%% R.E. Showalter -- Chapter 1
\chapter{Elements of Hilbert Space}
\pagenumbering{arabic}
\section{Linear Algebra}			% 1
\setcounter{equation}{0}


We begin with some notation. A function $F$ with domain $\dom (F) =A$ 
and range $\Rg (F)$ a subset of $B$ is denoted by $F:A\to B$. 
That a point $x\in A$ is mapped by $F$ to a point $F(x)\in B$ is 
indicated by $x\mapsto F(x)$. 
If $S$ is a subset of $A$ then the {\it image\/} of $S$ by $F$ is 
$F(S) = \{F(x):x\in S\}$. Thus $\Rg(F) = F(A)$. 
The  pre-image or inverse image of a set $T\subset B$ is $F^{-1}(T) = 
\{ x\in A :F(x)\in T\}$. 
A function is called {\it injective\/} if it is one-to-one, 
{\it surjective\/} if it is onto, 
and {\it bijective\/} if it is both injective and surjective. 
Then it is called, respectively, an {\it injection, surjection\/}, 
or {\it bijection\/}. 

$\KK$ will denote the field of scalars for our vector spaces and is always one 
of $\RR$ (real number system) or $\CC$ (complex numbers). 
The choice in most situations will be clear from the context or immaterial, 
so we usually avoid mention of it. 

The ``strong inclusion'' $K\subset\subset G$ between subsets of Euclidean 
space $\RR^n$ means $K$ is compact, $G$ is open, and $K\subset G$. 
If $A$ and $B$ are sets, their Cartesian product is given by $A\times B = 
\{ [a,b] :a\in A,\ b\in B\}$. 
If $A$ and $B$ are subsets of $\KK^n$ (or any other vector space) their set 
sum is $A+B = \{a+b :a\in A,\ b\in B\}$. 

\subsection{}  	%1.1
A {\it linear space\/} over the field $\KK$ is a non-empty set $V$ of 
vectors with a binary operation {\it addition\/} $+:V\times V\to V$ and 
a {\it scalar multiplication\/} $\cdot : \KK\times V\to V$ such that 
$(V,+)$ is an Abelian group, i.e., 
$$\begin{array}{ll}
(x+y) +z=x+(y+z)\ ,&x, y, z\in V\ ,\\
\mbox{there is a zero } \theta\in V :x+\theta =x\ ,&x\in V\ ,\\
\mbox{if $x\in V$,  there is } -x \in V:x+(-x)=\theta\ ,&\mbox{and}\\
x+y=y+x\ ,&x, y\in V\ ,
\end{array}$$
and we have 
$$\begin{array}{ll}
(\alpha+\beta) \cdot x = \alpha\cdot x+\beta\cdot x\ ,
	&\alpha \cdot (x+y) = \alpha\cdot x+\alpha \cdot y\ ,\\
\alpha\cdot (\beta\cdot x) = (\alpha\beta)\cdot x\ ,&1\cdot x=x\ ,
\qquad\qquad x,y\in V\ ,\ \alpha,\beta\in \KK\ .
\end{array}$$
We shall suppress the symbol for scalar multiplication since there is no 
need for it. 

\examples
(a) The set $\KK^n$ of $n$-tuples of scalars is a linear space over $\KK\,$. 
Addition and scalar multiplication are defined coordinatewise: 
$$\begin{array}{l}
(x_1,x_2,\ldots,x_n) + (y_1,y_2,\ldots,y_n) 
= (x_1+y_1,x_2+y_2,\ldots , x_n+y_n)\\
\alpha (x_1,x_2,\ldots,x_n) = (\alpha x_1,\alpha x_2,\ldots,\alpha x_n)\ .
\end{array}$$

\noindent (b) The set $\KK^X$ of functions $f:X\to \KK$ is a linear space, 
where $X$ is a non-empty set, and we define $(f_1+f_2)(x) = f_1 (x)+f_2(x)$, 
$(\alpha f)(x) = \alpha f(x)$, $x\in X$. 
\smallskip

\noindent (c) Let $G\subset \RR^n$ be open. 
The above pointwise  definitions of linear operations give a linear space 
structure on the set $C(G,\KK)$ of continuous $f:G\to \KK$. 
We normally shorten this to $C(G)$. 
\smallskip

\noindent (d) For each $n$-tuple $\alpha = (\alpha_1,\alpha_2,\ldots,
\alpha_n)$ of non-negative integers, we denote by $D^\alpha$ the 
{\it partial derivative} 
$${\partial^{|\alpha|} \over 
\partial x_1^{\alpha_1} \partial x_2^{\alpha_2} \cdots 
\partial x_n^{\alpha_n}}$$ 
of order $|\alpha| = \alpha_1 + \alpha_2 +\cdots + \alpha_n$. 
The sets $C^m(G) = \{f\in C(G): D^\alpha f\in C(G)$ for all $\alpha$, 
$|\alpha|\le m\}$, $m\ge 0$, and $C^\infty G = \bigcap_{m\ge1} C^m(G)$ are 
linear spaces with the operations defined above. 
We let $D^\theta$ be the identity, where $\theta = (0,0,\ldots,0)$, so 
$C^0(G) = C(G)$. 
\smallskip

\noindent (e) For $f\in C(G)$, the {\it support\/} of $f$ is the closure 
in $G$ of the set $\{x\in G:f(x)\ne0\}$ and we denote it by $\supp(f)$. 
$C_0(G)$ is the subset of those functions in $C(G)$ with compact support. 
Similarly, we define $C_0^m (G) = C^m(G) \cap C_0(G)$, $m\ge1$ and 
$C_0^\infty (G) = C^\infty (G) \cap C_0(G)$. 
\smallskip

\noindent (f) If $f:A\to B$ and $C\subset A$, we denote $f|_C$ the 
{\it restriction\/} of $f$ to $C$. We obtain useful linear spaces of 
functions on the closure $\bar G$ as follows: 
$$C^m(\bar G) = \bigl\{ f|_{\bar G} : f\in C_0^m (\RR^n)\bigr\}\quad ,\quad 
C^\infty (\bar G) = \bigl\{ f|_{\bar G} : f\in C_0^\infty (\RR^n)\bigr\}\ .$$ 
These spaces play a central role in our work below. 

\subsection{}   	%1.2
A subset $M$ of the linear space $V$ is a {\it subspace\/} of $V$ if it 
is closed under the linear operations. That is, $x+y\in M$ whenever $x,y\in M$ 
and $\alpha x\in M$ for each $\alpha\in\KK$ and $x\in M$. 
We denote that $M$ is a subspace of $V$ by $M\le V$. 
It follows that $M$ is then (and only then) a linear space with addition 
and scalar multiplication inherited from $V$. 

\examples 
We have three chains of subspaces given by 
$$\begin{array}{rcl}
C^j(G)&\le&C^k (G)\  \le\  \KK^G\ ,\\
C^j (\bar G)&\le&C^k(\bar G)\ ,\quad \mbox{and}\\
\{\theta\}\  \le\  C_0^j (G)&\le&C_0^k (G)\ ,\quad 0\le k\le j\le \infty\ .
\end{array}$$ 
Moreover, for each $k$ as above, we can identify $\varphi\in C_0^k(G)$ with 
that $\Phi\in C^k(\bar G)$ obtained by defining $\Phi$ to be equal to 
$\varphi$ on $G$ and zero on $\partial G$, the boundary of $G$. 
Likewise we can identify each $\Phi \in C^k(\bar G)$ with $\Phi|_G\in C^K(G)$. 
These identifications are ``compatible'' and we have 
$C_0^k (G) \le C^k(\bar G) \le C^k(G)$. 

\subsection{} 	% 1.3
We let $M$ be a subspace of $V$ and construct a corresponding {\it quotient 
space\/}. For each $x\in V$, define a {\it coset\/} $\hat x =\{y\in V: y-x
\in M\} = \{x+m:m\in M\}$. The set $V/M = \{\hat x:x\in V\}$ is the 
{\it quotient set\/}. 
Any $y\in \hat x$ is a {\it representative\/} of the coset $\hat x$ and 
we clearly have $y\in \hat x$ if and only if $x\in \hat y$ if and only if 
$\hat x= \hat y$. We shall define addition of cosets by adding a corresponding 
pair of representatives and similarly define scalar multiplication. 
It is necessary to first verify that this definition is unambiguous. 

\plainlemma
{If $x_1,x_2\in \hat x$, $y_1,y_2\in\hat y$, and $\alpha \in\KK$, then 
$\widehat{(x_1+y_1)} = \widehat{(x_2+y_2)}$ and $\widehat{(\alpha x_1)} 
= \widehat{(\alpha x_2)}$.}

\noindent The proof follows easily, since $M$ is closed under addition and 
scalar multiplication, and we can define $\hat x+\hat y = \widehat{(x+y)}$ 
and $\alpha\hat x =\widehat{(\alpha x)}$. 
These operations make $V/M$ a linear space. 

\examples 
(a) Let $V= \RR^2$ and $M= \{ (0,x_2): x_2\in\RR\}$. 
Then $V/M$ is the set of parallel translates of the $x_2$-axis, $M$, 
and addition of two cosets is easily obtained by adding their (unique) 
representatives on the $x_1$-axis. 
\smallskip

\noindent (b) 
Take $V= C(G)$. Let $x_0\in G$ and $M= \{\varphi \in C(G): \varphi(x_0)=0\}$. 
Write  each $\varphi\in V$ in the form $\varphi (x) = (\varphi (x)-\varphi 
(x_0)) + \varphi (x_0)$. 
This representation can be used to show that $V/M$ is essentially 
equivalent (isomorphic) to $\KK$. 
\smallskip

\noindent (c) 
Let $V= C(\bar G)$ and $M= C_0(G)$.
We  can describe $V/M$ as a space of ``boundary values.'' 
To do this, begin by noting that for each $K\subset\subset G$ there is a 
$\psi \in C_0(G)$ with $\psi =1$ on $K$. (Cf.\ Section II.1.1.) 
Then write a given $\varphi \in C(\bar G)$ in the form 
$$\varphi = (\varphi\psi) + \varphi (1-\psi)\ ,$$ 
where the first term belongs to $M$ and the second equals $\varphi$ in 
a neighborhood of $\partial G$. 

\subsection{} 	% 1.4
Let $V$ and $W$ be linear spaces over $\KK$. 
A function $T:V\to W$ is {\it linear\/} if 
$$T(\alpha x +\beta y) = \alpha T(x) + \beta T(y)\ ,\qquad 
\alpha,\beta\in \KK\ ,\ x,y\in V\ .$$ 
That is, linear functions are those which preserve the linear operations. 
An {\it isomorphism\/} is a linear bijection. 
The set $\{x\in V:Tx=0\}$ is called the {\it kernel\/} of the (not 
necessarily linear) function ${T:V\to W}$ and we denote it by $K(T)$. 

