Functional Analysis

0. Preface

Functional Analysis can mean many different things to different mathematicians. The core of the subject, however, is to study linear spaces with some topology which allows us to do analysis; ones like spaces of functions, spaces of operators acting on the space of functions, etc.

The book consists of two parts. The first part covers the basics of Banach spaces theory with the emphasis on its applications. The second part covers topological vector spaces, especially locally convex ones, generalization of banach spaces. In both parts, we give principal results e.g., the closed graph theorem, resulting in some repetition. One reason for doing this is that one often only needs a Banach-version of such results. Another reason is that the approach seems more pedagogically sound; the statement of the results in their full generality may (at first) look cryptical to students new to them.

Knowledge of measure theory will not be needed except for Chapter 3.4 5 (Lp spaces). On the other hand, solid knowledge in general topology is mandatory.

1. Preliminaries

In this chapter we gather some standard results, some of which are not necessarily in functional analysis proper yet will be used later, primarily to fix language and formulation. We prefer to omit proofs for them since they can be easily found somewhere else.

1. Theorem (Ascoli) Let $ℱ$ be a family of continuous real-valued functions on a compact space $K$. If

(i) For each $x\in K$, $\underset{f\in ℱ}{\sup }|f(x)|<\mathrm{\infty }$.

(ii) For each $ϵ>0$, every point $x\in K$ has a neighborhood $\omega$ such that $|f(y)-f(x)|<ϵ$ for every $y\in \omega$ and every $f\in ℱ$ (i.e., $ℱ$ is equicontinuous),

then

$ℱ$ is totally bounded in the metric $d(f,g)=\underset{K}{\sup }|f-g|$.

1. Theorem (Heine-Borel) A totally bounded subset of a complete metric space is relatively compact.

As a corollary, the conclusion of the Ascoli's theorem thus says the closure of $ℱ$ is compact since $C(K)$ is complete.

A subset of a topological space is said to be meager if it is a union of countably many subsets of the space that have dense complement.

1. Theorem (Baire) Every complete metric space is non-meager.
Proof: (TODO: we should provide one.)

By modifying the proof so to use compact sets instead of balls we can show that every locally compact (Hausdorff?) space is also non-meager, though we will not be needing this.

1. Exercise Show by contradiction that the set of real numbers is uncountable, using the theorem.

In particular, a dense set may be meager.

1. Exercise Give an example of a dense but meager set. (Hint: you can find such an example in the book.)

1. Theorem (Tychonoff) Every product space of a nonempty collection of compact spaces is compact.

An linear operator from a linear space to a scalar field is called a linear functional.

1. Theorem (Hahn-Banach) Let $𝒳$ be a real vector space and $p$ be a function on $𝒳$ such that $p(x+y)\le p(x)+p(y)$ and $p(tx)=tp(x)$ for any $x\in 𝒳$ and any $t>0$. If $ℳ\subset 𝒳$ is a closed subspace and $f$ is a linear functional on $ℳ$ such that $f\le p$, then $f$ admits a linear extension $F$ defined in $𝒳$ such that $F\le p$.
Proof: First suppose that $𝒳=\{x+tz;x\in ℳ,t\in ℝ\}$ for some $z\in ̸ℳ$. By hypothesis we have:

$f(x)+f(y)\le p(x-z)+p(y+z)$ for all $x,y\in ℳ$,

which is equivalent to:

$\sup \{f(x)-p(x-z);x\in M\}\le \inf \{p(y+z)-f(y);y\in M\}$.

Let $c$ be some number in between the sup and the inf. Define $F(x+tz)=f(x)+tc$ for $x\in ℳ,t>0$. It follows that $F$ is an desired extension. Indeed, $f=F$ on $ℳ$ being clear, we also have:

$F(x+tz)\le tp(\frac{x}{t}+z)$ if $t>0$

and

$F(x+tz)\le -tp(\frac{x}{-t}-z)$ if $t<0$.

Let $\mathrm{\Omega }$ be the collection of pairs $(H,g_H)$ where $H$ is linear space with $ℳ\subset \mathscr{H}\subset X$ and $g_H$ is a linear function on $ℳ$ that extends $f$ and is dominated by $p$. It can be shown that $\mathrm{\Omega }$ is partially ordered and the union of every totally ordered sub-collection of $\mathrm{\Omega }$ is in $\mathrm{\Omega }$ (TODO: need more details). Hence, by Zorn's Lemma, we can find the maximal element $(L,g_L)$ and by the early part of the proof we can show that $ℒ=𝒳$. $◻$

We remark that a different choice of $c$ in the proof results in a different extension. Thus, an extension given by the Hahn-Banach theorem in general is not unique.

1. Exercise State the analog of the theorem for complex vector spaces and prove that this version can be reduced to the real version. (Hint: $Ref(ix)=Imf(x)$)

(TODO: mention moment problem.)

1 Lemma Let $g,f_1,...,f_n$ be linear functionals on the same linear space. $g$ is a linear combination of $f_1,...,f_n$ if and only if $\underset{j=1}{\overset{n}{\cap }}\ker f_j\subset \ker g$.
Proof: The direct part is clear. We shall show the converse by induction. Suppose $n=1$. We may suppose that there is $y$ such that $f_1(y)=1$. For any $x$, since $x-f_1(x)y\in \ker f_1\subset \ker g$,

$g(x-f_1(x)y)=0$ or $g(x)=g(y)f_1(x)$.

The basic case is thus proven. Now, suppose the lemma holds for some $n-1$. As before, we may suppose that there is some $y$ such that

$f_n(y)=1$ while $f_1(y)=f_2(y)=...=f_{n-1}(y)=0$.

For any $x\in \underset{j=1}{\overset{n-1}{\cap }}\ker f_j$, since $f_k(x-f_n(x)y)=0$ for every $k=1,2,...,n$, $g(x-f_n(x)y)=0$. Hence, the application of the inductive hypothesis to the linear functional $x\mapsto g(x-f_n(x)y)$ gives:

$g(x-f_n(x)y)=\sum _{j=1}^{n-1}{a}_j{f}_j(x)$

for some scalars $a_1,...,a_n$. $◻$ (TODO: the proof can be clearer and simpler).

Part 1: Banach spaces

2. Normed space

Let $𝒳$ be a linear space. A norm is a real-valued function $f$ on $X$, with the notation $\Vert \cdot \Vert =f(\cdot )$, such that

• (i) $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$ (w:triangular inequality)
• (ii) $\Vert \lambda x\Vert =|\lambda |\Vert x\Vert$ for any scalar $\lambda$
• (iii) $\Vert x\Vert =0$ implies $x=0$.

(ii) implies that $\Vert 0\Vert =0$. This and (ii) then implies $0=\Vert x-x\Vert \le \Vert x\Vert +\Vert -x\Vert =2\Vert x\Vert$ for all $x$; that is, norms are always non-negative. With the metric $d(x,y)=\Vert x-y\Vert$ a normed space is a metric space. If only (i) and only (ii) hold, the function is called a seminorm.

A linear space with a norm is called a normed space. We define the operator norm of a continuous linear operator $f$ between normed spaces $𝒳$ and $𝒴$, denoted by $\Vert f\Vert$, by

$\Vert f\Vert =\underset{\Vert x{\Vert }_{𝒳}\le 1}{\sup }\Vert f(x){\Vert }_{𝒴}$

2 Theorem Let $T$ be a linear operator from a normed space $𝒳$ to a normed space $𝒴$.

• (i) $T$ is continuous if and only if there is a constant $C>0$ such that $\Vert Tx\Vert \le C\Vert x\Vert$ for all $x\in X$
  
• (ii) $\Vert f\Vert =\inf \{$ any $C$ as in (i) $\}$, and $\Vert f\Vert =\underset{\Vert x\Vert =1}{\sup }\Vert f(x)\Vert$ if $𝒳$ has nonzero element.

Proof: See w:bounded operator and see w:operator norm. $◻$

A complete normed space is called a Banach space. While there is seemingly no prototypical example of a Banach space, we still give one example of a Banach space: $𝒞(K)$, the space of all continuous functions on a compact space $𝒦$, can be identified with a Banach space by introducing the norm:

$\Vert \cdot \Vert =su{p}_K|\cdot |$

It is a routine exercise to check that this is indeed a norm. The completeness holds since, from real analysis, we know that a uniform limit of a sequence of continuous functions is continuous. In concrete spaces like this one, one can directly show the completeness. More often than that, however, we will see that the completeness is a necessary condition for some results (especially, reflexivity), and thus the space has to be complete. The matter will be picked up in the later chapter.

2 Theorem (completion) Every normed space can be identified with a dense subspace of some Banach space.

The completeness is essential, for example, in the next result.

2 Theorem Let $𝒳$ be a normed linear space, and $𝒴$ be a Banach space. Let $ℳ\subset 𝒳$ be a subspace. If $ℳ$ is dense, then a continuous linear operator $T:ℳ\to 𝒴$ admits a unique extension to $𝒳$.
Proof: For each $x\in 𝒳$, let $x_n\in ℳ$ be Cauchy such that $x_n\to x$. By continuity $\Vert T{x}_n-T{x}_m\Vert \le \Vert T\Vert \Vert x_n-x_m\Vert \to 0$ as $n,m\to \mathrm{\infty }$. Since $𝒴$ is complete, $\underset{n\to \mathrm{\infty }}{\lim }T{x}_n$ exists in $𝒴$, and we denote it by $\tilde{T}x$. From this definition It is clear that the extension $\tilde{T}$ is linear. Moreover, if there is another extension $S$, then $Sx$ is the limit of $T{x}_n$; whence, the extension is unique. Finally, $\tilde{T}$ is continuous since

$\Vert \tilde{T}x\Vert =\Vert \tilde{T}\underset{n\to \mathrm{\infty }}{\lim }{x}_n\Vert \le \underset{n\to \mathrm{\infty }}{\lim }\Vert T\Vert \Vert x_n\Vert =\Vert T\Vert \Vert x\Vert$. $◻$

2 Theorem (open mapping theorem) Let $𝒳,𝒴$ be Banach spaces. If $T:𝒳\to 𝒴$ is a continuous linear surjection, then it is an open mapping; i.e., it maps open sets to open sets.
Proof: Let $B(r)=\{x\in 𝒳;\Vert x\Vert <r\}$. Since $T$ is surjective, $\underset{n=1}{\overset{\mathrm{\infty }}{\cup }}T(B(n))=T(\underset{n=1}{\overset{\mathrm{\infty }}{\cup }}B(n))=T(X)=Y$. Then by Baire's Theorem

A linear operator from a normed space $𝒳$ to $𝒴$ is said to be closed if its graph, that is the set $\{(x,Tx);x\in \mathrm{dom}(𝒳)\}$, is closed in $𝒳\times 𝒴$.

