Elements of Duality Theory

Abstract

In this chapter we mainly deal with relationship between a space (locally convex or Banach) and its dual space. The contents of the chapter: the general notion of duality; weak topology; polars and bipolars; the adjoint operator; Alaoglu’s theorem; $w^*$-convergence in a dual Banach space; the second dual space; weak convergence criteria in classical Banach spaces; total and norming sets of functionals; metrizability conditions; the Eberlein–Smulian theorem; reflexive Banach spaces.

17.1 Duality in Locally Convex Spaces

17.1.1 The General Notion of Duality. Polars

Definition 1.

Let X, Y be linear spaces over the same field $\mathbb K= \mathbb R$ or $\mathbb C$. A mapping that to each pair of elements $(x, y) \in X\times Y$ assigns a number $\langle x, y \rangle \in \mathbb K$ is called a duality pairing (or duality mapping, or simply a duality) if :

• (a)   $\langle {x, y} \rangle$ is a bilinear form:

  $$\begin{aligned} \langle {a_1 x_1 + a_2 x_2 , y} \rangle = a_1 \langle {x_1, y} \rangle + a_2 \langle {x_2, y} \rangle ;\\ \langle {x, a_1 y_1 + a_2 y_2} \rangle = a_1 \langle {x, y_1} \rangle + a_2 \langle {x, y_2} \rangle ; \end{aligned}$$

• (b)   $\langle {x, y} \rangle$ satisfies the non-degeneracy condition:

  • — for every $x\in X\setminus \{0\}$ there exists an $y\in Y$ such that $\langle {x, y} \rangle \not =0$, and

  • — for every $y\in Y\setminus \{0\}$ there exists an $x\in X$ such that $\langle {x, y} \rangle \not = 0$.

A pair of spaces X, Y with a duality pairing given on them is called a dual pair, or a pair of spaces in duality .

As in the previous chapter, in order to avoid treating separately the real and complex cases, we will tacitly assume that $\mathbb K= \mathbb C$. The simpler case of real spaces differs from the complex one only by absence of the symbol $\mathrm{Re}$ when one applies the Hahn–Banach separation theorem.

For us the most important example of a dual pair will consist of a locally convex space X and its dual space $Y = X^*$, with the duality pairing given by the evaluation of functionals on elements: $\langle {x, y}\rangle = y(x)$. To a certain extent this example describes the general situation.

Definition 2.

Let (X, Y) be a dual pair. For each element $y \in Y$ we define its action on the elements of the space X by the rule $y(x) = \langle {x, y} \rangle$. With this definition every element $y \in Y$ becomes a linear functional on X, i.e., $Y \subset X'$. The weak topology on X is defined as the topology $\sigma (X, Y)$ introduced in Definition 1 of Subsection 16.3.2. That is, a neighborhood basis of zero in the topology $\sigma (X, Y)$ is given by the family of sets $\left\{ {x \in X: \max _{y \in G} \left| \langle {x, y} \rangle \right| < \varepsilon } \right\}$, where $\varepsilon > 0$ and G runs over all finite subsets of the space Y.

Axiom (b) of a dual pair guarantees that the weak topology is separated. By Theorem 1 of Subsection 16.3.2, $\left( {X,\sigma (X, Y)}\right) ^* = Y$, so that any dual pair can be regarded as a pair of the form $(X, X^*)$. Nevertheless, the general definition of a dual pair has its own merit: in that definition, the spaces X and Y play completely equivalent roles. In particular, we could equally successfully regard the elements of the space X as functionals on Y and consider the weak topology $\sigma (Y, X)$ already on the space Y. This equivalence of the roles played by the two spaces allows one to transfer, by symmetry, the properties of one of the spaces in a dual pair to the other space.

Recall that $\sigma (X, Y)$ is the weakest topology in which all functionals $y \in Y$, i.e., all functionals of the form $y(x) = \langle {x, y} \rangle$, are continuous. In particular, if X is a locally convex space and $Y = X^*$, then $\sigma (X, Y)$ is weaker (possibly not strictly) than the initial topology of the space X. This remark can be taken as a justification of the term “weak topology”.

Let us mention yet another important connection between the original topology of a locally convex space X and the weak topology $\sigma (X, X^*)$.

Theorem 1.

Any convex closed subset of a locally convex space X is also closed in the weak topology $\sigma (X, X^*)$. In particular, every closed linear subspace of a locally convex space X is $\sigma (X, X^*)$-closed.

