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\centerline{\bf FUNCTIONAL ANALYSIS\footnote{These lecture notes were prepared
for the instructor's personal use in teaching a half-semester course
on functional analysis at the beginning graduate level at Penn State, in
Spring 1997.  They are certainly not meant to replace a good text on the
subject, such as those listed on this page.}}
\medskip
\centerline{Douglas N.~Arnold\footnote{Department of Mathematics,
Penn State University, University Park, PA   16802. \hfil\break
Web: http://www.math.psu.edu/dna/.}}
\vfill
{\parindent=0pt
References:
\smallskip
John B.~Conway, {\it A Course in Functional Analysis}, 2nd Edition,
Springer-Verlag, 1990.

Gert K.~Pedersen, {\it Analysis Now}, Springer-Verlag, 1989.

Walter Rudin, {\it Functional Analysis}, 2nd Edition, McGraw Hill, 1991.

Robert J.~Zimmer, {\it Essential Results of Functional Analysis},
University of Chicago Press, 1990.}
\vfill
\goodbreak\head CONTENTS \endhead\nobreak
\roster
\item"I." Vector spaces and their topology\dotfill 2
\item""\qquad Subspaces and quotient spaces\dotfill 4
\item""\qquad Basic properties of Hilbert spaces\dotfill 5

\item"II." Linear Operators and Functionals\dotfill 9
\item""\qquad The Hahn--Banach Theorem\dotfill 10
\item""\qquad Duality\dotfill 10

\item"III." Fundamental Theorems\dotfill 14
\item""\qquad The Open Mapping Theorem\dotfill 14
\item""\qquad The Uniform Boundedness Principle\dotfill 15
\item""\qquad The Closed Range Theorem\dotfill 16

\item"IV." Weak Topologies\dotfill 18
\item""\qquad The weak topology\dotfill 18
\item""\qquad The weak* topology\dotfill 19

\item"V." Compact Operators and their Spectra\dotfill 22
\item""\qquad Hilbert--Schmidt operators\dotfill 22
\item""\qquad Compact operators\dotfill 23
\item""\qquad Spectral Theorem for compact self-adjoint operators\dotfill 26
\item""\qquad The spectrum of a general compact operator\dotfill 28

\item"VI." Introduction to General Spectral Theory\dotfill 31
\item""\qquad The spectrum and resolvent in a Banach algebra\dotfill 31
\item""\qquad Spectral Theorem for bounded self-adjoint operators\dotfill 35
\endroster
\vfill\eject

\goodbreak\head I. Vector spaces and their topology \endhead\nobreak

Basic definitions: (1) Norm and seminorm on vector spaces (real or complex).
A norm defines a Hausdorff topology on a vector space in which
the algebraic operations are continuous, resulting in a {\it normed
linear space}.
If it is complete it is called a Banach space.

(2) Inner product and semi-inner-product.  In the real case an inner
product is a positive definite, symmetric bilinear form on $X\x
X\to\R$.  In the complex case it is positive definite, Hermitian
symmetric, sesquilinear form $X\x X\to\C$.  An (semi) inner product
gives rise to a (semi)norm.  An inner product space is thus a special
case of a normed linear space.  A complete inner product space is a
Hilbert space, a special case of a Banach space.

The polarization identity expresses the norm of an inner product space
in terms of the inner product.  For real inner product spaces it is
$$
(x,y)=\frac14(\|x+y\|^2-\|x-y\|^2).
$$
For complex spaces it is
$$
(x,y)=\frac14(\|x+y\|^2+i\|x+iy\|^2-\|x-y\|^2-i\|x-iy\|^2).
$$
In inner product spaces we also have the parallelogram law:
$$
\|x+y\|^2 + \|x-y\|^2 = 2(\|x\|^2+\|y\|^2).
$$
This gives a criterion for a normed space to be an inner product space.
Any norm coming from an inner product satisfies the parallelogram
law and, conversely, if a norm satisfies the parallelogram law, we can
show (but not so easily) that the polarization identity defines an inner
product, which gives rise to the norm.

(3) A {\it topological vector space}\/ is a vector space endowed with a
Hausdorff topology such that the algebraic operations are continuous.
Note that we can extend the notion of Cauchy sequence, and therefore of
completeness, to a TVS: a sequence $x_n$ in a TVS is Cauchy if for every
neighborhood $U$ of $0$ there exists $N$ such that $x_m-x_n\in U$ for
all $m,n\ge N$.

A normed linear space is a TVS, but there is another, more general
operation involving norms which endows a vector space with a
topology.  Let $X$ be a vector space and suppose that a family
$\{\|\cdot\|_\alpha\}_{\alpha\in \Cal A}$ of seminorms on $X$ is
given which are {\it sufficient}\/ in the sense that $\bigcap_\alpha
\{\|x\|_\alpha=0\}=0$.  Then the topology generated by the sets
$\{\|x\|_\alpha<r\}$, $\alpha\in\Cal A$, $r>0$, makes $X$ a TVS.
A sequence (or net) $x_n$ converges to $x$ iff $\|x_n-x\|_\alpha\to0$
for all $\alpha$.  Note that, a fortiori,
$|\,\|x_n\|_\alpha-\|x\|_\alpha|\to0$, showing that each seminorm is
continuous.

If the number of seminorms is finite, we may add them to get a norm
generating the same topology.  If the number is countable, we may
define a metric
$$
d(x,y)=\sum_n2^{-n}\frac{\|x-y\|_n}{1+\|x-y\|_n},
$$
so the topology is metrizable.

Examples: (0) On $\R^n$ or $\C^n$ we may put the $l_p$ norm,
$1\le p\le \infty$, or the weighted $l_p$ norm with some arbitrary
positive weight.  All of these norms are equivalent (indeed all
norms on a finite dimensional space are equivalent), and generate
the same Banach topology.  Only for $p=2$ is it a Hilbert space.

(2) If $\Omega$ is a subset of $\R^n$ (or, more generally, any Hausdorff
space) we may define the space $C_b(\Omega)$ of bounded continuous
functions with the supremum norm.  It is a Banach space.  If $X$ is
compact this is simply the space $C(\Omega)$ of continuous functions on
$\Omega$.

(3)  For simplicity, consider the unit interval, and define
$C^n([0,1])$ and $C^{n,\alpha}([0,1])$, $n\in\N$, $\alpha\in(0,1]$.
Both are Banach spaces with the natural norms.
$C^{0,1}$ is the space of Lipschitz functions.
$C([0,1])\subset C^{0,\alpha}\subset C^{0,\beta}\subset C^1([0,1])$
if $0<\alpha\le\beta\le 1$.

(4) For $1\le p<\infty$ and $\Omega$ an open or closed subspace
of $\R^n$ (or, more generally, a $\sigma$-finite measure space),
we have the space $L^p(\Omega)$ of equivalence classes of
measurable $p$-th power integrable functions (with equivalence
being equality off a set of measure zero), and for $p=\infty$
equivalence classes of essentially bounded functions (bounded
after modification on a set of measure zero).  For $1<p<\infty$
the triangle inequality is not obvious, it is Minkowski's inequality.
Since we modded out the functions with $L^p$-seminorm zero, this
is a normed linear space, and the Riesz-Fischer theorem asserts
that it is a Banach space.  $L^2$ is a Hilbert space.  If
$\operatorname{meas}(\Omega)<\infty$, then $L^p(\Omega)\subset L^q(\Omega)$
if $1\le q\le p\le\infty$.

(5) The sequence space $l_p$, $1\le p\le\infty$ is an example
of (4) in the case where the measure space is $\N$ with the counting
measure.  Each is a Banach space.  $l_2$ is a Hilbert space.
$l_p\subset l_q$ if $1\le p\le q \le\infty$ (note the inequality
is reversed from the previous example).
The subspace $c_0$ of sequences tending to $0$ is a closed subspace
of $l_\infty$.

(6) If $\Omega$ is an open set in $\R^n$ (or any Hausdorff space),
we can equip $C(\Omega)$ with the norms $f\mapsto |f(x)|$ indexed
by $x\in\Omega$.  This makes it a TVS, with the topology being that
of pointwise convergence.  It is not complete (pointwise limit of
continuous functions may not be continuous).

(7) If $\Omega$ is an open set in $\R^n$ we can equip $C(\Omega)$ with
the norms $f\mapsto \|f\|_{L^\infty(K)}$ indexed by compact subsets of
$\Omega$, thus defining the topology of uniform convergence on compact
subsets.  We get the same toplogy by using only the countably many
compact sets
$$
K_n=\{x\in\Omega:|x|\le n,\ \operatorname{dist}(x,\partial\Omega)\ge 1/n\}.
$$
The topology is complete.

(8) In the previous example, in the case $\Omega$ is a region in $\C$,
and we take complex-valued functions, we may consider the subspace
$H(\Omega)$ of holomorphic functions.  By Weierstrass's theorem
it is a closed subspace, hence itself a complete TVS.

(9) If $f,g\in L^1(I)$, $I=(0,1)$ and
$$
\int_0^1 f(x)\phi(x)\,dx = -\int_0^1 g(x)\phi'(x)\,dx,
$$
for all infinitely differentiable $\phi$ with support contained
in $I$ (so $\phi$ is identically zero near $0$ and $1$), then
we say that $f$ is weakly differentiable and that $f'=g$.  We can then
define the {\it Sobolev space}\/ $W^1_p(I)=\{\,f\in L^p(I)\,:\, f'\in L^p(I)\,\}$,
with the norm
$$
\|f\|_{W^1_p(I)}=\left(\int_0^1 |f(x)|^p\,dx+\int_0^1
|f'(x)|^p\,dx\right)^{1/p}.
$$
This is a larger space than $C^1(\bar I)$, but still incorporates
first order differentiability of $f$.  The case $p=2$ is particularly useful,
because it allows us to deal with differentiability in a Hilbert space
context.  Sobolev spaces can be extended to measure any degree of
differentiability (even fractional), and can be defined on arbitrary domains
in $\R^n$.

\goodbreak\subhead Subspaces and quotient spaces \endsubhead\nobreak

If $X$ is a vector space and $S$ a subspace, we may define the
vector space $X/S$ of cosets.  If $X$ is normed, we may define
$$
\|u\|_{X/S}=\inf_{x\in u}\|x\|_X\text{, or equivalently }
\|\bar x\|_{X/S}=\inf_{s\in S}\|x-s\|_X.
$$
This is a seminorm, and is a norm iff $S$ is closed.

\proclaim{Theorem} If $X$ is a Banach space and $S$ is a closed subspace
then $S$ is a Banach space and $X/S$ is a Banach space.
\endproclaim
\demo{Sketch} Suppose $x_n$ is a sequence of elements of $X$ for which
the cosets $\bar x_n$ are Cauchy.  We can take a subsequence with
$\|\bar x_n-\bar x_{n+1}\|_{X/S}\le 2^{-n-1}$, $n=1,2,\ldots$.  Set
$s_1=0$, define $s_2\in S$ such that $\|x_1-(x_2+s_2)\|_X\le 1/2$,
define $s_3\in S$ such that $\|(x_2+s_2)-(x_3+s_3)\|_X\le 1/4$, \dots.
Then $\{x_n+s_n\}$ is Cauchy in $X$ \dots \qed
\enddemo

A converse is true as well (and easily proved).
\proclaim{Theorem} If $X$ is a normed linear space and $S$ is a closed
subspace such that $S$ is a Banach space and $X/S$ is a Banach space,
then $X$ is a Banach space.
\endproclaim

Finite dimensional subspaces are always closed (they're complete).
More generally:
\proclaim{Theorem} If $S$ is a closed subspace of a Banach space and $V$
is a finite dimensional subspace, then $S+V$ is closed.
\endproclaim
\demo{Sketch} We easily pass to the case $V$ is one-dimensional and
$V\cap S=0$.  We then have that $S+V$ is algebraically a direct sum
and it is enough to show that the projections $S+V\to S$ and $S+V\to V$
are continuous (since then a Cauchy sequence in $S+V$ will lead to
a Cauchy sequence in each of the closed subspaces, and so to a convergent
subsequence).  Now the projection $\pi:X\to X/S$ restricts to a 1-1 map
on $V$ so an isomorphism of $V$ onto its image $\bar V$.  Let $\mu:\bar V\to V$
be the continuous inverse.  Since $\pi(S+V)\subset\bar V$, we may form
the composition $\mu\circ \pi|_{S+V}:S+V\to V$ and it is continuous.
But it is just the projection onto $V$.  The projection onto $S$ is
$id-\mu\circ\pi$, so it is also continuous.
\qed
\enddemo

\remark{Note} The sum of closed subspaces of a Banach space need
not be closed.  For a counterexample (in a separable Hilbert space), let
$S_1$ be the vector space of all real sequences $(x_n)_{n=1}^\infty$
for which $x_n=0$ if $n$ is odd, and $S_2$ be the sequences for which
$x_{2n}=nx_{2n-1}$, $n=1,2,\ldots$. Clearly $X_1=l_2\cap S_1$ and
$X_2=l_2\cap S_2$ are closed subspaces of $l_2$, the space of square
integrable sequences (they are defined as the intersection of the null
spaces of continuous linear functionals).  Obviously every sequence can be
written in a unique way as sum of elements of $S_1$ and
$S_2$:
$$
(x_1,x_2,\ldots)
= (0,x_2-x_1,0,x_4-2x_3,0,x_6-3x_5,\ldots)
+(x_1,x_1,x_3,2x_3,x_5,3x_5,\ldots).
$$
If a sequence has all but finitely many terms zero, so do the two
summands.  Thus all such sequences belong to $X_1+X_2$, showing
that $X_1+X_2$ is dense in $l_2$.  Now consider the sequence
$(1,0,1/2,0,1/3,\ldots)\in l_2$.  Its only decomposition as elements of $\Cal
S_1$ and $S_2$ is
$$
(1,0,1/2,0,1/3,0,\ldots)=(0,-1,0,-1,0,-1,\ldots)+(1,1,1/2,1,1/3,1,\ldots),
$$
and so it does not belong to $X_1+X_2$.  Thus $X_1+X_2$ is not closed
in $l_2$.
\endremark


\goodbreak\subhead Basic properties of Hilbert spaces\endsubhead\nobreak

An essential property of Hilbert space is that the distance of
a point to a closed convex set is alway attained.

\proclaim{Projection Theorem} Let $X$ be a Hilbert space, $K$ a closed
convex subset, and $x\in X$.  Then there exists a unique $\bar x\in K$
such that
$$
\|x-\bar x\|=\inf_{y\in K}\|x-y\|.
$$
\endproclaim
\demo{Proof} Translating, we may assume that $x=0$, and so we must
show that there is a unique element of $K$ of minimal norm.  Let
$d=\inf_{y\in K}\|y\|$ and chose $x_n\in K$ with $\|x_n\|\to d$.
Then the parallelogram law gives
$$
\left\|\frac{x_n-x_m}2\right\|^2=\frac12\|x_n\|^2+\frac12\|x_m\|^2
-\left\|\frac{x_n+x_m}2\right\|^2 \le \frac12\|x_n\|^2+\frac12\|x_m\|^2-d^2,
$$
where we have used convexity to infer that $(x_n+x_m)/2\in K$.  Thus
$x_n$ is a Cauchy sequence and so has a limit $\bar x$, which must
belong to $K$, since $K$ is closed.  Since the norm is continuous,
$\|\bar x\|=\lim_n\|x_n\|=d$.