2 Corollary If $T$ is a continuous linear operator between Banach spaces with closed range, then there exists a $K>0$ such that if $y\in \mathrm{im}(T)$ then $\Vert x\Vert \le K\Vert y\Vert$ for some $x$ with $Tx=y$.
Proof: Since $T$ is an open mapping, there exists a $\delta >0$ such that

$B_{\delta }(T(0))\subset T(B_1(0))$

where by $B_r(x)$ we denote an open ball of radius $r$ about $x$. By replacing $\delta$ with $\delta /2$ we can also archive that

$\overline{B_{\delta }(0)}\subset T(B_1(0))$

Let $y\in \mathrm{im}(T)\ne 0$ be given. Then there is $z$ with $Tz=y$. Since $\Vert T\left(\frac{\delta }{\Vert y\Vert }z\right)\Vert =\delta$, there exists $z_0\in B_1(0)$ such that $T(z_0)=T\left(\frac{\delta }{\Vert y\Vert }z\right)$. Let $x=\frac{\Vert y\Vert }{\delta }{z}_0$. Then

$\Vert x\Vert =\frac{\Vert y\Vert }{\delta }\Vert z_0\Vert \le \frac{1}{\delta }\Vert y\Vert$ and $Tx=y$

Finally, if $y=0$, just take $x=0$. $◻$

2 Exercise Use a quotient map to simplify the proof. (Hint: the answer will be given in the chapter 4)

2 Corollary If $(𝒳,\Vert \cdot {\Vert }_1)$ and $(𝒳,\Vert \cdot {\Vert }_2)$ are Banach spaces, then the norms $\Vert \cdot {\Vert }_1$ and $\Vert \cdot {\Vert }_2$ are equivalent; i.e., each norm is dominated by the other.
Proof: Let $I:(𝒳,\Vert \cdot {\Vert }_1+\Vert \cdot {\Vert }_2)\to (𝒳,\Vert \cdot {\Vert }_1)$ be the identity map. Then we have:

$\Vert I\cdot {\Vert }_1=\Vert \cdot {\Vert }_1\le (\Vert \cdot {\Vert }_1+\Vert \cdot {\Vert }_2)$.

This is to say, $I$ is continuous. Since Cauchy sequences apparently converge in the norm $\Vert \cdot {\Vert }_1+\Vert \cdot {\Vert }_2$, the open mapping theorem says that the inverse of $I$ is also continuous, which means explicitly:

$\Vert \cdot {\Vert }_1+\Vert \cdot {\Vert }_2=\Vert I^{-1}\cdot {\Vert }_1+\Vert I^{-1}\cdot {\Vert }_2\le \Vert I^{-1}\Vert \Vert \cdot {\Vert }_1$.

By the same argument we can show that $\Vert \cdot {\Vert }_1+\Vert \cdot {\Vert }_2$ is dominated by $\Vert \cdot {\Vert }_2$ $◻$

The technique using the corollary will be exploited in the proof of the next result.

2 Theorem (closed graph theorem) Let $𝒳,𝒴$ be Banach spaces, and $T:𝒳\to 𝒴$ a linear operator. The following are equivalent.

• (i) $T$ is continuos.
• (ii) If $x_j\to 0$ and $T{x}_j$ is convergent, then $T{x}_j\to 0$.
• (iii) The graph of $T$ is closed.

Proof: That (i) implies (ii) is clear. To show (iii), suppose $(x_j,T{x}_j)$ is convergent in $X$. Then $x_j$ converges to some $x_0$ or $x_j-x_0\to 0$, and $T{x}_j-Tx$ is convergent. Thus, if (ii) holds, $T(x_j-x)\to 0$. Finally, to prove (iii) $\Rightarrow$ (i), we note that Corollary 2.something gives the inequality:

$\Vert \cdot \Vert +\Vert T\cdot \Vert \le K\Vert \cdot \Vert$

since by hypothesis the norm in the left-hand side is complete. Hence, if $x_j\to x$, then $T{x}_j\to Tx$. $◻$

Note that when the domain of a linear operator is not a Banach space (e.g., just dense in a Banach space), the condition (ii) is not sufficient for the graph of the operator to be closed. A function is said to be densely defined if its domain is dense.

2 Exercise Let $T$ be a closed densely defined operator and $S$ be a linear operator with $\mathrm{dom}(T)\subset \mathrm{dom}(S)$. If there are constants $a,b$ such that (i) $0\le a<1$ and $b>0$ and (ii) $\Vert S(u)\Vert \le a\Vert T(u)\Vert +b\Vert u\Vert$ for every $u\in \mathrm{dom}(T)$, then $T+S$ is a closed densely defined operator.

2 Corollary A linear functional $u$ is continuous if and only if it has closed kernel.
Proof: $\ker (u)=u^{-1}(\{0\})$ is closed if $u$ is continuous. To show the converse, suppose $x_j\to 0$ and $u(x_j)$ is convergent. Since the corollary is obvious when $u$ is identically zero, we may suppose that there is $z$ such that $u(z)=1$. Then the sequence $x_j-u(x_j)z$ has a limit in the kernel of $u$, since the kernel is closed. It follows:

$0=u(\underset{j\to \mathrm{\infty }}{\lim }{x}_j-u(x_j)z)=u(\underset{j\to \mathrm{\infty }}{\lim }{x}_j)+u(-\underset{j\to \mathrm{\infty }}{\lim }u(x_j)z)=-\underset{j\to \mathrm{\infty }}{\lim }u(x_j)$. $◻$

When $𝒳,𝒴$ are normed spaces, by $L(𝒳,𝒴)$ we denote the space of all continuous linear operators from $𝒳$ to $𝒴$.

2 Theorem If $𝒴$ is complete, then every Cauchy sequence $T_n$ in $L(𝒳,𝒴)$ converges to a limit $T$ and $\Vert T\Vert =\underset{n\to \mathrm{\infty }}{\lim }\Vert T_n\Vert$.
Proof: Let $T_n$ be a Cauchy sequence in operator norm. For each $x\in 𝒳$, since

$\Vert T_n(x)-T_m(x)\Vert \le \Vert T_n-T_m\Vert \Vert x\Vert$

and $𝒴$ is complete, there is a limit $y$ to which $T_n(x)$ converges. Define $T(x)=y$. $T$ is linear since the limit operations are linear. It is also continuous since $\Vert T(x)\Vert =\Vert \lim T_n(x)\Vert \le \lim \Vert T_n\Vert \Vert x\Vert$. Finally, $\lim \Vert T_n-T\Vert =\underset{\Vert x\Vert \le 1}{\sup }\Vert \lim T_n(x)-T(x)\Vert$ and $|\Vert T^n\Vert -\Vert T\Vert |\le \Vert T^n-T\Vert \to 0$ as $n\to \mathrm{\infty }$. $◻$

2 Theorem (uniform boundedness principle) Let $ℱ$ be a family of continuos functions $f:X\to Y$ where $Y$ is a normed linear space. Suppose that $M\subset X$ is non-meager and that:

$\sup \{\Vert f(x)\Vert :f\in ℱ\}<\mathrm{\infty }$ for each $x\in M$

It then follows: there is some $G\subset X$ open and such that

(a) $\sup \{\Vert f(x)\Vert :f\in ℱ,x\in G\}<\mathrm{\infty }$

If we assume in addition that each member of $ℱ$ is a linear operator and $X$ is a normed linear space, then

(b) $\sup \{\Vert f\Vert :f\in ℱ\}<\mathrm{\infty }$

Proof: Let $E_j={\cap }_{f\in ℱ}\{x\in X;\Vert f(x)\Vert \le j\}$ be a sequence. By hypothesis, $M\subset \bigcup _{j=1}^{\mathrm{\infty }}{E}_j$ and each $E_j$ is closed since $\{x\in X;\Vert f(x)\Vert >j\}$ is open by continuity. It then follows that some $E_N$ has an interior point $y$; otherwise, $M$ fails to be non-meager. Hence, (a) holds. To show (b), making additional assumptions, we can find an open ball $B=B(2r,y)\subset E_N$. It then follows: for any $f\in ℱ$ and any $x\in X$ with $\Vert x\Vert =1$,

$\Vert f(x)\Vert =r^{-1}\Vert f(rx+y)-f(y)\Vert \le 2r^{-1}N$. $◻$

A family $\mathrm{\Gamma }$ of linear operators is said to be equicontinuous if given any neighborhood $W$ of $0$ we can find a neighborhood $V$ of $0$ such that:

$f(V)\subset W$ for every $f\in \mathrm{\Gamma }$

The conclusion of the theorem, therefore, means that the family satisfying the hypothesis of the theorem is equicontinuous.

2 Corollary Let $𝒳,𝒴,𝒵$ be normed spaces. Let $T:𝒳\times 𝒴$ be a bilinear or sesquilinear operator. If $T$ is separately continuous (i.e., the function is continuous when all but one variables are fixed) and $𝒴$ is complete, then $T$ is continuous.
Proof: For each $y\in 𝒴$,

$\sup \{\Vert T(x,y){\Vert }_{𝒵};\Vert x{\Vert }_{𝒳}\le 1\}=\Vert T(\cdot ,y)\Vert$

where the right-hand side is finite by continuity. Hence, the application of the principle of uniform boundedness to the family $\{T(x,\cdot );\Vert x{\Vert }_{𝒳}\le 1\}$ shows the family is equicontinuous. That is, there is $K>0$ such that:

$\Vert T(x,y){\Vert }_{𝒵}\le K\Vert y{\Vert }_{𝒴}$ for every $\Vert x{\Vert }_{𝒳}le1$ and every $y\in 𝒴$.

The theorem now follows since $𝒳\times 𝒴$ is a metric space. $◻$

Since scalar multiplication is a continuous operation in normed spaces, the corollary says, in particular, that every linear operator on finite dimensional normed spaces is continuous. The next is one more example of the techniques discussed so far.

2. Theorem (Hahn-Banach) Let $(𝒳,\Vert \cdot \Vert )$ be normed space and $ℳ\subset 𝒳$ be a linear subspace. If $z$ is a linear functional continuous on $ℳ$, then there exists a continuos linear functional $w$ on $𝒳$ such that $z=w$ on $ℳ$ and $\Vert z\Vert =\Vert w\Vert$.
Proof: Apply the Hahn-Banach stated in Chapter 1 with $\Vert z\Vert \Vert \cdot \Vert$ as a sublinear functional dominating $z$. Then:

$\Vert z\Vert =\sup \{\Vert w(x)\Vert ;x\in ℳ,\Vert x\Vert \le 1\}\le \sup \{\Vert w(x)\Vert ;x\in 𝒳,\Vert x\Vert \le 1\}=\Vert w\Vert \le \Vert z\Vert$;

that is, $\Vert z\Vert =\Vert w\Vert$. $◻$

2. Corollary Let $ℳ$ be a subspace of a normed linear space $𝒳$. Then $x$ is in the closure of $ℳ$ if and only if $z(x)$ = 0 for any $z\in {𝒳}^{*}$ that vanishes on $ℳ$.
Proof: By continuity $z(\overline{ℳ})\subset \overline{z(ℳ)}$. Thus, if $x\in \overline{ℳ}$, then $z(x)\in \overline{z(ℳ)}=\{0\}$. Conversely, suppose $x\in ̸\overline{ℳ}$. Then there is a $\delta >0$ such that $\Vert y-x\Vert \ge \delta$ for every $y\in ℳ$. Define a linear functional $z(y+\lambda x)=\lambda$ for $y\in ℳ$ and scalars $\lambda$. For any $\lambda \ne 0$, since $-{\lambda }^{-1}y\in ℳ$,

$|z(y+\lambda x)|=|\lambda |{\delta }^{-1}\delta \le {\delta }^{-1}|\lambda ||{\lambda }^{-1}y+x\Vert =\Vert y+\lambda x\Vert$.