Proof.

Let $A \subset X$ be a convex closed subset. Let us pick an arbitrary point $x \in X\setminus A$ and show that x is not a $\sigma (X, X^*)$-limit point of the set A. Since A is closed, there exists an open neighborhood U of the point x such that $U\cap A=\emptyset$. Since the space X is locally convex, we may assume that the neighborhood U is convex. By the geometric form of the Hahn–Banach theorem, there exist a functional $f \in X^* \setminus \{0\}$ and a scalar $\theta \in \mathbb {R}$ such that $\mathrm{Re} \, f(u) < \theta$ for all $u \in U$ and $\mathrm{Re} f(a) \geqslant \theta$ for all $a \in A$. In particular, $\mathrm{Re}\, f(x) < \theta$. Since f, and together with it $\mathrm{Re}\, f$ are $\sigma (X, X^*)$-continuous functions, the point x, at which $\mathrm{Re}\, f(x) < \theta$, cannot be a $\sigma (X, X^*)$-limit point for the set A, on which $\mathrm{Re}\, f(a)\geqslant \theta$.   $\square$

Definition 3.

Let (X, Y) be a dual pair. The polar of the set $A \subset X$ is the set $A^\circ \subset Y$, defined by the following rule: $y \in A^\circ$ if $|\langle {x, y}\rangle | \leqslant 1$ for all $x \in A$. The polar $A^\circ \subset X$ of a set $A\subset Y$ is defined by symmetry.

The annihilator of the set $A \subset X$ is the set $A^\bot \subset Y$ consisting of all elements $y \in Y$ such that $\langle {x, y} \rangle = 0$ for all $x \in A$. Obviously, $A^\bot \subset A^\circ$, and, by Lemma 2 of Subsection 16.3.2, if A is a linear subspace, then $A^\bot = A^\circ$. Moreover, $A^\bot = (\mathrm{Lin}\, A)^\bot$.

Example 1.

Consider the pair $(X, X^*)$, where X is a Banach space. Then $({B_X})^\circ = \overline{B}_{X^*}$. Indeed,

$$ f \in {\overline{B}}_{X^*} \Longleftrightarrow \Vert f\Vert \leqslant 1 \Longleftrightarrow \sup \limits _{x \in B_X } |f(x)| \leqslant 1 \Longleftrightarrow f \in (B_X)^\circ . $$

Theorem 2.

Polars possess the following properties:

• (i) If $A \subset B$, then $A^\circ \supset B^\circ$.

• (ii) $\{0_X \}^\circ = Y$ and $X^\circ = \{0_Y \}$, where $0_X$ and $0_Y$ are the zero elements of the spaces X and Y, respectively.

• (iii)   $(\lambda A)^\circ = \frac{1}{\lambda }A^\circ$ for any $\lambda \ne 0$.

• (iv) $\left( {\bigcup _{A \in \mathfrak {E}} A } \right) ^\circ = \bigcap _{A \in \mathfrak {E}}{A^\circ }$ for any family $\mathfrak {E}$ of subsets of the space X. In particular, $(A_1 \cup A_2)^\circ = A_1^\circ \cap A_2^\circ$.

• (v) For any point $x \in X$, the set $\{x\}^\circ$ is a convex balanced $\sigma (Y, X)$-closed neighborhood of zero.

• (vi) The polar of any set is a convex balanced $\sigma (Y, X)$-closed set.

• (vii) The sets of the form $A^\circ$, where A runs over all finite subsets of the space X, constitute a neighborhood basis of zero in the topology $\sigma (Y, X)$ .

Proof.

Properties (i)–(iv) are obvious. The fact that the set

$$\begin{aligned} \{x\}^0 = \{y \in Y:\, |\langle {x, y} \rangle | \leqslant 1\}&= \{y \in Y: | {x(y)} | \leqslant 1\} \nonumber \\&= x^{-1}\left( \{\lambda \in \mathbb {C}: | \lambda | \leqslant 1\}\right) \end{aligned}$$

(1)

is convex and balanced is a consequence of the linearity of x as a functional on Y. Since $\mathbb {C}_1 = \{\lambda \in \mathbb {C}: | \lambda | \leqslant 1\}$ is a closed neighborhood of zero in $\mathbb {C}$, and the functional x is continuous in the $\sigma (Y, X)$-topology, relation (1) means that $\{x\}^\circ$ is a $\sigma (Y, X)$-closed neighborhood of zero. This establishes property (v). Property (vi) follows from (v) thanks to property (iv): $A^\circ = \bigcap _{x \in A}{\{x\}^\circ }$, and the intersection operation does not destroy the properties of convexity, closedness, and balancedness.

