Banach Spaces

Abstract

Let X be a vector space and let $\varTheta$ be its zero element.

Appendices

Problems and Complements

1c Normed Spaces

1.1.:

Let E be a bounded, open subset of $\mathbb {R}^N$. The space $C({\bar{E}})$ endowed with the norm of $L^p(E)$ is not a Banach space.

1.2.:

Every normed space is homeomorphic to its open unit ball.

1.3.:

A normed space $\{X;\Vert \cdot \Vert \}$ is complete if and only if the intersection of a countable family of nested, closed balls is nonempty.

1.4.:

The $L^2$-norm and the sup-norm on C[0, 1] are not equivalent. In particular C[0, 1] is not complete in $L^p[0,1]$ for all $1\le p <\infty$. The norms of $L^p(E)$ and $L^q(E)$ for $1\le q<p<\infty$, are not equivalent.

The next proposition provides a criterion for a normed space to be a Banach space.

Proposition 1.1c

A normed space $\{X;\Vert \cdot \Vert \}$ is complete if and only if every absolutely convergent series converges to an element of X.

Proof

Let $\{X;\Vert \cdot \Vert \}$ be complete and let $\mathop {\textstyle {\sum }}\limits x_n$ be absolutely convergent in $\{X;\Vert \cdot \Vert \}$, so that $\mathop {\textstyle {\sum }}\limits \Vert x_n\Vert \le M$ for some $M>0$. Then $\{\sum _{j=1}^n x_j\}$ is Cauchy and hence convergent to some $x\in X$.

Conversely, let $\{x_n\}$ be Cauchy in $\{X;\Vert \cdot \Vert \}$, so that for each $j\in \mathbb {N}$ there exists $n_j$ such that

$$\begin{aligned} \Vert x_n-x_m\Vert \le 2^{-j}\quad \text { for }\> n,m\ge n_j. \end{aligned}$$

Set

$$\begin{aligned} y_j=x_{n_{j+1}}-x_{n_j}. \end{aligned}$$

The series $\mathop {\textstyle {\sum }}\limits y_j$ is absolutely convergent and we let x denote its limit. Thus the subsequence $\{x_{n_j}\}\subset \{x_n\}$ converges to x. Since $\{x_n\}$ is Cauchy, the whole sequence converges to x. $\blacksquare$

In the context of $L^p(E)$ the criterion has been used in the proof of Theorem 5.1 of Chap. 6.

Proposition 1.2c

Every norm p on $\mathbb {R}^N$ is equivalent to the Euclidean norm $\Vert \cdot \Vert$.

Remark 1.1c

The statement is a particular case of Proposition 12.1 of Chap. 2. The proof below is more direct using that the topology of $\mathbb {R}^N$ is generated by a norm $p(\cdot )$.

Proof

(of Proposition 1.2c) Let $\{\mathbf {e}_1,\dots ,\mathbf {e}_N\}$ be a basis of $\mathbb {R}^N$. Then

$$\begin{aligned} x=\mathop {\textstyle {\sum }}\limits _{j=1}^N c_j \mathbf {e}_j\quad \Longrightarrow \quad p(x)\le \mathop {\textstyle {\sum }}\limits _{j=1}^N |c_j| p(\mathbf {e}_j) \le C\Vert x\Vert \end{aligned}$$

for a constant C independent of x. This implies that

$$\begin{aligned} |p(x)-p(y)|\le p(x-y)\le C\Vert x-y\Vert \quad \text { for all }\> x,y\in \mathbb {R}^N. \end{aligned}$$

Thus $p(\cdot )$ is continuous in the topology of the Euclidean norm. Its restriction to the Euclidean unit sphere $S_1$ is a continuous function pointwise strictly bounded below in $S_1$. Since the $S_1$ is compact in $\mathbb {R}^N$ with its Euclidean topology, $p(\cdot )$ attains its minimum $c>0$ there. From this

$$\begin{aligned} p(x)=p\Big (\Vert x\Vert \frac{x}{\Vert x\Vert }\Big ) =\Vert x\Vert p\Big (\frac{x}{\Vert x\Vert }\Big )\ge c\Vert x\Vert \quad \text { for all }\> x\in \mathbb {R}^N-\{0\}. \end{aligned}$$

$\blacksquare$

1.1 1.1c Semi-Norms and Quotients

1.5.:

The quantities $\mathcal {V}_f[a,b]$ in (1.2), and $[f]_{\alpha }$ in (1.3) are semi-norms in their respective spaces.

1.6.:

Let $\{p_{\alpha }\}$ be a collection of semi-norms on X such that

$$\begin{aligned} p_\infty (x)=\sup _{\alpha } p_\alpha (x) <\infty \quad \text { for all }\> x\in X. \end{aligned}$$

Then $p_\infty (\cdot )$ defines a semi-norm on X.

1.7.:

Let $\{p_j\}$ for $j=1,\dots ,n$ be a finite collection of semi-norms on X. Then

$$\begin{aligned} q_1(x)=\mathop {\textstyle {\sum }}\limits _{j=1}^n p_j(x);\quad q_2(x)=\sqrt{\mathop {\textstyle {\sum }}\limits _{j=1}^n p_j^2(x)};\quad q_\infty (x)=\max _{1\le j\le n} p_j(x), \end{aligned}$$

are also semi-norms.

1.8.:

Prove that a semi-norm p on X is convex. Conversely a convex function $f:X\rightarrow \mathbb {R}^+$ which is homogeneous of order 1 in X, defines a semi-norm in X. Hint: Homogeneous of order $\alpha \in \mathbb {R}$ means that $f(tx)=|t|^{\alpha } f(x)$ for all $t\in \mathbb {R}$.

2c Finite and Infinite Dimensional Normed Spaces

2.1.:

An infinite dimensional Banach space $\{X;\Vert \cdot \Vert \}$ cannot have a countable Hamel basis (16.9 of the Complements of Chap. 2).

2.2.:

Let $\ell _o$ be the collection of all sequences of real numbers $\{c_n\}$ with only finitely many nonzero elements. There is no norm on $\ell _o$ by which $\ell _o$ would be a Banach space. See also § 9.6c of the Complements of Chap. 2.

2.3.:

$L^p(E)$ and $\ell _p$ are of infinite dimension for all $1\le p\le \infty$ and their dimension is larger than $\aleph _o$.

2.4.:

Let $C^1[0,1]$ be the collections of all continuously differentiable functions in [0, 1] with norm

$$\begin{aligned} C^1[0,1]\ni f\rightarrow \Vert f\Vert _{C^1[0,1]}=\sup _{[0,1]}|f| +\sup _{[0,1]} |f^\prime |. \end{aligned}$$

Let $X_o$ be a closed subspace of $C^1[0,1]$ which is also closed in $L^2[0,1]$. Prove that

i.:

The norms $\Vert \cdot \Vert _{C^1[0,1]}$ and $\Vert \cdot \Vert _{L^2[0,1]}$ are equivalent on $X_o$, i.e., there exists two positive constants $c\le C$ such that

$$\begin{aligned} c\,\Vert f\Vert _{C^1[0,1]}\le \Vert f\Vert _{L^2[0,1]}\le C\,\Vert f\Vert _{C^1[0,1]}. \end{aligned}$$

ii.:

$X_o$ is a finite dimensional subspace of $L^2[0,1]$.

2.5.:

The dimension of C[0, 1] and BV[0, 1] is uncountably infinite. The collection $\{x^n\}$ is an infinite set of linearly independent elements of C[0, 1] and BV[0, 1].

2.6.:

An infinite dimensional Banach space $\{X;\Vert \cdot \Vert \}$ has an infinite dimensional non-closed subspace. Fix a Hamel basis $\{x_\alpha \}$, for X, where $\alpha$ is an uncountable index, select an infinite, countable collection $\{x_n\}\subset \{x_\alpha \}$ of linearly independent elements and set $Y=\{x_n\}$. Then Y is non-closed, for otherwise it would be an infinite dimensional Banach space with a countable Hamel basis.

3c Linear Maps and Functionals

3.1.:

A linear map T from $\{X;\Vert \cdot \Vert _X\}$ into $\{Y;\Vert \cdot \Vert _Y\}$ is continuous if and only if it maps sequences $\{x_n\}$ converging to $\varTheta _X$ into bounded sequences of $\{Y;\Vert \cdot \Vert _Y\}$.

3.2.:

Two normed spaces $\{X;\Vert \!\cdot \!\Vert _1\}$ and $\{X;\Vert \!\cdot \!\Vert _2\}$ are homeomorphic if and only if there exist positive constants $0<c_o\le 1\le c_1$ such that

$$\begin{aligned} c_o\Vert x\Vert _1\le \Vert x\Vert _2\le c_1\Vert x\Vert _1\quad \text { for all }\> x\in X. \end{aligned}$$

3.3.:

A linear functional on a finite dimensional normed space is continuous.

3.4.:

Let E be a bounded, open subset of $\mathbb {R}^N$. A linear map $T:C({\bar{E}})\rightarrow \mathbb {R}$ is a positive functional, if $T(f)\ge 0$ whenever $f\ge 0$. A positive linear functional on $C({\bar{E}})$ is bounded. Thus positivity implies continuity. Moreover any two of the conditions

$$\begin{aligned} \text {(i)}~~\ \Vert T\Vert =1\qquad \text {(ii)}~~\ T(1)=1\qquad \text {(iii)}~~\ T\ge 0 \end{aligned}$$

implies the remaining one.

3.5.:

Let $\{X;\Vert \cdot \Vert \}$ be an infinite dimensional Banach space. There exists a discontinuous, linear map $T:X\rightarrow X$.