Since the inequality holds for $\lambda =0$ as well, $z$ is continuous. Hence, in view of the Hahn-Banach theorem, $z\in 𝒳$ while we still have $z=0$ on $ℳ$ and $z(x)\ne 0$. $◻$

2 Corollary If $(𝒳,\Vert \cdot \Vert )$ is a normed space and $z(x)=z(y)$ for all $z\in {𝒳}^{*}$, then $x=y$
Proof: Suppose $x\ne y$, and define $z(\lambda (x-y))=\lambda \Vert x-y\Vert$ for scalars $\lambda$. Now, $z$ is continuous since its domain is finite-dimensional, and so by the Hahn-Banach theorem $z\in {𝒳}^{*}$. Thus, $|z(x)-z(y)|=\Vert x-y\Vert \ne 0$, or $z(x)\ne z(y)$. $◻$

Here is a classic application.

2 Theorem Let $𝒳,𝒴$ be Banach spaces, $T:𝒳\to 𝒴$ be a linear operator. If $x_n\to 0$ implies that $(z\circ T)x_n\to 0$ for every $z\in {𝒳}^{*}$, then $T$ is continuous.
Proof: Suppose $x_n\to 0$ and $T{x}_n\to y$. For every $z\in {𝒳}^{*}$, by hypothesis and the continuity of $z$,

$0=\underset{n\to \mathrm{\infty }}{\lim }z(T{x}_n)=z(y)$.

Now, by the preceding corollary $y=0$ and the continuity follows from the closed graph theorem. $◻$

2 Theorem Let $𝒳$ be a Banach space and $E\subset 𝒳$. Then $E$ is bounded if and only if $\underset{E}{\sup }|f|<\mathrm{\infty }$ for every $f\in {𝒳}^{*}$
Proof: By continuity,

$\sup \{|f(x)|;x\in E\}\le \Vert f\Vert \underset{E}{\sup }\Vert \cdot \Vert$.

This proves the direct part. For the converse, define $T_{x}f=f(x)$ for $x\in E,f\in {𝒳}^{*}$. By hypothesis

$|T_{x}f|\le \underset{E}{\sup }|f|$ for every $x\in E$.

Thus, by the principle of uniform boundedness, there is $K>0$ such that:

$|T_{x}f|\le K\Vert f\Vert$ for every $x\in E,f\in {𝒳}^{*}$

Hence, in view of Theorem 2.something, for $x\in E$,

$\Vert x\Vert =\sup \{|f(x)|;f\in {𝒳}^{*},\Vert f\Vert \le 1\}\le K$.

$◻$

2. Corollary Let $(𝒳,\Vert \cdot \Vert )$ be Banach, $f_j\in {𝒳}^{*}$ and $ℳ\subset 𝒳$ dense and linear. Then $f_j(x)\to 0$ for every if and only if (1) $\underset{j}{\sup }\Vert f_j\Vert <\mathrm{\infty }$ and $f_j(y)\to 0$ for every $y\in ℳ$.
Proof: Since $f_j$ is Cauchy, it is bounded. This shows the direct part. To show the converse, let $x\in 𝒳$. If $y_j\in ℳ$, then

$\begin{array}{cc}|f_j(x)|& \le |f_j(x-y_j)|+|f_j(y_j)|\\ & =\underset{j}{\sup }\Vert f_j\Vert \Vert x-y_j\Vert +|f_j(y_j)|\\ \end{array}$

By denseness, we can take $y_j$ so that $\Vert y_j-x\Vert \to 0$. $◻$

2 Exercise Show the corollary may fail if $f_j$ is not bounded.

2 Theorem Let $T$ be a continuous linear operator into a Banach space. If $\Vert I-T\Vert <1$ where $I$ is the identity operator, then the inverse $T^{-1}$ exists, is continuous and can be written by:

$T^{-1}(x)=\sum _{k=0}^{\mathrm{\infty }}{(I-T)}^k(x)$ for each $x$ in the range of $T$.

Proof: For $n\ge m$, we have:

$\Vert \sum _{k=m}^n{(I-T)}^k(x)\Vert \le \Vert x\Vert \sum _{k=m}^n{\Vert I-T\Vert }^k$.

Since the series is geometric by hypothesis, the right-hand side is finite. Let $S_n=\sum _{k=0}^n{(I-T)}^k$. By the above, each time $x$ is fixed, $S_n(x)$ is a Cauchy sequence and the assumed completeness implies that the sequence converges to the limit, which we denote by $S(x)$. Since for each $x$ $\underset{n\ge 1}{\sup }\Vert S_n(x)\Vert <\mathrm{\infty }$, it follows from the principle of uniform boundedness that:

$\underset{n\ge 1}{\sup }\Vert S_n\Vert \le \mathrm{\infty }$.

Thus, by the continuity of norms,

$\Vert S(x)\Vert =\underset{n\to \mathrm{\infty }}{\lim }\Vert S_n(x)\Vert \le (\underset{n\ge 1}{\sup }\Vert S_n\Vert )\Vert x\Vert$.

This shows that $S$ is a continuous linear operator since the linearity is easily checked. Finally,

$\Vert TS(x)-x\Vert =\Vert \underset{n\to \mathrm{\infty }}{\lim }-(I-T{)}^{n+1}(x)\Vert \le \Vert x\Vert \underset{n\to \mathrm{\infty }}{\lim }\Vert I-T{\Vert }^{n+1}=0$.

Hence, $S$ is the inverse to $T$. $◻$

2 Corollary The space of invertible continuous linear operators $𝒳$ is an open subspace of $L(𝒳,𝒳)$.
Proof: If $T\in L(𝒳,𝒳)$ and $\Vert S-T\Vert <\frac{1}{\Vert T^{-1}\Vert }$, then $S$ is invertible. $◻$

If $𝐅$ is a scalar field and $𝒳$ is a normed space, then $L(𝒳,𝐅)$ is called a dual of $𝒳$ and is denoted by ${𝒳}^{*}$. In view of Theorem 2.something, it is a Banach space.

3. Hilbert space

A Banach space is called a Hilbert space if for each ordered pair $(x,y)$ of elements in the space there is a unique complex (or real) number called an inner product of $x$ and $y$, denoted by $⟨x,y⟩$, subject to the following conditions:

• (i) The functional $f(x)=⟨x,y⟩$ is linear.
• (ii) $⟨x,y⟩=\overline{⟨y,x⟩}$
• (iii) $⟨x,x⟩=0$ implies that $x=0$

We define $\Vert x\Vert =⟨x,x{⟩}^{1/2}$ and this can be shown to be a norm. Indeed, it is clear that $\Vert \alpha x\Vert =|\alpha |\Vert x\Vert$ and (iii) is the reason that $\Vert x\Vert =0$ implies that $x=0$. Finally, the triangular inequality follows from the next lemma.

3 Lemma (Schwarz's inequality) $|⟨x,y⟩|\le \Vert x\Vert \Vert y\Vert$ where the equality holds if and only if we can write $x=\lambda y$ for some scalar $\lambda$.

If we assume the lemma for a moment, since $\mathrm{Re}(\alpha )\le |\alpha |$ for any complex number $\alpha$ it then follows:

$\Vert x+y{\Vert }^2$	$=\Vert x{\Vert }^2+2\mathrm{Re}⟨x,y⟩+\Vert y{\Vert }^2$
	$\le \Vert x{\Vert }^2+2|⟨x,y⟩|+\Vert y{\Vert }^2$
	$\le (\Vert x\Vert +\Vert y\Vert {)}^2$

Proof of Lemma: First suppose $\Vert x\Vert =1$. If $\alpha =\overline{⟨x,y⟩}$, it then follows:

$0\le \Vert \alpha x-y{\Vert }^2=|\alpha {|}^2-2\mathrm{Re}(\alpha ⟨x,y⟩)+\Vert y{\Vert }^2=-|\alpha {|}^2+\Vert y{\Vert }^2$

where the equation becomes $0$ if and only if $x=\lambda y$. Since we may suppose that $x\ne 0$, the general case follows easily. $◻$

3 Theorem A normed linear space is a pre-Hilbert space if and only if $\Vert x-y{\Vert }^2=2\Vert x{\Vert }^2+2\Vert y{\Vert }^2-\Vert x+y{\Vert }^2$ (polarization identity)
Proof: The direct part is clear. To show the converse, we define

$⟨x,y⟩=4^{-1}(\Vert x+y{\Vert }^2-\Vert x-y{\Vert }^2+i\Vert x+iy{\Vert }^2-i\Vert x-iy{\Vert }^2$.

It is then immediate that $⟨x,y⟩=\overline{⟨y,x⟩}$, $⟨-x,y⟩=-⟨x,y⟩$ and $⟨ix,y⟩=i⟨x,y⟩$. Moreover, since the calculation:

$\Vert x_1+x_2+y{\Vert }^2-\Vert x_1+x_2-y{\Vert }^2$	$=2\Vert x_1+y{\Vert }^2-2\Vert x_1-y{\Vert }^2-\Vert x_1-x_2+y{\Vert }^2-\Vert x_1-x_2-y{\Vert }^2$
	$=\sum _{j=1}^2\Vert x_j+y{\Vert }^2-\Vert x_j-y{\Vert }^2$,

we have: $⟨x_1+x_2,y⟩=⟨x_1,y⟩+⟨x_2,y⟩$. If $\alpha$ is a real scalar and ${\alpha }_j$ is a sequence of rational numbers converging to $\alpha$, then by continuity and the above, we get: $⟨\alpha x,y⟩=\underset{j\to \mathrm{\infty }}{\lim }⟨{\alpha }_{j}x,y⟩=\alpha ⟨x,y⟩.◻$

3 Lemma Let $\mathscr{H}$ be a pre-Hilbert. Then $x_j\to x$ in norm if and only if for any $y\in \mathscr{H}$ $\Vert x_j\Vert \to \Vert x\Vert$ and $⟨x_j-x,y⟩\to 0$ as $j\to \mathrm{\infty }$.
Proof: The direct part holds since:

$|\Vert x_j\Vert -\Vert x\Vert |+|⟨x_j-x,y⟩|\le \Vert x_j-x\Vert (1+\Vert y\Vert )\to 0$ as $j\to \mathrm{\infty }$.