Now let us turn to property (vii). If the subset $A\subset X$ is finite, then $A^\circ = \bigcap _{x \in A}{\{x\}^\circ }$ is a finite intersection of $\sigma (Y, X)$-neighborhoods. Hence, the polar of a finite set is a weak neighborhood. Further, by definition, every $\sigma (Y, X)$-neighborhood contains a set of the form $U_{G,\varepsilon }= \left\{ {y \in Y:\mathop {\max }_{g \in G} |{g(x)}| < \varepsilon } \right\}$, where $G = \{{g_1 , g_2 ,\ldots , g_n}\}\subset X$ and $\varepsilon > 0$. For $A = \frac{1}{2\varepsilon }G$ we have $U_{G,\varepsilon } \supset A^\circ$. That is, every $\sigma (Y, X)$-neighborhood contains a set of the form $A^\circ$, with $A \subset X$ finite.   $\square$

Corollary 1.

The annihilator of any set $A \subset X$ is a $\sigma (Y, X)$-closed linear subspace.

Proof.

The linearity is verified directly, while $\sigma (Y, X)$-closedness follows, for example, from property (vi) above and the relations $A^\bot = (\mathrm{Lin}\, A)^\bot = (\mathrm{Lin}\, A)^\circ$.   $\square$

Similarly to how with a convex sets one associates its convex hull, and with a subspace its linear hull (span), convex balanced sets lead to the absolute convex hull construction.

Definition 4.

Let X be a linear space. An absolutely convex combination of a finite collection of elements $\{x_k\}_{k = 1}^n \subset X$ is any sum $\sum _{k = 1}^n {\lambda _k x_k}$, where $\lambda _k$ are scalars with $\sum _{k = 1}^n{|\lambda _k|} \leqslant 1$. The absolutely convex hull of the set A in the linear space X is the set $\mathrm{aconv} A$ consisting of all absolutely convex combinations of elements of A. The closure of the set $\mathrm{aconv}\, A$ in the topology $\tau$ is denoted by $\tau$-$\overline{ \mathrm{aconv}} A$ or, if the context makes it clear, simply by $\overline{\mathrm{aconv}} A$.

Exercises

1.

Consider the real dual pair $({\mathbb {R}^2,\mathbb {R}^2})$ with the scalar product as the duality pairing. Construct in the plane the polars of the following sets:

• — $\{(0,1)\}$;

• — $\{(1,1)\}$;

• — $\{(1,1),(0,1)\}$;

• — $\{(x_1, x_2):\, | {x_1} | + | {x_2} | \leqslant 1\}$;

• — $\{(x_1, x_2):\, | {x_1} |^2 + | {x_2} |^2 \leqslant 1\}$.

2.

Consider the pair (X, Y), where $X = Y = C[0,1]$. Which of the expressions listed below give a duality on this pair?

• — $\langle {f, g} \rangle = \int _0^1 {f(t)g(t)dt}$;

• — $\langle {f, g} \rangle = \int _0^{1/2}{f(t)g(t)dt}$;

• — $\langle {f, g} \rangle = \int _0^{1/2}{f(t)g(t)dt} - \int _{1/2}^1 {f(t)g(t)dt}$;

• — $\langle {f, g} \rangle = \int _0^1 {f^2(t)g(t)dt}$;

• — $\langle {f, g} \rangle = \int _0^1 {f(t^2)g(t)dt}$;

• — $\langle {f, g} \rangle = f(0)g(0)$;

• — $\langle {f, g} \rangle = \int _0^1 {f(t)g(t)dt} + f(0)g(0)$.

3.

Let A be a subset of a linear space X. Then $\mathrm{aconv}\, A$ is a convex balanced set.

4.

Any convex balanced set containing the set A also contains $\mathrm{aconv} A$.

5.

$\mathrm{aconv} A$ is equal to the intersection of all convex balanced sets containing A.

6.

Let A be a subspace of the linear space X. Then $\mathrm{aconv} A = A$.

7.