Having fixed a Hamel basis $\{x_\alpha \}$ for X, after a possible renormalization, we may assume that $\Vert x_\alpha \Vert =1$ for all $\alpha$. Every element $x\in X$ can be represented in a unique way, as the finite linear combination of elements of $\{x_\alpha \}$, i.e., for every $x\in X$ there exists a unique m-tuple of real numbers $\{c_1,\dots ,c_m\}$ such that

$$\begin{aligned} x=\mathop {\textstyle {\sum }}\limits _{j=1}^m\,c_jx_{\alpha _j}. \end{aligned}$$

(3.1c)

Since X is of infinite dimension, the index $\alpha$ ranges over some set A such that $(A)\ge (\mathbb {N})$. Out of $\{x_\alpha \}$ select a countable collection $\{x_n\}\subset \{x_\alpha \}$. Then set

$$\begin{aligned} T(x_n)=n x_n \quad \text { and }\quad T(x_\alpha )=\varTheta \quad \text { if }\>\alpha \notin \mathbb {N}. \end{aligned}$$

For $x\in X$, having determined its representation (3.1c), set also

$$\begin{aligned} T(x)=\mathop {\textstyle {\sum }}\limits _{j=1}^m\,c_j T(x_{\alpha _j}). \end{aligned}$$

In view of the uniqueness of the representation (3.1c), this defines a linear map from X into X. Such a map, however, is discontinuous since $\Vert T(x_n)\Vert =n\Vert x_n\Vert \rightarrow \infty$ as $n\rightarrow \infty$.

3.6.:

Let $\{X;\Vert \cdot \Vert \}$ be an infinite dimensional Banach space. There exists a discontinuous, linear functional $T:X\rightarrow \mathbb {R}$.

3.7.:

Are the constructions of 3.5 and 3.6 possible if $\{X;\Vert \cdot \Vert \}$ is an infinite dimensional normed space?

Remark 3.1c

The conclusion of 3.6 is in general false for metric spaces. For example if X is a set, the discrete metric, generates the discrete topology on X. With respect to such a topology there exists no discontinuous maps $T:X\rightarrow \mathbb {R}$.

However there exist metric, not normed, spaces that admit discontinuous linear functionals. As an example consider $L^p(E)$ for $0<p<1$ endowed with the metric (3.1c) of § 3.4c of the Complements of Chap. 6. If E is a Lebesgue measurable subset of $\mathbb {R}^N$ and $\mu$ is the Lebesgue measure, then every nontrivial, linear functional on $L^p(E)$ is discontinuous.⁴

3.8.:

Let the unit ball in $\mathbb {R}^N$ in some norm $\Vert \cdot \Vert$, be $\prod _{j=1}^N[-1,1]$. Compute the unit ball of the dual space.

3.9.:

Let $X=\bigoplus _{j=1}^k X_j$ where $X_j$ for $j=1,\dots ,k$ are Banach spaces. Then $X^*=\bigoplus _{j=1}^k X_j^*$. In particular $(L^p(E)^k)^*=L^q(E)^k$ where $1\le p<\infty$ and p and q are Hölder conjugate.

6c Equibounded Families of Linear Maps

Let $\{X;\Vert \cdot \Vert _X\}$ and $\{Y;\Vert \cdot \Vert _Y\}$ be Banach spaces.

6.1.:

Let $T(x,y):X\times Y \rightarrow \mathbb {R}$ be a functional linear and continuous in each of the two variables. Then T is linear and bounded with respect to both variables.

6.2.:

Let $\{T_n\}$ be a sequence in $\mathcal {B}(X;Y)$, such that the limit of $\{T_n(x)\}$ exists for all $x\in X$. Then $T(x)=\lim T_n(x)$ defines a bounded, linear map from $\{X;\Vert \cdot \Vert _X\}$ into $\{Y;\Vert \cdot \Vert _Y\}$.

6.3.:

Let $\{T_n\}$ be a sequence in $\mathcal {B}(X;Y)$, such that $\Vert T_n\Vert \le C$ for some positive constant C and all $n\in \mathbb {N}$. Let $X_o$ be the set of $x\in X$ for which $\{T_n(x)\}$ converges. Then $X_o$ is a closed subspace of $\{X;\Vert \cdot \Vert _X\}$.

6.4.:

Let $T_1$ and $T_2$ be elements of $\mathcal {B}(X;X)$ and let $T_1T_2(\cdot )=T_1(T_2(\cdot ))$ be the composition map. Then $T_1T_2\in \mathcal {B}(X;X)$ and $\Vert T_1T_2\Vert \le \Vert T_1\Vert \,\Vert T_2\Vert$.

Let $T\in \mathcal {B}(X;X)$ satisfy $\Vert T\Vert <1$. Then $(I+T)^{-1}$ exists as an element of $\mathcal {B}(X;X)$ and

$$\begin{aligned} (I+T)^{-1}=I+\mathop {\textstyle {\sum }}\limits \,(-1)^n T^n. \end{aligned}$$

Hint: Since $\Vert T\Vert <1$, the series $\mathop {\textstyle {\sum }}\limits \Vert T^n\Vert \le \mathop {\textstyle {\sum }}\limits \Vert T\Vert ^n$ converges. Since $\mathcal {B}(X;X)$ is a Banach space, $\mathop {\textstyle {\sum }}\limits (-T)^n$ converges to an element $\mathcal {B}(X;X)$.

6.5.:

Let E be a bounded open set in $\mathbb {R}^N$. Fix $h\in L^\infty (E)$ and find $f\in L^\infty (E)$ such that

$$\begin{aligned} h=(I+T)f\qquad \text { i.e., formally }\qquad f=(I+T)^{-1}h \end{aligned}$$

(6.1c)

where T(f) is the Riesz potential introduced in (4.2). Verify that T is a bounded linear map from $L^\infty (E)$ into itself. Give conditions on E so that such a formal solution formula is actually justified and exhibit explicitly the solution f.

6.6.:

Let E be a Lebesgue measurable subset of $\mathbb {R}^N$ of finite measure and let $1\le p,q\le \infty$ be conjugate. Then if $q>p$ the space $L^q(E)$ is of first category in $L^p(E)$. Hint: $L^q(E)$ is the union of $\big [\Vert g\Vert _q\le n\big ]$.

8c The Open Mapping Theorem

$\{X;\Vert \cdot \Vert _X\}$ and $\{Y;\Vert \cdot \Vert _Y\}$ are Banach spaces:

8.1.:

$T\in \mathcal {B}(X;Y)$ is a homeomorphism if and only if it is onto, and there exist positive constants $c_1\le c_2$ such that

$$\begin{aligned} c_1\Vert x\Vert _X\le \Vert T(x)\Vert _Y\le c_2\Vert x\Vert _X\qquad \text { for all }\> x\in X. \end{aligned}$$

8.2.:

A bounded linear map T from a subspace $X_o\subset X$ into Y, has closed graph if and only if its domain is closed.

8.3.:

The sup-norm on C[0, 1] generates a strictly stronger topology than the $L^2$-norm.

8.4.:

Let $T\in \mathcal {B}(X;Y)$. If T(X) is of second category in Y, then T is onto.

9c The Hahn–Banach Theorem

Let X be a vector space over the complex field $\mathbb {C}$. The norm $\Vert \!\cdot \!\Vert$ is defined as in (i)–(iii) of § 1, except that $\lambda \in \mathbb {C}$. In such a case $|\lambda |$ is the modulus of $\lambda$ as element of $\mathbb {C}$.

Denote by $X_{\mathbb {R}}$ the vector space X when multiplication is restricted to scalars in $\mathbb {R}$. A linear functional $T:X\rightarrow \mathbb {C}$ is separated into its real and imaginary part by

$$\begin{aligned} T(x)=T_{\mathbb {R}}(x)+iT_i(x) \end{aligned}$$

(9.1c)

where the maps $T_{\mathbb {R}}$ and $T_i$ are functionals from $X_{\mathbb {R}}$ into $\mathbb {R}$. Since $T:X\rightarrow \mathbb {C}$ is linear, $T(i x)=i T(x)$ for all $x\in X$. From this compute

$$\begin{aligned} T(ix)=T_{\mathbb {R}}(ix)+iT_i(ix)\qquad iT(x)=iT_{\mathbb {R}}(x)-T_i(x). \end{aligned}$$

This implies

$$\begin{aligned} T_i(x)=-T_{\mathbb {R}}(ix)\qquad \text { for all }\> x\in X. \end{aligned}$$

(9.2c)

Thus $T:X\rightarrow \mathbb {C}$ is identified by its real part $T_{\mathbb {R}}$ regarded as a linear functional from $X_{\mathbb {R}}$ into $\mathbb {R}$. Vice versa any such real-valued functional $T_{\mathbb {R}}:X_{\mathbb {R}}\rightarrow \mathbb {R}$ identifies a linear functional $T:X\rightarrow \mathbb {C}$ by the formulae (9.1c)–(9.2c).