For uniqueness, note that if $\|\bar x\|=\|\tilde x\|=d$, then
$\|(\bar x+\tilde x)/2\|=d$ and the parallelogram law gives
$$
\|\bar x-\tilde x\|^2=2\|\bar x\|^2 +2 \|\tilde x\|^2 - \|\bar x+\tilde x\|^2
=2d^2+2d^2-4d^2=0.\qed
$$
\enddemo

The unique nearest element to $x$ in $K$ is often denoted $P_K x$,
and referred to as the projection of $x$ onto $K$.  It satisfies
$P_K\circ P_K=P_K$, the definition of a projection.
This terminology is especially used when $K$ is a closed linear subspace
of $X$, in which case $P_K$ is a linear projection operator.

\subsubhead Projection and orthogonality\endsubsubhead
If $S$ is any subset of a Hilbert space $X$, let
$$
S^\perp = \{\, x\in X \,:\, \<x,s\>=0\text{ for all $s\in S$}\,\}.
$$
Then $S^\perp$ is a closed subspace of $X$.  We obviously have
$S\cap S^\perp = 0$ and $S\subset S^{\perp\perp}$.

Claim: If $S$ is a closed subspace of $X$, $x\in X$, and $P_S x$ the
projection of $x$ onto $S$, then $x-P_S x\in S^\perp$.  Indeed,
if $s\in S$ is arbitrary and $t\in\R$, then
$$
\|x-P_S x\|^2 \le \|x-P_S x - t s\|^2=\|x-P_S x\|^2-2t(x-P_S x,s)+t^2\|s\|^2,
$$
so the quadratic polynomial on the right hand side has a minimum at $t=0$.
Setting the derivative there to 0 gives $(x-P_S x,s)=0$.

Thus we can write any $x\in X$ as $s+s^\perp$ with $s\in S$ and $s^\perp\in
S^\perp$ (namely $s=P_S x$, $s^\perp=x-P_S x$).  Such a decomposition
is certainly unique (if $\bar s+\bar s^\perp$ were another one
we would have $s-\bar s=\bar s^\perp-s^\perp\in S\cap S^\perp=0$.)
We clearly have $\|x\|^2=\|s\|^2+\|s^\perp\|^2$.

An immediate corollary is that $S^{\perp\perp}=S$ for $S$ a closed subspace,
since if $x\in S^{\perp\perp}$ we can write it as $s+s^\perp$, whence
$s^\perp\in S^\perp\cap S^{\perp\perp}=0$, i.e., $x\in S$.  We thus
see that the decomposition
$$
x=(I-P_S)x+P_S x
$$
is the (unique) decomposition of $x$ into elements of $S^\perp$ and
$S^{\perp\perp}$.  Thus $P_{S^\perp}=I-P_S$.  For any subset $S$ of $X$,
$S^{\perp\perp}$ is the smallest closed subspace containing $S$.

\subsubhead Orthonormal sets and bases in Hilbert space \endsubsubhead

Let $e_1$, $e_2$, \dots, $e_N$ be orthonormal elements of a Hilbert space
$X$, and let $S$ be their span.  Then $\sum_n\<x,e_n\>e_n\in S$ and
$x-\sum_n\<x,e_n\>e_n\perp S$, so $\sum_n\<x,e_n\>e_n=P_S x$.  But
$\|\sum_n\<x,e_n\>e_n\|^2=\sum_{n=1}^N \<x,e_n\>^2$, so
$$
\sum_{n=1}^N \<x,e_n\>^2 \le \|x\|^2
$$
(Bessel's inequality).  Now let $\Cal E$ be an orthonormal set of
arbitrary cardinality.  It follows from Bessel's inequality that
for $\epsilon>0$ and $x\in X$, $\{\,e\in \Cal E\,:\, \<x,e\>\ge \epsilon\,\}$
is finite, and hence that $\{\,e\in \Cal E\,:\, \<x,e\> > 0\,\}$ is countable.
We can thus extend Bessel's inequality to an arbitrary orthonormal set:
$$
\sum_{e\in \Cal E}\<x,e\>^2 \le \|x\|^2,
$$
where the sum is just a countable sum of positive terms.

It is useful to extend the notion of sums over sets of arbitrary cardinality.
If $\Cal E$ is an arbitary set and $f:\Cal E\to X$ a function mapping
into a Hilbert space (or any normed linear space or even TVS), we say
$$
\sum_{e\in \Cal E} f(e) = x
\tag{$\star$}
$$
if the net $\sum_{e\in\Cal F} f(e)$, indexed by the {\it finite}\/ subsets
$\Cal F$ of $\Cal E$, converges to $x$.  In other words, ($\star$) holds if,
for any neighborhood $U$ of the origin, there is a finite set $\Cal F_0\subset
\Cal E$ such that $x-\sum_{e\in \Cal F} f(e)\in U$ whenever
$\Cal F$ is a finite subset of $\Cal E$ containing $\Cal F_0$.  In the
case $\Cal E=\N$, this is equivalent to absolute convergence of a series.
Note that if $\sum_{e\in \Cal E} f(e)$ converges, then for all $\epsilon$
there is a finite $\Cal F_0$ such that if $\Cal F_1$ and $\Cal F_2$ are
finite supersets of $\Cal F_0$, then $\|\sum_{e\in\Cal F_1}f(e)-\sum_{e\in\Cal
F_2}f(e)\|\le \epsilon$.  It follows easily that each of the sets
$\{\,e\in \Cal E\,|\, \|f(e)\|\ge 1/n\,\}$ is finite, and hence,
$f(e)=0$ for all but countably many $e\in\Cal E$.

\proclaim{Lemma}  If $\Cal E$ is an orthonormal subset of a Hilbert
space $X$ and $x\in X$, then
$$
\sum_{e\in\Cal E} \<x,e\>e
$$
converges.
\endproclaim
\demo{Proof} We may order the elements $e_1$, $e_2$, \dots of $\Cal E$ for
which $\<x,e\>\ne 0$.  Note that
$$
\|\sum_{n=1}^N\<x,e_n\>e_n\|^2=\sum_{n=1}^N |\<x,e_n\>|^2\le \|x\|^2.
$$
This shows that the partial sums $s_N=\sum_{n=1}^N\<x,e_n\>e_n$ form a Cauchy
sequence, and so converge to an element $\sum_{n=1}^\infty\<x,e_n\>e_n$
of $X$.  As an exercise in applying the definition, we show that
$\sum_{e\in\Cal E} \<x,e\>e=\sum_{n=1}^\infty\<x,e_n\>e_n$.  Given
$\epsilon>0$ pick $N$ large enough that $\sum_{n=N+1}^\infty
|\<x,e_n\>|^2<\epsilon$.  If $M>N$ and $\Cal F$ is a finite subset of
$\Cal E$ containing $e_1$, \dots, $e_N$, then
$$
\|\sum_{n=1}^M\<x,e_n\>e_n-\sum_{e\in\Cal F} \<x,e\>e\|^2\le \epsilon.
$$
Letting $M$ tend to infinity,
$$
\|\sum_{n=1}^\infty\<x,e_n\>e_n-\sum_{e\in\Cal F} \<x,e\>e\|^2\le \epsilon,
$$
as required. \qed
\enddemo

Recall the proof that every vector space has a basis.  We consider the
set of all linearly independent subsets of the vector space ordered by
inclusions, and note that if we have a totally ordered subset of this
set, then the union is a linearly independent subset containing all its
members.  Therefore Zorn's lemma implies that there exists a maximal
linearly independent set.  It follows directly from the maximality
that this set also spans, i.e., is a basis.  In an inner product
space we can use the same argument to establish the existence
of an orthonormal basis.

In fact, while bases exist for all vector spaces, for infinite
dimensional spaces they are difficult or impossible to construct and
almost never used.  Another notion of basis is much more useful, namely
one that uses the topology to allow infinite linear combinations.
To distinguish ordinary bases from such notions, an ordinary basis
is called a Hamel basis.

Here we describe an orthonormal Hilbert space basis.  By definition
this is a maximal orthonormal set.  By Zorn's lemma, any orthonormal
set in a Hilbert space can be extended to a basis, and so orthonormal
bases exist.  If $\Cal E$ is such an orthonormal basis, and $x\in X$,
then
$$
x=\sum_{e\in \Cal E} \<x,e\>e.
$$
Indeed, we know that the sum on the right exists in $X$ and it is easy to check
that its inner product with any $e_0\in\Cal E$ is $\<x,e_0\>$.  Thus
$y:=x-\sum_{e\in \Cal E} \<x,e\>e$ is orthogonal to $\Cal E$, and if it weren't
zero, then we could adjoin $y/\|y\|$ to $\Cal E$ to get a larger
orthonormal set.

Thus we've shown that any element $x$ of $X$ can be expressed as $\sum c_e e$
for some $c_e\in\R$, all but countably many of which are $0$.  It is easily
seen that this determines the $c_e$ uniquely, namely $c_e=\<x,e\>$, and
that $\|x\|^2=\sum c_e^2$.

The notion of orthonormal basis allows us to define a Hilbert space dimension,
namely the cardinality of any orthonormal basis.  To know that this
is well defined, we need to check that any two bases have the same cardinality.
If one is finite, this is trivial.  Otherwise, let $\Cal E$ and $\Cal F$
be two infinite orthonormal bases.  For each $0\ne x\in X$, 
the inner product $\<x,e\>\ne 0$ for at least one $e\in\Cal E$.  Thus
$$
\Cal F\subset \bigcup_{e\in\Cal E} \{\,f\in \Cal F\,:\, \<f,e\>\ne 0\,\},
$$
i.e., $\Cal F$ is contained in the union of $\operatorname{card}\Cal
E$ countable sets.  Therefore $\operatorname{card}\Cal F\le
\aleph_0\operatorname{card}\Cal E=\operatorname{card}\Cal E$.

If $\Cal S$ is any set, we define a particular Hilbert space $l^2(\Cal S)$
as the set of functions $c:\Cal S\to \R$ which are zero off a countable
set and such that $\sum_{s\in\Cal S} c_s^2 <\infty$. We thus see that via
a basis, any Hilbert space can be put into a norm-preserving (and so
inner-product-preserving) linear bijection (or Hilbert space
isomorphism) with an $l^2(\Cal S)$.  Thus, up to isomorphism, there is
just one Hilbert space for each cardinality.  In particular there is
only one infinite dimensional separable Hilbert space (up to isometry).

Example: The best known example of an orthonormal basis in an infinite
Hilbert space is the set of functions $e_n=\exp(2\pi i n \theta)$ which
form a basis for complex-valued $L^2([0,1])$.  (They are obviously orthonormal,
and they are a maximal orthonormal set by the Weierstrass approximation
Theorem.  Thus an arbitrary $L^2$
function has an $L^2$ convergent Fourier series
$$
f(\theta)=\sum_{n=-\infty}^\infty \hat f(n) e^{2\pi i n \theta},
$$
with $\hat f(n)=\<f,e_n\>=\int_0^1 f(\theta)e^{-2\pi i n \theta}\,d\theta$.
Thus from the Hilbert space point of view, the theory of Fourier series
is rather simple.  More difficult analysis comes in when we consider convergence
in other topologies (pointwise, uniform, almost everywhere, $L^p$, $C^1$, \dots).

\subsubhead Schauder bases \endsubsubhead
An orthonormal basis in a Hilbert space is a special example of a Schauder
basis.  A subset $\Cal E$ of a Banach space $X$ is called a Schauder basis
if for every $x\in X$ there is a unique function $c:\Cal E\to\R$ such that
$x=\sum_{e\in\Cal E} c_e e$.  Schauder constructed a useful Schauder basis
for $C([0,1])$, and there is useful Schauder bases in many other separable
Banach spaces.  In 1973 Per Enflo settled a long-standing open question
by proving that there exist separable Banach spaces with no Schauder bases.

\goodbreak\head II. Linear Operators and Functionals \endhead\nobreak
%
$B(X,Y)$ = bounded linear operators between normed linear spaces $X$ and $Y$.
A linear operator is bounded iff it is bounded on every ball iff it is
bounded on some ball iff it is continuous at every point iff it is continuous
at some point.

\proclaim{Theorem} If $X$ is a normed linear space and $Y$ is a Banach space,
then $B(X,Y)$ is a Banach space with the norm
$$
\|T\|_{B(X,Y)}=\sup_{0\ne x\in X}\frac{\|Tx\|_Y}{\|x\|_X}.
$$
\endproclaim
\demo{Proof}  It is easy to check that $B(X,Y)$ is a normed linear space,
and the only issue is to show that it is complete.

Suppose that $T_n$ is a Cauchy sequence in $B(X,Y)$.  Then for each $x\in X$
$T_n x$ is Cauchy in the complete space $Y$, so there exists $Tx\in Y$
with $T_n x\to Tx$.  Clearly $T:X\to Y$ is linear.  Is it bounded?
The real sequence $\|T_n\|$ is Cauchy, hence bounded, say $\|T_n\|\le K$.
It follows that $\|T\|\le K$, and so $T\in B(X,Y)$.  To conclude
the proof, we need to show that $\|T_n-T\|\to 0$.  We have
$$
\multline
\|T_n-T\|=\sup_{\|x\|\le 1}\|T_n x-Tx\|=\sup_{\|x\|\le 1}\lim_{m\to\infty}
\|T_n x-T_mx\|
\\
=\sup_{\|x\|\le 1}\limsup_{m\to\infty}
\|T_n x-T_mx\|\le \limsup_{m\to\infty}\|T_n-T_m\|.
\endmultline
$$
Thus $\limsup_{n\to\infty}\|T_n-T\|=0$.
\qed
\enddemo

If $T\in B(X,Y)$ and $U\in B(Y,Z)$, then $UT=U\circ T\in B(X,Z)$ and
$\|UT\|_{B(X,Z)}\le \|U\|_{B(Y,Z)}\|T\|_{B(X,Y)}$.  In particular,
$B(X):=B(X,X)$ is a {\it Banach algebra}, i.e., it has an additional
``multiplication'' operation which makes it a non-commutative algebra,
and the multiplication is continuous.

The dual space is $X^*:=B(X,\R)$ (or $B(X,\C)$ for complex vector spaces).
It is a Banach space (whether $X$ is or not).

\goodbreak\subhead The Hahn--Banach Theorem \endsubhead\nobreak
%
A key theorem for dealing with dual spaces of normed linear spaces
is the Hahn-Banach Theorem.  It assures us that the dual space of
a nontrivial normed linear space is itself nontrivial.  (Note: the
norm is important for this.  There exist topological vector spaces,
e.g., $L^p$ for $0<p<1$, with no non-zero continuous linear functionals.)

\proclaim{Hahn-Banach} If $f$ is a bounded linear functional on a subspace
of a normed linear space, then $f$ extends to the whole space with preservation
of norm.
\endproclaim

Note that there are virtually no hypotheses beyond linearity and existence
of a norm.  In fact for some purposes a weaker version is useful.
For $X$ a vector space, we say that $p:X\to\R$ is sublinear if $p(x+y)\le p(x)+p(y)$
and $p(\alpha x)=\alpha p(x)$ for $x,y\in X$, $\alpha\ge 0$.