Let A be a subset of a topological space X. Then $\overline{\mathrm{aconv}} A$ is the smallest, with respect to inclusion, closed convex balanced set containing A.

8.

Let (X, Y) be a dual pair, $A \subset X$. Then $({\mathrm{aconv} A})^\circ = A^\circ$.

9.

Consider on X the weak topology $\sigma (X, Y)$. Then the polar of any set coincides with the polar of its closure. Further, $({\overline{\mathrm{aconv}} A})^\circ = A^\circ$ for any set $A \subset X$.

10.

For any dual pair $(X, X^*)$, where X is a Banach space, describe explicitly the neighborhoods of zero in the topologies $\sigma (X, X^*)$ and $\sigma (X^*, X)$.

11.

Is the open unit ball in a Banach space X a $\sigma (X, X^*)$-open set?

12.

Is the closed unit ball in a Banach space X a $\sigma (X, X^*)$-closed set?

Note that the number of elements in an absolutely convex combination can be arbitrarily large. Consequently, even for a compact set A the absolutely convex hull is not necessarily closed: $\overline{\mathrm{aconv}} A$ will contain, in particular, the infinite sums of the form $\sum _{k = 1}^\infty {\lambda _k} x_k$ with $\sum _{k = 1}^\infty {|\lambda _k|} \leqslant 1$.

13.

On the following example, convince yourself that $\overline{\mathrm{aconv}} A$ is not exhausted by the sums $\sum _{k = 1}^\infty {\lambda _k} x_k$ with $\sum _{k = 1}^\infty {|\lambda _k|} \leqslant 1$. Let $A = \{x_n = e_1 + e_{n + 1} :n \in \mathbb {N}\}$, where $e_n$ is an orthonormal system in the Hilbert space H. Then the vector $e_1$ lies in $\overline{\mathrm{aconv}} A$, but cannot be written as $\sum _{k = 1}^\infty {\lambda _k} x_k$.

14.

In a finite-dimensional space the absolutely convex hull of any closed bounded set is closed.

17.1.2 The Bipolar Theorem

Let (X, Y) be a dual pair, $A \subset X$. Then $A^\circ \subset Y$, and now one can consider the polar of $A^\circ$.

Definition 1.

The set $( A^\circ )^\circ \subset X$ is called the bipolar $A \subset X$ and is denoted by $A^{\circ \circ }$.

Theorem 1.

The bipolar $A^{\circ \circ }$ of a set $A \subset X$ coincides with the $\sigma (X, Y)$-closed absolutely convex hull of A .

Proof.

First, we note that $A^{\circ \circ } \supset A$. Indeed, if $x \in A$, then by the definition of the set $A^\circ$, $|{\langle {x, y} \rangle } | \leqslant 1$ for any $y \in A^\circ$. But this means precisely that x belongs to the polar $A^\circ$.

Further, the bipolar is a particular case of a polar. Therefore, by assertion (vi) in Theorem 2 of Subsection 17.1.1, $A^{\circ \circ }$ is a convex balanced $\sigma (X, Y)$-closed set. Hence, $A^{\circ \circ } \supset \overline{\mathrm{aconv}\,} A$. To prove the opposite inclusion, pick an arbitrary point $x_0 \in X\setminus \overline{\mathrm{aconv}} A$, and let us show that $x_0 \notin A^{\circ \circ }$. Indeed, since $x_0 \notin \overline{\mathrm{aconv}} A$ and $\overline{\mathrm{aconv}} A$ is a convex balanced $\sigma (X, Y)$-closed set, the Hahn–Banach theorem in the form indicated in Exercise 9 in Subsection 16.3.1 says that there exists a $\sigma (X, Y)$-continuous linear functional y on X such that

• I. $|y(x)| \leqslant 1$ for all $x \in \overline{\mathrm{aconv}} A$, and

• II. $|y(x_0)| > 1$.

But every $\sigma (X, Y)$-continuous linear functional is an element of the space Y. Condition I means that $y \in ({\overline{\text{ aconv }} A})^\circ \subset A^\circ$. But then condition II means that $x_0 \notin A^{\circ \circ }$, as we needed to show.    $\square$

Corollary 1.

If $A \subset X$ is a $\sigma (X, Y)$-closed convex and balanced set, then $A^{\circ \circ } = A$. In particular, $B^{\circ \circ \circ } = B^\circ$ for any set $B \subset Y$.