Conversely, we have:

$\Vert x_j-x{\Vert }^2=\Vert x_j{\Vert }^2-2\mathrm{Re}⟨x_j,x⟩+\Vert x{\Vert }^2\to 0$ as $j\to \mathrm{\infty }$

$◻$

3 Lemma Let $D$ be a convex closed subset of a Hilbert space. Then $D$ admits a unique element $z$ such that

$\Vert z\Vert =\inf \{\Vert x\Vert ;x\in D\}$.

Proof: For each $n=1,2,...$, there is some $x_n\in D$ such that $0\le \Vert x_n\Vert -\delta \le n^{-1}$. That is, $\delta =\underset{n\to \mathrm{\infty }}{\lim }\Vert x_n\Vert$. Since $D$ is convex,

$\frac{x_n+x_m}{2}\in D$ and so $\delta \le \frac{1}{2}\Vert x_n+x_m\Vert$.

It follows:

$\Vert x_n-x_m{\Vert }^2$	$=2\Vert x_n{\Vert }^2+2\Vert x_m{\Vert }^2-\Vert x_n+x_m{\Vert }^2$
	$\le 2\Vert x_n{\Vert }^2+2\Vert x_m{\Vert }^2-4{\delta }^2$
	$\to 2{\delta }^2+2{\delta }^2-4{\delta }^2=0$ as $n,m\to \mathrm{\infty }$

This is to say, $x_n$ is Cauchy. Since $D$ is a closed subset of a complete metric space, whence it is complete, there is a limit $z\in D$ with $\Vert z\Vert =\delta$. The uniqueness follows since if $\Vert w\Vert =\delta$ we have

$\Vert z-w{\Vert }^2=2\Vert z{\Vert }^2+2\Vert w{\Vert }^2-\Vert z+w{\Vert }^2$

where the right side is $\le 0$ for the same reason as before. $◻$

3 Theorem (orthogonal decomposition) Let $\mathscr{H}$ be a Hilbert space and $ℳ\subset \mathscr{H}$ be a subspace. For every $x\in H$ we can write

$x=y+z$

where $y\in ℳ$ and $z\in {ℳ}^{\mathrm{\perp }}$, and $y$ and $z$ are uniquely determined by $x$
Proof: Clearly $x-ℳ$ is convex, and it is also closed since the image of a closed set under translation is again closed. Now, the application of the above lemma gives an element $y\in ℳ$ such that $\Vert x-y\Vert =\inf \{\Vert x-w\Vert ;w\in ℳ\}$. Let $z=x-y$. This $z$ turns out to be orthogonal to $ℳ$. Indeed, for any $w\in ℳ$ with $\Vert w\Vert =1$, if $\lambda =⟨w,z⟩$, we have

$\Vert z{\Vert }^2\le \Vert x-(y+\lambda w){\Vert }^2=\Vert z-\lambda w{\Vert }^2=\Vert z{\Vert }^2-2\mathrm{Re}⟨z,\lambda w⟩+|\lambda {|}^2=\Vert z{\Vert }^2-|\lambda {|}^2$

since $y+\lambda w\in ℳ$. That is, $0\le -|\lambda {|}^2$ or $\lambda =0$. The uniqueness holds since if $x=y_1+y_2=y_2+z_2$ and $y_1,y_2\in ℳ$ and $z_1,z_2\in {ℳ}^{\mathrm{\perp }}$, then

$\Vert y_1-y_2{\Vert }^2=⟨y_1-y_2,y_1-y_2⟩=⟨y_1-y_2,z_1-z_2⟩$ (since $y_1-y_2=z_1-z_2$).

The right-hand side is equal to 0 since $z_1-z_2\in {ℳ}^{\mathrm{\perp }}$. Thus, $y_1=y_2$. $◻$

3 Corollary Let $ℳ$ be a subspace of a Hilbert space $\mathscr{H}$. Then

• (i) ${ℳ}^{\mathrm{\perp }}=\{0\}$ if and only if $ℳ$ is dense in $\mathscr{H}$.
• (ii) ${ℳ}^{\mathrm{\perp }\mathrm{\perp }}=\overline{ℳ}$.

Proof: Left as an exercise.

3 Theorem (representation theorem) Every continuous linear functional $f$ on a Hilbert space $\mathscr{H}$ has the form:

$f(x)=⟨x,y⟩$ with a unique $y\in ℳ$ and $\Vert f\Vert =\Vert y{\Vert }_{\mathscr{H}}$

Proof: Let $ℳ=f^{-1}(\{0\})$. Since $f$ is continuous and any finite set is closed in $ℂ$ or $ℝ$, $ℳ$ is closed. If $ℳ=\mathscr{H}$, then take $y=0$. If not, by the above corollary, there is a nonzero $z\in \mathscr{H}$ orthogonal to $ℳ$. By replacing $z$ with $z\Vert z{\Vert }^{-1}$ we may suppose that $\Vert z\Vert =1$. For any $x\in \mathscr{H}$, since $zf(x)-f(z)x$ is in the kernel of $f$ and thus is orthogonal to $z$, we have:

$0=⟨zf(x)-f(z)x,z⟩=⟨z,z⟩f(x)-⟨f(z)x,z⟩$

and so:

$f(x)=⟨x,\overline{f(z)}z⟩$

The uniqueness follows since $⟨x,y_1⟩=⟨x,y_2⟩$ for all $x\in \mathscr{H}$ means that $y_1-y_2\in {\mathscr{H}}^{\mathrm{\perp }}=\{0\}$. Finally, we have the identity:

$\Vert y\Vert =|⟨\frac{y}{\Vert y\Vert },y⟩|\le \Vert f\Vert \le \Vert y\Vert$

where the last inequality is Schwarz's inequality. $◻$

3 Exercise Using Lemma 1.6 give an alternative proof of the preceding theorem.

In view of Theorem 3.something, for each $x\in \mathscr{H}$, we can write: $x=y+z$ where $y\in ℳ$, a closed subspace of $\mathscr{H}$, and $z\in {ℳ}^{\mathrm{\perp }}$. Denote each $y$, which is uniquely determined by $x$, by $\pi (x)$. The function $\pi$ then turns out to be a linear operator. Indeed, for given $x_1,x_2\in \mathscr{H}$, we write:

$x_1=y_1+z_1,x_2=y_2+z_2$ and $x_1+x_2=y_3+z_3$

where $y_j\in ℳ$ and $z_j\in {ℳ}^{\mathrm{\perp }}$ for $j=1,2,3$. By the uniqueness of decomposition

$\pi (x_1)+\pi (x_2)=y_1+y_2=y_3=\pi (x_1+x_2)$.

The similar reasoning shows that $\pi$ commutes with scalars. Now, for $x=y+z\in \mathscr{H}$ (where $y\in ℳ$ and $z\in {ℳ}^{\mathrm{\perp }}$), we have:

$\Vert x{\Vert }^2=\Vert \pi (x){\Vert }^2+\Vert z{\Vert }^2\ge \Vert \pi (x){\Vert }^2$

That is, $\pi$ is continuous with $\Vert \pi \Vert \le 1$. In particular, when $ℳ$ is a nonzero space, there is $x_0\in ℳ$ with $\pi (x_0)=x_0$ and $\Vert x_0\Vert =1$ and consequently $\Vert pi\Vert =1$. Such $\pi$ is called an orthogonal projection.

The next theorem gives an alternative to the Hahn-Banach theorem.

3 Theorem Let $ℳ$ be a linear (not necessarily closed) subspace of a Hilbert space. Every continuous linear functional on $ℳ$ can be extended to a unique continuous linear functional on $\mathscr{H}$ that has the same norm and vanishes on ${ℳ}^{\mathrm{\perp }}$
Proof: Define $g=f\circ {\pi }_{\overline{ℳ}}$. By the same argument used in the proof of Theorem 2.something (Hahn-Banach) and the fact that $\Vert {\pi }_{ℱ}\Vert =1$, we obtain $\Vert f\Vert =\Vert g\Vert$. Since $g=0$ on ${ℳ}^{\mathrm{\perp }}$, it remains to show the uniqueness. For this, let $h$ be another extension with the desired properties. Since the kernel of $f-h$ is closed and thus contain $\overline{ℳ}$, $f=h$ on $\overline{ℳ}$. Hence, for any $x\in \mathscr{H}$,

$h(x)=(h\circ {\pi }_{\overline{ℳ}})x=(f\circ {\pi }_{\overline{ℳ}})x=g(x)$.

The extension $g$ is thus unique. $◻$

3 Theorem Let ${ℳ}_n$ be an increasing sequence of closed subspaces, and $ℳ$ be the closure of ${ℳ}_1\cup {ℳ}_2\cup ...$. If ${\pi }_{ℳ}$ is an orthogonal projection onto $ℳ$, then for every $x\in ℳ$ ${\pi }_{{ℳ}_n}(x)\to x$. Proof: Let $𝒩=\{x\in ℳ;{\pi }_{{ℳ}_n}(x)\to x(n\to \mathrm{\infty })\}$. Then $𝒩$ is closed. Indeed, if $x_j\in 𝒩$ and $x_j\to x$, then

$\Vert {\pi }_{{ℳ}_n}(x)-x\Vert \le 2\Vert x-x_j\Vert +\Vert {\pi }_{{ℳ}_n}(x_j)-x_j\Vert$

and so $x\in 𝒩$. Since $ℳ\subset \overline{𝒩}$, the proof is complete. $◻$

3 Lemma (Bessel's inequality) If $u_k$ is a sequence orthonormal in a Hilbert space $\mathscr{H}$, then

$\sum _{k=1}^{\mathrm{\infty }}|⟨x,u_k⟩{|}^2\le \Vert x{\Vert }^2$ for any $x\in \mathscr{H}$.

Proof: If $⟨x,y⟩=0$, then $\Vert x+y{\Vert }^2=\Vert x{\Vert }^2+\Vert y{\Vert }^2$. Thus,

$\Vert x-\sum _{k=1}^n⟨x,u_k⟩{\Vert }^2=\Vert x{\Vert }^2-2\mathrm{Re}\sum _{k=1}^n|⟨x,u_k⟩{|}^2+\sum _{k=1}^n|⟨x,u_k⟩{|}^2=\Vert x{\Vert }^2-\sum _{k=1}^n|⟨x,u_k⟩{|}^2$.

Letting $n\to \mathrm{\infty }$ completes the proof. $◻$.