1.1 9.1c The Complex Hahn–Banach Theorem

Theorem 9.1c

([17, 151]) Let X be a complex vector space and let $p:X\rightarrow \mathbb {R}$ be a semi-norm on X. Then, every linear functional $T_o:X_o\rightarrow \mathbb {C}$ defined on a subspace $X_o$ of X and satisfying $|T_o(x)|\le p(x)$ for all $x\in X_o$, admits an extension $T:X\rightarrow \mathbf {C}$ such that

$$\begin{aligned} |T(x)|\le p(x)\quad \text { for all }\>x\in X\quad \text { and }\quad T(x)=T_o(x)\quad \text { for }\> x\in X_o. \end{aligned}$$

Proof

Denote by $X_{o,\mathbb {R}}$ the real subspace of $X_o$ and by $T_{o,\mathbb {R}}:X_{o,\mathbb {R}}\rightarrow \mathbb {R}$ the real part of T. By the representation (9.1c)–(9.2c) it suffices to extend $T_{o,\mathbb {R}}$ into a linear map $T_{\mathbb {R}}:X_{\mathbb {R}}\rightarrow \mathbb {R}$. This follows from the Hahn–Banach theorem, since

$$\begin{aligned} T_{o,\mathbb {R}}(x)\le |T(x)|\le p(x)\qquad \text { for all } \> x\in X. \end{aligned}$$

$\blacksquare$

1.2 9.2c Linear Functionals in $L^\infty (E)$

The Riesz representation theorem for the bounded linear functionals in $L^p(E)$, fails for $p=\infty$. A counterexample can be constructed as follows.

On $C[-1,1]$ let $T_o(f)=f(0)$. This is bounded, linear functional on $C[-1,1]$. The boundedness of $T_o$ is meant in the sense of $L^\infty [-1,1]$, i.e.,

$$\begin{aligned} \Vert T_o\Vert =\sup _{\genfrac{}{}{0.0pt}{}{\varphi \in C[-1,1]}{\Vert \varphi \Vert _\infty =1}}\, |T_o(\varphi )|. \end{aligned}$$

Then, by the Hahn–Banach theorem, $T_o$ can be extended to a bounded linear functional T in $L^\infty [-1,1]$ coinciding with $T_o$ on $C[-1,1]$ and such that $\Vert T\Vert =\Vert T_o\Vert$. For such an extension there exists no function $g\in L^1[-1,1]$ such that

$$\begin{aligned} T(f)=\int _{-1}^1 fgdx\qquad \text { for all }\quad f\in L^\infty [-1,1]. \end{aligned}$$

11c Separating Convex Subsets of X

1.1 11.1c A Counterexample of Tukey [164]

The nonempty interior assumption is essential. We produce two disjoint, closed, convex sets $C_1$ and $C_2$ in $\ell _2$, such that $C_1-C_2$ is dense in $\ell _2$. This implies that $C_1-C_2$ cannot be separated from $\varTheta$ and hence $C_1$ and $C_2$ cannot be separated. Identify $x\in \ell _2$ with its corresponding sequence $\{x_n\}$ and define

$$\begin{aligned} \begin{aligned} C_1&=\Big \{x\in \ell _2\,\bigm |\, x_1>\big |n^2\big (x_n-{\textstyle \frac{1}{n}}\big )\big |, \text { for all }\> n>1\Big \}\\ C_2&=\big \{x\in \ell _2\,\bigm |\, x_n=0 \text { for all }\> n>1\Big \}. \end{aligned} \end{aligned}$$

One verifies that $C_1$ and $C_2$ are convex and closed. They are also disjoint. Indeed if $y\in C_1\cap C_2$

$$\begin{aligned} y_1\ge \big |n^2\big (0-{\textstyle \frac{1}{n}}\big )\big |=n \quad \text { for all }\> n\ge 2. \end{aligned}$$

Thus $y_1=\infty$ and hence $y\notin \ell _2$. To show that $C_1-C_2$ is dense in $\ell _2$, fix $z\in \ell _2$ and $\varepsilon >0$ and pick $n_\varepsilon$ so large that

$$\begin{aligned} \mathop {\textstyle {\sum }}\limits _{n\ge n_\varepsilon } n^{-2}+ \mathop {\textstyle {\sum }}\limits _{n\ge n_\varepsilon } z_n^2<\frac{\varepsilon ^2}{4}. \end{aligned}$$

Choose a number

$$\begin{aligned} x_1>\big |n^2\big (z_n-{\textstyle \frac{1}{n}}\big )\big | \qquad \text { for }\> 2\le n_\varepsilon <n \end{aligned}$$

and let $x=\{x_n\}\in C_1$ and $y=\{y_n\}\in C_2$ be defined by

$$\begin{aligned} x_n=\left\{ \begin{array}{ll} x_1\quad &{}\text { for }\> n=1;\\ z_n\quad &{}\text { for }\> 2\le n <n_\varepsilon ;\\ \frac{1}{n}\quad &{}\text { for }\> n \ge n_\varepsilon ; \end{array}\right. \quad \qquad y_n=\left\{ \begin{array}{ll} z_1+x_1\quad &{}\text { for }\> n=1;\\ 0&{}\text { for }\> n>1. \end{array}\right. \end{aligned}$$

By the triangle inequality $\Vert z-(x-y)\Vert <\varepsilon$.

1.1.1 11.1.1c A Variant of Tukey’s Counterexample

Corollary 11.1c

Every infinite dimensional Banach space X contains a 1-dimensional subspace $C_1$ and a closed, convex set $C_2$ that cannot be separated by a bounded linear functional.

Proof

Let $\{\mathbf {e}_n\}$ be a sequence of linearly independent elements on the unit sphere of X and set $C_2=\{\mathbf {e}_1\}$ and

$$\begin{aligned} C_1=\left\{ \genfrac{}{}{0.0pt}{}{x\in X\text { of the form }\> x=\mathop {\textstyle {\sum }}\limits x_n\mathbf {e}_n\> \text { with}}{x_1\ge n^3\big |x_n-n^{-2}\big |\>\text { for all }\> n\ge 2} \right\} \end{aligned}$$

$\blacksquare$

1.2 11.2c A Counterexample of Goffman and Pedrick [56]

Let X be a linear space with a countably infinite Hamel basis $\{x_n\}$. Let $C_1$ be the collections of all elements $x\in X$ whose last nonzero coefficient of its representations in terms of $\{x_n\}$ is positive. The set $C_1$ is convex and $\varTheta \notin C_1$. Any functional T that separates $\{\varTheta \}$ from $C_1$ must be either non-positive or nonnegative on $C_1$. Let then T be a linear functional on X, which is nonnegative on $C_1$. For every $x_j\in \{x_n\}$ and all $\alpha \in \mathbb {R}$ the element $\alpha x_j+x_{j+1}\in C_1$. Therefore

$$\begin{aligned} T(\alpha x_j+x_{j+1})=\alpha T(x_j)+T(x_{j+1}) \ge 0 \quad \text { for all }\>\alpha \in \mathbb {R}. \end{aligned}$$

Thus $T(x_j)=0$. Since $x_j\in \{x_n\}$ is arbitrary, T vanishes on all basis elements of X and therefore is the zero functional.

1.3 11.3c Extreme Points of a Convex Set

Let C be a convex set in a linear, normed space. A point $x\in C$ is an extreme point of C if there do not exist distinct points $u,v\in C$ and $t\in (0,1)$, such that $x=t u+(1-t)v$, that is, if no line segment in C has x as an interior point.

11.1.:

The extreme points of a convex, closed polyhedron in $\mathbb {R}^N$ are its vertices.

11.2.:

An open convex polyhedron in $\mathbb {R}^N$ has no extreme points.

11.3.:

A closed $\frac{1}{2}$-space in $\mathbb {R}^N$ has no extreme points.

11.4.:

The set of extreme points of the closed unit ball of a uniformly convex linear normed space $\{X;\Vert \cdot \Vert \}$ is the unit sphere. In particular the extreme points of $L^p(E)$ for $1<p<\infty$ are all those functions in $L^p(E)$ such that $\Vert f\Vert _p=1$.

11.5.:

The set of the extreme points of the closed unit ball in $L^\infty (E)$ are those $f\in L^\infty (E)$ such that $|f|=1$ a.e. in E.

11.6.:

The set of the extreme points of the closed unit ball in $L^1(E)$ is empty.

11.7.:

Let E be a bounded, connected, open set in $\mathbb {R}^N$. The closed unit ball of $C({\bar{E}})$ has no, non constant, extreme points. Examine the case when E has several connected components.

A set $E\subset C$ is an extremal subset of C if it satisfies the property:

$$\begin{aligned} \left\{ \genfrac{}{}{0.0pt}{}{\text {if there exist}\ u,v\in C\ \text {and}\ t\in (0,1)\ \text {such that}}{tu+(1-t)v\in E,\ \text {then}\ u,v\in E.} \right\} \end{aligned}$$

Extreme points are extremal sets. Convex portions of a face of a closed, convex polyhedron are extremal sets of that polyhedron. Prove the following:

Lemma 11.1c

If $E_1$ is an extremal subset of C and $E_2$ is an extremal subset of $E_1$, then $E_2$ is an extremal subset of C.

Lemma 11.2c

A nonempty, convex, compact subset C of a linear, normed space X, has extreme points.

Proof

Consider the collection $\mathcal {E}$ of all closed, extremal subsets of C, partially ordered by inclusion. Such a collection is not empty since $C\in \mathcal {E}$. Any linearly ordered subcollection $\mathcal {E}_1\subset \mathcal {E}$ has the finite intersection property (§ 5 of Chap. 2) and therefore

$$\begin{aligned} E_o=\mathop {\textstyle {\bigcap }}\limits \{E\,|\,E\in \mathcal {E}_1\}\quad \text { is not empty, closed and compact}. \end{aligned}$$

We claim that $E_o$ is a singleton which is extreme of C. If $E_o$ contains more than one point, pick $p,q\in E_o$ with $p\ne q$, let $T\in X^*$ be such that $T(p)>T(q)$ (Corollary 10.1), and set

$$\begin{aligned} E_o^\prime = \Big \{x\in E_o\,\big |\, T(x)=\inf _{y\in E_o}T(y)\Big \}. \end{aligned}$$

The set $E_o^\prime$ is well defined, since T is continuous and $E_o$ is compact, and is a proper subset of $E_o$, since $T(p)>T(q)$. The proof consists in verifying that $E_o^\prime$ is a convex, closed, extremal subset of $E_o$. Thus by Lemma 11.1c it would be a closed, extremal subset of C, properly contained in $E_o$, thereby contradicting the definition of $E_o$.