\plainlemma
{If $T:V\to W$ is linear, then $K(T)$ is a subspace of $V$, 
$\Rg(T)$ is a subspace of $W$, and $K(T) = \{\theta\}$ if and only if $T$ 
is an injection.} 

\examples 
(a) Let $M$ be a subspace of $V$. The identity $i_M:M\to V$ is a linear 
injection ${x\mapsto x}$ and its range is $M$. 
\smallskip

\noindent (b) 
The {\it quotient\/} map $q_M:V\to V/M$, $x\mapsto\hat x$, is a linear 
surjection with kernel $K(q_M)=M$. 
\smallskip

\noindent (c) 
Let $G$ be the open interval $(a,b)$ in $\RR$ and consider $D\equiv d/dx$: 
$V\to C(\bar G)$, where  $V$ is a subspace of $C^1 (\bar G)$. 
If $V= C^1 (\bar G)$, then $D$ is a linear surjection with $K(D)$ consisting 
of constant functions on $\bar G$. 
If $V= \{\varphi \in C^1 (\bar G): \varphi (a) =0\}$, then $D$ is an 
isomorphism. Finally, if $V= \{\varphi \in C^1(\bar G):\varphi (a)=
\varphi(b)=0\}$, 
then $\Rg  (D) = \{\varphi\in C(\bar G):\int_a^b \varphi =0\}$. 
\qed

Our next result shows  how each linear function can be factored into the 
product of a linear injection and an appropriate quotient map. 

\begin{theorem}\label{thm1-1A}
Let $T:V\to W$ be linear and $M$ be a subspace of $K(T)$. 
Then there is exactly one function ${\hatT:V/M\to W}$ for which 
${\hatT\circ q_M=T}$, and $\hatT$ is linear with $\Rg (\hatT)=\Rg(T)$. 
Finally, $\hatT$ is injective if and only if ${M=K(T)}$.
\end{theorem}  

\proof 
If $x_1,x_2\in\hat x$, then $x_1-x_2\in M\subset K(T)$, so $T(x_1)= T(x_2)$. 
Thus we can define a function as desired by the formula $\hatT(\hat x) 
= T(x)$. The uniqueness and linearity of $\hatT$ follow since $q_M$ is 
surjective and linear. 
The equality of the ranges follows, since $q_M$ is surjective, and the last 
statement follows from the observation that $K(T)\subset M$ if and only if 
${v\in V}$ and $\hatT(\hat x) =0$ imply ${\hat x=\hat 0}$. 
\qed

An immediate corollary is that each linear function ${T:V\to W}$ can be 
factored into a product of a surjection, an isomorphism, and an injection: 
${T=i_{\Rg(T)} \circ \hatT\circ q_{K(T)}}$. 

A function $T:V\to W$ is called {\it conjugate linear\/} if 
$$T(\alpha x+\beta y) = \bar \alpha T(x) +\bar \beta T(y)\ ,\qquad 
\alpha,\beta \in \KK\ ,\ x,y\in V\ .$$ 
Results similar to those above hold for such functions. 

\subsection{} 	% 1.5
Let $V$ and $W$ be linear spaces over $\KK$ and consider the set 
$L(V,W)$ of linear functions from $V$ to $W$. 
The set $W^V$ of all functions from $V$ to $W$ is a linear space under 
the pointwise definitions of addition and scalar multiplication 
(cf.\ Example 1.1(b)), and $L(V,W)$ is a subspace. 

We define $V^*$ to be the linear space of all conjugate linear functionals 
from ${V\to \KK}$. $V^*$ is called the {\it algebraic dual\/} of $V$. 
Note that there is a bijection ${f\mapsto \bar f}$ of $\L(V,\KK)$ onto 
$V^*$, where $\bar f$ is the functional defined by $\bar f(x)=\overline{f(x)}$ 
for ${x\in V}$ and is called the {\it conjugate\/} of the functional 
${f:V\to \KK}$. Such spaces provide a useful means of constructing large 
linear spaces containing a given class of functions. 
We illustrate this technique in a simple situation. 

\example 
Let $G$ be open in $\RR^n$ and $x_0\in G$. 
We shall imbed the space $C(G)$ in the algebraic dual of $C_0(G)$. 
For each ${f\in C(G)}$, define ${T_f\in C_0(G)^*}$ by 
$$T_f(\varphi) = \int_G f\bar \varphi\ ,\qquad \varphi\in C_0(G)\ .$$ 
Since $f\bar\varphi \in C_0(G)$, the Riemann integral is adequate here. 
An easy exercise shows that the function ${f\mapsto T_f:C(G)\to C_0(G)^*}$  
is a linear injection, so we may thus identify $C(G)$ with a subspace of 
$C_0(G)^*$. This linear injection is not surjective; we can exhibit 
functionals  on $C_0(G)$ which are not identified with functions in $C(G)$. 
In particular, the {\it Dirac functional\/} $\delta_{x_0}$ defined by 
$$\delta_{x_0} (\varphi) = \overline{\varphi (x_0)}\ ,\qquad 
\varphi\in C_0(G)\ ,$$ 
cannot be obtained as $T_f$ for any $f\in C(G)$. 
That is, ${T_f=\delta_{x_0}}$ implies that ${f(x)=0}$ for all ${x\in G}$, 
${x\ne x_0}$, and thus ${f=0}$, a contradiction. 

\section{Convergence and Continuity}	% 2
\setcounter{equation}{0}

The absolute value function on $\RR$ and modulus function on $\CC$ are 
denoted by $|\cdot|$, and each gives a notion of length or distance in 
the corresponding space and permits the discussion of convergence of sequences 
in that space or continuity of functions on that space. 
We shall extend these concepts to a general linear space. 

\subsection{} 	%2.1
A {\it seminorm\/} on the linear space $V$ is a function $p:V\to \RR$ 
for which $p(\alpha x) = |\alpha|p(x)$ and $p(x+y)\le p(x)+p(y)$ for all 
$\alpha\in\KK$ and $x,y\in V$. The pair $V,p$ is called a {\it seminormed 
space\/}. 
\smallskip

\begin{lemma}\label{lem1-2-1}
If $V,p$ is a seminormed space, then 
\begin{description}
\item[(a)] $|p(x)-p(y)| \le p(x-y)\ ,\qquad x,y\in V\ ,$
\item[(b)] $p(x)\ge0\ ,\qquad x\in V$\ , {\it and} 
\item[(c)]  the kernel $K(p)$  is a subspace of $V$. 
\item[(d)]  If $T\in L(W,V)$, then $p\circ T:W\to \RR$ 
is a seminorm on $W$. 
\item[(e)]  If $p_j$ is a seminorm on $V$ and $\alpha_j\ge0$, 
$1\le j\le n$, then $\sum_{j=1}^n \alpha_j p_j$ is a seminorm on $V$.
\end{description}
\end{lemma}

\proof 
We have $p(x) = p(x-y+y)\le p(x-y) + p(y)$ so $p(x) - p(y) \le p(x-y)$. 
Similarly, $p(y) -p(x)\le p(y-x) = p(x-y)$, so the result follows. 
Setting ${y=0}$ in (a) and noting ${p(0)=0}$, we obtain (b). 
The result (c) follows directly from the definitions, and (d) and (e) 
are straightforward exercises. 

If $p$ is a seminorm with the property that $p(x)>0$ for each $x\ne\theta$, 
we call it a {\it norm\/}. 

\examples 
(a) For $1\le k\le n$ we define seminorms on $\KK^n$ by $p_k(x)= 
\sum_{j=1}^k |x_j|$, $q_k(x) = (\sum_{j=1}^k |x_j|^2)^{1/2}$, 
and ${r_k(x) = \max \{|x_j| :1\le j\le k\}}$.  
Each of $p_n$, $q_n$ and $r_n$ is a norm. 

(b) If $J\subset X$ and $f\in \KK^X$, we define $p_J(f) 
= \sup \{|f(x)|: x\in J\}$. 
Then for each finite ${J\subset X}$, $p_J$ is a seminorm on $\KK^X$. 

(c)  For each $K\subset\subset G$, $p_K$ is a seminorm on $C(G)$. 
Also, ${p_{\bar G} = p_G}$ is a norm on $C(\bar G)$. 

(d) For each $j$, $0\le j\le k$, and $K\subset\subset G$ we can define a 
seminorm on $C^k(G)$ by $p_{j,K}(f)=\sup \{|D^\alpha f(x)| :x\in K$, 
$|\alpha| \le j\}$. 
Each such $p_{j,G}$ is a norm on $C^k(\bar G)$. 

\subsection{}	%2.2
Seminorms permit a discussion of convergence. 
We say the sequence $\{x_n\}$ in $V$ {\it converges\/} to ${x\in V}$ if 
$\lim_{n\to\infty} p(x_n-x)=0$; that is, if ${\{p(x_n-x)\}}$ is a sequence 
in $\RR$ converging to $0$. 
Formally, this means that for every ${\varep >0}$ there is an integer 
${N\ge 0}$ such that ${p(x_n-x)<\varep}$ for all ${n\ge N}$. 
We denote this by ${x_n\to x}$ in $V,p$ and suppress the mention of $p$ 
when it is clear what is meant. 

Let $S\subset V$. 
The {\it closure\/} of $S$ in $V,p$ is the set ${\bar S= \{x\in V: x_n\to x}$ 
in $V,p$ for some sequence $\{x_n\}$ in $S\}$, and $S$ is called 
{\it closed\/} if ${S=\bar S}$. 
The closure $\bar S$ of $S$ is the  smallest closed set containing 
${S:S\subset \bar S}$, ${\bar S = \bbarS}$, and if 
$S\subset K=\bar K$ then $\bar S\subset K$. 

\plainlemma{Let $V,p$ be a seminormed space and $M$ be a subspace of $V$. 
Then $\bar M$ is a subspace of $V$.} 

\proof 
Let $x,y\in \bar M$. 
Then there are sequences $x_n,y_n\in M$ such that ${x_n\to x}$ and 
${y_n\to y}$ in $V,p$. 
But $p((x+y)- (x_n+y_n)) \le p(x-x_n) + p(y-y_n) \to 0$  which shows 
that $(x_n+y_n)\to x+y$. 
Since $x_n+y_n\in M$, all $n$, this implies that $x+y\in \bar M$. 
Similarly, for $\alpha\in \KK$ we have $p(\alpha x-\alpha x_n)= 
|\alpha| p(x-x_n) \to0$, so $\alpha x \in \bar M$. 
\qed

\subsection{}	%2.3
Let $V,p$ and $W,q$ be seminormed spaces and $T:V\to W$ (not necessarily 
linear). Then $T$ is called {\it continuous at\/} ${x\in V}$ if for 
every ${\varep>0}$ there is a ${\delta>0}$ for which ${y\in V}$ and 
${p(x-y)<\delta}$ implies $q(T(x)-T(y))<\varep$. 
$T$ is {\it continuous\/} if it is continuous at every ${x\in V}$. 

\begin{theorem}\label{thm1-2A}
$T$ is continuous at $x$ if and only if $x_n\to x$ in $V,p$ 
implies ${Tx_n\to Tx}$ in $W,q$.
\end{theorem} 

\proof 
Let $T$ be continuous at $x$ and $\varep>0$. 
Choose ${\delta>0}$ as in the definition above and then $N$ such that 
${n\ge N}$ implies ${p(x_n-x)<\delta}$, where ${x_n\to x}$ in $V,p$ is given. 
Then ${n\ge N}$ implies ${q(Tx_n-Tx)<\varep}$, so ${Tx_n\to Tx}$ in  $W,q$. 