\proclaim{Generalized Hahn-Banach} Let $X$ be a vector space, $p:X\to\R$
a sublinear functional, $S$ a subspace of $X$, and $f:S\to\R$ a linear
function satisfying $f(x)\le p(x)$ for all $x\in S$, then $f
$ can be extended to $X$ so that the same inequality holds for all $x\in X$.
\endproclaim
\demo{Sketch} It suffices to extend $f$ to the space spanned by $S$ and
one element $x_0\in X\setminus S$, preserving the inequality, since
if we can do that we can complete the proof with Zorn's lemma.

We need to define $f(x_0)$ such that $f(tx_0+s)\le p(tx_0+s)$ for
all $t\in\R$, $s\in S$.  The case $t=0$ is known and it is easy to use
homogeneity to restrict to $t=\pm 1$.  Thus we need to find a value
$f(x_0)\in\R$ such that
$$
f(s)-p(-x_0+s)\le f(x_0)\le p(x_0+s)-f(s) \text{\quad for all $s\in S$}.
$$
Now it is easy to check that for any $s_1$, $s_2\in S$,
$f(s_1)-p(-x_0+s_1)\le p(x_0+s_2)-f(s_2)$, and so such an $f(x_0)$ exists.
\qed
\enddemo

\proclaim{Corollary} If $X$ is a normed linear space and $x\in X$, then
there exists $f\in X^*$ of norm 1 such that $f(x)=\|x\|$.
\endproclaim

\proclaim{Corollary} If $X$ is a normed linear space, $S$ a closed subspace,
and $x\in X$, then there exists $f\in X^*$ of norm 1 such that $f(x)=\|\bar
x\|_{X/S}$.
\endproclaim

\goodbreak\subhead Duality \endsubhead\nobreak
%
If $X$ and $Y$ are normed linear spaces and $T:X\to Y$, then we
get a natural map $T^*:Y^*\to X^*$ by $T^*f(x)=f(Tx)$ for all $f\in Y^*$,
$x\in X$.  In particular, if $T\in B(X,Y)$, then $T^*\in B(Y^*,X^*)$.
In fact, $\|T^*\|_{B(Y^*,X^*)}=\|T\|_{B(X,Y)}$.  To prove this,
note that $|T^*f(x)|=|f(Tx)|\le \|f\|\|T\|\|x\|$.  Therefore
$\|T^*f\|\le \|f\|\|T\|$, so $T^*$ is indeed bounded, with $\|T^*\|\le
\|T\|$.  Also,
given any $y\in Y$, we can find $g\in Y^*$ such that $|g(y)|=\|y\|$,
$\|g\|=1$.  Applying this with $y=Tx$ ($x\in X$ arbitrary), gives
$\|Tx\|=|g(Tx)|=|T^*g x|\le\|T^*\|\|g\|\|x\|=\|T^*\|\|x\|$.  This
shows that $\|T\|\le \|T^*\|$.  Note that if $T\in B(X,Y)$, $U\in B(Y,Z)$,
then $(UT)^*=T^* U^*$.

If $X$ is a Banach space and $S$ a subset, let
$$
S^a=\{\,f\in X^*\,|\,f(s)=0\quad\forall s\in S\,\}
$$
denote the {\it annihilator}\/ of $S$.  If $V$ is a subset of $X^*$, we similarly
set
$$
{}^aV=\{\,x\in X\,|\,f(x)=0\quad\forall f\in V\,\}.
$$
Note the distinction between $V^a$, which is a subset of $X^{**}$
and ${}^aV$, which is a subset of $X$.  All annihilators are closed
subspaces.

It is easy to see that $S\subset T\subset X$ implies that $T^a\subset
S^a$, and $V\subset W\subset X^*$ implies that ${}^aW\subset {}^aV$.
Obviously $S\subset {}^a(S^a)$ if $S\subset X$ and $V\subset ({}^aV)^a$
if $V\subset X^*$.  The Hahn-Banach theorem implies that $S={}^a(S^a)$
in case $S$ is a closed subspace of $X$ (but it can happen that
$V\subsetneq({}^aV)^a$ for $V$ a closed subspace of $X^*$.  For
$S\subset X$ arbitrary, ${}^a(S^a)$ is the smallest closed subspace of
$X$ containing the subset $S$, namely the closure of the span of $S$.

Now suppose that $T:X\to Y$ is a bounded linear operator between Banach
spaces.  Let $g\in Y^*$.  Then $g(Tx)=0\quad\forall x\in X$ $\iff$
$T^*g(x)=0\quad\forall x\in X$ $\iff$ $T^*g=0$.  I.e.,
$$
\Ra(T)^a=\Nu(T^*).
$$
Similarly, for $x\in X$, $Tx=0$ $\iff$ $f(Tx)=0\quad\forall f\in Y^*$
$\iff$ $T^*f(x)=0\quad\forall f\in Y^*$, or
$$
{}^a\Ra(T^*)=\Nu(T).
$$
Taking annihilators gives two more results:
$$
\overline{\Ra(T)}={}^a\Nu(T^*), \qquad
\overline{\Ra(T^*)}\subset\Nu(T)^a.
$$
In particular we see that $T^*$ is injective iff $T$ has dense range;
and $T$ is injective if $T^*$ has dense range.

Note: we will have further results in this direction once we introduce
the weak*-topology on $X^*$.  In particular, $({}^aS)^a$ is the weak*
closure of a subspace $S$ of $X^*$ and $T$ is injective iff $T^*$ has 
weak* dense range.

\subsubhead Dual of a subspace \endsubsubhead
An important case is when $T$ is the inclusion map $i:S\to X$, where
$S$ is a closed subspace of $X$.  Then $r=i^*:X^*\to S^*$ is just the
restriction map: $r f(s)=f(s)$.  Hahn-Banach tells us that $r$ is
surjective.  Obviously $\Nu(r)=S^a$.  Thus we have a canonical isomorphism
$\bar r:X^*/S^a\to S^*$.  In fact, the Hahn-Banach theorem shows
that it is an isometry.  Via this isometry one often identifies
$X^*/S^a$ with $S^*$.

\subsubhead Dual of a quotient space \endsubsubhead
Next, consider the projection map $\pi:X\to X/S$ where $S$ is a closed
subspace.  We then have $\pi^*:(X/S)^* \to X^*$.  Since $\pi$ is surjective,
this map is injective.  It is easy to see that the range is contained
in $S^a$.  In fact we now show that $\pi^*$ maps $(X/S)^*$ onto
$S^a$, hence provides a canonical isomorphism of $S^a$ with $(X/S)^*$.
Indeed, if $f\in S^a$, then we have a splitting $f=g\circ\pi$ with
$g\in (X/S)^*$ (just define $g(c)=f(x)$ where $x$ is any element
of the coset $c$).  Thus $f=\pi^*g$ is indeed in the range of $\pi^*$.
This correspondence is again an isometry.

\subsubhead Dual of a Hilbert space \endsubsubhead
The identification of dual spaces can be quite tricky.  The case of Hilbert
spaces is easy.  

\proclaim{Riesz Representation Theorem} If $X$ is a real Hilbert space, define $j:X\to X^*$
by $j_y(x)=\<x,y\>$.  This map is a linear isometry of $X$ onto $X^*$.
For a complex Hilbert space it is a conjugate linear isometry
(it satisfies $j_{\alpha y}=\bar\alpha j_y$).
\endproclaim
\demo{Proof}  It is easy to see that $j$ is an isometry of $X$ into $X^*$ and
the main issue is to show that any $f\in X^*$ can be written as $j_y$ for
some $y$.  We may assume that $f\ne0$, so $\Nu(f)$ is a proper closed subspace
of $X$.  Let $y_0\in[\Nu(f)]^\perp$ be of norm $1$ and set $y=(fy_0)y_0$.
For all $x\in X$, we clearly have that $(fy_0)x-(fx)y_0\in \Nu(f)$, so
$$
j_y(x)=\<x,(fy_0)y_0\>=\<(fy_0)x,y_0\>=\<(fx)y_0,y_0\>=fx.\qed
$$
\enddemo

Via the map $j$ we can define an inner product on $X^*$, so it is again
a Hilbert space.

Note that if $S$ is a closed subspace of $X$, then $x\in S^\perp$
$\iff$ $j_s\in S^a$.  The Riesz map $j$ is sometimes used to identify
$X$ and $X^*$.  Under this identification there is no distinction
between $S^\perp$ and $S^a$.

\subsubhead Dual of $C(\Omega)$ \endsubsubhead
Note: there are two quite distinct theorems referred to as the Riesz
Representation Theorem.  The proceeding is the easy one.  The hard one identifies
the dual of $C(\Omega)$ where $\Omega$ is a compact subset of $\R^n$
(this can be generalized considerably).  It states that there is an
isometry between $C(\Omega)^*$ and the space of finite signed measures
on $\Omega$.  (A finite signed measure is a set function of the form
$\mu=\mu_1-\mu_2$ where $\mu_i$ is a finite measure, and we view such as
a functional on $C(X)$ by $f\mapsto \int_\Omega f\,d\mu_1-\int_\Omega
f\,d\mu_2$.)  This is the real-valued case; in the complex-valued
case the isometry is with complex measures $\mu + i\lambda$ where
$\mu$ and $\lambda$ are finite signed measures.

\subsubhead Dual of $C^1$ \endsubsubhead
It is easy to deduce a representation for an arbitrary
element of the dual of, e.g., $C^1([0,1])$.  The map
$f\mapsto (f,f')$ is an isometry of $C^1$ onto a closed
subspace of $C\x C$.  By the Hahn-Banach Theorem, every
element of $(C^1)^*$ extends to a functional on $C\x C$,
which is easily seen to be of the form
$$
(f,g)\mapsto \int f\,d\mu + \int g\,d\nu
$$
where $\mu$ and and $\nu$ are signed measures ($(X\x Y)^*=X^*\x Y^*$
with the obvious identifications).  Thus any continuous linear functional
on $C^1$ can be written
$$
f\mapsto \int f\,d\mu + \int f'\,d\nu.
$$
In this representation the measures $\mu$ and $\lambda$ are {\it not}\/
unique.

\subsubhead Dual of $L^p$ \endsubsubhead
H\"older's inequality states that if $1\le p\le\infty$, $q=p/(p-1)$, then
$$
\int f g \le \|f\|_{L^p}\|g\|_{L^q}
$$
for all $f\in L^p$, $g\in L^q$.  This shows that the map
$g\mapsto \lambda_g$:
$$
\lambda_g(f)=\int f g ,
$$
maps $L^q$ linearly into $(L^p)^*$ with $\|\lambda_g\|_{(L^p)^*}\le
\|g\|_{L^q}$.  The choice $f=\sign(g)|g|^{q-1}$ shows
that there is equality.  In fact, if $p<\infty$, $\lambda$ is
a linear isometry of $L^q$ onto $(L^p)^*$. For $p=\infty$ it
is an isometric injection, but not in general surjective.
Thus the dual of $L^p$ is $L^q$ for $p$ finite.  The dual of
$L^\infty$ is a very big space, much bigger than $L^1$ and rarely
used.

\subsubhead Dual of $c_0$ \endsubsubhead
The above considerations apply to the dual of the sequence spaces
$l_p$.  Let us now show that the dual of $c_0$ is $l_1$.
For any $c=(c_n)\in c_0$ and $d=(d_n)\in l_1$, we define $\lambda_d(c)=\sum c_n
d_n$.  Clearly
$$
|\lambda_d(c)|\le \sup |c_n|\sum|d_n|=\|c\|_{c_0}\|d\|_{l_1},
$$
so $\|\lambda_d\|_{c_0^*}\le\|d\|_{l_1}$.  Taking
$$
c_n=\cases \sign(d_n),& n\le N\\ 0, & n>N,\endcases
$$
we see that equality holds.  Thus $\lambda:l_1\to c_0^*$ is an isometric
injection.  We now show that it is onto.  Given $f\in c_0^*$, define
$d_n=f(e^{(n)})$ where $e^{(n)}$ is the usual unit sequence
$e^{(n)}_m=\delta_{mn}$.  Let $s_n=\sign(d_n)$.  Then
$|d_n|=f(s_n e^{(n)})$, so
$$
\sum_{n=0}^N |d_n|=\sum_{n=0}^Nf(s_n e^{(n)})
=f(\sum_{n=0}^Ns_n e^{(n)})\le \|f\|.
$$
Letting $N\to\infty$ we conclude that $d\in l_1$.  Now by construction
$\lambda_d$ agrees with $f$ on all sequences with only finitely many
nonzeros.  But these are dense in $c_0$, so $f=\lambda_d$.

\subsubhead The bidual \endsubsubhead
If $X$ is any normed linear space, we have a natural map
$i:X\to X^{**}$ given by
$$
i_x(f)=f(x),\qquad x\in X, \quad f\in X^*.
$$
Clearly $\|i_x\|\le\|f\|$ and, by the Hahn-Banach
theorem, equality holds.  Thus $X$ may be identified as a subspace
of the Banach space $X^{**}$.  If we define $\tilde X$ as the closure
of $i(X)$ in $X^{**}$, then $X$ is isometrically embedded as a dense
subspace of the Banach space $\tilde X$.  This determines $\tilde X$ up
to isometry, and is what we define as the completion of $X$.  Thus any
normed linear space has a completion.

If $i$ is onto, i.e., if $X$ is isomorphic with $X^{**}$ via this identification,
we say that $X$ is reflexive (which can only happen is $X$ is complete).
In particular, one can check that if $X$ is a Hilbert space and $j:X\to X^*$
is the Riesz isomorphism,  and $j^*:X^*\to X^{**}$ the Riesz isomorphism
for $X^*$, then $i=j^*\circ j$, so $X$ is reflexive.

Similarly, the canonical isometries of $L^q$ onto $(L^p)^*$ and then
$L^p$ onto $(L^q)^*$ compose to give the natural map of $L^p$ into
its bidual, and we conclude that $L^p$ (and $l_p$) is reflexive for
$1<p<\infty$.  None of $L^1$, $l_1$, $L^\infty$, $l_\infty$, $c_0$,
or $C(X)$ are reflexive.

If $X$ is reflexive, then $i({}^aS)=S^a$ for $S\subset X^*$.  In other
words, if we identify $X$ and $X^{**}$, the distinction between the two
kinds of annihilators disappears.  In particular, for reflexive Banach
spaces, $\overline{\Ra(T^*)}=\Nu(T)^a$ and $T$ is injective iff $T^*$
has dense range.

\goodbreak\head III. Fundamental Theorems \endhead\nobreak

The Open Mapping Theorem and the Uniform Boundedness Principle
join the Hahn-Banach Theorem as the ``big three''.
These two are fairly easy consequence of the Baire Category
Theorem.

\proclaim{Baire Category Theorem} A complete metric space cannot
be written as a countable union of nowhere dense sets.
\endproclaim
\demo{Sketch of proof} If the statement were false, we could write
$M=\bigcup_{n\in\N} F_n$ with $F_n$ a closed subset which does not contain any
open set.  In particular, $F_0$ is a proper closed set, so there exists
$x_0\in M$, $\epsilon_0\in (0,1)$ such that $E(x_0,\epsilon_0)\subset
M\setminus F_0$.  Since no ball is contained in $F_1$, there exists
$x_1\in E(x_0,\epsilon_0/2)$ and $\epsilon_1\in(0,\epsilon_0/2)$ such
that $E(x_1,\epsilon_1)\subset M\setminus F_1$.  In this way we get
a nested sequence of balls such that the $n$th ball has radius at most
$2^{-n}$ and is disjoint from $F_n$.  It is then easy to check that
their centers form a Cauchy sequence and its limit, which must exist
by completeness, can't belong to any $F_n$.
\qed\enddemo

\goodbreak\subhead The Open Mapping Theorem \endsubhead\nobreak
%
The Open Mapping Theorem follows from the Baire Category Theorem and
the following lemma.