If there exists $u,v\in C$ and $t\in (0,1)$ such that $tu+(1-t)v\in E_o^\prime \subset E_o$, then $u,v\in E_o$ and

$$\begin{aligned} tT(u)+(1-t)T(v)=T(tu+(1-t)v)=\inf _{y\in E_o}T(y). \end{aligned}$$

Now

$$\begin{aligned} T(u)>\inf _{y\in E_o}T(y)\quad \text { implies }\quad T(v)<\inf _{y\in E_o}T(y) \end{aligned}$$

contradicting that $v\in E_o$. Thus

$$\begin{aligned} T(u)=T(v)=\inf _{y\in E_o}T(y) \quad \text { which implies }\> u,v\in E_o^\prime . \end{aligned}$$

$\blacksquare$

Theorem 11.1c

(Krein–Milman [86]) A compact, convex set C in a normed linear space is the closed convex hull of its extreme points.

Proof

Let E be the set of the extreme points of C. By the definition of convex hull, $\overline{c(E)}\subset C$. To prove the converse inclusion, proceed by contradiction, assuming that there exists $x_o\in C$ such that $x_o\notin \overline{c(E)}$. By Proposition 11.3 there exists $T\in X^*$ such that $T(x_o)<T(\overline{c(E)})$. Set

$$\begin{aligned} C_1=\Big \{x\in C\,\bigm |\, T(x)=\inf _{y\in C} T(y)\Big \}. \end{aligned}$$

Proceeding as in the proof of Lemma 11.2c one verifies that $C_1$ is a nonempty, convex, closed, extremal subset of C. Moreover $C_1\cap E=\emptyset$. Since $C_1$ is convex and compact it has at least one extreme point $y_o$, which by Lemma 11.1c is also an extreme point of C. $\blacksquare$

11.8.:

Give an example to show that the compactness assumption on C is essential.

1.4 11.4c A General Version of the Krein–Milman Theorem

Prove that Theorem 11.1c continues to hold, with essentially the same proof if X is a Hausdorff topological vector space on which $X^*$ separates points. Compactness of C is referred to the topology of X.

12c Weak Topologies

Proposition 12.1c

Let $\{X;\Vert \cdot \Vert \}$ be a normed space and let $\mathcal {W}$ denote its weak topology. Denote also by $\mathcal {O}$ a weak neighborhood of the origin of X. Then:

(i):

For every $V\in \mathcal {W}$ and $x_o\in V$, there exists $\mathcal {O}$ such that $x_o+\mathcal {O}\subset V$.

(ii):

For every $V\in \mathcal {W}$ and $x_o\in V$, there exists $\mathcal {O}$ such that $\mathcal {O}+\mathcal {O}+x_o\subset V$.

(iii) :

For every $\mathcal {O}$ there exists $\mathcal {O}^\prime$ such that $\lambda \mathcal {O}^\prime \subset \mathcal {O}$, for all $|\lambda |\le 1$.

Proof

(of (i)) An open weak neighborhood V of $x_o\in X$ contains an open set of the form

$$\begin{aligned} V_o=\mathop {\textstyle {\bigcap }}\limits _{j=1}^m T^{-1}_j (T_j(x_o)-\alpha _j,T_j(x_o)+\alpha _j) \end{aligned}$$

for some finite m and $\alpha _j>0$. Equivalently

$$\begin{aligned} V_o=\{x\in X\bigm | |T_j(x-x_o)|<\alpha _j \quad \text { for }\> j=1,\dots ,n\}. \end{aligned}$$

The neighborhood of the origin

$$\begin{aligned} \mathcal {O}=\{x\in X\bigm | |T_j(x)|<\alpha _j \quad \text { for }\> j=1,\dots ,n\big \} \end{aligned}$$

is such that $x_o+\mathcal {O}\subset V_o$. $\blacksquare$

The remaining statements are proved similarly.

12.1.:

In a finite dimensional normed linear space, the notions of weak and strong convergence coincide.

12.2.:

Construct examples and counterexamples for the following statements:

i.:

A weakly closed subset of a normed linear space is also strongly closed. The converse is false.

ii.:

A strongly sequentially compact subset of a normed linear space is also weakly sequentially compact. The converse is false.

12.3.:

A normed linear space is weakly complete if and only if it is complete in its strong topology. A weakly dense set in $\{X;\Vert \cdot \Vert \}$ is also strongly dense.

12.4.:

The weak topology on a normed linear space is Hausdorff (Proposition 11.3).

1.1 12.1c Infinite Dimensional Normed Spaces

Let $\{X;\Vert \cdot \Vert \}$ be an infinite dimensional Banach space. There exists a countable collection $\{X_n\}$ of infinite dimensional, closed subspaces of X such that $X_{n+1}\subset X_n$ with strict inclusion. For example one might take a nonzero functional $T_1\in X^*$ and set $X_1=\ker \{T_1\}$. Such a subspace is infinite dimensional (Hint: Proposition 5.1). Then select a nonzero functional $T_2\in X_1^*$, set $X_2=\ker \{T_2\}$, and proceed by induction.

For each n, select an element $x_n\in X_{n+1}-X_n$ so that $\Vert x_n\Vert =2^{-n}$. This generates a sequence $\{x_n\}$ of linearly independent elements of X whose span $X_o$ is isomorphic to $\ell _\infty$ by the representation

$$\begin{aligned} \ell _\infty \ni \{c_n\}\>\longrightarrow \>\mathop {\textstyle {\sum }}\limits \, c_n x_n\in X \end{aligned}$$

Since the dimension of $\ell _\infty$ is not less than the cardinality of $\mathbb {R}$ the dimension of an infinite dimensional Banach space is at least the cardinality of $\mathbb {R}$. Compare with 2.1.

12.5.:

A Banach space $\{X;\Vert \cdot \Vert \}$ is finite dimensional if and only if every linear subspace is closed (2.6). However an infinite dimensional Banach space X contains infinitely many, infinite dimensional, closed subspaces.

12.6.:

The weak topology of an infinite dimensional normed space X is not normable, i.e., there exists no norm $\Vert \cdot \Vert _w$ on X that generates the weak topology. This follows from Corollary 12.3. Give an alternate proof. Hint: If such $\Vert \cdot \Vert _w$ exists, the open unit ball, with respect to such a norm, would be an open neighborhood of the origin.

1.2 12.2c About Corollary 12.5

In the context of $L^p(E)$ spaces the corollary had been established in [14]. When $p=2$ the coefficients $\alpha _j$ can be given an elegant form as established by the following proposition.

Proposition 12.2c

Let $\{f_n\}$ be a sequence of functions in $L^2(E)$ weakly convergent to some $f\in L^2(E)$. There exists a subsequence $\{f_{n_j}\}$ such that setting

$$\begin{aligned} \varphi _m=\frac{f_{n_1}+f_{n_2}+\cdots +f_{n_m}}{m} \end{aligned}$$

the sequence $\{\varphi _m\}$ converges to f strongly in $L^2(E)$.

Proof

By possibly replacing $f_n$ with $f_n-f$, we may assume that $f=0$. Fix $n_1=1$. Since $\{f_n\}\rightarrow 0$ weakly in $L^2(E)$, there exists an index $n_2$ such that

$$\begin{aligned} \Big |\int _Ef_{n_1}f_{n_2}d\mu \Big |\le \frac{1}{2}. \end{aligned}$$

Then there is an index $n_3$ such that

$$\begin{aligned} \Big |\int _Ef_{n_1}f_{n_3}d\mu \Big | \le \frac{1}{3}\qquad \text { and }\qquad \Big |\int _Ef_{n_2}f_{n_3}d\mu \Big |\le \frac{1}{3}. \end{aligned}$$

Proceeding in this fashion we extract, out of $\{f_n\}$, a subsequence $\{f_{n_j}\}$, such that, for all $k\ge 2$

$$\begin{aligned} \Big |\int _Ef_{n_\ell }f_{n_k}d\mu \Big | \le \frac{1}{k}\quad \text { for all }\>\ell =1,\dots ,k-1. \end{aligned}$$

Denoting by M the upper bound of $\Vert f_n\Vert _2$, compute

$$\begin{aligned} \int _E\varphi _{m}^2d\mu&=\frac{1}{m^2}\int _E(f_{n_1}+f_{n_2}+\cdots f_{n_m})^2d\mu \\&\le \frac{1}{m^2}\Big (m M^2+2+4\frac{1}{2}+\cdots +2m\frac{1}{m}\Big )\\&\le \frac{M^2+2}{m}\>\longrightarrow \>0\quad \text { as }\> m\rightarrow \infty . \end{aligned}$$

$\blacksquare$

1.3 12.3c Weak Closure and Weak Sequential Closure

For a subset E of a normed space $\{X;\Vert \cdot \Vert \}$ denote by $\bar{E}_{w-\sigma }$ its is weak sequential closure, that is the set of all limit points of weakly convergent sequences in E. Equivalently, by $\bar{E}_{w-\sigma }$ is the set of all points $x\in X$, for which there exists a sequence $\{x_n\}\subset E$ weakly convergent to x. By the definition $\bar{E}_{w-\sigma }\subset \bar{E}_w$. The inclusion is in general strict, as shown by the following examples.