Corollary 2.

$A^{ \bot \bot } = \overline{\mathrm{Lin}\,} A$ for any set $A \subset X$. If A is a linear subspace, then $A^{ \bot \bot } = \overline{A}$. Finally, $B^{ \bot \bot \bot } = B^\bot$ for any $B \subset Y$.

Proof.

$A^{ \bot \bot } = (A^\bot )^\bot = ((\mathrm{Lin}\, A)^\bot )^\bot = (\mathrm{Lin}\, A)^{\circ \circ } = \overline{\mathrm{Lin}\,} A$.   $\square$

Corollary 3.

If $A_1, A_2 \subset X$ are $\sigma (X, Y)$-closed convex and balanced sets, then the equality $A_1 = A_2$ is equivalent to the equality $A_1^\circ = A_2^\circ$. Moreover, if $A_1$, $A_2$ are subspaces, then the equality $A_1 = A_2$ is equivalent to the equality $A_1^\bot = A_2^\bot$.

Proof.

If two sets are equal, then their polars are also equal, so the implication $A_1 = A_2 \Longrightarrow A_1^\circ = A_2^\circ$ is obvious with no supplementary restrictions on the sets. Conversely, if $A_1^\circ = A_2^\circ$, then $A_1^{\circ \circ } = A_2^{\circ \circ }$, and it remains to use the bipolar theorem.

   $\square$

Recall that the spaces figuring in a dual pair (X, Y) play equivalent roles, and all assertions concerning the polars and bipolars of subsets of the space X are valid, upon exchanging the roles of the spaces in the dual pair, for the subsets of the space Y.

Theorem 2.

Let (X, Y) be a dual pair and A be a subset of Y. Then the following conditions are equivalent:

• (i) the set of functionals $A \subset Y$ separates the points of the space X;

• (ii) $A^\bot = \{0\}$;

• (iii) $A^{ \bot \bot } = Y$;

• (iv) the linear span of the set A is $\sigma (Y, X)$-dense in Y.

Proof.

(i) $\Longrightarrow$ (ii). The inclusion $A^\bot \supset \{0\}$ holds always. If now $x \in X\setminus \{0\}$, then according to (i), there exists an $y \in A$ such that $\langle {x, y} \rangle \ne 0$. Consequently, $x \notin A^\bot$.

(ii) $\Longrightarrow$ (i). Let $x \in X\setminus \{0\}$. Then $x \notin A^\bot$, and so there exists an $y \in A$ such that $\langle {x, y} \rangle \ne 0$.

(ii) $\Longleftrightarrow$ (iii). Since $A^\bot$ and $\{0\}$ are $\sigma (X, Y)$-closed subspaces, Corollary 3 applies.

(iii) $\Longleftrightarrow$ (iv). By Corollary 2, $A^{ \bot \bot } = \overline{\mathrm{Lin}}\, A$.

Exercises

1.

Construct in the plane the bipolars of the sets from Exercise 1 of Subsection 17.1.1.

2.

Consider the real dual pair (X, Y), where $X = Y = \mathbb {R}^2$, with the duality pairing $\langle {\mathbf {x},\mathbf {y}} \rangle = x_1 y_2 - 2x_2 y_1$. Construct the polars and bipolars of the following subsets of the space X:

• — $\{(0,1)\}$;

• — $\{(1,1)\}$;

• — $\{(1,1),(0,1)\}$;

• — $\{(x_1, x_2): | {x_1} | + | {x_2} | \leqslant 1\}$;

• — $\{(x_1, x_2): | {x_1} |^2 + | {x_2} |^2 \leqslant 1\}$.

3.

Do the polars of the above sets change if they are regarded as subsets of Y? Do the bipolars change?

4.

Consider the real dual pair (X, Y), where $X = C[0,1]$ and $Y = L_1[0,1]$, with the duality pairing $\langle {f, g} \rangle = \int _0^1{f(t)g(t)dt}$. Construct the polars and bipolars of the following subsets of the space X:

• — $\{{f \in C[0,1]:f(t) > 0\ \mathrm{for\ all}\ t \in [0,1]}\}$;

• — $\{{f \in C[0,1]:0 \leqslant f(t) \leqslant 1\ \mathrm{for\ all }\ t \in [0,1]} \}$;

• — $\{{f \in C[0,1]:f(t) = 0\ \mathrm{for\ all}\ t \in [0,1/2]}\}$.