\proclaim{Lemma} Let $T:X\to Y$ be a bounded linear operator between
Banach spaces.  If $E(0_Y,r)\subset \overline{T(E(0_X,1))}$ for some
$r>0$, then $E(0_Y,r)\subset T(B(0_X,2))$.
\endproclaim
\demo{Proof} Let $U=T(E(0_X,1))$.  Let $y\in Y$, $\|y\|<r$.  There exists
$y_0\in U$ with $\|y-y_0\|\le r/2$.  By homogeneity, there exists
$y_1\in \frac12U$ such that $\|y-y_0-y_1\|\le r/4$, $y_2\in \frac14U$ such
that $\|y-y_0-y_1-y_2\|\le r/8$, etc.  Take $x_n\in\frac1{2^n}U$ such
that $Tx_n=y_n$, and let $x=\sum_n x_n\in X$.  Then $\|x\|\le 2$ and
$Tx=\sum y_n=y$.
\qed\enddemo

\remark{Remark} The same proof works to prove the statement with $2$
replaced by any number greater than $1$.  With a small additional argument,
we can even replace it with $1$ itself.  However the statement above
is sufficient for our purposes.
\endremark

\proclaim{Open Mapping Theorem} A bounded linear surjection between
Banach spaces is open.
\endproclaim
\demo{Proof} It is enough to show that the image under $T$ of a ball about 0 contains
some ball about 0.  The sets $T(E(0,n))$ cover $Y$, so the closure of one
of them must contain an open ball.  By the previous result, we can
dispense with the closure.  The theorem easily follows using the linearity
of $T$.
\qed\enddemo

There are two major corollaries of the Open Mapping Theorem, each of which is
equivalent to it.

\proclaim{Inverse Mapping Theorem or Banach's Theorem} The inverse
of an invertible bounded linear operator between Banach spaces is
continuous.
\endproclaim
\demo{Proof} The map is open, so its inverse is continuous.\qed\enddemo

\proclaim{Closed Graph Theorem} A linear operator between Banach spaces
is continuous iff its graph is closed.
\endproclaim

A map between topological spaces is called closed if its graph is
closed.  In a general Hausdorff space, this is a weaker property than
continuity, but the theorem asserts that for linear operators between
Banach spaces it is equivalent.  The usefulness is that a direct proof
of continuity requires us to show that if $x_n$ converges to $x$ in $X$
then $Tx_n$ converges to $Tx$.  By using the closed graph theorem, we
get to assume as well that $Tx_n$ is converging to some $y$ in $Y$ and
we need only show that $y=Tx$.
\demo{Proof} Let $G=\{\,(x,Tx)\,|\,x\in X\,\}$ denote the graph. Then
the composition $G\subset X\times Y\to X$ is a bounded linear operator
between Banach spaces given by $(x,Tx)\mapsto x$.  It is clearly one-to-one
and onto, so the inverse is continuous by Banach's theorem.  But the composition
$X\to G\subset X\times Y \to Y$ is simply the $T$, so $T$ is continuous.
\qed
\enddemo

Banach's theorem leads immediately to this useful characterization of
closed imbeddings of Banach spaces.

\proclaim{Theorem} Let $T:X\to Y$ be a bounded linear map between Banach
spaces.  Then $T$ is one-to-one and has closed range if and only if
there exists a positive number $c$ such that
$$
\|x\|\le c\|Tx\|\qquad\forall x\in X.
$$
\endproclaim
\demo{Proof}  If the inequality holds, then $T$ is clearly one-to-one,
and if $Tx_n$ is a Cauchy sequence in $\Ra(T)$, then $x_n$ is Cauchy,
and hence $x_n$ converges to some $x$, so $Tx_n$ converges to $Tx$.
Thus the inequality implies that $\Ra(T)$ is closed.

For the other direction, suppose that $T$ is one-to-one with closed
range and consider the map $T^{-1}:\Ra(T)\to X$.  It is the inverse of a
bounded isomorphism, so is itself bounded.  The inequality follows immediately
(with $c$ the norm of $T^{-1})$.
\qed
\enddemo

Another useful corollary is that if a Banach space admits a second weaker
or stronger norm under which it is still Banach, then the two norms are
equivalent.  This follows directly from Banach's theorem applied
to the identity.

\goodbreak\subhead The Uniform Boundedness Principle \endsubhead\nobreak
%
The Uniform Boundedness Principle (or the Banach-Steinhaus Theorem)
also comes from the Baire Category Theorem.
\proclaim{Uniform Boundedness Principle} Suppose that $X$ and $Y$ are
Banach spaces and $\Cal S\subset B(X,Y)$.  If $\sup_{T\in\Cal S}\|T(x)\|_Y<\infty$
for all $x\in X$, then $\sup_{T\in\Cal S}\|T\|<\infty$.
\endproclaim
\demo{Proof} One of the closed sets $\{\,x\,|\,|f_n(x)|\le N\quad\forall
n\,\}$ must contain $E(x_0,r)$ for some $x_0\in X$, $r>0$.  Then, if
$\|x\|<r$, $|f_n(x)|\le |f_n(x+x_0)-f_n(x_0)|\le N+\sup|f_n(x_0)|=M$,
with $M$ independent of $n$.  This shows that the $\|f_n\|$ are uniformly
bounded (by $M/r$).
\qed
\enddemo

In words: a set of linear operators between Banach spaces which is bounded
pointwise is norm bounded.

The uniform boundedness theorem is often a way to generate counterexamples.
A typical example comes from the theory of Fourier series.
For $f:\R\to\C$ continuous and $1$-periodic the $n$th partial sum of
the Fourier series for $f$ is
$$
f_n(s)=\sum_{k=-n}^n\int_{-1}^{1} f(t)e^{-2\pi ikt}\,dt \, e^{2\pi iks}
=\int_{-1}^{1} f(t) D_n(s-t)\,dt
$$
where
$$
D_n(x)=\sum_{k=-n}^n e^{2\pi ikx}.
$$
Writing $z=e^{2\pi is}$, we have
$$
D_n(s)=\sum_{k=-n}^n
z^k=z^{-n}\frac{z^{2n}-1}{z-1}=\frac{z^{n+1/2}-z^{-n-1/2}}{z^{1/2}-z^{-1/2}}
=\frac{\sin(2n+1)\pi x}{\sin \pi x}.
$$
This is the Dirichlet kernel, a $C^\infty$ periodic function.  In particular,
the value of the $n$th partial sum of the Fourier series of $f$ at $0$ is
$$
T_nf:=f_n(0)=\int_{-1}^1 f(t) D_n(t)\,dt.
$$
We think of $T_n$ as a linear functional on the Banach space of $1$-periodic
continuous function endowed with the sup norm.  Clearly
$$
\|T_n\|\le C_n:=\int_{-1}^{1} |D_n(t)|\,dt.
$$
In fact this is an equality.  If $g(t)=\sign D_n(t)$, then $\sup|g|=1$
and $T_n g = C_n$.  Actually, $g$ is not continuous, so to make this
argument correct, we approximate $g$ by a continuous functions,
and thereby prove the norm equality.  Now one can calculate that
$\int|D_n|\to\infty$ as $n\to\infty$.  By the uniform boundedness
theorem we may conclude that there exists a continuous periodic function
for whose Fourier series diverges at $t=0$.

\goodbreak\subhead The Closed Range Theorem\endsubhead\nobreak
%
We now apply the Open Mapping Theorem to better understand
the relationship between $T$ and $T^*$.  The property of having
a closed range is significant to the structure of an operator
between Banach spaces.  If $T:X\to Y$ has a closed range $Z$
(which is then itself a Banach space), then $T$ factors as the
projection $X\to X/\Nu(T)$, the isomorphism $X/\Nu(T)\to Z$,
and the inclusion $Z\subset Y$.  The Closed Range Theorem
says that $T$ has a closed range if and only if $T^*$ does.

\proclaim{Theorem} Let $T:X\to Y$ be a bounded linear operator between
Banach spaces.  Then $T$ is invertible iff $T^*$ is.
\endproclaim
\demo{Proof} If $S=T^{-1}:Y\to X$ exists, then $ST=I_X$ and $TS=I_Y$,
so $T^*S^*=I_{X^*}$ and $S^*T^*=I_{Y^*}$, which shows that $T^*$ is
invertible.

Conversely, if $T^*$ is invertible, then it is open, so there is
a number $c>0$ such that $T^* B_{Y^*}(0,1)$ contains $B_{X^*}(0,c)$.
Thus, for $x\in X$
$$
\align
\|Tx\|&=\sup_{f\in B_{Y^*}(0,1)} |f(Tx)|=\sup_{f\in B_{Y^*}(0,1)}|(T^*f)x|
\\ &\ge \sup_{g\in B_{X^*}(0,c)}|g(x)|=c\|x\|.
\endalign
$$
The existence of $c>0$ such that $\|Tx\|\ge c\|x\|$ $\forall x\in X$
is equivalent to the statement that $T$ is injective with closed range.
But since $T^*$ is injective, $T$ has dense range.
\qed
\enddemo

\proclaim{Lemma} Let $T:X\to Y$ be a linear map between
Banach spaces such that $T^*$ is an injection with
closed range.  Then $T$ is a surjection.
\endproclaim
\demo{Proof} Let $E$ be the closed unit ball of $X$ and $F=\overline{TE}$.
It suffices to show that $F$ contains a ball around the origin,
since then, by the lemma used to prove the Open Mapping Theorem,
$T$ is onto.

There exists $c>0$ such that $\|T^*f\|\ge c\|f\|$ for all $f\in Y^*$.
We shall show that $F$ contains the ball of radius $c$ around the
origin in $Y$.  Otherwise there exists $y\in Y$, $\|y\|\le c$,
$y\notin F$.  Since $F$ is a closed convex set we can find a functional
$f\in Y^*$ such that $|f(Tx)|\le \alpha$ for all $x\in E$ and $f(y)>\alpha$.
Thus $\|f\|>\alpha/c$, but
$$
\|T^*f\|=\sup_{x\in E}|T^*f(x)|=\sup_{x\in E}|f(Tx)|\le \alpha.
$$
This is a contradiction.\qed
\enddemo

\proclaim{Closed Range Theorem} Let $T:X\to Y$ be a bounded linear
operator between Banach spaces.  Then T has closed range if and
only if $T^*$ does.
\endproclaim
\demo{Proof}  1) $\Ra(T)$ closed $\implies$ $\Ra(T^*)$ closed.

Let $Z=\Ra(T)$.  Then $\bar T:X/\Nu(T)\to Z$ is an isomorphism (Inverse
Mapping Theorem).  The diagram
$$
\CD
X @>T>> Y \\
@V\pi VV  @AA\cup A \\
X/\Nu(T) @>\cong>\bar T> Z.
\endCD
$$
commutes.  Taking adjoints,
$$
\CD
X^* @<T^*<< Y^* \\
@A\cup AA  @VV\pi V \\
\Nu(T)^a @<\cong<\bar T^*< Y^*/Z^a.
\endCD
$$
This shows that $\Ra(T^*)=\Nu(T)^a$.

2) $\Ra(T^*)$ closed $\implies$ $\Ra(T)$ closed.

Let $Z=\overline{\Ra(T)}$ (so $Z^a=\Nu(T^*)$) and let $S$ be the
range restriction of $T$, $S:X\to Z$.  The
adjoint is $S^*:Y^*/Z^a\to X^*$, the lifting of $T^*$ to
$Y^*/Z^a$.  Now $\Ra(S^*)=\Ra(T^*)$, is closed,
and $S^*$ is an injection.  We wish to show that $S$
is onto $Z$.  Thus the theorem follows from the preceding
lemma.
\qed
\enddemo

\goodbreak\head IV. Weak Topologies \endhead\nobreak
%
\goodbreak\subhead The weak topology \endsubhead\nobreak
%
Let $X$ be a Banach space.  For each $f\in X^*$ the map $x\mapsto |f(x)|$
is a seminorm on $X$, and the set of all such seminorms, as $f$ varies over
$X^*$, is sufficient by the Hahn-Banach Theorem.  Therefore we can endow
$X$ with a new TVS structure from this family of seminorms.  This is
called the weak topology on $X$.  In particular, $x_n\to x$ weakly
(written $x_n @>w>> x$)
iff $f(x_n)\to f(x)$ for all $f\in X^*$.  Thus the weak topology is
weaker than the norm topology, but all the elements of $X^*$ remain
continuous when $X$ is endowed with the weak topology (it is by definition
the weakest topology for which all the elements of $X^*$ are continuous).

Note that the open sets of the weak topology are rather big.  If
$U$ is an weak neighborhood of $0$ in an infinite dimensional
Banach space then, by definition, there exists $\epsilon>0$ and
finitely many functionals $f_n\in X^*$ such that
$\{\,x\,|\,|f_n(x)|<\epsilon\quad\forall N\,\}$ is contained in $U$.  Thus
$U$ contains the infinite dimensional closed subspace
$\Nu(f_1)\cap\ldots\cap\Nu(f_n)$.

If $x_n @>w>> x$ weakly, then, viewing the $x_n$ as linear functionals
on $X^*$ (via the canonical embedding of $X$ into $X^{**}$), we see that
the sequence of real numbers obtained by applying the $x_n$ to any $f\in
X^*$ is convergent and hence bounded uniformly in $n$.  By the Uniform
Boundedness Principle, it follows that the $x_n$ are bounded.  
\proclaim{Theorem} If a sequence of elements of a Banach space converges
weakly, then the sequence is norm bounded.
\endproclaim

On the other hand, if the $x_n$ are small in norm, then their weak limit
is too.
\proclaim{Theorem} If $x_n@>w>>x$ in some Banach space, then
$\|x\|\le \displaystyle\liminf_{n\to\infty}\|x_n\|$.
\endproclaim
\demo{Proof} Take $f\in X^*$ of norm 1 such that $f(x)=\|x\|$.
Then $f(x_n)\le \|x_n\|$, and taking the $\liminf$ gives the result.
\qed
\enddemo

For convex sets (in particular, for subspaces) weak closure coincides
with norm closure:
\proclaim{Theorem} 1) The weak closure of a convex set is equal
to its norm closure.

2) A convex set is weak closed iff it is
normed closed.

3) A convex set is weak dense iff it is norm dense.
\endproclaim
\demo{Proof} The second and third statement obviously follow from the
first, and the weak closure obviously contains the norm closure.  So it
remains to show that if $x$ does not belong to the norm closure
of a convex set $E$, then there is a weak neighborhood of $x$ which
doesn't intersect $E$.  This follows immediately from the following convex
separation theorem.
\qed
\enddemo

\proclaim{Theorem} Let $E$ be a nonempty closed convex subset of a Banach
space $X$ and $x$ a point in the complement of $E$.  Then there exists
$f\in X^*$ such that $f(x)<\inf_{y\in E} f(y)$.
\endproclaim

In fact we shall prove a stronger result:
\proclaim{Theorem} Let $E$ and $F$ be disjoint, nonempty, convex subsets
of a Banach space $X$ with $F$ open. Then there exists
$f\in X^*$ such that $f(x)<\inf_{y\in E} f(y)$ for all $x\in F$.
\endproclaim
(The previous result follows by taking $F$ to be any ball about
$x$ disjoint from $E$.)
\demo{Proof}  This is a consequence of the generalized Hahn-Banach Theorem.
Pick $x_0\in E$ and $y_0\in F$ and set $z_0=x_0-y_0$ and $G=F-E+z_0$.
Then $G$ is a convex open set containing $0$ but not containing $z_0$.
(The convexity of $G$ follows directly from that of $E$ and $F$; the
fact that $G$ is open follows from the representation of
$G=\bigcup_{y\in E}F-y+z_0$ as a union of open sets; obviously
$0=y_0-x_0+z_0\in G$, and $z_0\notin G$ since $E$ and $F$ and disjoint.)