Conversely, if $T$ is not continuous at $x$, then there is an ${\varep>0}$ 
such that for every ${n\ge1}$ there is an ${x_n\in V}$ with 
${p(x_n-x)<1/n}$ and ${q(Tx_n-Tx)\ge\varep}$. 
That is, ${x_n\to x}$ in $V,p$ but $\{Tx_n\}$ does not converge to $Tx$ 
in $W,q$. 
\qed

We record the facts that our algebraic operations and seminorm are always 
continuous. 

\plainlemma
{If $V,p$ is a seminormed space, the functions $(\alpha,x)\mapsto \alpha x: 
\KK\times V\to V$, $(x,y)\mapsto x+y: V\times V\to V$, and ${p:V\to\RR}$ are 
all continuous.} 

\proof 
The estimate $p(\alpha x-\alpha_nx_n)\le |\alpha-\alpha_n| p(x) +|\alpha_n| 
p(x-x_n)$ implies the continuity of scalar multiplication. 
Continuity of addition follows from an estimate in the preceding Lemma, 
and continuity of $p$ follows from the Lemma of 2.1. 
\qed

Suppose $p$ and $q$ are seminorms on the linear space $V$. 
We say $p$ is {\it stronger\/} than $q$ (or $q$ is {\it weaker\/} than $p$) 
if for any sequence $\{x_n\}$ in $V$, ${p(x_n)\to 0}$ implies 
${q(x_n)\to0}$. 
\medskip

\begin{theorem}\label{thm1-2B}
The following are equivalent: 
\begin{description}
\item[(a)] $p$ is stronger than $q$, 
\item[(b)] the identity $I:V,p\to V,q$ is continuous, and 
\item[(c)] there is a constant $K\ge 0$ such that 
$$q(x) \le Kp(x)\ ,\qquad x\in V\ .$$ 
\end{description} 
\end{theorem}

\proof 
By Theorem \ref{thm1-2A}, (a) is equivalent to having the identity 
${I:V,p\to V,q}$ continuous at $0$, so (b) implies (a). 
If (c) holds, then $q(x-y)\le Kp(x-y)$, ${x,y\in V}$, so (b) is true. 

We claim now that (a) implies (c). 
If (c) is false, then for every integer ${n\ge1}$ there is an ${x_n\in V}$ 
for which $q(x_n)>np(x_n)$. 
Setting $y_n= (1/q(x_n))x_n$, ${n\ge1}$, we have obtained a sequence 
for which ${q(y_n)=1}$ and ${p(y_n)\to0}$, thereby contradicting (a). 
\medskip

\begin{theorem}\label{thm1-2C}
Let $V,p$ and $W,q$ be seminormed spaces and $T\in L(V,W)$. 
The following are equivalent:
\begin{description}
\item[(a)] $T$ is continuous at $\theta\in V$\ ,
\item[(b)] $T$ is continuous, and 
\item[(c)] there is a constant $K\ge 0$ such that 
$$q\bigl(T(x)\bigr) \le Kp(x)\ ,\qquad x\in V\ .$$
\end{description}
\end{theorem}

\proof 
By Theorem \ref{thm1-2B}, each of these is equivalent to requiring that 
the seminorm $p$ be stronger than the seminorm ${q\circ T}$ on $V$. 
\qed

\subsection{}	%2.4
If $V,p$ and $W,q$ are seminormed spaces, we denote by $\L(V,W)$ the set of 
continuous linear functions from $V$ to $W$. 
This is a subspace of $L(V,W)$ whose elements are frequently called the 
{\it bounded\/} operators from $V$ to $W$ (because of Theorem \ref{thm1-2C}). 

Let $T\in \L(V,W)$ and consider 
\begin{eqnarray*}
\lambda&\equiv&\sup\{ q\bigl( T(x)\bigr) :x\in V\ ,
\quad p(x)\le 1\}\ ,\\
\mu&\equiv&\inf \{ K>0: q\bigl( T(x)\bigr) \le Kp(x)\ \mbox{ for all }\ 
x\in V\}\ .
\end{eqnarray*}
If $K$ belongs to the set defining $\mu$, then for every $x\in V:p(x)\le1$ 
we have $q(T(x))\le K$, hence ${\lambda\le K}$. 
This holds for all such $K$, so ${\lambda\le\mu}$. 
If ${x\in V}$ with ${p(x)>0}$, then $y\equiv (1/p(x))x$ satisfies 
${p(y)=1}$, so $q(T(y))\le\lambda$. 
That is $q(T(x))\le\lambda p(x)$ whenever ${p(x)>0}$. 
But by Theorem \ref{thm1-2C}(c) this last inequality is trivially satisfied 
when ${p(x)=0}$, so we have ${\mu\le\lambda}$. 
These remarks prove the first part of the following result; the remaining 
parts are straightforward. 

\begin{theorem}\label{thm1-2D}
Let $V,p$ and $W,q$ be seminormed spaces. 
For each $T\in\L(V,W)$ we define a real number by $|T|_{p,q}\equiv 
\sup \{q(T(x)):x\in V$, $p(x)\le 1\}$. 
Then we have $|T|_{p,q} = \sup \{q(T(x)):x\in V$, $p(x)=1\} = \inf\{ K>0: 
q(T(x))\le Kp(x)$ for all $x\in V\}$ and $|\cdot|_{p,q}$ is a seminorm 
on $\L(V,W)$. Furthermore, $q(T(x))\le |T|_{p,q}\cdot p(x)$, ${x\in V}$, 
and $|\cdot|_{p,q}$ is a norm whenever $q$ is a norm. 
\end{theorem} 

\definitions 
The {\it dual\/} of the seminormed space $V,p$ is the linear space 
$V'= \{f\in V^* : f$ is continuous$\}$ with the norm 
$$\|f\|_{V'} = \sup \bigl\{ |f(x)| :x\in V\ ,\quad p(x)\le 1\bigr\}\ .$$ 
If $V,p$ and $W,q$ are seminormed spaces, then $T\in \L(V,W)$ is called a 
{\it contraction\/} if $|T|_{p,q}\le1$, and $T$ is called an 
{\it isometry\/} if $|T|_{p,q}=1$. 
\qed 

\section{Completeness}		%3
\setcounter{equation}{0}

\subsection{}			%3.1
A sequence $\{x_n\}$ in a seminormed space $V,p$ is called {\it Cauchy\/} 
if $\lim_{m,n\to\infty} p(x_m\break -x_n)=0$, that is, if for every 
${\varep>0}$ there is an integer $N$ such that $p(x_m-x_n)<\varep$ for all 
${m,n\ge N}$. 
Every convergent sequence is Cauchy. 
We call $V,p$ {\it complete\/} if every Cauchy sequence is convergent. 
A complete normed linear space is a {\it Banach space\/}. 

\examples 
Each of the seminormed spaces of Examples 2.1(a-d) is complete. 

(e) Let $G= (0,1)\subset \RR^1$ and consider $C(\bar G)$ with the norm 
$p(x) = \int_0^1 |x(t)|\,dt$. 
Let $0<c<1$ and for each $n$ with $0<c-1/n$ define $x_n\in C(\bar G)$ by 
$$x_n(t)=\cases{
1\ ,&$c\le t\le 1$\cr
n(t-c)+1\ ,&$c-1/n<t<c$\cr
0\ ,&$0\le t\le c-1/n$\cr}$$
For $m\ge n$ we have $p(x_m-x_n)\le 1/n$, so $\{x_m\}$ is Cauchy. 
If $x\in C(\bar G)$, then 
$$p(x_n-x) \ge \int_0^{c-1/n} |x(t)|\,dt 
+ \int_c^1 |1-x(t)|\, d(t)\ .$$ 
This shows that if $\{x_n\}$ converges to $x$ then $x(t)=0$ for 
$0\le t<c$ and $x(t)=1$ for $c\le t\le 1$, a contradiction. 
Hence $C(\bar G)$, $p$ is not complete. 

\subsection{}		%3.2
We consider the problem of extending a given function to a larger domain. 

\plainlemma
{Let $T:D\to W$ be given, where $D$ is a subset of the seminormed space 
$V,p$ and $W,q$ is a normed linear space. 
There is at most one continuous $\bar T:\bar D\to W$ for which 
${\bar T|_D=T}$.} 

\proof 
Suppose $T_1$ and $T_2$ are continuous functions  from $\bar D$ to $W$ which 
agree with $T$ on $D$. 
Let $x\in \bar D$. 
Then there are $x_n\in D$ with $x_n\to x$ in $V,p$. 
Continuity of $T_1$ and $T_2$ shows $T_1x_n\to T_1x$ and $T_2x_n\to T_2x$. 
But $T_1x_n=T_2x_n$ for all $n$, so $T_1x= T_2x$ by the uniqueness of 
limits in the normed space $W,q$. 

\begin{theorem}\label{thm1-3A}
Let $T\in \L(D,W)$, where $D$ is a subspace of the seminormed space $V,p$ 
and $W,q$ is a Banach space.  
Then there exists a unique $\bar T\in \L(\bar D,W)$ such that 
$\bar T|_D=T$, and $|\bar T|_{p,q} = |T|_{p,q}$.
\end{theorem}

\proof Uniqueness follows from the preceding lemma. 
Let $x\in\bar D$. 
If ${x_n\in D}$ and $x_n\to x$ in $V,p$, then $\{x_n\}$ is Cauchy 
and the estimate 
$$q\bigl( T(x_m) - T(x_n)\bigr) \le Kp(x_m-x_n)$$ 
shows $\{T(x_n)\}$ is Cauchy in $W,q$, hence, convergent to some $y\in W$. 
If $x'_n\in D$ and $x'_n\to x$ in $V,p$, then $Tx'_n\to y$, so we can 
define $\bar T:\bar D\to W$ by $T(x)=y$. 
The linearity of $T$ on $D$ and the continuity of addition and scalar 
multiplication imply that $\bar T$ is linear. 
Finally, the continuity of seminorms and the estimates 
$$q\bigl(T(x_n)\bigr) \le |T|_{p,q}\, p(x_n)$$ 
show $\bar T$ is continuous on $|\bar T|_{p,q} = |T|_{p,q}$. 