12.7.:

Let $E=\{\sqrt{n}\mathbf {e}_n\}\subset \ell _2$, where $\mathbf {e}_n$ is the infinite sequence consisting of zeroes except the nth entry which is one. Denoting by $\mathbf {0}$ the zero element of $\ell _2$

$$\begin{aligned} \mathbf {0}\in \overline{\{\sqrt{n}\mathbf {e}_n\}}_{w} \qquad \text { but }\qquad \mathbf {0}\notin \overline{\{\sqrt{n}\mathbf {e}_n\}}_{w-\sigma }. \end{aligned}$$

(12.1c)

To prove the first of these, pick $T\in \ell _2^*$ and for a fixed $\alpha >0$, consider the weak, open neighborhood of the origin

$$\begin{aligned} \mathcal {O}_{T,\alpha }=\{x\in \ell _2\,\bigm |\, |T(x)|<\alpha \}. \end{aligned}$$

By the Riesz representation theorem any such T is identified by some $\mathbf {t}\in \ell _2$ acting on elements $\mathbf {a}\in \ell _2$ by the formula

$$\begin{aligned} T(\mathbf {a})=\mathbf {t}\cdot \mathbf {a}=\mathop {\textstyle {\sum }}\limits t_na_n. \end{aligned}$$

Since $\mathbf {t}\in \ell _2$, for all $M>0$, there exists $n>M$ such that $|t_n|\le \alpha /\sqrt{n}$. Indeed otherwise

$$\begin{aligned} \Vert \mathbf {t}\Vert ^2\ge \mathop {\textstyle {\sum }}\limits _{n>M} t_n^2>\alpha \mathop {\textstyle {\sum }}\limits \frac{1}{n}. \end{aligned}$$

For such an index, $|T(\sqrt{n}\mathbf {e}_n)|<\alpha$, and hence $\sqrt{n}\mathbf {e}_n\in \mathcal {O}_{T,\alpha }$. Let now

$$\begin{aligned} \mathcal {O}=\mathop {\textstyle {\bigcap }}\limits _{j=1}^kT_j^{-1}(-\alpha _j,\alpha _j), \quad \text { for }\>\alpha _j>0\quad \text { for some }\>k\in \mathbb {N}. \end{aligned}$$

(12.2c)

Prove that n can be chosen so large that $\sqrt{n}\mathbf {e}_n\in \mathcal {O}$. Therefore any weak, open neighborhood of the origin intersects $\{\sqrt{n}\mathbf {e}_n\}$.

To prove the second of (12.1c) we establish that there exists no subsequence $\{\sqrt{m}\mathbf {e}_m\}\subset \{\sqrt{n}\mathbf {e}_n\}$, weakly convergent to $\mathbf {0}$. Having picked such a subsequence assume first that some fixed index $j\in \mathbb {N}$ occurs infinitely many times in $\{\sqrt{m}\mathbf {e}_m\}$. Pick $T_j\in \ell _2^*$ corresponding to $\mathbf {e}_j\in \ell _2$. For such a choice the sequence $\{T_j(\sqrt{m}\mathbf {e}_m)\}$ contains the value $\sqrt{j}$ infinitely many times, and thus it cannot converge to zero. If no index $j\in \mathbb {N}$ occurs infinitely many times in $\{\sqrt{m}\mathbf {e}_m\}$ pick a subsequence

$$\begin{aligned} \{\sqrt{m_j}\mathbf {e}_{m_j}\}\subset \{\sqrt{m}\mathbf {e}_m\}\quad \text { so that }\quad m_j\ge 2^j \end{aligned}$$

and set

$$\begin{aligned} \mathbf {t}=\mathop {\textstyle {\sum }}\limits _{j\in \mathbb {N}}\frac{1}{\sqrt{m_j}}\mathbf {e}_{m_j}\quad \text { satisfying }\quad \Vert \mathbf {t}\Vert ^2=\mathop {\textstyle {\sum }}\limits _{j\in \mathbb {N}}\frac{1}{m_j}<\infty . \end{aligned}$$

Therefore $\mathbf {t}\in \ell _2$ and we let $T\in \ell _2^*$ be its corresponding functional. For such a functional $T(\sqrt{m_j}\mathbf {e}_{m_j})=1$. Therefore the sequence $\{T(\sqrt{m}\mathbf {e}_m)\}$ contains 1 infinitely many times and thus it cannot converge to zero.

Remark 12.1c

The set E is countable, so that countability alone is not sufficient to identifty weak closure with weak sequential closure.

Remark 12.2c

The set $\{\sqrt{n}\mathbf {e}_n\}$ is unbounded in $\ell _2$. However the strict inclusion $\bar{E}_{w-\sigma }\subset \bar{E}_w$ continues to hold even for bounded sets, as shown by the next example.

12.8.:

Consider the set $E\subset \ell _1$

$$\begin{aligned} E=\mathop {\textstyle {\bigcup }}\limits _{n,m\in \mathbb {N}; m\ne n} \{\mathbf {e}_m-\mathbf {e}_n\}. \end{aligned}$$

We claim that E is bounded in $\ell _1$, its weak closure contains $\mathbf {0}$ but its weak sequential closure does not contain $\mathbf {0}$.

To establish the first claim let $\mathcal {O}$ be a weakly open set of the form (12.2c). Every $T_j\in \ell _1^*$ is identified with an element $\mathbf {t}_j\in \ell _\infty$ acting on elements $\mathbf {a}\in \ell _1$ by the formula

$$\begin{aligned} T_j(\mathbf {a})=\mathbf {t}_j\cdot \mathbf {a}=\mathop {\textstyle {\sum }}\limits t_{j,n}a_n. \end{aligned}$$

Since $\mathbf {t}_j\in \ell _\infty$ for $j=1,\dots ,k$, as n ranges over $\mathbb {N}$, the k-tuple $(t_{1,n},\dots , t_{k,n})$ ranges over a bounded set $K\subset \mathbb {R}^k$, which, without loss of generality we may assume to be a closed cube with faces parallel to the coordinate planes. Pick $\varepsilon =\min \{\alpha _1,\dots ,\alpha _k\}$ and subdivide K into no less than $\varepsilon ^{-k}$ homotetic closed sub-cubes $K_\varepsilon$ of edge not exceeding $\varepsilon$. At least one of these sub-cubes must contain infinitely many k-tuples $(t_{1,n},\dots , t_{k,n})$. Select one such cube and relabel the corresponding k-tuples as $\{(t_{1,n_i},\dots , t_{k,n_i})\}_{i\in \mathbb {N}}$. For the element $\mathbf {e}_{m_i}-\mathbf {e}_{n_i}\in \{\mathbf {e}_m-\mathbf {e}_n\}$ compute

$$\begin{aligned} |T_j(\mathbf {e}_{m_i}-\mathbf {e}_{n_i})|=|t_{j,m_i}-t_{j,n_i}|\le \varepsilon \le \alpha _j \quad \text { for }\> j=1,\dots k. \end{aligned}$$

Therefore $\mathbf {e}_{m_i}-\mathbf {e}_{n_i}\in \mathcal {O}$. Thus every weak open neighborhood of $\mathbf {0}$ intersects $\{\mathbf {e}_m-\mathbf {e}_n\}$ and hence $\mathbf {0}\in \overline{\{\mathbf {e}_m-\mathbf {e}_n\}}_w$.

To show that no subsequence $\{\mathbf {e}_{m_j}-\mathbf {e}_{n_j}\}\subset \{\mathbf {e}_m-\mathbf {e}_n\}$ converges weakly to $\mathbf {0}$ it suffices for any such subsequence, to construct a functional $T\in \ell _1^*$ such that $\{T(\mathbf {e}_{m_j}-\mathbf {e}_{n_j})\}$ does not converge to zero. Construct one such T by examining separately the following cases

i.:

Some pair $(m_j,n_j)$ occurs infinitely many times;

ii.:

Some index $m_j$ occurs infinitely many times and $n_j\rightarrow \infty$, or vice versa.

iii.:

Both $m_j, n_j\rightarrow \infty$.

12.9.:

Let $E=\ell _2$ be defined by (von Neuman [113])

$$\begin{aligned} E=\{\mathbf {e}_n+n\mathbf {e}_m\}\quad \text { for }\>0\le n<m. \end{aligned}$$

Prove that E is strongly closed (Hint: all the sequences in E, Cauchy in $\ell _2$ are constant).

Prove that the origin of $\ell _2$ is in the weak closure but not in the weak sequential closure of E.

12.10.:

Prove that $\ell _2$ and $\ell _1$ equipped with their weak topology do not satisfy the first axiom of countability.

14c Weak Compactness

14.1.:

A weakly compact subset of a normed linear space is weakly closed. See also Proposition 5.1 of Chap. 2.

14.2. :

A weakly compact, convex subset of a normed linear space is strongly closed.

1.1 14.1c Linear Functionals on Subspaces of $C({\bar{E}})$

Let E be a bounded, open set in $\mathbb {R}^N$ and let $C({\bar{E}})$ denote the space of the continuous functions in ${\bar{E}}$ equipped with the sup-norm.