Since $G$ is open and convex and contains $0$, for each
$x\in X$,
$\{\,t>0\,|\,t^{-1}x\in G\,\}$ is a nonempty open semi-infinite interval.  Define
$p(x)\in [0,\infty)$ to be the left endpoint of this interval.  By
definition $p$ is positively homogeneous.  Since $G$ is convex,
$t^{-1}x\in G$ and $s^{-1}y\in G$ imply that
$$
(t+s)^{-1}(x+y)=\frac t{s+t} t^{-1}x + \frac s{s+t} s^{-1}y\in G,
$$
whence $p$ is subadditive.  Thus $p$ is a sublinear functional.  Moreover,
$G=\{\,x\in X\,|\, p(x)<1\,\}$.

Define a linear functional $f$ on $X_0:=\R z_0$ by $f(z_0)=1$.  Then
$f(tz_0)=t\le t p(z_0)=p(t z_0)$ for $t\ge 0$ and $f(tz_0)<0\le p(t z_0)$ for
$t<0$.  Thus $f$ is a linear functional on $X_0$ satisfying $f(x)\le
p(x)$ there.  By Hahn-Banach we can extend $f$ to a linear functional on
$X$ satisfying the same inequality.  This implies that $f$ is bounded
(by 1) on the open set $G$, so $f$ belongs to $X^*$.

If $x\in F$, $y\in E$, then $x-y+z_0\in G$, so
$f(x)-f(y)+1=f(x-y+z_0)<1$, or $f(x)<f(y)$.  Therefore $\sup_{x\in
F}f(x)\le\inf_{y\in E}f(y)$.  Since $f(F)$ is an open interval,
$f(\tilde x)<\sup_{x\in F}f(x)$ for all $\tilde x\in F$, and so we have the
theorem.
\qed
\enddemo

\goodbreak\subhead The weak* topology \endsubhead\nobreak
%
On the dual space $X^*$ we have two new topologies.  We may endow it
with the weak topology, the weakest one such that all functionals in
$X^{**}$ are continuous, or we may endow it with the topology generated
by all the seminorms $f\mapsto f(x)$, $x\in X$.  (This is obviously
a sufficient family of functionals.)  The last is called the weak*
topology and is a weaker topology than the weak topology.  If $X$ is
reflexive, the weak and weak* topologies coincide.

Examples of weak and weak* convergence:
1) Consider weak convergence in $L^p(\Omega)$ where $\Omega$ is a
bounded subset of $\R^n$.  From the characterization of the dual of
$L^p$ we see that
$$
f_n @>w*>> f \text{ in $L^\infty$}
\implies
f_n @>w>> f \text{ weak in $L^p$}
\implies
f_n @>w>> f \text{ weak in $L^q$}
$$
whenever $1\le q\le p<\infty$.  In particular we claim that the complex
exponentials $e^{2\pi i n x} @>w*>> 0$ in $L^\infty([0,1])$ as
$n\to\infty$.  This is simply the statement that
$$
\lim_{n\to\infty}\int_0^1 g(x)e^{2\pi i n x}\,dx=0,
$$
for all $g\in L^1([0,1])$, i.e., that the Fourier coefficients of an
$L^1$ tend to $0$, which is known as the Riemann--Lebesgue Lemma.
(Proof: certainly true if $g$ is a trigonometric polynomial.  The trig
polynomials are dense in $C([0,1])$ by the Weierstrass Approximation
Theorem, and $C([0,1])$ is dense in $L^1([0,1])$.)  This is one common
example of weak convergence which is not norm convergence, namely
weak vanishing by oscillation.

2) Another common situation is weak vanishing to infinity.  As a very
simple example, it is easy to see that the unit vectors in $l_p$
converge weakly to zero for $1<p<\infty$ (and weak* in $l_\infty$,
but not weakly in $l_1$).  As a more interesting example, let $f_n\in
L^p(\R)$ be a sequence of function which are uniformly bounded in $L^p$,
and for which $f_n|_{[-n,n]}\equiv0$.  Then we claim that $f_n\to 0$
weakly in $L^p$ if $1< p<\infty$.  Thus we have to show that
$$
\lim_{n\to\infty}\int_{\R} f_ng\,dx=0
$$
for all $g\in L^q$.
Let $S_n=\{\,x\in\R\,|\,|x|\ge n\,\}$.  Then $\lim_n\int_{S_n}
|g|^q\,dx=0$ (by the dominated convergence theorem).  But
$$
|\int_{\R}
f_ng\,dx|=|\int_{S_n} f_ng\,dx|\le \|f_n\|_{L^p}\|g\|_{L^q(S_n)} \le
C\|g\|_{L^q(S_n)}\to 0.
$$
The same proof shows that if the $f_n$ are
uniformly bounded they tend to $0$ in $L^\infty$ weak*.  Note that the
characteristic functions $\chi_{[n,n+1]}$ do {\it not}\/ tend to zero
weakly in $L^1$ however.

3) Consider the measure $\phi_n=2n\chi_{[-1/n,1/n]}dx$.  Formally
$\phi_n$ tends to the delta function $\delta_0$ as $n\to\infty$.
Using the weak* topology on $C([-1,1])$ this convergence becomes precise:
$\phi_n @>w*>> \delta_0$.

\proclaim{Theorem (Alaoglu)} The unit ball in $X^*$ is weak* compact.
\endproclaim
\demo{Proof} For $x\in X$, let $I_x=\{\,t\in\R\,:\, |t|\le\|x\|\,\}$,
and set $\Omega=\Pi_{x\in X} I_x$.  Recall that this Cartesian product
is nothing but the set of all functions $f$ on $X$ with $f(x)\in I_x$
for all $x$.  This set is endowed with the Cartesian product topology,
namely the weakest topology such that for all $x\in X$,
the functions $f\mapsto f(x)$ (from $\Omega$ to $I_x$) are continuous.
Tychonoff's Theorem states that $\Omega$ is compact with this topology.

Now let $E$ be the unit ball in $X^*$.  Then $E\subset\Omega$ and
the topology thereby induced on $E$ is precisely the weak* topology.
Now for each pair $x,y\in X$ and each $c\in R$, define
$F_{x,y}(f)=f(x)+f(y)-f(x+y)$, $G_{x,c}=f(c x)-cf(x)$.  These are
continuous functions on $\Omega$ and
$$
E=\bigcap_{x,y\in X}F_{x,y}^{-1}(0)\cap\bigcap\Sb x\in X\\c\in\R\endSb 
G_{c,x}^{-1}(0).
$$
Thus $E$ is a closed subset of a compact set, and therefore compact itself.
\qed
\enddemo

\proclaim{Corollary}  If $f_n@>w^*>>f$ in $X^*$, then
$\|f\|\le\displaystyle\liminf_{n\to\infty} \|f_n\|_{X^*}$.
\endproclaim
\demo{Proof} Let $C=\liminf\|f_n\|$ and let $\epsilon>0$ be arbitrary.
Then there exists a subsequence (also denoted $f_n$) with
$\|f_n\|\le C+\epsilon$.  The ball of radius $C+\epsilon$ being
weak* compact, and so weak* closed, $\|f\|<C+\epsilon$.  Since
$\epsilon$ was arbitrary, this gives the result.
\qed
\enddemo

On $X^{**}$ the weak* topology is that induced by the functionals
in $X^*$.

\proclaim{Theorem} The unit ball of $X$ is weak* dense
in the unit ball of $X^{**}$.
\endproclaim
\demo{Proof} Let $z$ belong to the unit ball of $X^{**}$.  We need to show
that for any $f_1,\ldots,f_n\in X^*$ of norm $1$, and any $\epsilon>0$, the set
$$
\{\,w\in X^{**}\,|\,|(w-z)(f_i)|<\epsilon,\quad i=1,\ldots,n\,\}
$$
contains a point of the unit ball of $X$.  (Since any neighborhood of $z$
contains a set of this form.)

It is enough to show that there exists $y\in X$
with $\|y\|<1+\epsilon$ such that $(y-z)(f_i)=0$ for each $i$.  Because
then $y/(1+\epsilon)$ belongs to the closed unit ball of $X$, and
$$
|((1+\epsilon)^{-1}y-z)(f_i)|=|((1+\epsilon)^{-1}y-y)(f_i)|
\le \|((1+\epsilon)^{-1}y-y\|=\|y\|\frac{\epsilon}{1+\epsilon}<\epsilon.
$$

Let $S$ be the span of the $f_i$ in $X^*$.  Since $S$ is finite
dimensional the canonical map $X\to S^*$ is surjective.  (This is equivalent
to saying that if the null space of a linear functional $g$ contains the
intersection of the null spaces of a finite set of linear functionals $g_i$,
then $g$ is a linear combination of the $g_i$, which is a simple, purely
algebraic result. [Proof: The nullspace of the map $(g_1,\ldots,g_n):X\to\R^n$
is contained in the nullspace of $g$, so $g=T\circ(g_1,\ldots,g_n)$ for
some linear $T:\R^n\to\R$.])  Consequently $X/{}^aS$ is isometrically
isomorphic to $S$.

In particular $z|_S$ concides with $y+{}^aS$ for some $y\in X$.  Since
$\|z\|_{S^*}\le 1$, and we can choose the coset representative $y$
with $\|y\|\le 1+\epsilon$ as claimed.
\qed
\enddemo

\proclaim{Corollary} The closed unit ball of a Banach space $X$ is weakly
compact if and only if $X$ is reflexive.
\endproclaim
\demo{Proof}  If the closed unit ball of $X$ is weakly compact, then
it is weak* compact when viewed as a subset of $X^{**}$.  Thus the ball
is weak* closed, and so, by the previous theorem, the embedding of
the the unit ball of $X$ contains the ball of $X^{**}$.  It follows
that the embedding of $X$ is all of $X^{**}$.

The reverse direction is immediate from the Alaoglu theorem.\qed
\enddemo

\goodbreak\head V. Compact Operators and their Spectra \endhead\nobreak
\subhead Hilbert--Schmidt operators \endsubhead\nobreak
%
\proclaim{Lemma} Suppose that $\{\,e_i\,\}$ and $\{\,\tilde e_i\,\}$ are two
orthonormal bases for a separable Hilbert space $X$, and $T\in B(X)$.  Then
$$
\sum_{i,j}|\langle Te_i,e_j\rangle|^2 = \sum_{i,j}|\langle T\tilde e_i,\tilde e_j\rangle|^2.
$$
\endproclaim
\demo{Proof} For all $w\in X$, $\sum_j|\langle w,e_j\rangle|^2=\|w\|^2$, so
$$
\sum_{i,j}|\langle Te_i,e_j\rangle|^2 =\sum_i\|Te_i\|^2=\sum_j\|T^*e_j\|^2.
$$
But
$$
\sum_i\|T^*e_i\|^2=\sum_{i,j}|\langle T^*e_i,\tilde e_j\rangle|^2=\sum_j\|T\tilde e_j\|^2.
\qed
$$
\enddemo

\definition{Definition} If $T\in B(X)$ define $\|T\|_2$ by
$$
\|T\|_2^2=\sum_{i,j}|\langle Te_i,e_j\rangle|^2=\sum_i\|Te_i\|^2
$$
where $\{\,e_i\,\}$ is
any orthonormal basis for $X$.  $T$ is called a Hilbert--Schmidt operator
if $\|T\|_2<\infty$, and $\|T\|_2$ is called the Hilbert--Schmidt norm of
$T$.
\enddefinition

We have just seen that if $T$ is Hilbert--Schmidt, then so is $T^*$ and
their Hilbert--Schmidt norms coincide.

\proclaim{Proposition} $\|T\|\le \|T\|_2$.
\endproclaim
\demo{Proof}
Let $x=\sum c_ie_i$ be an arbitrary element of $X$.  Then
$$
\|Tx\|^2=\sum_i\Bigl|\sum_j c_j\langle T e_j,e_i\rangle\Bigr|^2.
$$
By Cauchy--Schwarz
$$
|\sum_j c_j\langle T e_j,e_i\rangle|^2
\le \sum_j c_j^2 \cdot \sum_j|\langle T e_j,e_i\rangle|^2
=\|x\|^2\sum_j|\langle T e_j,e_i\rangle|^2.
$$
Summing on $i$ gives the result.
\qed
\enddemo

\proclaim{Proposition} Let $\Omega$ be an open subset of $\R^n$
and $K\in L^2(\Omega\x\Omega)$.  Define
$$
T_K u(x)=\int_\Omega K(x,y)u(y)\,dy,\qquad\text{for all $x\in \Omega$}.
$$
Then $T_K$ defines a Hilbert--Schmidt operator on $L^2(\Omega)$ and
$\|T_K\|_2 = \|K\|_{L^2}$.
\endproclaim
\demo{Proof}  For $x\in\Omega$, set $K_x(y)=K(x,y)$.  By Fubini's theorem,
$K_x\in L^2(\Omega)$ for almost all $x\in\Omega$, and
$$
\|K\|_{L^2}^2=\int\|K_x\|^2\,dx.
$$
Now, $T_K u(x)=\langle K_x,u\rangle$, so, if
$\{\,e_i\,\}$ is an orthonormal basis, then
$$
\multline
\|T_K\|_2^2=\sum_i\|T_Ke_i\|^2=\sum_i\int |(T_Ke_i)(x)|^2\,dx
=\sum_i\int |\langle K_x,e_i\rangle|^2\,dx
\\
=\int\sum_i|\langle K_x,e_i\rangle|^2\,dx
=\int\|K_x\|^2\,dx=\|K\|_{L^2}^2.
\endmultline
$$
\qed
\enddemo

\goodbreak\subhead Compact operators \endsubhead\nobreak
%
\definition{Definition} A bounded linear operator between
Banach spaces is called compact if it maps the unit
ball (and therefore every bounded set) to a precompact set.
\enddefinition

For example, if $T$ has finite rank ($\dim\Ra(T)<\infty$), then
$T$ is compact.

Recall the following characterization of precompact sets in
a metric space, which is often useful.