2 Lemma A normed space $(𝒳,\Vert \cdot \Vert )$ is complete (in the sense of metric spaces) if and only if $\sum _{j=1}^{\mathrm{\infty }}\Vert x_j\Vert <\mathrm{\infty }$ implies $\sum _{j=1}^{\mathrm{\infty }}{x}_j$ exists.
Proof: If $\sum _{j=1}^{\mathrm{\infty }}\Vert x_j\Vert <\mathrm{\infty }$, then since

$\Vert \sum _{j=1}^n{x}_j-\sum _{j=1}^m{x}_j\Vert =\Vert \sum _{j=m+1}^n{x}_j\Vert \le \sum _{j=m+1}^n\Vert x_j\Vert \to 0$ as $n>m$ and $n,m\to \mathrm{\infty }$,

$\sum _{j=1}^{\mathrm{\infty }}{x}_j$ exists by completeness. Conversely, suppose $x_j$ is a Cauchy sequence. Thus, for each $j=1,2,...$, there exists an index $k_j$ such that $\Vert x_n-x_m\Vert <2^{-j}$ for any $n,m\ge k_j$. Let $x_{k_0}=0$. Then $\sum _{j=0}^{\mathrm{\infty }}\Vert x_{k_{j+1}}-x_{k_j}\Vert <2$. Hence, by assumption we can get the limit $x=\sum _{j=0}^{\mathrm{\infty }}{x}_{k_{j+1}}-x_{k_j}$, and since

$\Vert x_{n_k}-x\Vert =\Vert \sum _{j=1}^n{x}_{k_{j+1}}-x_{k_j}-x\Vert \to 0$ as $n\to \mathrm{\infty }$,

we conclude that $x_j$ has a subsequence converging to $x$; thus, it converges to $x$. $◻$

3 Theorem (Parseval) Let $u_k$ be a orthonormal sequence in a Hilbert space $\mathscr{H}$. Then the following are equivalent:

• (i) $\mathrm{span}\{u_1,u_2,...\}$ is dense in $\mathscr{H}$.
• (ii) For each $x\in \mathscr{H}$, $x=\sum _{k=1}^{\mathrm{\infty }}⟨x,u_k⟩u_k$.
• (iii) For each $x,y\in \mathscr{H}$, $⟨x,y⟩=\sum _{k=1}^{\mathrm{\infty }}⟨x,u_k⟩\overline{⟨y,u_k⟩}$.
• (iv) $\Vert x{\Vert }^2=\sum _{k=1}^{\mathrm{\infty }}|⟨x,u_k⟩{|}^2$ (the Parseval equality).

Proof: Let $ℳ=\mathrm{span}\{u_1,u_2,...\}$. If $v\in ℳ$, then it has the form: $v=\sum _{k=1}^{\mathrm{\infty }}{\alpha }_k{u}_k$ for some scalars ${\alpha }_k$. Since $⟨v,u_j⟩=\sum _{k=1}^{\mathrm{\infty }}{a}_j⟨u_k,u_j⟩=a_j$ we can also write: $v=\sum _{k=1}^{\mathrm{\infty }}⟨v,u_k⟩u_k$. Let $y=\sum _{k=1}^{\mathrm{\infty }}⟨x,u_k⟩u_k$. Bessel's inequality and that $\mathscr{H}$ is complete ensure that $y$ exists. Since

$⟨y,v⟩=\sum _{k=1}^{\mathrm{\infty }}⟨x,u_k⟩⟨u_k,v⟩=\sum _{k=1}^{\mathrm{\infty }}⟨x,⟨v,u_k⟩u_k⟩=⟨x,\sum _{k=1}^{\mathrm{\infty }}⟨v,u_k⟩u_k⟩=⟨x,v⟩$

for all $v\in ℳ$, we have $x-y\in {ℳ}^{\mathrm{\perp }}=\{0\}$, proving (i) $\Rightarrow$ (ii). Now (ii) $\Rightarrow$ (iii) follows since

$|⟨x,y⟩-\sum _{k=1}^n⟨x,u_k⟩\overline{⟨y,u_k⟩}|=|⟨x,y-\sum _{k=1}^n⟨y,u_k⟩u_k⟩|\to 0$ as $n\to \mathrm{\infty }$

To get (iii) $\Rightarrow$ (iv), take $x=y$. To show (iv) $\Rightarrow$ (i), suppose that (i) is false. Then there exists a $z\in (\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\mathrm{\{}{\mathrm{u}}_{\mathrm{1}},{\mathrm{u}}_{\mathrm{2}},\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{\}}{)}^{\mathrm{\perp }}$ with $z\ne 0$. Then

$\sum _{k=1}^{\mathrm{\infty }}|⟨z,u_k⟩{|}^2=0<\Vert z{\Vert }^2$.

Thus, (iv) is false.

3 Theorem (adjoint) Let $𝒳,𝒴$ be Hilbert spaces. Let $T$ be a linear operator from a subspace of $𝒳$ to $𝒴$. For each $y\in 𝒴$, $⟨Tx,y{⟩}_{𝒴}$ is continuous for all $x$ in the domain of $T$ if and only if there exists some $z\in 𝒳$ such that:

$⟨Tx,y{⟩}_{𝒴}=⟨x,z{⟩}_{𝒳}$ for all $x$ in the domain of $T$.

Here $z$ is uniquely determined by $y$ if and only if the domain of $T$ is dense.
Proof: The existence follows from the Hahn-Banach theorem (or the projection argument for Hilbert spaces) and the Riesz representation theorem. (FIXME: need more detail). Suppose the uniqueness of $z$. It then follows: if $x\in \mathrm{dom}T$ and $w\in (\mathrm{dom}T{)}^{\mathrm{\perp }}$, then:

$⟨Tx,y⟩=⟨x,z⟩=⟨x,z+w⟩$

and the uniqueness shows that $z=z+w$ and $w=0$. Conversely, if $\mathrm{dom}T$ is dense, then

$⟨Tx,y⟩=⟨x,z⟩=⟨x,w⟩$ for $x\in \mathrm{dom}T$ implies that $z-w\in (\mathrm{dom}T{)}^{\mathrm{\perp }}=0$.

When the domain of $T$ is dense, for each $y$ satisfying the continuity condition, we denote $z$ in the theorem by $T^{*}y$. $T^{*}$ can be shown to be linear (since $T$ is densely defined but not because of linearlity), and we call it the adjoint of $T$.

3 Theorem Let $𝒳,𝒴$ be Hilbert spaces and $T:𝒳\to 𝒴$ be a closed densely defined (i.e., having dense domain and closed graph) operator. Then:

• (i) $T^{*}$ is a closed densely defined.
• (ii) $T^{**}=T$.

Proof: Let $z\in (\mathrm{dom}{T}^{*}{)}^{\mathrm{\perp }}$. If $u\oplus y$ is orthogonal to the graph of $T$, then

$0=⟨u\oplus y,x\oplus Tx⟩=⟨u,x⟩+⟨y,Tx⟩$ for all $x$ in the domain of $T$;

that is, $y\in (\mathrm{dom}{T}^{*}{)}^{\mathrm{\perp }}$. Thus, :

$0=⟨0,u⟩+⟨z,y⟩$ for $u\oplus y\in (\mathrm{gra}T{)}^{\mathrm{\perp }}$.

This is to say: $0\oplus z\in (\mathrm{gra}T{)}^{\mathrm{\perp }\mathrm{\perp }}=\mathrm{gra}T$ and $z=T(0)=0$. (ii) (i) implies the existence of $T^{**}$ and the equality holds since the following are equivalent:

• (a) $w\oplus z\in \mathrm{gra}{T}^{**}$
• (b) $0=⟨-T^{*}y,w⟩+⟨y,z⟩=⟨(-T^{*}y)\oplus y,w\oplus z⟩$ for $u\oplus y\in \mathrm{gra}(T{)}^{\mathrm{\perp }}$.
• (c) $0=⟨u\oplus y,w\oplus z⟩$ for $u\oplus y\in \mathrm{gra}(T{)}^{\mathrm{\perp }}$.
• (d) $w\oplus z\in \mathrm{gra}{T}^{\mathrm{\perp }\mathrm{\perp }}$.

Note that the equivalence of (b) and (c) follows, as in the proof of (i), from that $y$ is in the domain of $T^{*}$ and so $-T^{*}y=u$. We conclude that $T=T^{**}$, and (by applying (ii)) that $T^{*}$ has closed graph. $◻$

3 Corollary Let ${\mathscr{H}}_1,{\mathscr{H}}_2$ be Hilbert spaces, and $T:H_1\to H_2/math>acloseddenselydefinedlinearoperator.<math>u\in \mathrm{dom}$ if and only if there is some $K>0$ such that:

$\Vert ⟨T^{*}f,u\Vert \le K\Vert f\Vert$ for every $f\in {\mathrm{dom}}^{*}$

Proof: If $Tu$ is defined, then take $K=\Vert Tu{\Vert }_{{\mathscr{H}}_2}$. As for the converse, the estimate means that the map $f\to ⟨T^{*}f,u$ is continuous; hence, $u\in {\mathrm{dom}}^{*}*=\mathrm{dom}$. $◻$

3 Theorem Let ${\mathscr{H}}_1,{\mathscr{H}}_2,{\mathscr{H}}_3$ be Hilbert spaces, and $T_j:H_j\to H_{j+1}(j=1,2)$ densely defined linear operators. Suppose $\mathrm{ran}{T}_1\subset \mathrm{dom}{T}_2$. Then $\mathrm{gra}{T}_1*\circ T_2^{*}\subset \mathrm{gra}(T_2\circ T_1{)}^{*}$ where the equality holds if $T_j^{*}*=T_j(j=1,2)$
Proof: Let $z\in \mathrm{dom}{T}_1^{*}\circ T_2^{*}$ be given. Then

$⟨(T_2\circ T_1)u,z⟩=⟨T_{1}u,T_2^{*}z⟩==⟨u,T_1^{*}\circ T_2^{*}z⟩$ for every $u\in \mathrm{dom}{T}_1$. But, by definition, $(T_1\circ T_2{)}^{*}$ denotes an element such as $T_1^{*}\circ T_2^{*}z⟩$.

Hence, $(T_1\circ T_2{)}^{*}=T_1^{*}\circ T_2^{*}$ in $\mathrm{dom}{T}_1^{*}\circ T_2^{*}$. Finally, if $T_j^{*}*=T_j(j=1,2)$, then, by the fact we have just proved

$\mathrm{gra}{T}_1^{*}\circ T_2^{*}\subset \mathrm{gra}(T_2\circ T_1{)}^{*}=\mathrm{gra}(T_2^{*}*\circ T_1^{*}*{)}^{*}\subset \mathrm{gra}{T}_1^{*}\circ T_2^{*}$. $◻$

3 Theorem Let $ℳ$ be a closed linear subspace of a Hilbert space $\mathscr{H}$. $P$ is an orthogonal projection onto $ℳ$ if and only if $P=P^{*}=P^2$ and the range of $P$ is $ℳ$.
Proof: The direct part is clear except for $P=P^{*}$. To see this, note that for any $x\in \mathscr{H}$ $⟨Px,x⟩$ is a real number since $⟨Px,x⟩=⟨Px,x-Px⟩+⟨Px,Px⟩$. Hence,

$⟨Px,x⟩=⟨x,Px⟩=⟨P^{*}x,x⟩$ for every $x\in \mathscr{H}$.