\subsection{}	%3.3 
A {\it completion\/} of the seminormed space $V,p$ is a complete seminormed 
space $W,q$ and a linear injection ${T:V\to W}$ for which $\Rg(T)$ is dense 
in $W$  and $T$ preserves seminorms: 
$q(T(x))= p(x)$ for all ${x\in V}$. 
By identifying $V,p$ with $\Rg(T),q$, we may visualize $V$ as being dense 
and contained in a corresponding space that is complete. 
The completion of a normed space is a Banach space and linear injection 
as above. 
If two Banach spaces are completions of a given normed space, then we can 
use Theorem \ref{thm1-3A} to construct a linear norm-preserving bijection 
between them, so the completion of a normed space is essentially unique. 

We first construct a completion of a given seminormed space $V,p$. 
Let $W$ be the set of all Cauchy sequences in $V,p$. 
From the estimate $|p(x_n) - p(x_m)| \le p(x_n-x_m)$ it follows that 
$\bar p (\{x_n\}) = \lim_{n\to\infty} p(x_n)$ defines a function 
$\bar p:W\to \RR$ and it easily follows that $\bar p$ is a seminorm on $W$.  
For each ${x\in V}$, let $Tx= \{x,x,x,\ldots\}$, the indicated constant 
sequence. 
Then $T:V,p\to W,\bar p$ is a linear seminorm-preserving injection. 
If $\{x_n\}\in W$, then for any ${\varep>0}$ there is an integer $N$ 
such that $p(x_n-x_N) <\varep/2$ for ${n\ge N}$, and we have 
$\bar p(\{x_n\} - T(x_N))\le \varep/2 <\varep$. 
Thus, $\Rg (T)$ is dense in $W$. 
Finally, we verify that $W,\bar p$ is complete. 
Let $\{\bar x_n\}$ be a Cauchy sequence in $W,\bar p$ and for each 
${n\ge1}$ pick ${x_n\in V}$ with $\bar p(\bar x_n-T(x_n))<1/n$. 
Define $\bar x_0 = \{x_1,x_2,x_2,\ldots\}$. 
From the estimate 
$$p(x_m-x_n) = \bar p(Tx_m-Tx_n) \le 1/m + \bar p(\bar x_m-\bar x_n)+1/n$$ 
it follows that $\bar x_0 \in W$, and from 
$$\bar p(\bar x_n-\bar x_0) \le \bar p (\bar x_n-Tx_n) 
+ \bar p(Tx_n-\bar x_0) < 1/n + \lim_{m\to\infty} p(x_n-x_m)$$ 
we deduce that $\bar x_n\to \bar x_0$ in $W,\bar p$. 
Thus, we have proved the following. 

\begin{theorem}\label{thm1-3B} 
Every seminormed space has a completion. 
\end{theorem}

\subsection{}		%3.4 
In order to obtain from a normed space a corresponding normed completion, 
we shall identify those elements of $W$ which have the same limit by 
factoring $W$ by the kernel of $\bar p$. 
Before describing this quotient space, we consider quotients in a seminormed 
space. 
\medskip

\begin{theorem}\label{thm1-3C} 
Let $V,p$ be a seminormed space, $M$ a subspace of $V$ and define 
$$\hat p(\hat x) = \inf \bigl\{ p(y) : y\in\hat x\bigr\}\ ,\qquad 
\hat x\in V/M\ .$$ 
\begin{description}
\item[(a)] $V/M, \hat p$ is a seminormed space and the quotient map 
$q:V\to V/M$ has $(p,\hat p)$-seminorm $=1$. 
\item[(b)] If $D$ is dense in $V$, then $\hat D=\{\hat x:x\in D\}$ is 
dense in $V/M$. 
\item[(c)] $\hat p$ is a norm if and only if $M$ is closed. 
\item[(d)] If $V,p$ is complete, then $V/M,\hat p$ is complete.  
\end{description}
\end{theorem} 

\proof 
We leave (a) and (b) as exercises. 
Part~(c) follows from the observation that $\hat p(\hat x)=0$ if and only if 
${x\in \bar M}$. 

To prove (d), we recall that a Cauchy sequence converges if it has a 
convergent subsequence so we need only consider a sequence $\{\hat x_n\}$ 
in $V/M$ for which $\hat p(\hat x_{n+1} - \hat x_n) < 1/2^n$, ${n\ge1}$. 
For each ${n\ge1}$ we pick ${y_n\in \hat x_n}$ with 
$p(y_{n+1}-y_n) < 1/2^n$. 
For ${m\ge n}$ we obtain 
$$p(y_m-y_n) \le \sum_{k=0}^{m-1-n} p(y_{n+1+k} - y_{n+k} ) 
< \sum_{k=0}^\infty 2^{-(n+k)} = 2^{1-n}\ .$$ 
Thus $\{y_n\}$ is Cauchy in $V,p$ and part (a) shows $\hat x_n\to \hat x$ 
in $V/M$, where $x$ is the limit of $\{y_n\}$ in $V,p$. 
\qed 

Given $V,p$ and the completion $W,\bar p$ constructed for 
Theorem \ref{thm1-3B}, we consider the quotient space $W/K$ and its 
corresponding seminorm $\hat p$, where $K$ is the kernel of $\bar p$. 
The continuity of $\bar p:W\to \RR$ implies that $K$ is closed, so 
$\hat p$ is a norm on $W/K$. 
$W,\bar p$ is complete, so $W/K$, $\hat p$ is a Banach space. 
The quotient map ${q:W\to W/K}$ satisfies $\hat p(q(x)) = \hat p(\hat x) 
= \bar p(y)$ for all ${y\in q(x)}$, so $q$ preserves the seminorms. 
Since $\Rg(T)$ is dense in $W$ it follows that the linear map 
${q\circ T: V\to W/K}$ has a dense range  in $W/K$. 
We have $\hat p((q\circ T)x) = \hat p(\widehat{Tx})= p(x)$ for ${x\in V}$, 
hence $K(q\circ T) \le K(p)$. 
If $p$ is a norm this shows that ${q\circ T}$ is injective and proves 
the following. 

\begin{theorem}\label{thm1-3D}
Every normed space has a completion. 
\end{theorem} 

\subsection{}		%3.5
We briefly consider the vector space $\L(V,W)$. 

\begin{theorem}\label{thm1-3E}
If $V,p$ is a seminormed space and $W,q$ is a Banach space, then $\L(V,W)$ 
is a Banach space. 
In particular, the dual $V'$ of a seminormed space is complete.  
\end{theorem} 

\proof 
Let $\{T_n\}$ be a Cauchy sequence in $\L(V,W)$. 
For each ${x\in V}$, the estimate 
$$q(T_mx - T_nx) \le |T_m-T_n| p(x)$$ 
shows that $\{T_nx\}$ is Cauchy, hence convergent to a unique ${T(x)\in W}$. 
This defines ${T:V\to W}$ and the continuity of addition  and scalar 
multiplication in $W$ will imply that ${T\in L(V,W)}$. 
We have 
$$q\bigl( T_n(x)\bigr) \le |T_n| p(x)\ ,\qquad x\in V\ ,$$ 
and $\{|T_n|\}$ is Cauchy, hence, bounded in $\RR$, so the continuity 
of $q$ shows that $T\in \L(V,W)$ with $|T|\le K\equiv \sup\{|T_n|:n\ge1\}$. 

To show $T_n\to T$ in $\L(V,W)$, let $\varep>0$ and choose $N$ so large that 
${m,n\ge N}$ implies ${|T_m-T_n|<\varep}$. 
Hence, for ${m,n\ge N}$, we have 
$$q\bigl( T_m (x) - T_n(x)\bigr) < \varep p(x)\ ,\qquad x\in V\ .$$ 
Letting $m\to\infty$ shows that for $n\ge N$ we have 
$$q\bigl( T(x) - T_n(x)\bigr) \le \varep p(x)\ ,\qquad x\in V\ ,$$ 
so $|T-T_n|\le\varep$. 

\section{Hilbert Space}			% 4
\setcounter{equation}{0}

\subsection{}		%4.1 
A {\it scalar product\/} on the vector space $V$ is a function 
$V\times V\to \KK$ whose value at $x,y$ is denoted by $(x,y)$ and which 
satisfies 
(a)~$x\mapsto (x,y): V\to \KK$ is linear for every ${y\in V}$, 
(b)~$(x,y)=\overline{(y,x)}$, ${x,v\in V}$, and 
(c)~$(x,x)>0$ for each ${x\ne0}$. 
From (a) and (b) it follows that for each ${x\in V}$, the function 
$y\mapsto (x,y)$ is conjugate-linear, i.e., $(x,\alpha y) = \bar\alpha (x,y)$. 
The pair $V,(\cdot,\cdot)$ is called a {\it scalar product space\/}. 
\medskip

\begin{theorem}\label{thm1-4A}
If $V,(\cdot,\cdot)$ is a scalar product space, then 
\begin{description}
\item[(a)] $|(x,y)|^2 \le (x,x)\cdot (y,y)\ ,\qquad x,y\in V\ ,$ 
\item[(b)] $\|x\| \equiv (x,x)^{1/2}$ defines a norm $\|\cdot\|$ on $V$ 
for which 
$$\|x+y\|^2 + \|x-y\|^2 = 2\bigl( \|x\|^2 + \|y\|^2\bigr)\ ,\qquad 
x,y\in V\ ,\ \mbox{ and}$$ 
\item[(c)] the scalar product is continuous from $V\times V$ to $K$.
\end{description}
\end{theorem} 

\proof 
Part (a) follows from the computation 
$$0\le (\alpha x+ \beta y, \alpha x+\beta y) = 
\beta\bigl(\beta (y,y) - |\alpha|^2\bigr)$$ 
for the scalars $\alpha = -\overline{(x,y)}$ and $\beta = (x,x)$. 
To prove (b), we use (a) to verify 
$$\|x+y\|^2 \le \|x\|^2 + 2|(x,y)| + \|y\|^2 \le 
\bigl( \|x\| + \|y\|\bigr)^2\ .$$ 
The remaining norm axioms are easy and the indicated identity is easily 
verified. Part (c) follows from the estimate 
$$|(x,y) - (x_n,y_n)| \le \|x\|\, \|y-y_n\| + \|y_n\|\, \|x-x_n\|$$ 
applied to a pair of sequences, $x_n\to x$ and 
$y_n\to y$ in $V,\|\cdot\|$. 
\qed 

A {\it Hilbert space\/} is a scalar product space for which the 
corresponding normed space is complete. 