\proclaim{Proposition} Let $M$ be a metric space.  Then
the following are equivalent:
\roster
\item $M$ is precompact.
\item For all $\epsilon>0$ there exist finitely many sets of
diameter at most $\epsilon$ which cover $M$.
\item Every sequence contains a Cauchy subsequence.
\endroster
\endproclaim
\demo{Sketch of proof} (1) $\implies$ (2) and (3) $\implies$ (1) are
easy.  For (2) $\implies$ (3) use a Cantor diagonalization argument
to extract a Cauchy subsequence.\qed
\enddemo

\proclaim{Theorem} Let $X$ and $Y$ be Banach spaces and $B_c(X,Y)$
the space of compact linear operators from $X$ to $Y$.  Then
$B_c(X,Y)$ is a closed subspace of $B(X,Y)$.
\endproclaim
\demo{Proof} Suppose $T_n\in B_c(X,Y)$, $T\in B(X,Y)$, $\|T_n-T\|\to 0$.
We must show that $T$ is compact.  Thus we must show that $T(E)$ is precompact
in $Y$, where $E$ is the unit ball in $X$.  For this, it is enough to show
that for any $\epsilon>0$ there are finitely many balls $U_i$ of radius $\epsilon$
in $Y$ such that
$$
T(E)\subset \bigcup_i U_i.
$$
Choose $n$ large enough that $\|T-T_n\|\le\epsilon/2$, and let $V_1,V_2,\ldots,
V_n$ be finitely many balls of radius $\epsilon/2$ which cover $T_nE$.
For each $i$ let $U_i$ be the ball of radius $\epsilon$ with the same
center as $V_i$.\qed
\enddemo

It follows that closure of the finite rank operators in $B(X,Y)$ is
contained in $B_c(X,Y)$.  In general, this may be a strict inclusion,
but if $Y$ is a Hilbert space, it is equality.  To prove this,
choose an orthonormal basis for $Y$, and consider the finite rank
operators of the form $PT$ where $P$ is the orthogonal projection
of $Y$ onto the span of finitely many basis elements.  Using the fact
that $TE$ is compact ($E$ the unit ball of $X$) and that $\|P\|=1$,
we can find for any $\epsilon>0$, an operator $P$ of this form with
$\sup_{x\in E}\|(PT-T)x\|\le\epsilon$.

The next result is obvious but useful.
\proclaim{Theorem}  Let $X$ and $Y$ be Banach spaces and $T\in B_c(X,Y)$.
If $Z$ is another Banach space and $S\in B(Y,Z)$ then $ST$ is compact.
If $S\in B(Z,X)$, then $TS$ is compact.  If $X=Y$, then $B_c(X):=B_c(X,X)$
is a two-sided ideal in $B(X)$.
\endproclaim

\proclaim{Theorem} Let $X$ and $Y$ be Banach spaces and $T\in B(X,Y)$.
Then $T$ is compact if and only if $T^*$ is compact.
\endproclaim
\demo{Proof} Let $E$ be the unit ball in $X$ and $F$ the unit ball in
$Y^*$.  Suppose that $T$ is compact.  Given $\epsilon>0$ we must exhibit
finitely many sets of diameter at most $\epsilon$ which cover $T^*F$.
First choose $m$ sets of diameter at most $\epsilon/3$ which cover
$TE$, and let $Tx_i$ belong to the $i$th set.  Also, let $I_1,\ldots,I_n$
be $n$ intervals of length $\epsilon/3$
which cover the interval $[-\|T\|,\|T\|]$.  For any $m$-tuple
$(j_1,\ldots,j_m)$ of integers with $1\le j_i\le n$ we define the
set
$$
\{\,f\in F\,|\,f(Tx_i)\in I_{j_i},\quad i=1,\ldots,m\,\}.
$$
These sets clearly cover $F$, so there images under $T^*$ cover $T^*F$,
so it suffices to show that the images have diameter at most $\epsilon$.
Indeed, if $f$ and $g$ belong to the set above, and $x$ is any element
of $E$, pick $i$ such that $\|Tx-Tx_i\|\le\epsilon/3$.  We know
that $\|f(Tx_i)-g(Tx_i)\|\le \epsilon/3$.  Thus
$$
\multline
|(T^*f-T^*g)(x)|=|(f-g)(Tx)|
\\
\le |f(Tx)-f(Tx_i)|+|g(Tx)-g(Tx_i)|+|(f-g)(Tx_i)|
\le \epsilon.
\endmultline
$$

This shows that $T$ compact $\implies$ $T^*$ compact.  Conversely,
suppose that $T^*:Y^*\to X^*$ is compact.  Then $T^{**}$ maps the unit
ball of $X^{**}$ into a precompact subset of $Y^{**}$.  But the unit
ball of $X$ may be viewed as a subset of the unit ball of its bidual,
and the restriction of $T^{**}$ to the unit ball of $X$ coincides with
$T$ there.  Thus $T$ maps the unit ball of $X$ to a precompact set.
\qed
\enddemo

\proclaim{Theorem} If $T$ is a compact operator from a Banach space to itself,
then $\Nu(\id -T)$ is finite dimensional and $\Ra(\id-T)$ is closed.
\endproclaim
\demo{Proof} $T$ is a compact operator that restricts to the identity
on $\Nu(\id -T)$.  Hence the closed unit ball in $\Nu(\id -T)$ is compact,
whence the dimension of $\Nu(\id -T)$ is finite.

Now any finite dimensional subspace is complemented (see below), so there
exists a closed subspace $M$ of $X$ such that $\Nu(\id -T)+M=X$ and
$\Nu(\id -T)\cap M=0$.  Let $S=(\id-T)|_M$, so $S$ is injective and
$\Ra(S)=\Ra(\id-T)$.
We will show that for some $c>0$, $\|Sx\|\ge c\|x\|$ for all $x\in M$,
which will imply that $\Ra(S)$ is closed.  If the desired inequality
doesn't hold for any $c>0$, we can choose $x_n\in M$ of norm 1 with
$Sx_n\to 0$.  After passing to a subsequence we may arrange that also $Tx_n$
converges to some $x_0\in X$.  It follows that $x_n\to x_0$, so $x_0\in M$
and $Sx_0=0$.  Therefore $x_0=0$ which is impossible (since $\|x_n\|=1$).
\qed
\enddemo

In the proof we used the first part of the following lemma.  We say that
a closed subspace $N$ is complemented in a Banach space $X$ if there
is another closed subspace such that $M\oplus N=X$.

\proclaim{Lemma} A finite dimensional or finite codimensional closed subspace
of a Banach space is complemented.
\endproclaim
\demo{Proof}  If $M$ is a finite dimensional subspace, choose a basis
$x_1,\ldots,x_n$ and define a linear functionals $\phi_i:M\to\R$ by
$\phi_i(x_j)=\delta_{ij}$.  Extend the $\phi_i$ to be bounded linear
functionals on $X$.  Then we can take $N=\Nu(\phi_1)\cap\ldots\cap\Nu(\phi_n)$.

If $M$ is finite codimensional, we can take $N$ to be the span
of a set of nonzero coset representatives.
\qed
\enddemo

A simple generalization of the theorem will be useful when we study the
spectrum of compact operators.
\proclaim{Theorem} If $T$ is a compact operator from a Banach space to itself,
$\lambda$ a non-zero complex number, and $n$ a positive integer, then
$\Nu[(\lambda\id-T)^n]$ is finite dimensional and
$\Ra[(\lambda\id-T)^n]$ is closed.
\endproclaim
\demo{Proof} Expanding we see that $(\lambda\id-T)^n=\lambda^n(\id-S)$
for some compact operator $S$, so the result reduces to the previous
one.
\qed
\enddemo

We close the section with a good source of examples of compact operators,
which includes, for example, any matrix operator on $l_2$ for which
the matrix entries are square-summable.

\proclaim{Theorem} A Hilbert--Schmidt operator on a separable Hilbert space is
compact.
\endproclaim
\demo{Proof}  Let $\{\,e_i\,\}$ be an orthonormal basis.  Let $T$ be a given
Hilbert--Schmidt operator (so $\sum_i \|Te_i\|^2<\infty$).  Define
$T_n$ by $T_ne_i=Te_i$ if $i\le n$, $T_ne_i=0$ otherwise.  Then
$\|T-T_n\|\le \|T-T_n\|_2=\sum_{i=n+1}^\infty\|Te_i\|^2\to 0$.\qed
\enddemo

\goodbreak\subhead Spectral Theorem for compact self-adjoint operators \endsubhead\nobreak
%
In this section we assume that $X$ is a {\it complex}\/ Hilbert space.
If $T:X\to X$ is a bounded linear operator, we view $T^*$ as a map
from $X\to X$ via the Riesz isometry between $X$ and $X^*$.  That is,
$T^*$ is defined by
$$
\langle T^*x,y\rangle=\langle x,Ty\rangle.
$$
In the case of a finite dimensional complex Hilbert space, $T$ can be represented
by a complex square matrix, and $T^*$ is represented by its Hermitian
transpose.

Recall that a Hermitian symmetric matrix has real eigenvalues
and an orthonormal basis of eigenvectors.  For a self-adjoint operator
on a Hilbert space, it is easy to see that any eigenvalues are real,
and that eigenvectors corresponding to distinct eigenvalues are orthogonal.
However there may not exist an orthonormal basis of eigenvectors,
or even any nonzero eigenvectors at all.  For example,
let $X=L^2([0,1])$, and define $Tu(x)=x\,u(x)$ for $u\in L^2$.  Then
$T$ is clearly bounded and self-adjoint.  But it is easy to see
that $T$ does not have any eigenvalues.

\proclaim{Spectral Theorem for Compact Self-Adjoint Operators in Hilbert Space}
Let $T$ be a compact self-adjoint operator in a Hilbert space $X$.  Then
there is an orthonormal basis consisting of eigenvectors of $T$.
\endproclaim

Before proceeding to the proof we prove one lemma.

\proclaim{Lemma} If $T$ is a self-adjoint operator on a Hilbert space,
then
$$
\|T\|=\sup_{\|x\|\le1}|\langle Tx,x\rangle|.
$$
\endproclaim
\demo{Proof} Let $\alpha=\sup_{\|x\|\le1}|\langle Tx,x\rangle|$.  It is
enough to prove that
$$
|\langle Tx,y\rangle|\le\alpha\|x\|\|y\|
$$
for all $x$ and $y$.  We can obviously assume that $x$ and $y$
are nonzero.  Moreover, we may multiply $y$ by a complex number of
modulus one, so we can assume that $\langle Tx,y\rangle\ge0$.
Then
$$
\langle T(x+y),x+y\rangle - \langle T(x-y),x-y\rangle
=4\Re\langle Tx,y\rangle=4|\langle Tx,y\rangle|.
$$
so
$$
|\langle Tx,y\rangle|\le \frac\alpha4(\|x+y\|^2+\|x-y\|^2)
=\frac\alpha2(\|x\|^2+\|y\|^2).
$$
Now apply this result with $x$ replaced by $\sqrt{\|y\|/\|x\|}\,x$
and $y$ replaced by $\sqrt{\|x\|/\|y\|}\,y$.
\qed
\enddemo

\demo{Proof of spectral theorem for compact self-adjoint operators}
We first show that $T$ has a nonzero eigenvector.  If $T=0$, this
is obvious, so we assume that $T\ne0$.  Choose
a sequence $x_n\in X$ with $\|x_n\|=1$ so that $|\langle Tx_n,x_n\rangle|
\to \|T\|$.  Since $T$ is self-adjoint, $\langle Tx_n,x_n\rangle\in\R$,
so we may pass to a subsequence (still denoted $x_n$), for which
$\langle Tx_n,x_n\rangle\to\lambda=\pm\|T\|$.
Since $T$ is compact we may pass to a further subsequence
and assume that $Tx_n\to y\in X$.  Note that $\|y\|\ge|\lambda|>0$.

Using the fact that $T$ is self-adjoint and $\lambda$ is real, we get
$$
\align
\|Tx_n-\lambda x_n\|^2&=\|Tx_n\|^2-2\lambda\langle Tx_n,x_n\rangle
+\lambda^2\|x_n\|^2
\\
&\le 2\|T\|^2-2\lambda\langle Tx_n,x_n\rangle\to 2\|T\|^2-2\lambda^2=0.
\endalign
$$
Since $Tx_n\to y$ we infer that $\lambda x_n\to y$ as well, or $x_n\to
y/\lambda\ne0$.  Applying $T$ we have $Ty/\lambda=y$, so $\lambda$ is
indeed a nonzero eigenvalue.

To complete the proof, consider the set of all orthonormal subsets
of $X$ consisting of eigenvectors of $T$.  By Zorn's lemma, it
has a maximal element $S$.  Let $W$ be the closure of the span
of $S$.  Clearly $TW\subset W$, and it follows
directly (since $T$ is self-adjoint), that $TW^\perp\subset W^\perp$.
Therefore $T$ restricts to a self-adjoint operator on $W^\perp$ and
thus, unless $W^\perp=0$, $T$ has an eigenvector in $W^\perp$.  But
this clearly contradicts the maximality of $S$ (since we can
adjoin this element to $S$ to get a larger orthonormal set of
eigenvectors).  Thus $W^\perp=0$, and $S$ is an orthonormal basis.
\qed
\enddemo

The following structure result on the set of eigenvalues is generally
considered part of the spectral theorem as well.

\proclaim{Theorem}  If $T$ is a compact self-adjoint operator on a Hilbert space,
then the set of nonzero eigenvalues of $T$ is either a finite set or a
sequence approaching $0$ and the corresponding eigenspaces are all finite
dimensional.
\endproclaim
\remark{Remark} $0$ may or may not be an eigenvalue, and its eigenspace
may or may not be finite.
\endremark
\demo{Proof} Let $e_i$ be an orthonormal basis of eigenvectors, with
$Te_i=\lambda_ie_i$.  Here $i$ ranges over some index set $I$.
It suffices to show that
$S=\{\,i\in I\,|\,|\lambda_i|\ge\epsilon\,\}$ is finite for all $\epsilon>0$.
Then if $i,j\in I$
$$
\|Te_i-Te_j\|^2=\|\lambda_ie_j-\lambda_je_j\|^2=|\lambda_i|^2+|\lambda_j|^2,
$$
so if $i,j\in S$, then $\|Te_i-Te_j\|^2\ge 2\epsilon^2$.  If $S$ were infinite,
we could then choose a sequence of unit elements in $X$ whose
image under $T$ has no convergent subsequence, which violates the
compactness of $T$.
\qed
\enddemo

Suppose, for concreteness, that $X$ is an infinite dimensional
separable Hilbert space and that $\{\,e_n\,\}_{n\in\N}$ is an
orthonormal basis adapted to a compact self-adjoint operator $T$ on
$X$.  Then the map $U:X\to l_2$ given by
$$
U(\sum_n c_ne_n)=(c_0,c_1,\ldots),
$$
is an isometric isomorphism.  Moreover, when we use this map to transfer
the action of $T$ to $l_2$, i.e., when we consider the operator
$UTU^{-1}$ on $l_2$, we see that this operator is simply multiplication
by the bounded sequence $(\lambda_0,\lambda_1,\ldots)\in l_\infty$.
Thus the spectral theorem says that every compact self-adjoint $T$ is
unitarily equivalent to a multiplication operator on $l_2$.  (An
isometric isomorphism of Hilbert spaces is also called a unitary
operator.  Note that it is characterized by the property $U^*=U^{-1}$.)

A useful extension is the spectral theorem for commuting self-adjoint
compact operators.