By Lemma 3.something $P=P^{*}$. For the converse we only have to verify $x-Px\in {ℳ}^{\mathrm{\perp }}$. But it holds since $\ker P=\ker P^{*}=(\mathrm{ran}P{)}^{\mathrm{\perp }}$ and for any $x$ $P(x-Px)=Px-P^{2}x=0$. $◻$

As an application we shall prove:

3 Theorem Let ${\mathscr{H}}_1,{\mathscr{H}}_2$ be Hilbert spaces, $T:{\mathscr{H}}_1\to {\mathscr{H}}_2$ be a closed densely defined linear operator. Then $T$ is surjective if and only if there is a $K>0$ such that

$\Vert f{\Vert }_{{\mathscr{H}}_2}\le K\Vert T^{*}f{\Vert }_{{\mathscr{H}}_1}$ for every $f\in \mathrm{dom}{T}^{*}$.

Proof: Suppose $T$ is surjective. It suffices to show that

$E=\{f;f\in \mathrm{dom}{T}^{*},\Vert T^{*}f{\Vert }_{{\mathscr{H}}_1}\le 1\}$

is bounded. If $g\in {\mathscr{H}}_2$, then we can write $g=Tu$ and so:

$|⟨f,g⟩{|}_{{\mathscr{H}}_2}=|⟨T^{*}f,u⟩{|}_{{\mathscr{H}}_1}\le \Vert u\Vert$ for every $f\in E$.

That $E$ is bounded now follows from Theorem 2.something. To show the converse, let $g\in {\mathscr{H}}_2$ be given. Since $T^{*}$ is injective, we can define a linear functional $L$ by $L(T^{*}f)=⟨f,g{⟩}_{{\mathscr{H}}_2}$ for $f\in {\mathscr{H}}_2$. For every $f\in \mathrm{dom}{T}^{*}$,

$|L(T^{*}f)|=|⟨f,g{⟩}_{{\mathscr{H}}_2}|\le K\Vert T^{*}f\Vert$.

Thus, $L$ is continuous on the range of $T^{*}$. It follows from the Hahn-Banach theorem that we may assume that $L$ is defined and continuous on ${\mathscr{H}}_1$. Thus, by Theorem 3.something, we can write $L(\cdot )=⟨\cdot ,u{⟩}_{{\mathscr{H}}_1}$ in ${\mathscr{H}}_1$. Then:

$L(T^{*}f)=⟨f,g{⟩}_{{\mathscr{H}}_2}=⟨T^{*}f,u{⟩}_{{\mathscr{H}}_1}=⟨f,Tu{⟩}_{{\mathscr{H}}_2}$ for every $f\in {\mathscr{H}}_2$,

and so $Tu=g$. $◻$

3 Corollary Let $T,{\mathscr{H}}_1,{\mathscr{H}}_2$ be as given in the preceding theorem. Then $\mathrm{ran}T$ is closed if and only if $\mathrm{ran}{T}^{*}$ is closed.
Proof: Suppose $\mathrm{ran}T$ is closed. Since $T^{*}$ restricted to $(\ker T^{*}{)}^{\mathrm{\perp }}$ has the same range as $T^{*}$, we only have to show this operator has the closed range. Thus, suppose $T^{*}{f}_j$ is convergent and $f_j\in (\ker T^{*}{)}^{\mathrm{\perp }}=\mathrm{ran}T$. Applying the preceding theorem with $ℱ=\mathrm{ran}T$, we get:

$\Vert f_j-f_k{\Vert }_{{\mathscr{H}}_2}\le K\Vert T^{*}(f_j-f_k){\Vert }_{{\mathscr{H}}_1}\to 0$ as $j,k\to \mathrm{\infty }$.

Thus, there is a limit $f$ of $f_j$ and since $T^{*}$ is a closed operator, $T^{*}{f}_j\to T^{*}f$. Hence, $\mathrm{ran}{T}^{*}$ is closed. The converse holds since $T^{**}=T$. $◻$

3 Lemma Let $S,T$ be linear operators from a Hilbert space $\mathscr{H}$ over $ℂ$ to itself. If $⟨Tx,x⟩=⟨Sx,x⟩$ for $x\in \mathscr{H}$, then $S=T$.
Proof: Let $R=T-S$. We have $0=⟨R(x+y),x+y⟩=⟨Rx,y⟩+⟨Ry,x⟩$ and $0=i⟨R(x+iy),x+iy⟩=⟨Rx,y⟩+i^2⟨Ry,x⟩$. Summing the two we get: $0=2⟨Rx,y⟩$ for $x,y\in \mathscr{H}$. Taking $y=Rx$ gives $0=\Vert Rx{|}^2$ for all $x\in \mathscr{H}$ or $R=0$. $◻$

3 Lemma Let $T$ be a continuous linear operator on a Hilbert space $\mathscr{H}$. Then $T{T}^{*}=T^{*}T$ if and only if $\Vert Tx\Vert =\Vert T^{*}x\Vert$ for all $x\in \mathscr{H}$.

Continuous linear operators with the above equivalent conditions are said to be normal. For example, an orthogonal projection is normal. See w:normal operator for additional examples and the proof of the above lemma.

3 Theorem Let $N$ be a normal operator. If $\alpha$ and $\beta$ are distinct eigenvalues of $N$, then the respective eigenspaces of $\alpha$ and $\beta$ are orthogonal to each other.
Proof: Let $I$ be the identity operator, and $x,y$ be arbitrary eigenvectors for $\alpha ,\beta$, respectively. Since the adjoint of $\alpha I$ is $\overline{\alpha }I$, we have:

$0=\Vert (N-\alpha I)x\Vert =\Vert (N-\alpha I{)}^{*}x\Vert =\Vert N^{*}x-\overline{\alpha }x\Vert$.

That is, $N^{*}x=\overline{\alpha }x$, and we thus have:

$\overline{\alpha }⟨x,y⟩=⟨N^{*}x,y⟩=⟨x,Ny⟩=\overline{\beta }⟨x,y⟩$

If $⟨x,y⟩$ is nonzero, we must have $\alpha =\beta$. $◻$

A sequence $x_n$ in a Hilbert space is said to converge weakly to $x$ if $⟨x_n-x,y⟩\to 0$

3 Theorem Let $x_k$ be an orthogonal sequence in a Hilbert space $\mathscr{H}$. The series $\sum _{k=1}^{\mathrm{\infty }}{x}_k$ converges if and only if the series $\sum _{k=1}^{\mathrm{\infty }}⟨x_k$ converges weakly. Proof: (TODO: write one using the uniform boundedness principle and completeness.)

3 Exercise Let $\mathscr{H}$ be a Hilbert space with orthogonal basis $e_1,e_2,...$, and $x_n$ be a sequence with $\Vert x_n\Vert \le K$. Prove that there is a subsequence of $x_n$ that converges weakly to some $x$ and that $\Vert x\Vert \le K$. (Hint: Since $⟨x_n,e_k⟩$ is bounded, by Cantor's diagonal argument, we can find a sequence $x_{n_k}$ such that $⟨x_{n_k},e_k⟩$ is convergent for every $k$.)

We will obtain a more general version of this in the next chapter.

4 Geometry of Banach spaces

The major topic of the chapter is the notion of reflexibility of Banach spaces, which, as it turns out, is a geometric property.

If $𝒳$ is a Banach space, we can introduce a topology called the "weak topology" in addition to the original norm one, as follows.

From the definitions follows immediately:

4. Corollary

• (i) $x_n$ converges weakly (i.e., in the terms of a weak topology) to $x$ in $𝒳$ if and only if $f(x_n)$ converges to $f(x)$ for every $f\in {𝒳}^{*}$.
• (ii) $f_n$ converges weak-* to $f$ in ${𝒳}^{*}$ if and only if $f_n(x)$ converges to $f(x)$ for every $x\in 𝒳$.

In particular, when $𝒳$ is a Hilbert space, in view of Theorem 3.something the above discussion remains valid if we replace each occurrence of $f(x)$ by $⟨x,y⟩$ for some $y$.

We remark this easy-to-remember characterization does not sufficiently tell the whole story.

4. Corollary The continuous dual in terms of a weak topology is the same as the continuous dual in terms of a norm topology.
Proof: Let ${\tau }_1$ be the weak topology and $ta{u}_2$ be the norm topology. By definition ${\tau }_1\subset {\tau }_2$ and so: $(E,{\tau }_2{)}^{*}\subset (E,{\tau }_1{)}^{*}$. Conversely, if $f\in (E,{\tau }_1{)}^{*}$, then $f$ is norm-continuous; thus, $f\in (E,{\tau }_2{)}^{*}$. $◻$

4. Theorem Let $𝒳$ be a Banach space. A linear functional $z$ is continuous on ${𝒳}^{*}$ if and only if it is continuous on the closed unit ball of ${𝒳}^{*}$

4 Theorem Every finite dimensional Banach space is reflexive (thus, complete).
Proof: (TODO)

A linear operator $T$ is said to be a compact operator if the image of the open unit ball under $T$ is relatively compact. We recall that if a linear operator between normed spaces maps bounded sets to bounded sets, then it is continuous. Thus, every compact operator is continuous.

4. Theorem Let $𝒳$ be a reflexive Banach space and $𝒴$ be a Banach space. Then a linear operator $T:𝒳\to 𝒴$ is a compact operator if and only if $T$ sends weakly convergent sequence to norm convergent ones.
Proof: Let $x_n$ converges weakly to $0$, and suppose $T{x}_n$ is not convergent. That is, there is an $ϵ>0$ such that $T{x}_n\ge ϵ$ for infinitely many $n$. Denote this subsequence by $y_n$. By hypothesis we can then show (TODO: do this indeed) that it contains a subsequence $y_{n_k}$ such that $T{y}_{n_k}$ converges in norm, which is a contradiction. To show the converse, let $E$ be a bounded set. Then since $𝒳$ is reflexive every countable subset of $E$ contains a sequence $x_n$ that is Cauchy in the weak topology and so by the hypothesis $T{x}_n$ is a Cauchy sequence in norm. Thus, $T(E)$ is contained in a compact subset of $𝒴$. $◻$

4 Lemma Let $r>0$. A normed space $𝒳$ is finite-dimensional if and only if its closed ball of radius $r$ is compact.
Proof: If $𝒳$ is not finite dimensional, using w:Riesz's_lemma, we can construct a sequence $x_j$ such that:

$1=\Vert x_j\Vert \le \Vert x_j-\sum _{k=1}^{j-1}{a}_k{x}_k\Vert$ for any sequence of scalars $a_k$.