\examples 
(a) Let $V= \KK^N$ with vectors $x= (x_1,x_2,\ldots,x_N)$ and define 
$(x,y) = \sum_{j=1}^N x_j \bar y_j$. 
Then $V, (\cdot,\cdot)$  is a Hilbert space (with the norm 
$\|x\| = (\sum_{j=1}^N |x_j|^2)^{1/2}$) which we refer to as 
Euclidean space. 

(b) We define $C_0(G)$ a scalar product by 
$$(\varphi,\psi) = \int_G \varphi \bar \psi$$ 
where $G$ is open in $\RR^n$ and the Riemann integral is used. 
This scalar product space is not complete. 

(c) On the space $L^2(G)$ of (equivalence classes of) Lebesgue 
square-summable $\KK$-valued functions we define the scalar product as 
in (b) but with the Lebesgue integral. 
This gives a Hilbert space in which $C_0(G)$ is a dense subspace. 
\qed 

Suppose $V,(\cdot,\cdot)$ is a scalar product space and let 
$B,\|\cdot\|$ denote the completion of $V,\|\cdot\|$. 
For each ${y\in V}$, the function $x\mapsto (x,y)$ is linear, hence has 
a unique extension to $B$, thereby extending the definition of $(x,y)$ 
to ${B\times V}$.  
It is easy to verify that for each ${x\in B}$, the function  $y\mapsto (x,y)$ 
is in $V'$ and we can similarly extend it to define $(x,y)$ on $B\times B$. 
By checking that (the extended) function $(\cdot,\cdot)$ is a scalar 
product on $B$, we have proved the following result. 

\begin{theorem}\label{thm1-4B} 
Every scalar product space has a (unique) completion which is a Hilbert 
space and whose scalar product is the extension by continuity of the given 
scalar product. 
\end{theorem} 

\example 
$L^2(G)$ is the completion of $C_0(G)$ with the scalar product given above. 

\subsection{}		%4.2 
The scalar product gives us a notion of angles between vectors. 
(In particular, recall the formula $(x,y) = \|x\|\, \|y\|\cos(\theta)$ 
in Example~(a) above.) 
We call the vectors $x,y$ {\it orthogonal\/} if $(x,y)=0$. 
For a given subset $M$ of the scalar product space $V$, we define the 
{\it orthogonal complement\/} of $M$ to be the set 
$$M^\bot = \bigl\{ x\in V : (x,y)=0\ \mbox{ for all }\  y\in M\bigr\}\ .$$ 

\plainlemma
{$M^\bot$ is a closed subspace of $V$ and $M\cap M^\bot = \{0\}$.} 

\proof 
For each $y\in M$, the set $\{x\in V: (x,y)=0\}$ is a closed subspace and so 
then is the intersection of all these for ${y\in M}$. 
The only vector orthogonal to itself is the zero vector, so the second 
statement follows. 
\qed 

A set $K$ in the vector space $V$ is {\it convex\/} if for ${x,y\in K}$ and 
$0\le\alpha\le1$, we have $\alpha x+ (1-\alpha) y\in K$. 
That is, if a pair of vectors is in $K$, then so also is the line segment 
joining them. 

\begin{theorem}\label{thm1-4C}
A non-empty closed convex subset $K$ of the Hilbert space $H$ has an 
element of minimal norm.
\end{theorem} 

\proof 
Setting $d\equiv \inf\{\|x\|:x\in K\}$, we can find a sequence $x_n\in K$ 
for which $\|x_n\|\to d$. 
Since $K$ is convex we have $(1/2)(x_n+x_m)\in K$ for $m,n\ge1$, hence 
$\|x_n+x_m\|^2\ge 4d^2$. 
From Theorem \ref{thm1-4A}(b) we obtain the estimate $\|x_n-x_m\|^2 \le 
2(\|x_n\|^2 + \|x_m\|^2)-4d^2$. 
The right side of this inequality converges to 0, so $\{x_n\}$ is Cauchy,  
hence, convergent to some ${x\in H}$. 
$K$ is closed, so ${x\in K}$, and the continuity of the norm shows that 
$\|x\|= \lim_n \|x_n\|=d$. 
\qed 

We note that the element with minimal norm is unique, for if ${y\in K}$ 
with $\|y\|=d$, then $(1/2)(x+y)\in K$ and Theorem \ref{thm1-4A}(b) 
give us, respectively, $4d^2 \le \|x+y\|^2 = 4d^2 - \|x-y\|^2$. 
That is, $\|x-y\|=0$. 

\begin{theorem}\label{thm1-4D}
Let $M$ be a closed subspace of the Hilbert space $H$. 
Then for every ${x\in H}$ we have ${x=m+n}$, where ${m\in M}$ and 
${n\in M^\bot}$ are uniquely determined by $x$. 
\end{theorem} 

\proof 
The uniqueness follows easily, since if $x=m_1+n_1$ with $m_1\in M$, 
$n_1\in M^\bot$, then $m_1-m=n-n_1\in M\cap M^\bot = \{\theta\}$. 
To establish the existence of such a pair, define $K=\{x+y:y\in M\}$ 
and use Theorem \ref{thm1-4C} to find $n\in K$ with 
$\|n\|= \inf \{\|x+y\| :y\in M\}$.  
Then  set $m=x-n$. 
It is clear that $m\in M$ and we need only to verify that $n\in M^\bot$. 
Let $y\in M$. 
For each $\alpha\in \KK$, we have $n-\alpha y\in K$, hence 
$\|n-\alpha y\|^2\ge \|n\|^2$. 
Setting $\alpha=\beta(n,y)$, $\beta>0$, gives  us 
$|(n,y)|^2 (\beta\|y\|^2-2) \ge0$, and this can hold 
for all $\beta>0$ only if $(n,y)=0$. 
\qed 

\subsection{}		%4.3 
From Theorem \ref{thm1-4D} it follows that for each closed subspace $M$ 
of a Hilbert space $H$ we can define a function ${P_M :H\to M}$ by 
$P_M: x=m+n\mapsto m$, where ${m\in M}$ and ${n\in M^\bot}$ as above. 
The linearity of $P_M$ is immediate and the computation 
$$\|P_Mx\|^2 \le \|P_Mx\|^2 + \|n\|^2 = \|P_Mx+n\|^2 = \|x\|^2$$ 
shows $P_M\in \L(H,H)$ with $\|P_M\|\le 1$. 
Also, $P_Mx=x$ exactly when ${x\in M}$, so $P_M\circ P_M = P_M$. 
The operator $P_M$ is called the {\it projection on\/} $M$. 

If $P\in \L(B,B)$ satisfies $P\circ P=P$, then $P$ is called a 
{\it projection\/} on the Banach space $B$. 
The result of Theorem \ref{thm1-4D} is a guarantee of a rich supply of 
projections in a Hilbert space.  

\subsection{}		%4.4 
We recall that the (continuous) dual of a seminormed space is a Banach space. 
We shall show there is a natural correspondence between a Hilbert space $H$ 
and its dual $H'$. 
Consider for each fixed ${x\in H}$ the function $f_x$ defined by the scalar 
product: $f_x(y) = (x,y)$, $y\in H$. 
It is easy to check that $f_x\in H'$ and $\|f_x\|_{H'} = \|x\|$. 
Furthermore,  the map $x\mapsto f_x: H\to H'$ is linear: 
\begin{eqnarray*}
f_{x+z} &=& f_x+f_z\ ,\qquad x,z\in H\ , \cr 
f_{\alpha x} &=& \alpha f_x\ ,\qquad \alpha\in \KK\ ,\ x\in H\ .
\end{eqnarray*} 
Finally, the function $x\mapsto f_x:H\to H'$ is a norm preserving and linear 
injection. 
The above also holds in any scalar product space, but for Hilbert spaces this 
function is also surjective. 
This follows from the next result. 

\begin{theorem}\label{thm1-4E}
Let $H$ be a Hilbert space and $f\in H'$. 
Then there is an element ${x\in H}$ (and only one) for which 
$$f(y) = (x,y)\ ,\qquad y\in H\ .$$ 
\end{theorem}

\proof 
We need only verify the existence of $x\in H$. 
If $f=\theta$ we take $x=\theta$, so assume $f\ne\theta$ in $H'$. 
Then the kernel of $f$, $K= \{x\in H:f(x)=0\}$ is a closed subspace 
of $H$ with $K^\bot \ne \{\theta\}$. 
Let $n\in K^\bot$ be chosen with $\|n\|=1$. 
For each $z\in K^\bot$ it follows that $\overline{f(n)}z- \overline{f(z)}n 
\in K\cap K^\bot = \{\theta\}$, so $z$ is a scalar multiple of $n$. 
(That is, $K^\bot$ is one-dimensional.) 
Thus, each $y\in H$ is of the form $y=P_K(y) +\lambda n$ where 
$(y,n)=\lambda (n,n)=\lambda$. 
But we also have $f(y)=\bar\lambda f(n)$, since $P_K(y)\in K$, and thus 
$f(y) = (f(n)n,y)$ for all ${y\in H}$.  
\smallskip

The function $x\mapsto f_x$ from $H$ to $H'$ will occur frequently in 
our later discussions and it is called the {\it Riesz map\/} and is 
denoted by $R_H$. 
Note that it depends on the scalar product as well as the space. 
In particular, $R_H$ is an isometry of $H$ onto $H'$ defined by 
$$R_H (x)(y) = (x,y)_H\ ,\qquad x,y\in H\ .$$ 

\section{Dual Operators; Identifications}	%5
\setcounter{equation}{0}

\subsection{}		%5.1 
Suppose $V$ and $W$ are linear spaces and $T\in L(V,W)$. 
Then we define the {\it dual operator\/} $T'\in L(W^*,V^*)$ by 
$$T'(f) = f\circ T\ ,\qquad f\in  W^*\ .$$ 

\begin{theorem}\label{thm1-5A}
If $V$ is a linear space, $W,q$ is a seminorm space, and $T\in L(V,W)$ has 
dense range, then $T'$ is injective on $W'$. 
If $V,p$ and $W,q$ are seminorm spaces and $T\in \L(V,W)$, then the 
restriction of the dual $T'$ to $W'$ belongs to $\L(W',V')$ and it satisfies 
$$\|T'\|_{\L(W',V')} \le |T|_{p,q}\ .$$ 
\end{theorem} 

\proof 
The first part follows from Section 3.2. 
The second is obtained from the estimate 
$$|T'f(x)| \le \|f\|_{W'} |T|_{p,q}\, p(x)\ ,\qquad f\in W'\ ,\ x\in V\ .$$ 

We give two basic examples. 
Let $V$ be a subspace of the seminorm space $W,q$ and let ${i:V\to W}$ be 
the identity. 
Then $i'(f) = f\circ i$ is the restriction of $f$ to the subspace $V$; 
$i'$ is injective on $W'$ if (and only if) $V$ is dense in $W$. 
In such cases we may actually identify $i'(W')$ with $W'$, and we denote 
this identification by $W'\le V^*$. 