Let $X_o$ be a subspace of $C({\bar{E}})$ closed in the topology of $L^2(E)$. Prove that there exist positive constants $C_o\le C_1$, such that

$$\begin{aligned} C_o\Vert f\Vert _\infty \le \Vert f\Vert _2\le C_1\Vert f\Vert _\infty . \end{aligned}$$

Let $T_{o,y}\in X_o^*$ be the evaluation map at y

$$\begin{aligned} T_{o,y}(f)=f(y)\qquad \text { for all }\> f\in X_o. \end{aligned}$$

By the Hahn–Banach theorem there exists a functional $T_y\in L^2(E)^*$ such that $T_y=T_{o,y}$ on $X_o$. By the Riesz representation of the bounded linear functionals in $L^2(E)$, there exists a function $K(\cdot ,y)\in L^2(E)$, such that

$$\begin{aligned} f(y)=\int _EK(x,y)f(x)d\mu \qquad \text { for all }\> f\in X_o. \end{aligned}$$

(14.1c)

Proposition 14.1c

The unit ball of $X_o$ is compact.

Proof

The closed unit ball $\bar{B}_{o,1}$ of $X_o$ is also weakly closed. Since $L^2(E)$ is reflexive, also $X_o$ is reflexive. Therefore $\bar{B}_{o,1}$ is bounded and weakly closed and hence sequentially compact. In particular every sequence $\{f_n\}\subset \bar{B}_{o,1}$ contains in turn a subsequence $\{f_{n^\prime }\}$ weakly convergent to some $f\in \bar{B}_{o,1}$. By (14.1c) such a sequence converges pointwise to f and

$$\begin{aligned} \Vert f_n\Vert _\infty \le (\text {const})\Vert f_n\Vert _2\le (\text {const})^\prime \end{aligned}$$

for a constant independent of n. Therefore, by the Lebesgue dominated convergence theorem, $\{f_{n^\prime }\}\rightarrow f$ strongly in $L^2(E)$.

Thus every sequence $\{f_n\}\subset \bar{B}_{o,1}$ contains in turn a strongly convergent subsequence. This implies that $\bar{B}_{o,1}$ is compact in the strong topology inherited form $L^2(E)$. $\blacksquare$

Corollary 14.1c

Every subspace $X_o\subset C({\bar{E}})$, closed in $L^2(E)$ is finite dimensional.

1.2 14.2c Weak Compactness and Boundedness

14.3.:

Give a different proof of Corollary 14.1 by means of the uniform boundedness principle. Hint: For all $T\in X^*$, the image T(E) is a compact and hence bounded subset of $\mathbb {R}$.

15c The Weak${}^*$ Topology of $X^*$

1.1 15.1c Total Sets of X

The linear span $X_*$ of a collection of linear functionals on a linear vector space X is a total set of X, if it separates its points, that is, if for any two distinct elements $x,y\in X$, there exists $T\in X_*$ such that $T(x)\ne T(y)$. The dual $X^*$ of a normed space $\{X;\Vert \cdot \Vert \}$ separates the points of X and therefore is a total set of X. However the notion of total set does not require a topology being placed on X nor the continuity of the linear maps in $X_*$.

If $X_*$ is total for X, one might endow X with the weakest topology $\mathcal {W}_*$, for which all the elements of $X_*$ are continuous. A $\mathcal {W}_*$-open base neighborhood of the origin is of the form

$$\begin{aligned} \mathcal {O}_* =\left\{ \genfrac{}{}{0.0pt}{}{x\in X\,\bigm |\,|T_j(x)|<\alpha _j\> \text { for }\> T_j\in X_*\>\text { and }\>\alpha _j\in \mathbb {R}^+}{\text {for }\> j=1,\dots ,n\> \text { for some finite }\> n.}\right\} \end{aligned}$$

(15.1c)

The construction is in all similar to the construction of the weak topology of a normed space $\{X;\Vert \cdot \Vert \}$. Since $X_*$ is total, the $\mathcal {W}_*$ topology on X is Hausdorff and one verifies that the operations of sum and product by scalars are continuous in the indicated topology. Thus $\{X;\mathcal {W}_*\}$ is a Hausdorff, topological vector space. By construction the elements of $X_*$ are continuous from $\{X;\mathcal {W}_*\}$ into $\mathbb {R}$. The next proposition asserts that these are the only bounded linear functional on $\{X;\mathcal {W}_*\}$.

Proposition 15.1c

Let $T:\{X;\mathcal {W}_*\}\rightarrow \mathbb {R}$ be nonidentically zero, linear and continuous. Then $T\in X_*$.

Proof

By the continuity of T, there exists a $\mathcal {W}_*$-open set of the form (15.1c) such that $\mathcal {O}_*\subset T^{-1}(-1,1)$. Let $T_j\in X_*$ for $j=1,\dots ,n$ be the functionals in $X_*$ that identify $\mathcal {O}_*$. If $x\in {\mathop {\textstyle {\bigcap }}\limits }_{j=1}^n\ker \{T_j\}$ then $x\in T^{-1}(-\varepsilon ,\varepsilon )$ for all $\varepsilon >0$. Therefore T vanishes on ${\mathop {\textstyle {\bigcap }}\limits }_{j=1}^n\ker \{T_j\}$, and by Proposition 11.2 of Chap. 2, T is a linear combination of the $T_j$. $\blacksquare$

15.1.:

Let $\{X;\Vert \cdot \Vert \}$ be a normed space. Then $X_{**}$ as defined in (13.3) separates the points in $X^*$ and is total for $X^*$. The elements of $X_{**}$ are the only bounded linear functionals on $X^*$ equipped with its $\text {weak}^*$ topology. In particular every $T_*\in X^{**}-X_{**}$ is discontinuous with respect to the $\text {weak}^*$ topology of $X^*$,

15.2.:

Let $\{X;\Vert \cdot \Vert \}$ be a normed space. Then $X^*$ equipped with its $\text {weak}^*$ topology is a topological vector space whose dual is isometrically isomorphic to X.

15.3.:

Let $\{X;\Vert \cdot \Vert \}$ be a normed space. Then every convex, $\text {weak}^*$ compact subset of $X^*$ is the $\text {weak}^*$ closed convex hull of its extreme points (§ 11.4c).

1.2 15.2c Metrization Properties of Weak${}^*$ Compact Subsets of $X^*$

Proposition 15.2c

Let $\{X;\Vert \cdot \Vert \}$ be a Banach space. Then the weak$^{*}$ topology of a $\text {weak}^*$ compact set $K\subset X^*$ is metric if and only if $\{X;\Vert \cdot \Vert \}$ is separable.

Proof

(sufficient condition) Assume first that X is separable and let $\{x_n\}\subset X$ be a sequence dense in X. For $T_1,T_2\in X^*$ set

$$\begin{aligned} d(T_1,T_2)=\mathop {\textstyle {\sum }}\limits \frac{1}{2^n} \frac{|(T_1-T_2)(x_n)|}{1+|(T_1-T_2)(x_n)|}. \end{aligned}$$

Keeping in mind that X as a subset of $X^{**}$ separates the points of $X^*$, one verifies that

$$\begin{aligned} d:X^*\times X^*\rightarrow \mathbb {R}^+ \end{aligned}$$

is a metric which generates a metric topology on $X^*$ (§ 13.11c of the Complements of Chap. 2). One also verifies that every ball $B_\varepsilon (T)\subset X^*$ of center $T\in X^*$ and radius $\varepsilon >0$ in the metric $d(\cdot ,\cdot )$ contains a weak${}^*$ open neighborhood of T. Therefore the identity map from $X^*$ equipped with its weak${}^*$ topology onto $X^*$ equipped with the metric topology of $d(\cdot ,\cdot )$, is continuous. Let $\{K;\text {weak}^*\}$ and $\{K;d\}$ be the topological spaces formed by K equipped with the topologies inherited respectively from the weak${}^*$ and metric topologies of $X^*$. The identity map from the compact space $\{K;\text {weak}^*\}$ onto $\{K;d\}$ is continuous and one-to-one. To show that it is a homeomorphism we appeal to Proposition 5.1 of Chap. 2. A $\text {weak}^*$-closed set $E\subset \{K;\text {weak}^*\}$ is $\text {weak}^*$ compact. Since the identity map from $\{K;\text {weak}^*\}$ onto $\{K;d\}$ is continuous, the image of E in $\{K;d\}$ is compact and hence closed in the metric topology of $\{K;d\}$, since the latter is Hausdorff. $\blacksquare$

Remark 15.1c

The identity map from $\{K;\text {weak}^*\}$ onto $\{K;d\}$, being a homeomorphism, identifies the two topologies. For this is essential that $\{K;\text {weak}^*\}$ be $\text {weak}^*$ compact. In particular, the proposition does not imply that the $\text {weak}^*$ topology of the whole $X^*$ is metrizable.

Proof

(necessary condition) Assume conversely that the weak$^{*}$ topology of $K\subset X^*$ is metric. Up to a translation, may assume that $\varTheta \in K$. There exists a countable collection $\{\mathcal {O}^*_n\}$ of weak${}^*$ open neighborhoods of $\varTheta ^*$ such that $\mathop {\textstyle {\bigcap }}\limits \mathcal {O}^*_n=\varTheta ^*$. Without loss of generality the $\mathcal {O}^*_n$ are of the form

$$\begin{aligned} \mathcal {O}^*_n=\left\{ \genfrac{}{}{0.0pt}{}{T\in X^*\bigm | |x_{j_n}(T)|<\alpha _{j_n}\>\text { for }\> \alpha _{j_n}>0\> \text { and }\> x_{j_n}\in X}{\text {for }\> j_n\in \{j_{n,1},\dots ,j_{n,k}\}\subset \mathbb {N}}\right\} . \end{aligned}$$

The collection of $\{x_{j_n}\}$ is countable and its closed linear span is separable. We claim that $\overline{\{x_{j_n}\}}=X$. If not, there exists a nontrivial $T\in X^*$ vanishing on $\overline{\{x_{j_n}\}}$ (Corollary 10.3). However if $T(x_{j_n})=0$ for all $x_{j_n}$ then $T\in \mathcal {O}_n^*$ for all n and thus $T=\varTheta ^*$.$\blacksquare$

16c The Alaoglu Theorem

16.1.:

Let E be a bounded open set in $\mathbb {R}^N$. Prove that there is no linear normed space $\{X;\Vert \cdot \Vert \}$ whose dual is $L^1(E)$ with respect to the Lebesgue measure. Hint: Problems 11.6, and 15.3.