5.

Consider the same sets of continuous functions as subsets of the space $Y = L_1 [0,1]$. Construct their polars and bipolars with respect to the duality from the preceding exercise.

17.1.3 The Adjoint Operator

Definition 1.

Let $X_1, X_2$ be linear spaces, and let $T:X_1 \rightarrow X_2$ be a linear operator. The algebraic adjoint (or algebraic conjugate, or algebraic dual) of the operator T is the operator ${T}':{Y}' \rightarrow {X}'$ acting as ${T}' f = f \circ T$.

Further, let $(X_1, Y_1)$, $(X_2, Y_2)$ be dual pairs, and $T: X_1 \rightarrow X_2$ a linear operator. We say that an adjoint operator $T^* :Y_2 \rightarrow Y_1$ exists for T if for any $y \in Y_2$ there exists an element $T^* y \in Y_1$, such that $\langle {Tx, y} \rangle = \langle {x, T^* y}\rangle$ for all $x \in X_1$.

Treating the elements of the spaces $Y_1$ and $Y_2$ as functionals on the space $X_1$ and $X_2$, respectively, we see that $T^* y = y \circ T$, which explains the correctness of the definition and the linearity of the operator $T^*$. Clearly, an adjoint operator exists for the operator T if and only if $T'(Y_2)\subset Y_1$, and in this case $T^*$ is the restriction of the algebraic adjoint ${T}'$ to $Y_2$. For dual pairs $(X_1, X_1^*)$, $(X_2, X_2^*)$, where $X_1$ and $X_2$ are Banach spaces, the new definition of adjoint (conjugate, or dual) operator agrees with the already familiar definition of the adjoint (conjugate, or dual) of an operator in Banach spaces.

Let us mention a few simple facts.

Theorem 1.

Let $X_1$, $X_2$ be locally convex spaces, and let $T:X_1 \rightarrow X_2$ be a continuous linear operator. Then there exists the adjoint operator $T^*:X_2^* \rightarrow ~X_1^*$ for T.

Proof.

Let $f \in X_2^*$. Then the functional ${T}'\, f = f \circ T$ is continuous, as the composition of two continuous mappings. Therefore, ${T}'(X_2^*) \subset X_1^*$.   $\square$

Theorem 2.

Let $(X_1 , Y_1)$, $(X_2 , Y_2)$ be dual pairs, and let $T:X_1 \rightarrow X_2$ be a linear operator, and $T^* :Y_2 \rightarrow Y_1$ its adjoint. Then for any set $A \subset Y_2$,

$$\begin{aligned} T^{-1}(A^\circ ) = (T^* A)^\circ . \end{aligned}$$

(2)

Proof.

We have

$$ x \in T^{-1}(A^\circ )\,\Longleftrightarrow \, Tx \in A^\circ \Longleftrightarrow \forall y \in A, \, | \langle {Tx, y} \rangle | \leqslant 1 $$

$$ \Longleftrightarrow \, \forall y \in A, \, | \langle {x, T^* y} \rangle | \leqslant 1 \, \Longleftrightarrow \forall z \in T^* A, \, | \langle {x, z} \rangle | \leqslant 1\, \Longleftrightarrow \, x \in (T^* A)^\circ . \qquad {\square } $$

Theorem 3.

Let $(X_1,Y_1)$, $(X_2 , Y_2)$ be dual pairs, and let $T:X_1 \rightarrow X_2$ be a linear operator. Then the following conditions are equivalent:

• (a) the operator T has an adjoint;

• (b) T is weakly continuous, i.e., continuous as an operator acting

  from $({X_1 ,\sigma (X_1 , Y_1)})$ to $({X_2 ,\sigma (X_2 , Y_2)})$.

Proof.

(a) $\Longrightarrow$ (b). In view of the linearity, it suffices to verify the continuity of the operator at zero. Recall (item (vii) Theorem 2 in Subsection 17.1.1) that a neighborhood basis of zero in the topology $\sigma (X_2 , Y_2)$ is provided by the polars of the finite subsets $A \subset Y_2$. By relation (2), the preimage $T^{-1}(A^\circ )$ of the neighborhood $A^\circ$ is itself the polar $(T^* A)^\circ$ of the finite set $T^* A \subset Y_1$. That is, $T^{-1}(A^\circ )$ is a neighborhood of zero in $\sigma (X_1 , Y_1)$.