Consider the quotient map $q:W\to W/V$ where $V$ and $W,q$ are given as above. 
It is clear that if $g\in (W/V)^*$ and $f=q'(g)$, i.e., $f=g\circ q$, 
then $f\in W^*$ and $V\le K(f)$. 
Conversely, if $f\in W^*$ and $V\le K(f)$, then Theorem \ref{thm1-1A} 
shows there is a $g\in (W/V)^*$ for which $q'(g)=f$. 
These remarks show that $\Rg (q') = \{f\in W^*:V\le K(f)\}$. 
Finally, we note by Theorem \ref{thm1-3C} that $|q|_{q,\hat q}=1$, 
so it follows that $g\in (W,V)'$ if and only if $q'(g)\in W'$. 

\subsection{}		%5.2 
Let $V$ and $W$ be Hilbert spaces and $T\in \L(V,W)$. 
We define the {\it adjoint\/} of $T$ as follows:  
if $u\in W$, then the functional $v\mapsto (u,Tv)_W$ belongs to $V'$, 
so Theorem \ref{thm1-4E} shows that there is a unique $T^*u\in V$ such that 
$$(T^* u,v)_V = (u,Tv)_W\ ,\qquad u\in W\ ,\ v\in V\ .$$ 

\begin{theorem}\label{thm1-5B}
If $V$ and $W$ are Hilbert spaces and $T\in\L(V,W)$, then 
$T^* \in \L(W,V)$, $\Rg(T)^\bot = K(T^*)$ and $\Rg (T^*)^\bot = K(T)$. 
If $T$ is an isomorphism with $T^{-1} \in \L(W,V)$, then $T^*$ is an 
isomorphism and $(T^*)^{-1} = (T^{-1})^*$.
\end{theorem} 

We leave the proof as an exercise and proceed to show that dual operators 
are essentially equivalent to the corresponding adjoint. 
Let $V$ and $W$ be Hilbert spaces and denote by $R_V$ and $R_W$ the 
corresponding Riesz maps (Section~4.4) onto their respective dual spaces. 
Let $T\in \L(V,W)$ and consider its dual $T'\in \L(W',V')$ and its adjoint 
$T^*\in \L(W,V)$. 
For ${u\in W}$ and ${v\in V}$ we have $R_V\circ T^*(u)(v) = (T^*u,v)_V 
= (u,Tv)_W = R_W(u)(Tv) = (T'\circ R_Wu)(v)$. 
This shows that $R_V\circ T^* = T'\circ R_W$, so the Riesz maps permit 
us to study either the dual or the adjoint and deduce information on both. 
As an example of this we have the following.  
\medskip

\begin{corollary}\label{cor1-5C}
If $V$ and $W$ are Hilbert spaces, and $T\in \L(V,W)$, then $\Rg(T)$ is dense 
in $W$ if and only if $T'$ is injective, and $T$ is injective if and only if 
$\Rg (T')$ is dense in $V'$.  
If $T$ is an isomorphism with $T^{-1}\in \L(W,V)$, then $T'\in\L(W',V')$ 
is an isomorphism with continuous inverse. 
\end{corollary} 

\subsection{}		%5.3
It is extremely useful to make certain identifications between various 
linear spaces and we shall discuss a number of examples which will appear 
frequently in the following. 

First, consider the linear space $C_0(G)$ and the Hilbert space $L^2(G)$. 
Elements of $C_0(G)$ are functions while elements of $L^2(G)$ are 
{\it equivalence classes\/} of functions. 
Since each $f\in C_0(G)$ is square-summable on $G$, it belongs to exactly 
one such equivalence class, say $i(f)\in L^2(G)$. 
This defines a linear injection $i:C_0(G)\to L^2(G)$ whose range is dense 
in $L^2(G)$. 
The dual $i':L^2 (G)'\to C_0(G)^*$ is then a linear injection which is just  
restriction to $C_0(G)$. 

The Riesz map $R$ of $L^2(G)$ (with the usual scalar product) onto $L^2(G)'$ 
is defined as in Section~4.4. 
Finally, we have a linear injection $T:C_0(G)\to C_0(G)^*$ given in 
Section~1.5 by  
$$(Tf)(\varphi) = \int_G f(x)\bar\varphi (x)\,dx\ ,\qquad f,\varphi\in C_0(G)
\ .$$ 
Both $R$ and $T$ are possible identifications of (equivalence classes of) 
functions with conjugate-linear functionals. 
Moreover we have the important identity 
$$T= i'\circ R\circ i\ .$$ 
This shows that all four injections may be used simultaneously to identify 
the various pairs as subspaces. That is, we identify 
$$C_0(G) \le L^2 (G) = L^2 (G)'\le C_0(G)^*\ ,$$ 
and thereby reduce each of $i,R,i'$ and $T$ to the identity function from a 
subspace to the whole space. 
Moreover, once we identify $C_0(G) \le L^2 (G)$, $L^2(G)'\le C_0(G)^*$, and 
$C_0(G) \le C_0(G)^*$, by means of $i,i'$, and $T$, respectively, then it 
follows that the identification of $L^2(G)$ with $L^2(G)'$ through the 
Riesz map $R$ is possible (i.e., compatible with the three preceding) 
{\it only if\/} the $R$ corresponds to the standard scalar product on $L^2(G)$. 
For example, suppose $R$ is defined through the (equivalent)  
scalar-product 
$$(Rf)(g) = \int_G a(x) f(x) \overline{g(x)}\, dx\ ,\qquad f,g\in L^2(G)\ ,$$ 
where $a(\cdot)\in L^\infty (G)$ and $a(x)\ge c>0$, $x\in G$. 
Then, with the three identifications above, $R$ corresponds to 
multiplication by the function $a(\cdot)$. 
Other examples will be given later. 

\subsection{}		%5.4 
We shall find the concept of a sesquilinear form is as important to us as 
that of a linear operator. 
The theory of sesquilinear forms is analogous to that of linear operators 
and we discuss it briefly. 

Let $V$ be a linear space over the field $\KK$. 
A {\it sesquilinear form\/} on $V$ is a $\KK$-valued function $a(\cdot,\cdot)$ 
on the product ${V\times V}$ such that $x\mapsto a(x,y)$ is linear for 
every $y\in V$ and $y\mapsto a(x,y)$ is conjugate linear for every $x\in V$. 
Thus, each sesquilinear form $a(\cdot,\cdot)$ on $V$ corresponds to a 
unique $\A\in L(V,V^*)$ given by  
\begin{equation}\label{eq151}	
a(x,y) = \A x(y)\ ,\qquad x,y\in V\ .
\end{equation} 
Conversely, if $\A \in \L(V,V^*)$ is given, then 
Equation (\ref{eq151}) defines a sesquilinear form on $V$. 
\medskip

\begin{theorem}\label{thm1-5D}
Let $V,p$ be a normed linear space and $a(\cdot,\cdot)$ a sesquilinear 
form on $V$. The following are equivalent: 
\begin{description}
\item[(a)] $a(\cdot,\cdot)$ is continuous at $(\theta,\theta)$, 
\item[(b)] $a(\cdot,\cdot)$ is continuous on $V\times V$, 
\item[(c)] there is a constant $K\ge 0$ such that 
\begin{equation}\label{eq152}
|a(x,y)| \le Kp(x)p(y)\ ,\qquad x,y\in V\ , 
\end{equation}
\item[(d)] $\A\in \L (V,V')$.
\end{description}
\end{theorem} 

\proof 
It is clear that (c) and (d) are equivalent, (c) implies (b), and (b) 
implies (a). 
We shall show that (a) implies (c). 
The continuity of $a(\cdot,\cdot)$ at $(\theta,\theta)$ implies that there 
is a $\delta>0$ such that $p(x)\le\delta$ and $p(y)\le\delta$ imply 
$|a(x,y)| \le1$. 
Thus, if $x\ne0$ and $y\ne0$ we obtain Equation (\ref{eq152}) with 
$K= 1/\delta^2$. 

When we consider real spaces (i.e., $\KK=\RR$) there is no distinction 
between linear and conjugate-linear functions. 
Then a sesquilinear form is linear in both variables and we call it 
{\it bilinear\/}. 

\section{Uniform Boundedness; Weak Compactness}			% 6
\setcounter{equation}{0}

A sequence $\{x_n\}$ in the Hilbert space $H$  is called {\it weakly 
convergent\/} to $x\in H$ if $\lim_{n\to\infty} (x_n,v)_H = (x,v)_H$ 
for every $v\in H$. 
The weak limit $x$ is clearly unique. 
Similarly, $\{x_n\}$ is {\it weakly bounded\/} if $|(x_n,v)_H|$ is 
bounded for every $v\in H$. 

Our first result is a simple form of the {\it principle of uniform 
boundedness\/}. 

\begin{theorem}\label{thm1-6A}
A sequence $\{x_n\}$ is weakly bounded if and only if it is bounded. 
\end{theorem} 

\proof 
Let $\{x_n\}$ be weakly bounded. 
We first show that on some sphere, $s(x,r) = \{y\in H:\|y-x\|<r\}$, 
$\{x_n\}$ is uniformly bounded: 
there is a $K\ge 0$ with $|(x_n,y)_H|\le K$ for all $y\in s(x,r)$. 
Suppose not. 
Then there is an integer $n_1$ and $y_1\in s(0,1)$: 
$|(x_{n_1},y_1)_H|>1$. 
Since $y\mapsto (x_{n_1},y)_H$ is continuous, there is an $r_1<1$ 
such that $|(x_{n_1},y)_H|>1$ for $y\in s(y_1,r_1)$. 
Similarly, there is an integer $n_2>n_1$ and $\overline{s(y_2,r_2)} 
\subset s(y_1,r_1)$ such that $r_2<1/2$ and $|(x_{n_2},y)_H|>2$ for 
$y\in s(y_2,r_2)$. 
We inductively define $\overline{s(y_j,r_j)} \subset s(y_{j-1},r_{j-1})$ 
with $r_j<1/j$ and $|(x_{n_j},y)_H|>j$ for $y\in s(y_j,r_j)$. 
Since $\|y_m-y_n\|<1/n$ if $m>n$ and $H$ is complete, $\{y_n\}$ converges 
to some $y\in H$. But then $y\in s(y_j,r_j)$, hence $|(x_{n_j},y)_H|>j$ 
for all $j\ge1$, a contradiction. 

Thus $\{x_n\}$ is uniformly bounded on some sphere $s(y,r):|(x_n,y+rz)_H|
\le K$ for all $z$ with $\|z\|\le 1$. If $\|z\|\le1$, then 
$$|(x_n,z)_H| = (1/r) |x_n,y+rz)_H - (x_n,y)_H| \le 2K/r\ ,$$ 
so $\|x_n\| \le 2K/r$ for all $n$. 
\qed 

We next show that bounded  sequences have weakly convergent subsequences. 