16.2.:

Let E be a bounded open set in $\mathbb {R}^N$. Prove that $C({\bar{E}})$ is not the dual of any linear normed space. Hint: Problems 11.7, and 15.3.

16.3.:

The weak${}^*$ topology of the closed unit ball of $L^\infty (E)$ and $\ell _\infty$ are metrizable.

16.4.:

A bounded and weakly closed subset E of a Banach space X, need not be weakly compact. In particular the closed unit balls of $L^1(E)$, $L^\infty (E)$, $\ell _1$ and $\ell _\infty$ are weakly closed but not weakly compact.

16.5.:

Let $\{X;\Vert \cdot \Vert \}$ be a normed space and let

$$\begin{aligned} B_1=\{x\in X\bigm | \Vert x\Vert \le 1\},\qquad S_1=\{x\in X\bigm | \Vert x\Vert = 1\} \end{aligned}$$

be respectively the unit ball and the unit sphere in X. Prove or disprove by a counterexample the following statements:

i.:

$B_1$ is strongly and weakly bounded.

ii.:

$B_1$ is weakly sequentially closed.

iii.:

$S_1$ is weakly sequentially closed.

iv.:

If $\{X;\Vert \cdot \Vert \}$ is reflexive then $S_1$ is weakly sequentially closed.

v.:

If $\{X;\Vert \cdot \Vert \}$ is reflexive then $B_1$ is weakly sequentially compact.

vi.:

If $\{X;\Vert \cdot \Vert \}$ is reflexive then it is Banach.

vii.:

If $\{X;\Vert \cdot \Vert \}$ is Banach then it is reflexive.

viii.:

The weak topology of $\{X;\Vert \cdot \Vert \}$ is Hausdorff.

1.1 16.1c The Weak${}^*$ Topology of $X^{**}$

Denote by $X^{***}$ the dual of $X^{**}$, that is the collection of all bounded linear functionals $f:X^{**}\rightarrow \mathbb {R}$. It is a Banach space by the norm

$$\begin{aligned} \Vert f\Vert =\sup _{x^{**}\in X^{**}; x^{**}\ne 0} \frac{f(x^{**})}{\Vert x^{**}\Vert }= \sup _{x^{**}\in X^{**}; \Vert x^{**}\Vert =1}f(x^{**}). \end{aligned}$$

(16.1c)

Every element $x^*\in X^*$ identifies an element $f_{x^*}\in X^{***}$ by the formula

$$\begin{aligned} X^{**}\ni x^{**}\longrightarrow f_{x^*}(x^{**})= x^{**}(x^*). \end{aligned}$$

(16.2c)

Let $X_{***}$ denote the collection of all such functionals, that is

$$\begin{aligned} X_{***}=\left\{ \genfrac{}{}{0.0pt}{}{\text {the collection of all functionals}\ f_{x^*}\in X^{***}}{\text {of the form (13.2) as}\ x^*\ \text {ranges over}\ X^*}\right\} . \end{aligned}$$

From Corollary 10.2 and (16.1c) it follows that $\Vert f_{x^*}\Vert =\Vert x^*\Vert$. Therefore the injection map

$$\begin{aligned} X^*\ni x^*\longrightarrow f_{x^*}\in X_{***}\subset X^{***} \end{aligned}$$

(16.3c)

is an isometric isomorphism between $X^*$ and $X_{***}$. In general, not all the bounded linear functionals $f:X^{**}\rightarrow \mathbb {R}$ are derived from the injection map (16.3c); otherwise said, the inclusion $X_{***}\subset X^{***}$ is in general strict. The set $X_{***}$ is total for $X^{**}$ and it generates a topology $\mathcal {W}_{**}$ on $X^{**}$ which is the $\text {weak}^*$ topology generated by $X^*$ on $X^{**}$ (§ 15.1c). By this topology, $\{X^{**};\mathcal {W}_{**}\}$ turns into a Hausdorff, linear, topological vector space (§ 15.1c), and for such a space the separation Proposition 11.3 holds. In particular $\mathcal {W}_{**}$-closed sets are separated from points by a bounded linear functional on $\{X^{**};\mathcal {W}_{**}\}$. By Proposition 15.1c, any such a functional is of the form (16.2c). Let

$$\begin{aligned} \begin{array}{ll} B\>&{}\text {the unit ball of}\ X\ \text {closed in the norm of}\ \{X;\Vert \cdot \Vert _X\}\\ B^{**}\>&{}\text {the unit ball of}\ X^{**}\ \text {closed in the norm of}\ \{X^{**};\Vert \cdot \Vert _{**}\}. \end{array} \end{aligned}$$

Since the natural injection $X\ni x\rightarrow f_x\in X^{**}$ defined by (14.2) is an isometric isomorphism, the ball B when regarded as a subset of $B^{**}$ is a closed subset of $B^{**}$ and the inclusion is proper unless X is reflexive. If $B^{**}$ is given the $\mathcal {W}_{**}$ topology, by the Alaoglu Theorem is weak${}^*$ compact, and since $\mathcal {W}_{**}$ is Hausdorff, $B^{**}$ is both $\Vert \cdot \Vert _{**}$-closed and $\mathcal {W}_{**}$-closed (Proposition 5.1-(iii) of Chap. 2). Set

$$\begin{aligned} B_{**}=\left\{ \genfrac{}{}{0.0pt}{}{\text {the}\ \mathcal {W}_{**}\text {-closure of norm-closed unit ball of}\ X}{\text {regarded as a subset of}\ \{X^{**};\mathcal {W}_{**}\}} \right\} \end{aligned}$$

The next proposition asserts that while B with its norm topology is a subset $B^{**}$, with in general strict inclusion, when it is given the $\mathcal {W}_{**}$ topology, it is actually dense in $B^{**}$.

Proposition 16.1c

([57]) $B_{**}=B^{**}$.

Proof

$B_{**}$ is a closed and convex subset of $B^{**}$ (10.1-(iii) of the Complements of Chap. 2). If $B_{**}\subset B^{**}$ with proper inclusion, select and fix $x^{**}\in B^{**}-B_{**}$. By Proposition 11.3 there exists a nontrivial, bounded, linear functional f on $\{X^{**};\mathcal {W}_{**}\}$ and real numbers $\alpha <\beta$ such that

$$\begin{aligned} f(y)\le \alpha <\beta \le f(x^{**})\quad \text { for all }\>y\in B_{**}. \end{aligned}$$

By Proposition 15.1c, such a functional f is of the form (16.2c). Therefore, there exists $x^*\in X^*$ such that

$$\begin{aligned} x^*(y)\le \alpha <\beta \le x^{**}(x^*)\quad \text { for all }\>y\in B. \end{aligned}$$

(16.4c)

Now $y\in B$ implies $-y\in B$. Therefore

$$\begin{aligned} \Vert x^*\Vert =\sup _{\Vert y\Vert =1}x^*(y)\le \alpha . \end{aligned}$$

On the other hand, since $x^{**}\in B^{**}$

$$\begin{aligned} \alpha <\beta \le x^{**}(x^*)\le \Vert x^{**}\Vert \,\Vert x^*\Vert \le \alpha . \end{aligned}$$

This contradicts (16.4c) and proves the proposition. $\blacksquare$

Corollary 16.1c

The Banach space $\{X;\Vert \cdot \Vert \}$ when regarded as a subspace of $X^{**}$ and equipped with the weak${}^*$ topology of $X^{**}$, is dense in $X^{**}$.

1.2 16.2c Characterizing Reflexive Banach Spaces

The next statements follow from the previous proposition, upon observing that the $\mathcal {W}_{**}$ topology of X, as a subset of $X^{**}$, is precisely the weak topology generated on X by $X^*$.

Corollary 16.2c

A Banach space $\{X;\Vert \cdot \Vert \}$ is reflexive if and only if its closed unit ball is weakly compact.

Corollary 16.3c

A Banach space $\{X;\Vert \cdot \Vert \}$ is reflexive if and only if a bounded and weakly closed set is weakly compact.

Thus in some sense, reflexive Banach spaces are those for which a version of the Heine–Borel Theorem holds (Proposition 6.4 of Chap. 2).

16.6.:

Some of the following statements contain fallacies. Identify and disprove them, by an argument or a counterexample.

i.:

The closed unit ball B of a normed space $\{X;\Vert \cdot \Vert \}$, is convex and hence weakly closed (Proposition 12.2).

ii.:

Regarding B as a subset of $B^{**}$ its $\mathcal {W}_{**}$ topology coincides precisely with its weak topology.

iii.:

Therefore B as a subset of $B^{**}$ is $\mathcal {W}_{**}$ closed and hence $\mathcal {W}_{**}$-compact, since it is a closed subset of a compact set.

iv.:

Since the $\mathcal {W}_{**}$ topology on $B\subset B^{**}$ coincides with its weak topology, B is weakly compact.