By Theorem 1 of Subsection 16.3.2, which describes the dual space of a space equipped with the weak topology, the implication (b) $\Longrightarrow$ (a) is a particular case of Theorem 1.   $\square$

Theorems 1 and 3 admit the following consequence.

Corollary 1.

Let $X_1$, $X_2$ be locally convex spaces, and let $T:X_1 \rightarrow X_2$ be a continuous linear operator. Then T is weakly continuous, i.e., it is continuous in the topologies $\sigma (X_1, X_1^*)$ and $\sigma (X_2 , X_2^*)$.

Switching the roles of spaces in dual pairs, we obtain the following statement.

Theorem 4.

(I) Let $(X_1 , Y_1)$, $(X_2 , Y_2)$ be dual pairs, and let $T:X_1 \rightarrow ~X_2$ be a weakly continuous linear operator. Then the adjoint operator $T^* :Y_2 \rightarrow Y_1$ is continuous in the weak topologies $\sigma (Y_2 , X_2)$, $\sigma (Y_1, X_1)$.

(II) For the operator $R:Y_2 \rightarrow Y_1$ to be the adjoint of some weakly continuous operator acting from $X_1$ to $X_2$ it is necessary and sufficient that R is continuous in the topologies $\sigma (Y_2, X_2)$, $\sigma (Y_1, X_1)$.

Proof.

(I) The formula $\langle {Tx,y} \rangle = \langle {x, T^* y} \rangle$ says that the operator $T^*$ has an adjoint: $({T^*})^* = T$. By Theorem 3, applied to $T^*$ instead of T, the operator $T^*$ is weakly continuous.

(II) The necessity of the condition follows from assertion (I). Now suppose R is continuous in the topologies $\sigma (Y_2 , X_2)$ and $\sigma (Y_1, X_1)$. Then R has an adjoint $R^* :X_1 \rightarrow X_2$. By (I), the operator $R^*$ is weakly continuous. Hence, there exists the adjoint $({R^*})^* :Y_2 \rightarrow Y_1$. Since $\langle {x, ({R^*})^* y} \rangle = \langle {R^*x,y} \rangle = \langle {x, Ry} \rangle$, we see that $R = ({R^*})^*$, and so we have proved that R is an adjoint operator, more precisely, the adjoint of $R^*$.   $\square$

Recall that in Subsection 9.4.1 we established the following result. Let $X_1, X_2$ be Banach spaces, and $T:X_1 \rightarrow X_2$ a continuous operator. Then $T^* (X_2^*) \subset (\mathrm{Ker}\, T)^\bot$. In particular, if the operator $T^*$ is surjective, then T is injective. We are now ready to make this result more precise.

Theorem 5.

Let $X_1$, $X_2$ be locally convex spaces, and let $T:X_1 \rightarrow X_2$ be a continuous linear operator. Then:

• (a) $\left( {T^* (X_2^*)} \right) ^\bot = \mathrm{Ker}\, T$.

• (b) $(\mathrm{Ker}\, T)^\bot$ coincides with the $\sigma (X_1^* , X_1)$-closure of the subspace $T^* (X_2^*)$.

• (c) The operator T is injective if and only if the image of the operator $T^*$ is $\sigma (X_1^* , X_1)$-dense in $X_1^*$.

Proof.

(a) Using Theorem 2 and the equality of the polars and annihilators of subspaces, we have

$$ ({T^* (X_2^*)})^\bot = ({T^* (X_2^*)})^{\circ } = T^{-1} ({({X_2^*})^{\circ }}) = T^{-1}(0) = \mathrm{Ker}\, T. $$

(b) follows from (a) via a direct application of Corollary 2 of the bipolar theorem.

Finally, to prove (c), we remark that injectivity means that $\mathrm{Ker}\, T = \{0\}$. Both sides of this equality are closed (and hence, by Theorem 1 of Subsection 17.1.1, also weakly closed) subspaces of $X_1$. By Corollary 3, $\mathrm{Ker}\, T = \{0\}$ if and only if $(\mathrm{Ker}\, T)^\bot = X_1^*$, which in view of (b) is equivalent to $T^* (X_2^*)$ being $\sigma (X_1^* , X_1)$-dense in $X_1^*$.   $\square$

Exercises

1.

Apply Theorem 2 of Subsection 17.1.2 to prove Theorem 5.