Thus, in particular, $\Vert x_j-x_k\Vert \ge 1$ for all $j,k$. (TODO: fill gaps) $◻$

4 Corollary

• (i) Every finite-rank linear operator $T$ (i.e., a linear operator with finite-dimensional range) is a compact operator.
• (ii) Every linear operator $T$ with the finite-dimensional domain is continuous.

Proof: (i) is clear, and (ii) follows from (i) since the range of a linear operator has dimension less than that of the domain. $◻$

4 Theorem The set of all compact operators into a Banach space forms a closed subspace of the set of all continuous linear operators in operator norm.
Proof: Let $T$ be a linear operator and $\omega$ be the open unit ball in the domain of $T$. If $T$ is compact, then $T(\bar{\omega })$ is bounded (try scalar multiplication); thus, $T$ is continuous. Since the sum of two compacts sets is again compact, the sum of two compact operators is again compact. For the similar reason, $\alpha T$ is compact for any scalar $\alpha$. We conclude that the set of all compact operators, which we denote by $E$, forms a subspace of continuous linear operators. To show the closedness, suppose $S$ is in the closure of $E$. Let $ϵ>0$ be given. Then there is some compact operator $T$ such that $\Vert S-T\Vert <ϵ/2$. Also, since $T$ is a compact operator, we can cover $T(\omega )$ by a finite number of open balls of radius $ϵ/2$ centered at $z_1,z_2,...z_n$, respectively. It then follows: for $x\in \omega$, we can find some $j$ so that $\Vert Tx-z_j\Vert <ϵ/2$ and so $\Vert Sx-Tx\Vert \le \Vert Sx-z_j\Vert +\Vert z_j-Tx\Vert <ϵ$. This is to say, $S(\omega )$ is totally bounded and since the completeness its closure is compact. $◻$

4. Corollary If $T_n$ is a sequence of compact operators which converges in operator norm, then its limit is a compact operator.

4 Theorem (transpose) Let $𝒳,𝒴$ be Banach spaces, and $u:𝒳\to 𝒴$ be a continuous linear operator. Define ${}^{t}u:{𝒴}^{*}\to {𝒳}^{*}$ by the identity ${}^{t}u(z)(x)=u(z(x))$. Then ${}^{t}u$ is continuous both in operator norm and the weak-* topology, and $\Vert {}^{t}u\Vert =\Vert u\Vert$.
Proof: For any $z\in {𝒴}^{*}$

$\Vert {}^{t}u(z)\Vert =\underset{\Vert x\Vert \le 1}{\sup }|(u\circ z)(x)|\le \Vert u\Vert \Vert z\Vert$

Thus, $\Vert {}^{t}u\Vert \le \Vert u\Vert$ and ${}^{t}u$ is continuous in operator norm. To show the opposite inequality, let $ϵ>0$ be given. Then there is $x_0\in 𝒳$ with $(1-ϵ)\Vert u\Vert \le |u(x_0)|$. Using the Hahn-Banach theorem we can also find $\Vert z_0\Vert =1$ and $z_0(u(x_0))=|u(x_0)|$. Hence,

$\Vert {}^{t}u\Vert =\underset{\Vert z\Vert \le 1}{\sup }\Vert {}^{t}u(z)\Vert \ge \Vert {}^{t}u(z_0)\Vert =|z_0(u(x))|=|u(x_0)|\ge (1-ϵ)\Vert u\Vert$.

We conclude $\Vert {}^{t}u\Vert =\Vert u\Vert$. To show weak-* continuity let $V$ be a neighborhood of $0$ in ${𝒳}^{*}$; that is, $V=\{z;z\in {𝒳}^{*},|z(x_1)|<ϵ,...,|z(x_n)|<ϵ\}$ for some $ϵ>0,x_1,...,x_n\in 𝒳$. If we let $y_j=u(x_j)$, then

${}^{t}u(\{z;z\in {𝒴}^{*},|z(y_1)|<ϵ,...,|z(y_n)|<ϵ\})\subset V$

since $z(y_j)={}^{t}u(z)(x_j)$. This is to say, ${}^{t}u$ is weak-* continuous. $◻$

4 Exercise State and prove the analog of Theorem 3.something using the transpose instead of a Hilbert adjoint operator.

5 Lp spaces

5 Theorem (Minkowski's inequality) If $f,g\in L^p(d\mu )$ and $p>1$, then:

$\Vert f+g{\Vert }_p\le \Vert f{\Vert }_p+\Vert g{\Vert }_p$.

Proof (See Chapter 2 of An Introduction to Analysis for more details): The theorem is a consequence of Hölder's inequality:

$\int |fgd\mu |\le {(\int |f{|}^{p}d\mu )}^{1/p}{(\int |f{|}^{q}d\mu )}^{1/q}$ (where we let $q$ by $1/p+1/q=1$)

which is then a simple consequence of the following inequality:

$a^{\frac{1}{p}}{b}^{\frac{1}{q}}=\underset{t>0}{\inf }(\frac{1}{p}{t}^{-\frac{1}{q}}a+\frac{1}{q}{t}^{\frac{1}{p}}b)$ (where a, b > 0)

To simplify notation we denote $|f|,|g|$ by $f$ and $g$, respectively. First we have:

$\int (f+g{)}^{p}d\mu =\int f(f+g{)}^{p-1}+g(f+g{)}^{p-1}d\mu$

Then Hölder's inequality followed by division gives:

$\int (f+g{)}^{p}d\mu {(\int (f+g{)}^{(p-1)q}d\mu )}^{-1/q}\le {(\int f^{p}d\mu )}^{1/p}+{(\int g^{p}d\mu )}^{1/p}$

where $\frac{1}{p}+\frac{1}{q}=1$. The desired inequality follows after noting $(p-1)q=p$ and $1-\frac{1}{q}=\frac{1}{p}$. $◻$

By letting $\mu$ be a counting measure we also obtain the analog for the series:

5 Corollary $(\sum _{k=1}^{\mathrm{\infty }}|f_k+g_k{|}^p{)}^{1/p}\le (\sum _{k=1}^{\mathrm{\infty }}|f_k{|}^p{)}^{1/p}+(\sum _{k=1}^{\mathrm{\infty }}|g_k{|}^p{)}^{1/p}$

5 Theorem Let $1\le p\le \mathrm{\infty }$, $1\le q\le \mathrm{\infty }$, and $ℳ\subset L^p\cap L^q$ a linear subspace. If $ℳ$ is closed in both $L^p$ and $L^q$, then the topology for $ℳ$ inherited from $L^p$ and $L^q$ coincide.
Proof: The identity map $I:(E,\Vert \cdot {\Vert }_{L^q})\to (E,\Vert \cdot {\Vert }_{L^p\cap L^q)}$ is continuous. Since $I$ is a bijection between Banach spaces, the open mapping theorem says $I$ is a homeomorphism. $◻$

(Remark: the theorem is just an exercise 2.something.)

5 Theorem $L^p$ spaces are uniformly convex (thus, complete and reflexive).
Proof: (TODO)

5 Theorem (lifting property of $l^1$) Let ${ℬ}_1,{ℬ}_2$ be Banach spaces, and $Q:{ℬ}_1\to {ℬ}_2$ be a linear surjection. If $T:l^1\to {ℬ}_2$ is a a continuous linear operator, there exists a continuous linear operator $\tilde{T}:l^1\to {ℬ}_1$ such that $Q\circ \tilde{T}=T$.
Proof: Let $e_n$ be a canonical basis for $l^1$. Using Corollary 2.something to the open mapping theorem we obtain a sequence $x_n$ and constant $K$ so that:

$Q{x}_n=T{e}_n$, and $\Vert x_n\Vert \le K\Vert Q{x}_n\Vert$.

Then $\underset{n\ge 1}{\sup }\Vert x_n\Vert \le K\underset{n\ge 1}{\sup }\Vert Q{x}_n\Vert \le K\Vert T\Vert \underset{n\le 1}{\sup }\Vert e_n\Vert <\mathrm{\infty }$. Define $\tilde{T}$ by $\tilde{T}{e}_n=x_n$. Then $Q\circ \tilde{T}=T$, and $\tilde{T}$ is continuous since $\underset{n\ge 1}{\sup }\Vert \tilde{T}{e}_n\Vert <\mathrm{\infty }$. $◻$

Part 2: Topological vector spaces

Topological vector space

A vector space endowed by a topology that makes translations (i.e., $x+y$) and dilations (i.e., $\alpha x$) continuous is called a topological vector space or TVS for short.

A subset $E$ of a TVS is said to be:

• bounded if for every neighborhood $V$ of $0$ there exist $s>0$ such that $E\subset tV$ for every $t>s$
• balanced if $\lambda E\subset E$ for every scalar $\lambda$ with $|\lambda |\le 1$
• convex if ${\lambda }_{1}x+{\lambda }_{2}y\in E$ for any $x,y\in E$ and any ${\lambda }_1,{\lambda }_2\ge 0$ with ${\lambda }_{1}x+{\lambda }_{2}y=1$.

1 Corollary $(s+t)E=sE+tE$ for any $s,t>0$ if and only if $E$ is convex.
Proof: Supposing $s+t=1$ we obtain $sx+ty\in E$ for all $x,y\in E$. Conversely, if $E$ is convex,

$\frac{s}{s+t}x+\frac{t}{s+t}y\in E$, or $sx+sy\in (s+t)E$ for any $x,y\in E$.

Since $(s+t)E\subset sE+tE$ holds in general, the proof is complete.$◻$

Define $f(\lambda ,x)=\lambda x$ for scalars $\lambda$, vectors $x$. If $E$ is a balanced set, for any $|\lambda |\le 1$, by continuity,

$f(\lambda ,\bar{E})\subset \overline{f(\lambda ,E)}\subset \bar{E}$.

Hence, the closure of a balanced set is again balanced. In the similar manner, if $E$ is convex, for $s,t>0$

$f((s+t),\bar{E})=\overline{f((s+t),E)}=\overline{sE+tE}$,

meaning the closure of a convex set is again convex. Here the first equality holds since $f(\lambda ,\cdot )$ is injective if $\lambda \ne 0$. Moreover, the interior of $E$, denoted by $E^{\circ }$, is also convex. Indeed, for ${\lambda }_1,{\lambda }_2\ge 0$ with ${\lambda }_1+{\lambda }_2=1$

${\lambda }_1{E}^{\circ }+{\lambda }_2{E}^{\circ }\subset E$,

and since the left-hand side is open it is contained in $E^{\circ }$. Finally, a subspace of a TVS is a subset that is simultaneously a linear subspace and a topological subspace. Let $ℳ$ be a subspace of a TVS. Then $\overline{ℳ}$ is a topological subspace, and it is stable under scalar multiplication, as shown by the argument similar to the above. Let $g(x,y)=x+y$. If $ℳ$ is a subspace of a TVS, by continuity and linearity,

$g(\overline{ℳ},\overline{ℳ})\subset \overline{g(ℳ,ℳ)}=\overline{ℳ}$.