1.3 16.3c Metrization Properties of the Weak Topology of the Closed Unit Ball of a Banach Space

Proposition 16.2c

Let $\{X;\Vert \cdot \Vert \}$ be a Banach space. Then the weak topology of the closed unit ball $B\subset X$ is metric, if and only if the dual $X^*$ is separable.

Proof

If $X^*$ is separable, the weak${}^*$ topology of $B^{**}\subset X^{**}$ is metric. Identify X as a subset of $X^{**}$ by the natural injection map $X\ni x\rightarrow f_x\in X^{**}$ and observe that the weak${}^*$ topology inherited by X as a subset of $X^{**}$ is precisely the weak topology of X.

Assume next that the weak topology of the closed unit ball $B\subset X$ is metric. There exists a countable collection $\{\mathcal {O}_n\}$ of weakly open neighborhoods of the origin such that every weak neighborhood of the origin contains some $\mathcal {O}_n$. Without loss of generality the $\mathcal {O}_n$ can be taken of the form

$$\begin{aligned} \mathcal {O}_n=\left\{ \genfrac{}{}{0.0pt}{}{x\in X\bigm | |T_{j_n}(x)|<\varepsilon _n\>\text { for } \> \varepsilon _n>0\> \text { and }\> T_{j_n}\in X^*}{\text {for }\> j_n\in \{j_{n,1},\dots ,j_{n,k}\}\subset \mathbb {N}}\right\} \end{aligned}$$

The collection of $\{T_{j_n}\}$ is countable and its closed linear span is separable. We claim that $\overline{\{T_{j_n}\}}=X^*$. If not pick $\eta ^*\in X^*-\overline{\{T_{j_n}\}}$ at distance $\delta \in (0,1)$ from $\overline{\{T_{j_n}\}}$ and determine a nontrivial, bounded linear functional $x^{**}\in X^{**}$, of norm $\Vert x^{**}\Vert \le 1$, vanishing on $\overline{\{T_{j_n}\}}$ and such that $x^{**}(\eta ^*)=\delta$ (Proposition 10.3). The set

$$\begin{aligned} \mathcal {O}_\delta =\big \{x\in X\bigm | |\eta ^*(x)|<{\textstyle \frac{1}{2}}\delta \big \} \end{aligned}$$

is a weak neighborhood of the origin of X and hence $\mathcal {O}_n\subset \mathcal {O}_\delta$ for some n. Since $x^{**}\in B^{**}$, by the density Proposition 16.1c, there exists $x\in B$ such that

$$\begin{aligned} |x^{**}(T_{j_n})-T_{j_n}(x)|<\varepsilon _n\>\text { for all }\> j_n\in \{j_{n,1},\dots ,j_{n,k}\} \end{aligned}$$

and simultaneously

$$\begin{aligned} |\delta -\eta ^*(x)|=|x^{**}(\eta ^*)-\eta ^*(x)|<\varepsilon _n. \end{aligned}$$

By picking $\varepsilon _n$ sufficiently small we may insure that $\varepsilon _n<\frac{1}{2}\delta$. Then, since $x^{**}$ vanishes on $\overline{\{T_{j_n}\}}$, the first of these gives,

$$\begin{aligned} |T_{j_n}(x)|<\varepsilon _n\quad \text { for all }\> j_n\in \{j_{n,1},\dots ,j_{n,k}\} \end{aligned}$$

which implies that $x\in \mathcal {O}_n$. However the second of these implies $|\eta ^*(x)|>\frac{1}{2}\delta$. That is $x\notin \mathcal {O}_\delta \subset \mathcal {O}_n$. $\blacksquare$

16.7.:

The weak topology of the closed unit ball of $L^1(E)$ and $\ell _1$ are not metrizable.

1.4 16.4c Separating Closed Sets in a Reflexive Banach Space

Proposition 16.3c

Let $C_1$ and $C_2$ be two nonempty, disjoint, closed subsets of a reflexive Banach space $\{X;\Vert \cdot \Vert \}$. Assume that at least one of them is bounded. There exists a bounded linear functional T on X and real numbers $0<\alpha <\beta$ such that (11.2) holds. Equivalently, any two closed subsets of a reflexive Banach space, can be strictly separated, provided at least one of them is bounded.

Proof

If $C_1$ is closed and bounded it is $\text {weak}^*$ compact. The statement then follows from Proposition 11.3. $\blacksquare$

Remark 16.1c

The requirement that at least one of the two closed sets $C_1$ and $C_2$ be bounded is essential, in view of the counterexample in § 11.1c

17c Hilbert Spaces

1.1 17.1c On the Parallelogram identity

The parallelogram identity is equivalent to the existence of an inner product on a vector space X in the following sense.

A scalar product $\langle \cdot ,\cdot \rangle$ on a vector space X generates a norm $\Vert \!\cdot \!\Vert$ on X that satisfies the parallelogram identity. Vice versa, let $\{X;\Vert \cdot \Vert \}$ be a normed space whose norm $\Vert \cdot \Vert$ satisfies the parallelogram identity. Then setting

$$\begin{aligned} 4\langle x,y\rangle =\Vert x+y\Vert ^2-\Vert x-y\Vert ^2 \end{aligned}$$

defines a scalar product in X.

17.2.:

If $p\ne 2$ then $L^p(E)$ is not a Hilbert space.

17.3.:

Let $\{x_n\}$ and $\{y_n\}$ be Cauchy sequences in a Hilbert space H. Then $\{\langle x_n,y_n\rangle \}$ is a Cauchy sequence in $\mathbb {R}$.

18c Orthogonal Sets, Representations and Functionals

18.1.:

For an n-tuple $\{x_1,x_2,\dots ,x_n\}$ of orthogonal elements in a Hilbert space H

$$\begin{aligned} \Big \Vert \mathop {\textstyle {\sum }}\limits _{i=1}^n x_i\Big \Vert ^2= \mathop {\textstyle {\sum }}\limits _{i=1}^n\Vert x_i\Vert ^2\qquad \text {(Pythagora's theorem)} \end{aligned}$$

18.2.:

Let E be a subset of H. Then $E^\perp$ is a linear subspace of H and $\left( E^\perp \right) ^\perp$ is the smallest, closed, linear subspace of H containing E.

18.3.:

Every closed convex set of H has a unique element of least norm. More generally, let C be a weakly closed subset of a reflexive Banach space. Then the functional $h(x)=\Vert x\Vert$ takes its minimum on C, i.e., there exists $x_o\in C$ such that $\inf _{x\in C}\Vert x\Vert =\Vert x_o\Vert$.

18.4.:

Let E be a bounded open set in $\mathbb {R}^N$ and let $f\in C({\bar{E}})$. Denote by $\mathcal {P}_n$ the collections of polynomials of degree at most n in the coordinate variables. There exists a unique $P_o\in \mathcal {P}_n$ such that

$$\begin{aligned} \int _E|f-P|^2dx\ge \int _E|f-P_o|^2dx \qquad \text { for all }\> P\in \mathcal {P}_n. \end{aligned}$$

19c Orthonormal Systems

19.1.:

Let H be a Hilbert space and let $\mathcal {S}$ be an orthonormal system in H. Then for any pair x, y of elements in H

$$\begin{aligned} \mathop {\textstyle {\sum }}\limits _{\mathbf {u}\in \mathcal {S}}|\langle \mathbf {u},x\rangle | |\langle \mathbf {u},y\rangle |\le \Vert x\Vert \Vert y\Vert . \end{aligned}$$

19.2.:

Let $\mathcal {S}$ be an orthonormal system in H and denote by $H_o$ the closure of the linear span of $\mathcal {S}$. The projection of an element $x\in H$ into $H_o$ is defined by

$$\begin{aligned} x_o=\mathop {\textstyle {\sum }}\limits _{\mathbf {u}\in \mathcal {S}}\,\langle x,\mathbf {u}\rangle \mathbf {u}. \end{aligned}$$

Such a formula defines $x_o$ uniquely. Moreover $x_o\in H_o$ and $(x-x_o)\perp H_o$.

19.3.:

A Non Separable Hilbert Space: In $L^2_{{\text {loc}}}(\mathbb {R})$ with respect to the Lebesgue measure, define

$$\begin{aligned} L^2_{{\text {loc}}}(\mathbb {R}) \ni f,g \rightarrow \langle f,g \rangle \buildrel {\text {def}}\over {=} \lim _{\rho \rightarrow \infty } \frac{1}{\rho } \int _{-\rho }^{\rho } fgdx. \end{aligned}$$

Set

$$\begin{aligned} \begin{aligned} H_o&=\{f\in L^2_{{\text {loc}}}(\mathbb {R})\bigm | \Vert f\Vert =0\}\\ H_1&=\{ f\in L^2_{{\text {loc}}}(\mathbb {R})\bigm |\Vert f\Vert <\infty \}. \end{aligned} \end{aligned}$$

Since $H_o$ contains nonzero elements, $\Vert \cdot \Vert$ is not a norm. Introduce the quotient space

$$\begin{aligned} H=\frac{H_1}{H_o}\quad \text { of equivalence classes } \> f+H_o\quad \text { for }\> f\in H_1. \end{aligned}$$

Verify that $\langle \cdot ,\cdot \rangle$ is an inner product and $\Vert \cdot \Vert$ is a norm on H. The system

$$\begin{aligned} \big \{\sin \alpha x+H_o\big \}_{\alpha \in \mathbb {R}} \end{aligned}$$

is orthonormal in H, and uncountable. Therefore H is non separable, by Proposition 19.2.