\plainlemma 
{If $\{x_n\}$ is bounded in $H$ and $D$ is a dense subset of $H$, then 
$\lim_{n\to\infty} (x_n,v)_H= (x,v)_H$ for all $v\in D$ (if and) only if 
$\{x_n\}$ converges weakly to $x$.} 

\proof 
Let $\varep>0$ and $v\in H$. 
There is a $z\in D$ with $\|v-z\|<\varep$ and we obtain 
\begin{eqnarray*} 
|(x_n -x,v)_H| &\le&|(x_n,v-z)_H| + |(z,x_n-x)_H| +|(x,v-z)_H|\\ 
&<& \varep\|x_n\| + |(z,x_n-x)_H| + \varep \|x\|\ .\end{eqnarray*} 
Hence, for all $n$ sufficiently large (depending on $z$), we have 
$|(x_n-x,v)_H| < 2\varep \sup \{ \|x_m\| :m\ge1\}$. 
Since $\varep >0$ is arbitrary, the result follows. 

\begin{theorem}\label{thm1-6B}
Let the Hilbert space $H$ have a countable dense subset $D=\{y_n\}$. 
If $\{x_n\}$ is a bounded sequence in $H$, then it has a weakly convergent 
subsequence. 
\end{theorem} 

\proof 
Since $\{(x_n,y_1)_H\}$ is bounded in $\KK$, there is a subsequence 
$\{x_{1,n}\}$ of $\{x_n\}$ such that $\{(x_{1,n},y_1)_H\}$ converges. 
Similarly, for each $j\ge 2$ there is a subsequence $\{x_{j,n}\}$ of 
$\{x_{j-1,n}\}$ such that $\{(x_{j,n},y_k)_H\}$ converges in $\KK$ 
for $1\le k\le j$. 
It follows that $\{x_{n,n}\}$ is a subsequence of $\{x_n\}$ for which 
$\{(x_{n,n},y_k)_H\}$ converges for every $k\ge 1$. 

From the preceding remarks, it suffices to show that if $\{(x_n,y)_H\}$ 
converges in $\KK$ for every $y\in D$, then $\{x_k\}$ has a weak limit. 
So, we define $f(y) = \lim_{n\to\infty} (x_n,y)_H$, $y\in\langle D\rangle$, 
where $\langle D\rangle$ is the subspace of all linear combinations 
of elements of $D$. 
Clearly $f$ is linear; $f$ is continuous, since $\{x_n\}$ is bounded, and 
has by Theorem \ref{thm1-3A} a unique extension $f\in H'$. 
But then there is by Theorem \ref{thm1-4E} an $x\in H$ such that 
$f(y) = (x,y)_H$, $y\in H$. 
The Lemma above shows that $x$ is the weak limit of $\{x_n\}$. 

Any seminormed space which has a countable and dense subset is called 
{\it separable\/}. 
Theorem \ref{thm1-6B} states that any bounded set in a separable Hilbert 
space is {\it relatively sequentially weakly compact\/}. 
This result holds in any reflexive Banach space, but all the function 
spaces which we shall consider are separable Hilbert spaces, so 
Theorem \ref{thm1-6B} will suffice for our needs. 

\section{Expansion in Eigenfunctions}		%7 
\setcounter{equation}{0}

\subsection{}		%7.1
We consider the Fourier series of a vector in the scalar product space $H$ 
with respect to a given set of orthogonal vectors.  
The sequence $\{v_j\}$ of vectors in $H$ is called {\it orthogonal\/} 
if $(v_i,v_j)_H=0$ for each pair $i,j$ with $i\ne j$. 
Let $\{v_j\}$ be such a sequence of non-zero vectors and let $u\in H$. 
For each $j$ we define the {\it Fourier coefficient\/} of $u$ with respect 
to $v_j$ by $c_j=(u,v_j)_H/(v_j,v_j)_H$. 
For each $n\ge1$ it follows that $\sum_{j=1}^n c_jv_j$ is the projection 
of $u$ on the subspace $M_n$ spanned by $\{v_1,v_2,\ldots,v_n\}$. 
This follows from Theorem \ref{thm1-4D} by noting that $u-\sum_{j=1}^nc_jv_j$ 
is orthogonal to each $v_i$, $1\le j\le n$, hence belongs to $M_n^\bot$. 
We call the sequence of vectors {\it orthonormal\/} if they are orthogonal 
and if $(v_j,v_j)_H=1$ for each $j\ge1$. 

\begin{theorem}\label{thm1-7A} 
Let $\{v_j\}$ be an orthonormal sequence in the scalar product space $H$ 
and let $u\in H$. 
The Fourier coefficients of $u$ are given by $c_j = (u,v_j)_H$ and satisfy 
\begin{equation}\label{eq171}
\sum_{j=1}^\infty |c_j|^2 \le \|u\|^2\ .
\end{equation} 
Also we have $u=\sum_{j=1}^\infty c_jv_j$ if and only if equality holds 
in {\rm (\ref{eq171})}.
\end{theorem}

\proof 
Let $u_n \equiv \sum_{j=1}^n c_j v_j$, $n\ge1$. 
Then $u-u_n\perp u_n$ so we obtain 
\begin{equation}\label{eq172} 
\|u\|^2 = \|u-u_n\|^2 + \|u_n\|^2\ ,\qquad n\ge 1\ .
\end{equation} 
But $\|u_n\|^2 = \sum_{j=1}^n |c_j|^2$ follows since the set 
$\{v_i,\ldots,v_n\}$ is orthonormal, so we obtain $\sum_{j=1}^n |c_j|^2 
\le \|u\|^2$ for all $n$, hence (\ref{eq171}) holds. 
It follows from (\ref{eq172}) that $\lim_{n\to\infty} 
\|u-u_n\|-0$ if and only if equality holds in (\ref{eq171}).  

The inequality (\ref{eq171}) is {\it Bessel's inequality\/} and 
the corresponding equality is called {\it Parseval's equation\/}. 
The series $\sum_{j=1}^\infty c_jv_j$ above is the {\it Fourier series\/} 
of $u$ with respect to the orthonormal sequence $\{v_j\}$. 

\begin{theorem}\label{thm1-7B} 
Let $\{v_j\}$ be an orthonormal sequence in the scalar product space $H$. 
Then every element of $H$ equals the sum of its Fourier series if and only if 
$\{v_j\}$ is a {\em basis} for $H$, that is, its linear span is dense in $H$. 
\end{theorem}

\proof 
Suppose $\{v_j\}$ is a basis and let $u\in H$ be given. 
For any $\varep>0$, there is an $n\ge1$ for which the linear span $M$ of the 
set $\{v_1,v_2,\ldots,v_n\}$ contains an element which approximates $u$ 
within $\varep$. 
That is, $\inf \{\|u-w\| : w\in M\}<\varep$. 
If $u_n$ is given as in the proof of Theorem \ref{thm1-7A}, then we have 
$u-u_n\in M^\bot$. 
Hence, for any $w\in M$ we have 
$$\|u-u_n\|^2 = (u-u_n,u-w)_H \le \|u-u_n\|\, \|u-w\|\ ,$$ 
since $u_n-w\in M$. 
Taking the infimum over all $w\in M$ then gives 
\begin{equation}\label{eq173} 
\|u-u_n\|\le \inf \{\|u-w\| : w\in M\} < \varep\ .
\end{equation} 
Thus, $\lim_{n\to\infty} u_n = u$. 
The converse is clear. 

\subsection{}		%7.2 
Let $T\in \L(H)$. 
A non-zero vector $v\in H$ is called an {\it eigenvector\/} of $T$ if 
$T(v)=\lambda v$ for some $\lambda \in \KK$. 
The number $\lambda$ is the {\it eigenvalue\/} of $T$ corresponding to $v$. 
We shall show that certain operators possess a rich supply of eigenvectors. 
These eigenvectors form an orthonormal sequence to which we can apply the 
preceding Fourier series expansion techniques. 

An operator $T\in\L(H)$ is called {\it self-adjoint\/} if $(Tu,v)_H=
(u,Tv)_H$ for all $u,v\in H$. 
A self-adjoint $T$ is called {\it non-negative\/} if $(Tu,u)_H\ge0$ for 
all $u\in H$. 

\begin{lemma}\label{lem1-7-1}
\quad If $T\in \L(H)$ is non-negative self-adjoint, then 
$\|Tu\| \le \break  \|T\|^{1/2} (Tu,u)_H^{1/2}$, $u\in H$. 
\end{lemma} 

\proof 
The sesquilinear form $[u,v] \equiv (Tu,v)_H$ satisfies the first two 
scalar-product axioms and this is sufficient to obtain 
\begin{equation}\label{eq174}
|[u,v]|^2 \le[u,u][v,v]\ ,\qquad u,v\in H\ .
\end{equation} 
(If either factor on the right side is strictly positive, this follows from 
the proof of Theorem \ref{thm1-4A}. 
Otherwise, $0\le [u+tv,u+tv]= 2t[u,v]$ for all $t\in \RR$, hence, both 
sides of (\ref{eq174}) are zero.) 
The desired result follows by setting $v=T(u)$ in (\ref{eq174}). 