\proclaim{Theorem} If $T$ and $S$ are self-adjoint compact operators in
a Hilbert space $H$ and $TS=ST$, then there is an orthonormal basis of
$X$ whose elements are eigenvectors for both $S$ and $T$.
\endproclaim
\demo{Proof} For an eigenvalue $\lambda$ of $T$, let $X_\lambda$ denote
the corresponding eigenspace of $T$.  If $x\in X_\lambda$, then
$TSx=STx=\lambda Sx$, so $Sx\in X_\lambda$.  Thus $S$ restricts to
a self-adjoint operator on $X_\lambda$, and so there is an orthonormal
basis of $S$--eigenvectors for $X_\lambda$.  These are $T$--eigenvectors
as well.  Taking the union over all the eigenvalues $\lambda$ of $T$
completes the construction. \qed
\enddemo

Let $T_1$ and $T_2$ be any two self-adjoint operators and set
$T=T_1+iT_2$.  Then $T_1=(T+T^*)/2$ and $T_2=(T-T^*)/(2i)$.
Conversely, if $T$ is any element of $B(X)$, then we can define two
self-adjoint operators from these formulas and have $T=T_1+iT_2$.  Now
suppose that $T$ is compact and also {\it normal}, i.e., that $T$ and
$T^*$ commute.  Then $T_1$ and $T_2$ are compact and commute, and hence
we have an orthonormal basis whose elements are eigenvectors for both
$T_1$ and $T_2$, and hence for $T$.  Since the real and imaginary parts
of the eigenvalues are the eigenvalues of $T_1$ and $T_2$, we again see
that the eigenvalues form a sequence tending to zero and all have finite
dimensional eigenspaces.

We have thus shown that a compact normal operator admits an orthonormal
basis of eigenvectors.  Conversely, if $\{\,e_i\,\}$ is an orthonormal
basis of eigenvectors of $T$, then $\langle T^*e_i,e_j\rangle=0$
if $i\ne j$, which implies that each $e_i$ is also an eigenvector for
$T^*$.  Thus $T^*Te_i=TT^*e_i$ for all $i$, and it follows easily that
$T$ is normal.  We have thus shown:

\proclaim{Spectral Theorem for compact normal operators}
Let $T$ be a compact operator on a Hilbert space $X$.  Then there exists an
orthonormal basis for $X$ consisting of eigenvectors of $T$ if and only
if $T$ is normal.  In this case, the set of nonzero eigenvalues form
a finite set or a sequence tending to zero and the eigenspaces corresponding
to the nonzero eigenvalues are finite dimensional.  The eigenvalues are
all real if and only if the operator is self-adjoint.
\endproclaim

\goodbreak\subhead The spectrum of a general compact operator \endsubhead\nobreak
%
In this section we derive the structure of the spectrum of a compact operator
(not necessarily self-adjoint or normal) on a complex Banach space $X$.

For any operator $T$ on a complex Banach space, the {\it resolvent
set}\/ of $T$, $\rho(T)$ consists of those $\lambda\in\C$ such
that $T-\lambda\id$ is invertible, and the spectrum $\sigma(T)$
is the complement.  If $\lambda\in\sigma(T)$, then $T-\lambda\id$
may fail to be invertible in several ways.  (1) It may be that
$\Nu(T-\lambda\id)\ne0$, i.e., that $\lambda$ is an eigenvalue of $T$.
In this case we say that $\lambda$ belongs to the {\it point spectrum}\/
of $T$, denoted $\sigma_p(T)$.  (2) If $T-\lambda\id$ is injective, it
may be that its range is dense but not closed in $X$.  In this case
we say that $\lambda$ belongs to the {\it continuous spectrum}\/ of
$T$, $\sigma_c(T)$.  Or (3) it may be that $T-\lambda\id$ is injective
but that its range is not even dense in $X$.  This is the {\it residual
spectrum}, $\sigma_r(T)$.  Clearly we have a decomposition of $\C$
into the disjoint sets $\rho(T)$, $\sigma_p(T)$, $\sigma_c(T)$, and
$\sigma_r(T)$.  As an example of the continuous spectrum, consider
the operator $Te_n=\lambda_ne_n$ where the $e_n$ form an orthonormal
basis of a Hilbert space and the $\lambda_n$ form a positive sequence
tending to $0$.  Then $0\in\sigma_c(T)$.  If $Te_n=\lambda_ne_{n+1}$,
$0\in\sigma_r(T)$.

Now if $T$ is compact and $X$ is infinite dimensional, then
$0\in\sigma(T)$ (since if $T$ were invertible, the image of the unit ball
would contain an open set, and so couldn't be precompact).  From the
examples just given, we see that $0$ may belong to the point spectrum,
the continuous spectrum, or the residual spectrum.  However, we shall
show that all other elements of the spectrum are eigenvalues, i.e., that
$\sigma(T)=\sigma_p(T)\cup\{0\}$, and that, as in the normal case, the
point spectrum consists of a finite set or a sequence approaching zero.

The structure of the spectrum of a compact operator will be deduced from
two lemmas.  The first is purely algebraic.  To state it we need
some terminology: consider a linear operator $T$ from a
vector space $X$ to itself, and consider the chains of subspaces
$$
0=\Nu(\id)\subset\Nu(T)\subset\Nu(T^2)\subset\Nu(T^3)\subset\cdots.
$$
Either this chain is strictly increasing forever, or there is a least
$n\ge0$ such that $\Nu(T^n)=\Nu(T^{n+1})$, in which case only the first $n$
spaces are distinct and all the others equal the $n$th one.  In the latter
case we say that the kernel chain for $T$ {\it stabilizes}\/ at $n$.
In particular, the kernel chain stabilizes at $0$ iff $T$ is injective.
Similarly we may consider the chain
$$
X=\Ra(\id)\supset\Ra(T)\supset\Ra(T^2)\supset\Ra(T^3)\supset\cdots,
$$
and define what it means for the range chain to stabilize at $n>0$.
(So the range stabilizes at $0$ iff $T$ is surjective.)
It could happen that neither or only one of these chains stabilizes.
However:

\proclaim{Lemma} Let $T$ be a linear operator from a vector space $X$
to itself.  If the kernel chain stabilizes at $m$ and the range chain
stabilizes at $n$, then $m=n$ and $X$ decomposes as the direct sum of
$\Nu(T^n)$ and $\Ra(T^n)$.
\endproclaim
\demo{Proof}   Suppose $m$ were less than $n$.  Since the range chain
stabilizes at $n$, there exists $x$ with $T^{n-1}x\notin\Ra(T^n)$, and then
there exists $y$ such that $T^{n+1}y=T^nx$.  Thus $x-Ty\in\Nu(T^n)$, and,
since kernel chain stabilizes at $m<n$, $\Nu(T^n)=\Nu(T^{n-1})$.
Thus $T^{n-1}x=T^ny$, a contradiction.  Thus $m\ge n$.  A similar argument,
left to the reader, establishes the reverse inequality.

Now if $T^nx\in\Nu(T^n)$, then $T^{2n}x=0$, whence $T^nx=0$.  Thus
$\Nu(T^n)\cap\Ra(T^n)=0$.  Given $x$, let $T^{2n}y=T^nx$, so $x$ decomposes
as $T^ny\in\Ra(T^n)$ and $x-T^ny\in\Nu(T^n)$.
\qed
\enddemo

The second lemma brings in the topology of compact operators.

\proclaim{Lemma}  Let $T:X\to X$ be a compact operator on a
Banach space and $\lambda_1,\lambda_2,\ldots$ a sequence of complex
numbers with $\inf|\lambda_n|>0$.  Then the following is {\it impossible}:
There exists a strictly increasing chain of closed subspaces
$S_1\subset S_2\subset\cdots$ with $(\lambda_n\id-T)S_n\subset S_{n-1}$
for all $n$.
\endproclaim
\demo{Proof}  Suppose such a chain exists.  Note that each $TS_n\subset S_n$
for each $n$.  Since $S_n/S_{n-1}$ contains an element of norm $1$,
we may choose $y_n\in S_n$ with $\|y_n\|\le 2$,
$\operatorname{dist}(y_n,S_{n-1})=1$.  If $m<n$, then
$$
z:=\frac{Ty_m-(\lambda_n\id-T)y_n}{\lambda_n}\in S_{n-1},
$$
and
$$
\|Ty_m-Ty_n\|=|\lambda_n|\|y_n-z_n\|\ge |\lambda_n|.
$$
This implies that the sequence $(Ty_n)$ has no Cauchy subsequence,
which contradicts the compactness of $T$.\qed
\enddemo

We are now ready to prove the result quoted at the beginning of the subsection.
\proclaim{Theorem} Let $T$ be a compact operator on a Banach space $X$.  Then
any nonzero element of the spectrum of $T$ is an eigenvalue.  Moreover
$\sigma(T)$ is either finite or a sequence approaching zero.
\endproclaim
\demo{Proof} Consider the subspace chains $\Nu[(\lambda\id-T)^n]$ and
$\Ra[(\lambda\id-T)^n]$ (these are closed subspaces by a previous
result).  Clearly $\lambda\id-T$ maps $\Nu[(\lambda\id-T)^n]$
into $\Nu[(\lambda\id-T)^{n-1}]$, so the previous lemma
implies that the kernel chain stabilizes, say at $n$.  Now
$\Ra[(\lambda\id-T)^n]={}^a\Nu[(\lambda\id-T^*)^n]$ (since the range
is closed), and since these last stabilize, the range chain stabilizes
as well.

Thus we have $X=\Nu[(\lambda\id-T)^n]\oplus\Ra[(\lambda\id-T)^n]$.
Thus
$$
\Ra(\lambda\id-T)\ne X \implies
\Ra(\lambda\id-T)^n\ne X  \implies
\Nu(\lambda\id-T)^n\ne 0 \implies
\Nu(\lambda\id-T)\ne 0.
$$
In other words $\lambda\in\sigma(T) \implies\lambda\in\sigma_p(T)$.

Finally we prove the last statement.  If it were false we could find
a sequence of eigenvalues $\lambda_n$ with $\inf|\lambda_n|>0$.
Let $x_1,x_2,\ldots$ be corresponding nonzero eigenvectors and
set $S_n=\operatorname{span}[x_1,\ldots,x_n]$.  These form a
strictly increasing chain of subspaces (recall that eigenvectors
corresponding to distinct eigenvalues are linearly independent) and
$(\lambda_n\id-T)S_n\subset S_{n-1}$, which contradicts the lemma.
\qed
\enddemo

The above reasoning also gives us the Fredholm alternative:
\proclaim{Theorem} Let $T$ be a compact operator on a Banach space $X$
and $\lambda$ a nonzero complex number.  Then either (1) $\lambda\id-T$ is an isomorphism,
or (2) it is neither injective nor surjective.
\endproclaim
\demo{Proof}  Since the kernel chain and range chain for $S=\lambda\id-T$
stabilize, either they both stabilize at $0$, in which case $S$ is injective
and surjective, or neither does, in which case it is neither.\qed
\enddemo

We close this section with a result which is fundamental to the
study of Fredholm operators.
\proclaim{Theorem} Let $T$ be a compact operator on a Banach space $X$
and $\lambda$ a nonzero complex number.  Then
$$
\dim\Nu(\lambda\id-T)=\dim\Nu(\lambda\id-T^*)
=\operatorname{codim}\Ra(\lambda\id-T)=\operatorname{codim}\Ra(\lambda\id-T^*).
$$
\endproclaim
\demo{Proof}  Let $S=\lambda\id-T$.  Since $\Ra(S)$ is closed
$$
[X/\Ra(S)]^*\cong \Ra(S)^a=\Nu(S^*).
$$
Thus $[X/\Ra(S)]^*$ is finite dimensional, so $X/\Ra(S)$ is finite dimensional,
and these two spaces are of the same dimension.
Thus $\operatorname{codim}\Ra(S)=\dim\Nu(S^*)$.

For a general operator $S$ we only have $\overline{\Ra(S^*)}\subset\Nu(S)^a$,
but, as we now show, when $\Ra(S)$ is closed, $\Ra(S^*)=\Nu(S)^a$.  Indeed,
$S$ induces an isomorphism of $X/\Nu(S)$ onto $\Ra(S)$, and for any
$f\in \Nu(S)^a$, $f$ induces a map $X/\Nu(S)$ to $\R$.  It follows
that $f=gS$ for some bounded linear operator $g$ on $\Ra(S)$, which
can be extended to an element of $X^*$ by Hahn-Banach.  But
$f=gS$ simply means that $f=S^*g$, showing that $\Nu(S)^a\subset\Ra(S^*)$
(and so equality holds) as claimed.

Thus
$$
\Nu(S)^*\cong X^*/\Nu(S)^a = X^*/\Ra(S^*),
$$
so $\operatorname{codim}\Ra(S^*)=\dim\Nu(S)^*=\dim\Nu(S)$.

We complete the theorem by showing that
$\dim\Nu(S)\le\operatorname{codim}\Ra(S)$ and
$\dim\Nu(S^*)\le\operatorname{codim}\Ra(S^*)$.  Indeed,
since $\Ra(S)$ is closed with finite codimension, it is complemented
by a finite dimensional space $M$ (with $\dim M=\operatorname{codim}\Ra(S)$.
Since $\Nu(S)$ is finite dimensional, it is complemented by a space $N$.
Let $P$ denote the projection of $X$ onto $\Nu(S)$ which is a bounded
map which to the identity on $\Nu(S)$ and to zero on $N$.
Now if $\operatorname{codim}\Ra(S)<\dim\Nu(S)$, then there is a linear
map of $\Nu(S)$ onto $M$ which is not injective.  But then $T-fP$ is
a compact operator and $\lambda\id-T+fP$ is easily seen to be surjective.
By the Fredholm alternative, it is injective as well.  This implies
that $f$ is injective, a contradiction.  We have thus shown that
$\dim\Nu(S)\le\operatorname{codim}\Ra(S)$.  Since $T^*$ is compact, the
same argument shows that $\dim\Nu(S^*)\le\operatorname{codim}\Ra(S^*)$.
This completes the proof.
\qed
\enddemo

\goodbreak\head VI. Introduction to General Spectral Theory \endhead\nobreak

In this section we skim the surface of the spectral theory for a
general (not necessarily compact) operator on a Banach space,
before encountering a version of the Spectral Theorem for
a bounded self-adjoint operator in Hilbert space.  Our first
results don't require the full structure of in the algebra
of operators on a Banach space, but just an arbitrary
Banach algebra structure, and so we start there.

\goodbreak\subhead The spectrum and resolvent in a Banach algebra\endsubhead\nobreak
Let $X$ be a Banach algebra with an identity element denoted $\id$.  We
assume that the norm in $X$ has been normalized so that $\|\id\|=1$.
The two main examples to bear in mind are (1) $B(X)$, where $X$ is some
Banach space; and (2) $C(G)$ endowed with the sup norm, where $G$ is
some compact topological space, the multiplication is just pointwise
multiplication of functions, and $\id$ is the constant function $1$.

In this set up the resolvent set and spectrum may be defined as before:
$\rho(x)=\{\,\lambda\in\C\,|\,x-\lambda\id \text{ is invertible}\,\}$,
$\sigma(x)=\C\setminus\rho(x)$. 
The {\it spectral radius}\/ is defined to be $r(x)=\sup|\sigma(x)|$.
For $\lambda\in\rho(x)$, the resolvent
is defined as $R_x(\lambda)=(x-\lambda\id)^{-1}$.