Hence, $\overline{ℳ}$ is a linear subspace. We conclude that the closure of a subspace is a subspace.

Let $V$ be a neighborhood of $0$. By continuity there exists a $\delta >0$ and a neighborhood $W$ of $0$ such that:

$f(\{\lambda ;|\lambda |<\delta \},W)\subset V$

It follows that the set $\{\lambda ;|\lambda |<\delta \}W$ is a union of open sets, contained in $V$ and is balanced. In other words, every TVS admits a local base consisting of balanced sets.

1 Theorem Let $𝒳$ be a TVS, and $E\subset 𝒳$. The following are equivalent.

• (i) $E$ is bounded.
• (ii) Every countable subset of $E$ is bounded.
• (iii) for every balanced neighborhood $V$ of $0$ there exists a $t>0$ such that $E\subset tV$.

Proof: That (i) implies (ii) is clear. If (iii) is false, there exists a balanced neighborhood $V$ such that $E\subset ̸nV$ for every $n=1,2,...$. That is, there is a unbounded sequence $x_1,x_2,...$ in $E$. Finally, to show that (iii) implies (i), let $U$ be a neighborhood of 0, and $V$ be a balanced open set with $0\in V\subset U$. Choose $t$ so that $E\subset tV$, using the hypothesis. Then for any $s>t$, we have:

$E\subset tV=s\frac{t}{s}V\subset sV\subset sU$

$◻$

1 Corollary Every Cauchy sequence and every compact set in a TVS are bounded.
Proof: If the set is not bounded, it contains a sequence that is not Cauchy and does not have a convergent subsequence. $◻$

1 Lemma Let $f$ be a linear operator between TVSs. If $f(V)$ is bounded for some neighborhood $V$ of $0$, then $f$ is continuous.
$◻$

1 Theorem Let $f$ be a linear functional on a TVS $𝒳$.

• (i) $f$ has either closed or dense kernel.
• (ii) $f$ is continuous if and only if $\ker f$ is closed.

Proof: To show (i), suppose the kernel of $f$ is not closed. That means: there is a $y$ which is in the closure of $\ker f$ but $f(y)\ne 0$. For any $x\in 𝒳$, $x-\frac{f(x)}{f(y)}y$ is in the kernel of $f$. This is to say, every element of $𝒳$ is a linear combination of $y$ and some other element in $\ker$. Thus, $\ker f$ is dense. (ii) If $f$ is continuous, $\ker f=f^{-1}(\{0\})$ is closed. Conversely, suppose $\ker f$ is closed. Since $f$ is continuous when $f$ is identically zero, suppose there is a point $y$ with $f(y)=1$. Then there is a balanced neighborhood $V$ of $0$ such that $y+V\subset (\ker f{)}^c$. It then follows that $\underset{V}{\sup }|f|<1$. Indeed, suppose $|f(x)|\ge 1$. Then

$y-\frac{x}{f(x)}\in \ker (f)\cap (y+V)$ if $x\in V$, which is a contradiction.

The continuity of $f$ now follows from the lemma. $◻$

1 Lemma Let $V_0,V_1,...$ be a sequence of subsets of a a linear space containing $0$ such that $V_{n+1}+V_{n+1}\subset V_n$ for every $n\ge 0$. If $x\in V_{n_1}+...+V_{n_k}$ and $2^{-n_1}+...+2^{-n_k}\le 2^{-m}$, then $x\in V_m$.
Proof: We shall prove the lemma by induction over $k$. The basic case $k=1$ holds since $V_n\subset V_n+V_n$ for every $n$. Thus, assume that the lemma has been proven until $k-1$. First, suppose $n_1,...,n_k$ are not all distinct. By permutation, we may then assume that $n_1=n_2$. It then follows:

$x\in V_{n_1}+V_{n_2}+...+V_{n_k}\subset V_{n_2-1}+...+V_{n_k}$ and $2^{-n_1}+...2^{-n_k}=2^{-(n_2-1)}+...+2^{-n_k}\le 2^m$.

The inductive hypothesis now gives: $x\in V_m$. Next, suppose $n_1,...,n_k$ are all distinct. Again by permutation, we may assume that $n_1<n_2<...n_k$. Since no carry-over occurs then and $m<n_1$, $m+1<n_2$ and so:

$2^{-n_2}+...+2^{-n_k}\le 2^{-(m+1)}$.

Hence, by inductive hypothesis, $x\in V_{n_1}+V_{m+1}\subset V_m$. $◻$

1 Theorem Let $𝒳$ be a TVS.

• (i) If $𝒳$ is Hausdorff and has a countable local base, $𝒳$ is metrizable with the metric $d$ such that

$d(x,y)=d(x+z,y+z)$ and $d(\lambda x,0)\le d(x,0)$ for every $|\lambda |\le 1$

• (ii) For every neighborhood $V\subset 𝒳$ of $0$, there is a continuous function $g$ such that

$g(0)=0$, $g=1$ on $V^c$ and $g(x+y)\le g(x)+g(y)$ for any $x,y$.

Proof: To show (ii), let $V_0,V_1,...$ be a sequence of neighborhoods of $0$ satisfying the condition in the lemma and $V=V_0$. Define $g=1$ on $V^c$ and $g(x)=\inf \{2^{-n_1}+...+2^{-n_k};x\in V_{n_1}+...+V_{n_k}\}$ for every $x\in V$. To show the triangular inequality, we may assume that $g(x)$ and $g(y)$ are both $<1$, and thus suppose $x\in V_{n_1}+...+V_{n_k}$ and $y\in V_{m_1}+...+V_{m_j}$. Then

$x+y\in V_{n_1}+...+V_{n_k}+V_{m_1}+...+V_{m_j}$

Thus, $g(x+y)\le 2^{-n_1}+...+2^{-n_k}+2^{-m_1}+...+2^{-m_j}$. Taking inf over all such $n_1,...,n_k$ we obtain:

$g(x+y)\le g(x)+2^{-m_1}+...+2^{-m_j}$

and do the same for the rest we conclude $g(x+y)\le g(x)+g(y)$. This proves (ii) since $g$ is continuous at $0$ and it is then continuous everywhere by the triangular inequality. Now, to show (i), choose a sequence of balanced sets $V_0,V_1,...$ that is a local base, satisfies the condition in the lemma and is such that $V_0=𝒳$. As above, define $f(x)=\inf \{2^{-n_1}+...+2^{-n_k};x\in V_{n_1}+...+V_{n_k}\}$ for each $x\in 𝒳$. For the same reason as before, the triangular inequality holds. Clearly, $f(0)=0$. If $f(x)\le 2^{-m}$, then there are $n_1,...,n_k$ such that $2^{-n_1}+...+2^{-n_k}\le 2^{-m}$ and $x\in V_{n_1}+...+V_{n_k}$. Thus, $x\in V_m$ by the lemma. In particular, if $f(x)\le 2^{-m}$ for "every" $m$, then $x=0$ since $𝒳$ is Hausdorff. Since $V_n$ are balanced, if $|\lambda |\le 1$,

$\lambda x\in V_{n_1}+...+V_{n_k}$ for every $n_1,...,n_k$ with $x\in V_{n_1}+...+V_{n_k}$.

That means $f(\lambda x)\le f(x)$, and in particular $f(x)=f(-(-x))\le f(-x)\le f(x)$. Defining $d(x,y)=f(x-y)$ will complete the proof of (i). In fact, the properties of $f$ we have collected shows the function $d$ is a metric with the desired properties. The lemma then shows that given any $m$, $\{x;f(x)<\delta \}\subset V_m$ for some $\delta \le 2^m$. That is, the sets $\{x;f(x)<\delta \}$ over $\delta >0$ forms a local base for the original topology. $◻$

The second property of $d$ in (i) implies that open ball about the origin in terms of this $d$ is balanced, and when $𝒳$ has a countable local base consisting of convex sets it can be strengthened to:$d(\lambda x,y)\le \lambda d(x,y)$, which implies open balls about the origin are convex. Indeed, if $x,y\in V_{n_1}+...+V_{n_k}$, and if ${\lambda }_1\ge 0$ and ${\lambda }_2\ge 0$ with ${\lambda }_1+{\lambda }_2$, then

${\lambda }_{1}x+{\lambda }_{2}y\in V_{n_1}+...+V_{n_k}$

since the sum of convex sets is again convex. This is to say,

$f({\lambda }_{1}x+{\lambda }_{2}y)\le \min \{f(x),f(y)\}\le \frac{f(x)+f(y)}{2}$

and by iteration and continuity it can be shown that $f(\lambda x)\le \lambda f(x)$ for every $|\lambda |\le 1$.

Corollary For every neighborhood $V$ of some point $x$, there is a neighborhood of $x$ with $\bar{W}\subset V$
Proof: Since we may assume that $x=0$, take $W=\{x;g(x)<2^{-1}\}$. $◻$

Corollary If every finite set of a TVS $𝒳$ is closed, $𝒳$ is Hausdorff.
Proof: Let $x,y$ be given. By the preceding corollary we find an open set $V\subset \bar{V}\subset \{y{\}}^c$ containing $x$. $◻$

Locally convex spaces

A TVS with a local base consisting of convex sets is said to be locally convex. Since in this book we will never study non-Hausdorff locally convex spaces, we shall assume tacitly that every finite subset of every locally convex is closed, hence Hausdorff in view of Theorem something.

Lemma Let $𝒳$ be locally convex. The convex hull of a bounded set is bounded.

Distributions

The support of $u$, denoted by $\mathrm{supp}(u)$, is the complement of the union of all the open sets $\omega$ such that $u(w)=0$ for every $w\in 𝒟(\mathrm{\Omega })$.

Theorem Let $u\in {𝒟}^{\text{'}}(\mathrm{\Omega })$. Then $u(w)=0$ for every $w\in 𝒟((\mathrm{supp}(u){)}^c)$

Theorem A distribution $u\in {𝒟}^{\text{'}}(\mathrm{\Omega })$ has compact support if and only if it can be extended to an continuous linear functional on ${𝒞}^{\mathrm{\infty }}(\mathrm{\Omega })$.

Theorem If $u\in L^2({ℝ}^n)$, then

$\Vert \hat{u}{\Vert }_{L^2}=\Vert u{\Vert }_{L^2}$

Proof: When $u\in 𝒟({ℝ}^n)$, the theorem is a special case of Theorem something. Using Theorem something we can find a sequence $u_j\in 𝒟({ℝ}^n)$ such that $u_j\to u$ in $L^2$-norm. It follows:

$\Vert \widehat{u_j}-\widehat{u_k}{\Vert }_{L^2}=\Vert u_j-u_k{\Vert }_{L^2}\to 0$.

Hence, there is a limit $v\in L^2({ℝ}^n)$ such that $\widehat{u_j}\to v$ in $L^2$-norm and $v=\hat{u}$. $◻$