The operators we shall consider are the compact operators. 
If $V,W$ are seminormed spaces, then $T\in\L(V,W)$ is called {\it compact\/} 
if for any bounded sequence $\{u_n\}$ in $V$ its image $\{Tu_n\}$ has a 
subsequence which converges in $W$. 
The essential fact we need is the following. 

\begin{lemma}\label{lem1-7-2} 
If $T\in \L(H)$ is self-adjoint and compact, then there exists a vector 
$v$ with $\|v\|=1$ and $T(v)=\mu v$, where $|\mu| = \|T\|_{\L(H)} >0$. 
\end{lemma} 

\proof 
If $\lambda$ is defined to be $\|T\|_{\L(H)}$, it follows from 
Theorem \ref{thm1-2D} that there is a sequence $u_n$ in $H$ with $\|u_n\|=1$ 
and $\lim_{n\to\infty} \|Tu_n\|=\lambda$. 
Then $((\lambda^2 - T^2)u_n,u_n)_H= \lambda^2 - \|Tu_n\|^2$ converges  
to zero. 
The operator $\lambda^2-T^2$ is non-negative self-adjoint so 
Lemma \ref{lem1-7-1} implies $\{ (\lambda^2 -T^2)u_n\}$ converges 
to zero. 
Since $T$ is compact we may replace $\{u_n\}$ by an appropriate 
subsequence for which $\{Tu_n\}$ converges to some vector $w\in H$. 
Since $T$ is continuous there follows $\lim_{n\to\infty} (\lambda^2u_n) 
= \lim_{n\to\infty} T^2 u_n = Tw$, so $w=\lim_{n\to\infty} Tu_n = 
\lambda^{-2}T^2 (w)$. 
Note that $\|w\|=\lambda$ and $T^2 (w) = \lambda^2 w$. 
Thus, either $(\lambda+T)w\ne0$ and we can choose 
$v= (\lambda+T)w/\|(\lambda + T)w\|$, or  
$(\lambda+T)w=0$, and we can then choose $v=w/\|w\|$. 
Either way, the desired result follows. 

\begin{theorem}\label{thm1-7C} 
Let $H$ be a scalar product space and let $T\in \L(H)$ be self-adjoint 
and compact. Then there is an orthonormal sequence $\{v_j\}$ of 
eigenvectors of $T$ for which the corresponding sequence of eigenvalues 
$\{\lambda_j\}$ converges to zero and the eigenvectors are a basis for 
$\Rg (T)$. 
\end{theorem} 

\proof 
By Lemma \ref{lem1-7-2} it follows that there is a vector $v_1$ with 
$\|v_1\|=1$ and $T(v_1)=\lambda_1v_1$ with $|\lambda_1|=\|T\|_{\L(H)}$. 
Set $H_1 = \{v_1\}^\bot$ and note $T\{H_1\} \subset H_1$. 
Thus, the restriction $T|_{H_1}$ is self-adjoint and compact so 
Lemma \ref{lem1-7-2} implies the existence of an eigenvector 
$v_2$ of $T$ of unit length in $H_1$ with eigenvalue $\lambda_2$ 
satisfying $|\lambda_2| = \|T\|_{\L(H_1)} \le |\lambda_1|$. 
Set $H_2 = \{v_1,v_2\}^\bot$ and continue this procedure to obtain 
an orthonormal sequence $\{v_j\}$ in $H$ and sequence $\{\lambda_j\}$ 
in $\RR$ such that $T(v_j) = \lambda_j v_j$ and 
$|\lambda_{j+1}| \le |\lambda_j|$ for $j\ge1$. 

Suppose the sequence $\{\lambda_j\}$ is eventually zero; 
let $n$ be the first integer for which $\lambda_n=0$. 
Then $H_{n-1} \subset K(T)$, since $T(v_j)=0$ for $j\ge n$. 
Also we see $v_j\in \Rg (T)$ for $j<n$, so $\Rg (T)^\bot \subset 
\{v_1,v_2,\ldots,v_{n-1}\}^\bot = H_{n-1}$ and from 
Theorem \ref{thm1-5B} follows $K(T) = \Rg(T)^\bot \subset H_{n-1}$. 
Therefore $K(T) = H_{n-1}$ and $\Rg(T)$ equals the linear span of 
$\{v_1,v_2,\ldots,v_{n-1}\}$. 

Consider hereafter the case where each $\lambda_j$ is different from zero. 
We claim that $\lim_{j\to\infty} (\lambda_j)=0$. 
Otherwise, since $|\lambda_j|$ is decreasing we would have all 
$|\lambda_j| \ge\varep$ for some $\varep>0$. 
But then 
$$\|T(v_i) - T(v_j)\|^2 = \|\lambda_i v_i -\lambda_jv_j\|^2 
= \|\lambda_i v_i\|^2 + \|\lambda_j v_j\|^2 \ge 2\varep^2$$ 
for all $i\ne j$, so $\{ T(v_j)\}$ has no convergent subsequence, 
a contradiction. 
We shall show $\{v_j\}$ is a basis for $\Rg(T)$. 
Let $w\in \Rg(T)$ and $\sum b_jv_j$ the Fourier series of $w$. 
Then there is a $u\in H$ with $T(u)=w$ and we let $\sum c_jv_j$ 
be the Fourier series of $u$. 
The coefficients are related by 
$$b_j = (w,v_j)_H = (Tu,v_j)_H = (u,Tv_j)_H = \lambda_j c_j\ ,$$ 
so there follows $T(c_jv_j) = b_jv_j$, hence, 
\begin{equation}\label{eq175}	
w- \sum_{j=1}^n b_j v_j = T \left( u- \sum_{j=1}^n c_jv_j\right)\ ,
\qquad n\ge 1\ .
\end{equation} 
Since $T$ is bounded by $|\lambda_{n+1}|$ on $H_n$, and since 
$\|u-\sum_{j=1}^n c_jv_j\| \le \|u\|$ by (\ref{eq172}), 
we obtain from (\ref{eq175}) the estimate
\begin{equation}\label{eq176}
\left\| w-\sum_{j=1}^n b_ju_j\right\| \le |\lambda_{n+1}|\cdot \|u\|\ ,
\qquad n\ge 1\ .
\end{equation} 
Since $\lim_{j\to\infty} \lambda_j=0$, we have 
$w= \sum_{j=1}^\infty b_j v_j$ as desired. 

\exercises 

\begin{description}
\item[1.1.] 
Explain what ``compatible'' means in the Examples of Section 1.2.
\item[1.2.]
Prove the Lemmas of Sections 1.3 and 1.4. 
\item[1.3.] 
In Example (1.3.b), show $V/M$ is isomorphic to $\KK$. 
\item[1.4.] 
Let $V= C(\bar G)$ and $M= \{\varphi \in C(\bar G): \varphi|_{\partial G} 
=0\}$. Show $V/M$ is isomorphic to $\{\varphi|_{\partial G} :\varphi\in C
(\bar G)\}$, the space of ``boundary values'' of functions in $V$.
\item[1.5.] 
In Example (1.3.c), show $\hat\varphi_1=\hat\varphi_2$ if and only if 
$\varphi_1$ equals $\varphi_2$ on a neighborhood of $\partial G$. 
Find a space of functions isomorphic to $V/M$. 
\item[1.6.] 
In Example (1.4.c), find $K(D)$ and $\Rg(D)$ when 
$V= \{\varphi \in C^1(\bar G) :\varphi (a) = \varphi(b)\}$. 
\item[1.7.] 
Verify the last sentence in the Example of Section 1.5. 
\item[1.8.] 
Let $M_\alpha \le V$ for each $\alpha\in A$; show $\cap \{M_\alpha :\alpha 
\in A\}\le V$.
\medskip
\item[2.1.] 
Prove parts (d) and (e) of Lemma \ref{lem1-2-1}. 
\item[2.2.] 
If $V_1,p_1$ and $V_2,p_2$ are seminormed spaces, show $p(x)\equiv p_1(x_1) 
+ p_2(x_2)$ is a seminorm on the product $V_1\times V_2$. 
\item[2.3.] 
Let $V,p$ be a seminormed space. Show limits are unique if and only if 
$p$ is a norm. 
\item[2.4.] 
Verify all Examples in Section 2.1. 
\item[2.5.] 
Show $\cap_{\alpha\in A}\bar S_\alpha =\overline{\cap_{\alpha\in A}S_\alpha}$. 
Verify $\bar S=$ smallest closed set containing $S$. 
\item[2.6.] 
Show $T:V,p\to W,q$ is continuous if and only if $S$ closed in $W,q$ 
implies $T(S)$ closed in $V,p$. 
If $T\in L(V,W)$, then $T$  continuous if and only if $K(T)$ is closed. 
\item[2.7.] 
The composition of continuous functions is continuous; 
$T\in \L(V,W)$, $S\in \L(U,V)\Rightarrow T\circ S\in \L(U,W)$ and 
$|T\circ S| \le |T|\ |S|$. 
\item[2.8.] 
Finish proof of Theorem \ref{thm1-2D}. 
\item[2.9.]
Show $V'$ is isomorphic to $\L(V,\KK)$; they are equal  only if $\KK=\RR$.
\medskip
\item[3.1.] 
Show that a closed subspace of a seminormed space is complete. 
\item[3.2.] 
Show that a complete subspace of a normed space is closed. 
\item[3.3.] 
Show that a Cauchy sequence is convergent if and only if it has a 
convergent subsequence. 
\item[3.4.] 
Let $V,p$ be a seminormed space and $W,q$ a Banach space. 
Let the sequence $T_n\in \L(V,W)$ be given {\it uniformly bounded\/}: 
$|T_n|_{p,q} \le K$ for all $n\ge 1$. 
Suppose that $D$ is a dense subset of $V$ and $\{T_n(x)\}$ converges in $W$ 
for each $x\in D$. Then show $\{T_n(x)\}$ converges in $W$ for each 
$x\in V$ and $T(x) = \lim T_n (x)$ defines $T\in \L(V,W)$. 
Show that completeness of $W$ is necessary above. 
\item[3.5.] 
Let $V,p$ and $W,q$ be as given above. Show $\L(V,W)$ is isomorphic to 
$\L(V/\Ker (p),W)$. 
\item[3.6.] 
Prove the remark in Section 3.3 on uniqueness of a completion. 
\medskip
\item[4.1.] 
Show that the norms $p_2$ and $r_2$ of Section 2.1 are not obtained from 
scalar products. 
\item[4.2.] 
Let $M$ be a subspace of the scalar product space $V(\cdot,\cdot)$. 
Then the following are equivalent: 
$M$ is dense in $V$, $M^\bot =\{\theta\}$, and 
$\|f\|_{V'} = \sup \{|(f,v)_V|: v\in M\}$ for every $f\in V'$. 
\item[4.3.] 
Show $\lim x_n=x$ in $V$, $(\cdot,\cdot)$ if and only if $\lim \|x_n\|=\|x\|$ 
and $\lim f(x_n) = f(x)$ for all $f\in V'$. 
\item[4.4.] 
If $V$ is a scalar product space, show $V'$ is a Hilbert space. 
Show that the Riesz map of $V$ into $V'$ is surjective only if $V$ is 
complete. 
\medskip
\item[5.1.] 
Prove Theorem \ref{thm1-5B}. 
\item[5.2.] 
Prove Corollary \ref{cor1-5C}.  
\item[5.3.] 
Verify $T= i'\circ R\circ i$ in Section 5.3. 
\item[5.4.] 
In the situation of Theorem \ref{thm1-5B}, prove the following are 
equivalent: $\Rg (T)$ is closed, $\Rg (T^*)$ is closed, 
$\Rg(T) = K(T^*)^\bot$, and $\Rg (T^*) = K(T)^\bot$. 
\medskip
\item[7.1.] 
Let $G= (0,1)$ and $H= L^2(G)$. Show that the sequence $v_n(x)=2\sin (n\pi x)$, 
$n\ge1$ is orthonormal in $H$. 
\item[7.2.] 
In Theorem \ref{thm1-7A}, show that $\{u_n\}$ is a Cauchy sequence. 
\item[7.3.] 
Show that the eigenvalues of a non-negative self-adjoint operator are all 
non-negative. 
\item[7.4.] 
In the situation of Theorem \ref{thm1-7C}, show $K(T)$ is the 
orthogonal complement of the linear span of $\{v_1,v_2,v_3,\ldots\}$. 
\end{description}