\proclaim{Lemma} If $x,y\in X$ with $x$ invertible and $\|x^{-1}y\|<1$, then
$x-y$ is invertible,
$$
(x-y)^{-1}=\sum_{n=0}^\infty (x^{-1}y)^nx^{-1},
$$
and $\|(x-y)^{-1}\|\le \|x^{-1}\|/(1-\|x^{-1}y\|)$.
\endproclaim
\demo{Proof}
$$
\|\sum(x^{-1}y)^nx^{-1}\|\le\|x^{-1}\|\sum\|x^{-1}y\|^n
\le\|x^{-1}\|/(1-\|x^{-1}y\|),
$$
so the sum converges absolutely and the norm bound holds.  Also
$$
\sum_{n=0}^\infty (x^{-1}y)^nx^{-1}(x-y)
=\sum_{n=0}^\infty (x^{-1}y)^n-\sum_{n=0}^\infty (x^{-1}y)^{n+1}=\id,
$$
and similarly for the product in the reverse order.\qed
\enddemo

As a corollary, we see that if $|\lambda|>\|x\|$, then $\lambda \id-x$ is
invertible, i.e., $\lambda\in\rho(x)$.  In other words:

\proclaim{Proposition} $r(x)\le\|x\|$.\endproclaim

We also see from the lemma that $\lim_{\lambda\to\infty}\|R_x(\lambda)\|=0$.
Another corollary is that if $\lambda\in \rho(x)$ and
$|\mu|<\|R_x(\lambda)\|^{-1}$, then $\lambda-\mu\in\rho(x)$ and
$$
R_x(\lambda-\mu)=\sum_{n=0}^\infty R_x(\lambda)^{n+1}\mu^n.
$$

\proclaim{Theorem} The resolvent $\rho(x)$ is always open and contains
a neighborhood of $\infty$ in $\C$ and the spectrum is always non-empty
and compact.
\endproclaim
\demo{Proof} The above considerations show that the resolvent is open, and
so the spectrum is closed.  It is also bounded, so it is compact.

To see that the spectrum is non-empty, let $f\in X^*$ be arbitrary
and define $\phi(\lambda)=f[R_x(\lambda)]$.  Then $\phi$
maps $\rho(x)$ into $\C$, and it is easy to see that it is holomorphic
(since we have the power series expansion
$$
\phi(\lambda-\mu)=\sum_{n=0}^\infty f[R(\lambda)^{n+1}]\mu^n
$$
if $\mu$ is sufficiently small).  If $\sigma(x)=\emptyset$,
then $\phi$ is entire.  It is also bounded (since it tends to $0$ at infinity),
so Liouville's theorem implies that it is identically zero.  Thus for any
$f\in X^*$, $f[(\lambda \id-x)^{-1}]=0$ .  This implies that
$(\lambda \id-x)^{-1}=0$, which is clearly impossible.
\qed
\enddemo

\proclaim{Corollary (Gelfand--Mazur)} If $X$ is a complex Banach
division algebra, then $X$ is isometrically isomorphic to $\C$.
\endproclaim
\demo{Proof}  For each $0\ne x\in X$, let $\lambda\in\sigma(x)$.
Then $x-\lambda \id$ is not invertible, and since $X$ is a division
algebra, this means that $x=\lambda \id$.  Thus $X=\C \id$.
\qed
\enddemo

Now we turn to a bit of ``functional calculus.''  Let $x\in X$ and let
$f$ be a complex function of a complex variable which is holomorphic
on the closed disk of radius $\|x\|$ about the origin.  Then we make
two claims: (1) plugging $x$ into the power series expansion of $f$
defines an element $f(x)\in X$; and (2) the complex function $f$ maps
the spectrum of $x$ into the spectrum of $f(x)$.  (In fact onto, as
we shall show later in the case $f$ is polynomial.)  To prove these
claims, note that, by assumption, the radius of convergence of the
power series for $f$ about the origin exceeds $\|x\|$, so we can expand
$f(z)=\sum_{n=0}^\infty a_n z^n$ where $\sum |a_n|\|x\|^n<\infty$.
Thus the series $\sum a_n x^n$ is absolutely convergent in the Banach
space $X$; we call its limit $f(x)$.  (This is the {\it definition}\/
of $f(x)$.  It is a suggestive abuse of notation to use $f$ to denote
the this function, which maps a subset of $X$ into $X$, as well as the
original complex-valued function of a complex variable.)  Now suppose
that $\lambda\in\sigma(x)$.  Then
$$
f(\lambda)\id-f(x)=\sum_{n=1}^\infty a_n(\lambda^n \id -x^n)
=(\lambda \id - x)\sum_{n=1}^\infty a_n P_n
=\sum_{n=1}^\infty a_n P_n(\lambda \id - x),
$$
where
$$
P_n=\sum_{k=0}^{n-1} \lambda^k x^{n-k-1}.
$$
Note that $\|P_n\|\le n \|x\|^{n-1}$, so $\sum_{n=1}^\infty a_n P_n$
converges to some $y\in X$.  Thus
$$
f(\lambda)\id-f(x)=(\lambda \id - x)y=y(\lambda \id-x).
$$
Now $f(\lambda)\id-f(x)$ can't be invertible, because these formulas would
then imply that $\lambda \id-x$ would be invertible as
well, but $\lambda\in\sigma(x)$.  Thus we have verified that
$f(\lambda)\in\sigma\bigl(f(x)\bigr)$ for all $\lambda\in\sigma(x)$.

\proclaim{Theorem (Spectral Radius Formula)}
$r(x)=\lim_{n\to\infty}\|x^n\|^{1/n}=\inf_n\|x^n\|^{1/n}$.
\endproclaim
\demo{Proof}  If $\lambda\in \sigma(x)$, then $\lambda^n\in\sigma(x^n)$
(which is also evident algebraically), so $|\lambda^n|\le\|x^n\|$.
This shows that $r(x)\le \inf_n\|x^n\|^{1/n}$.

Now take $f\in X^*$, and consider
$$
\phi(\lambda)=f[(\lambda I-x)^{-1}]=\sum_{n=0}^\infty \lambda^{-n-1}f(x^n).
$$
Then $\phi$ is clearly holomorphic for $\lambda>\|x\|$, but we know it extends
holomorphically to $\lambda>r(x)$ and tends to $0$ as $\lambda$ tends
to infinity.  Let $\psi(\lambda)=\phi(1/\lambda)$.  Then $\psi$ extends
analytically to zero with value zero and defines an analytic function on
the open ball of radius $1/r(x)$ about zero, as does, therefore,
$$
\psi(\lambda)/\lambda=\sum_{n=0}^\infty f(\lambda^n x^n) .
$$
This shows that for each $|\lambda|<1/r(x)$ and each $f\in X^*$,
$f(\lambda^n x^n)$ is bounded.  By the uniform boundedness principle,
the set of elements $\lambda^n x^n$ are bounded in $X$, say by $K$.
Thus $\|x^n\|^{1/n}\le K^{1/n}/|\lambda|\to 1/|\lambda|$.  This is true for all 
$|\lambda|<1/r(x)$, so $\limsup\|x^n\|^{1/n}\le r(x)$.
\qed
\enddemo

\proclaim{Corollary} If $H$ is a Hilbert space and $T\in B(H)$ a
normal operator, then $r(T)=\|T\|$.
\endproclaim
\demo{Proof}
$$
\|T\|^2=\sup_{\|x\|\le1}\langle Tx,Tx\rangle=\sup_{\|x\|\le1}\langle T^*Tx,x\rangle
=\|T^*T\|,
$$
since $T^*T$ is self-adjoint.  Using the normality of $T$ we also get
$$
\align
\|T^*T\|^2=\sup_{\|x\|\le1}\langle T^*Tx,T^*Tx\rangle
&=\sup_{\|x\|\le1}\langle TT^*Tx,Tx\rangle
=\sup_{\|x\|\le1}\langle T^*T^2x,Tx\rangle
\\
&=\sup_{\|x\|\le1}\langle T^2x,T^2x\rangle
=\|T^2\|^2.
\endalign
$$
Thus $\|T\|^2=\|T^2\|$.  Replacing $T$ with $T^2$ gives,
$\|T\|^4=\|T^4\|$, and similarly for all powers of $2$.  The result
thus follows from the spectral radius formula.
\qed
\enddemo

As mentioned, we can now show that $p$ maps $\sigma(x)$ {\it onto}\/
$\sigma\bigl(p(x)\bigr)$ if $p$ is a polynomial.

\proclaim{Spectral Mapping Theorem} Let $X$ be a complex Banach algebra with
identity, $x\in X$, and let $p$ a polynomial in one variable with complex
coefficients.  Then $p\bigl(\sigma(x)\bigr)=\sigma\bigl(p(x)\bigr)$.
\endproclaim
\demo{Proof} We have already shown that
$p\bigl(\sigma(x)\bigr)\subset\sigma\bigl(p(x)\bigr)$.  Now suppose that
$\lambda\in\sigma\bigl(p(x)\bigr)$.  By the Fundamental Theorem of Algebra
we can factor $p-\lambda$, so
$$
p(x)-\lambda\id=a\Pi_{i=1}^n(x-\lambda_i\id),
$$
for some nonzero $a\in\C$ and some roots $\lambda_i\in\C$.
Since $p(x)-\lambda\id$ is not invertible, it follows that $x-\lambda_i\id$
is not invertible for at least one $i$.  In orther words,
$\lambda_i\in\sigma(x)$, so $\lambda=p(\lambda_i)\in p\bigl(\sigma(x)\bigr)$.
\qed
\enddemo

\goodbreak\subhead Spectral Theorem for bounded self-adjoint operators in
Hilbert space \endsubhead\nobreak
We now restrict to self-adjoint operators on Hilbert space and
close with a version of the Spectral Theorem for this class of
operator.  We follow Halmos's article ``What does the Spectral Theorem
Say?'' (American Mathematical Monthly 70, 1963) both in the relatively
elementary statement of the theorem and the outline of the proof.

First we note that self-adjoint operators have real spectra (not just
real eigenvalues).

\proclaim{Proposition} If $H$ is a Hilbert space and $T\in B(H)$ is
self-adjoint, then $\sigma(T)\subset\R$.
\endproclaim
\demo{Proof}
$$
|\langle (\lambda\id-T)x,x\rangle|\ge|\Im\langle (\lambda\id-T)x,x\rangle|
=|\Im\lambda|\|x\|^2,
$$
so if $\Im\lambda\ne0$, $\lambda\id-T$ is injective with closed range.
The same reasoning shows that $(\lambda\id-T)^*=\bar\lambda\id-T$
is injective, so $\Ra(\lambda\id-T)$ is dense.  Thus $\lambda\in\rho(T)$.
\qed
\enddemo

\proclaim{Spectral Theorem for self-adjoint operators in Hilbert space}
If $H$ is a complex Hilbert space and $T\in B(H)$ is self-adjoint, then there
exists a measure space $\Omega$ with measure $\mu$, a bounded measurable
function $\phi:\Omega\to\R$, and an isometric isomorphism $U:
L^2\to H$ such that
$$
U^{-1}TU=M_\phi
$$
where $M_\phi:L^2\to L^2$ is the operation of multiplication by $\phi$.
(Here $L^2$ means $L^2(\Omega,\mu;\C)$, the space of complex-valued
functions on $\Omega$ which are square integrable with respect to the
measure $\mu$.)
\endproclaim
\demo{Sketch of proof} Let $x$ be a nonzero element of $H$, and
consider the smallest closed subspace $M$ of $H$ containing $T^nx$ for
$n=0,1,\ldots$, i.e.,  $M=\overline{\{\,p(T)x\,|\,p\in\Bbb P_{\C}\,\}}$.
Here $\Bbb P_{\C}$ is the space of polynomials in one variable with
complex coefficients.  Both $M$ and its orthogonal complement are
invariant under $T$ (this uses the self-adjointness of $T$).  By a
straightforward application of Zorn's lemma we see that $H$ can be
written as a Hilbert space direct sum of $T$ invariant spaces of the
form of $M$.  If we can prove the theorem for each of these subspaces,
we can take direct products to get the result for all of $H$.  Therefore
we may assume from the start that $H=\overline{\{\,p(T)x\,|\,p\in\Bbb
P_{\C}\,\}}$ for some $x$.  (In other terminology, that $T$ has a {\it
cyclic vector}\/ $x$.)

Now set $\Omega=\sigma(T)$, which is a compact subset of the real line, and
consider the space $C=C(\Omega,\R)$, the space of all continuous
real-valued functions on $\Omega$.  The subspace of real-valued
polynomial functions is dense in $C$ (since any continuous function
on $\Omega$ can be extended to the interval $[-r(T),r(T)]$ thanks to
Tietze's extension theorem and then approximated arbitrarily closely by
a polynomial thanks to the Weierstrass approximation theorem).  For such
a polynomial function, $p$, define $Lp=\langle p(T)x,x\rangle\in\R$.
Clearly $L$ is linear and
$$
|Lp|\le \|p(T)\|\|x\|^2=r\bigl(p(T)\bigr)\|x\|^2,
$$
by the special form of the spectral radius formula for self-adjoint operators
in Hilbert space.  Since $\sigma\bigl(p(T)\bigr)=p\bigl(\sigma(T)\bigr)$,
we have $r\bigl(p(T)\bigr)=\|p\|_{L^\infty(\Omega)}=\|p\|_C$, and thus,
$$
|Lp|\le\|x\|^2\|p\|_C.
$$
This shows that $L$ is a bounded linear functional on a dense subspace of $C$
and so extends uniquely to define a bounded linear functional on $C$.

Next we show that $L$ is positive in the sense that $Lf\ge0$ for all
non-negative functions $f\in C$.  Indeed, if $f=p^2$ for some polynomial,
then
$$
Lf=\langle p(T)^2x,x\rangle=\langle p(T)x,p(T)x\rangle\ge0.
$$
For an arbitrary non-negative $f$, we can approximate $\sqrt f$ uniformly
by polynomials $p_n$, so $f=\lim p_n^2$ and $Lf=\lim Lp_n^2\ge0$.

We now apply the Riesz Representation Theorem for the representation
of the linear functional $L$ on $C$.  It state that there exists a
finite measure on $\Omega$ such that $Lf=\int f\,d\mu$ for $f\in\C$
(it is a positive measure since $L$ is positive).  In particular,
$\langle p(T)x,x\rangle=\int p\,d\mu$ for all $p\in\Bbb P_{\R}$.

We now turn to the space $L^2$ of complex-valued
functions on $\Omega$ which are square integrable with respect to the
measure $\mu$.  The subspace of complex-valued polynomial functions is
dense in $L^2$ (since the measure is finite, the $L^2$ norm is dominated
by the supremum norm).  For such a polynomial function, $q$, define
$Uq=q(T)x$.  Then
$$
\|Uq\|^2=\|q(T)x\|^2=\langle q(T)x,q(T)x\rangle
=\langle \bar q(T)q(T)x,x\rangle=\int|q|^2\,d\mu=\|q\|_{L^2}^2.
$$
Thus $U$ is an isometry of a dense subspace of $L^2$ into $H$ and
so extends to an isometry of $L^2$ onto a closed subspace of $H$.
In fact, $U$ is onto $H$ itself, since, by the assumption that $x$ is a cyclic
vector for $T$, the range of $U$ is dense.

Finally, define $\phi:\Omega\to\R$ by $\phi(\lambda)=\lambda$.
If $q$ is a complex polynomial, then $(M_\phi q)(\lambda)=\lambda q(\lambda)$,
which is also a polynomial.  Thus
$$
U^{-1}TUq=U^{-1}Tq(T)x=U^{-1}\bigl((M_\phi q)(T)\bigr)x=M_\phi q.
$$
Thus the bounded operators $U^{-1}TU$ and $M_\phi$ coincide on a dense
subset of $L^2$, and hence they are equal.
\qed
\enddemo

For a more precise description of the measure space and the extension
to normal operators, see Zimmer.

\end