2.

Under the assumptions of Theorem 5, the operator T is injective if and only if the image of the operator $T^*$ separates the points of $X_1$.

3.

Consider the pair of spaces (X, Y), where X is the space of infinitely differentiable functions f on the interval [0, 1] that obey the conditions $f(0) = f(1) = 0$, and $Y = C^\infty [0,1]$. Equip this pair with the duality pairing $\langle {f, g}\rangle = \int _0^1 {f(t)g(t)dt}$. Let $T:X \rightarrow X$ be the differentiation operator: $Tf = {df}/{dt}$. Does T admit an adjoint operator $T^* :Y \rightarrow Y$? Is T a weakly continuous operator?

4.

In the preceding exercise, replace Y by C[0, 1]. Do the conclusions change?

5.

Endow the pair (X, Y) considered in Exercise 3 above with an another duality pairing, given by $\langle {f, g} \rangle = \int _0^1 f(t)g(t^2)dt$. What is $T^*$ equal to in this case?

17.1.4 Alaoglu’s Theorem

Let X be a linear space. The Cartesian power $\mathbb {C}^X$ is the linear space of all complex-valued functions on X, and $X'$ is the space of linear functionals on X. Every functional is a complex-valued function, so ${X}'$ can be regarded as a subspace of $\mathbb {C}^X$. Equip $\mathbb {C}^X$ with the Tikhonov product topology. Then $\mathbb {C}^X$ becomes a locally convex space with the topology given by the neighborhood basis of zero consisting of the sets $U_{G,\varepsilon } = \{f \in \mathbb {C}^X: \max _{x \in G} |f(x)| < \varepsilon \}$, where G runs over all finite subsets of the space X and $\varepsilon > 0$.

Theorem 1.

$\left( {{X}',\sigma ({X}', X)} \right)$ is a closed subspace of the topological vector space $\mathbb {C}^X$. In other words,

• (i) ${X}'$ is closed in $\mathbb {C}^X$, and

• (ii) the topology induced by $\mathbb {C}^X$ on ${X}'$ coincides with the weak topology $\sigma ({X}', X)$.

Proof.

(i) For any $x_1, x_2 \in X$ and $a_1, a_2 \in \mathbb {C}$, define the function $F_{x_1, x_2, a_1, a_2} :\mathbb {C}^X \rightarrow \mathbb {C}$ by the formula

$$ F_{x_1, x_2, a_1, a_2} (f) = f(a_1 x_1 + a_2 x_2) - a_1 f(x_1) - a_2 f(x_2). $$

An element $f \in \mathbb {C}^X$ is a linear functional if and only if $F_{x_1, x_2, a_1, a_2} (f) = 0$ for all $x_1, x_2 \in X$ and all $a_1, a_2 \in \mathbb {C}$. In other words,

$$ {X}' = \bigcap _{x_1, x_2, a_1, a_2}{\mathrm{Ker}\, F_{x_1, x_2, a_1, a_2}}. $$

All the functionals $F_{x_1, x_2, a_1, a_2}$ are continuous, being linear combinations of coordinate projectors. Hence, their kernels are closed, and so their intersection ${X}'$ is also closed.

(ii) It suffices to recall that the neighborhoods of zero in the topology $\sigma ({X}', X)$ are defined by means of the duality pairing $\langle {f, x} \rangle = f(x)$ on the dual pair $({X}', X)$. In other words, a neighborhood basis of zero in the topology $\sigma ({X}', X)$ is given by the sets $\{{f \in {X}':\, \max _{x \in G} | {\, \langle {f, x} \rangle \,} | < \varepsilon }\} = U_{G,\varepsilon } \cap {X}'$.    $\square$

Theorem 2.

Let U be an absorbing subset of the linear space X. Then the polar $U^\circ \subset {X}'$ is $\sigma ({X}', X)$-compact.

Proof.

By Theorem 1, it suffices to establish the compactness of $U^\circ$ as a subset of the Tikhonov power $\mathbb {C}^X$. First, note that the polar $U^\circ$ is $\sigma ({X}', X)$-closed in ${X}'$, and ${X}'$, in turn, is closed in $\mathbb {C}^X$ by Theorem 1. Hence, $U^\circ$ is closed as a subset of the Tikhonov product $\mathbb {C}^X$. Next, for any $x \in X$ denote by n(x) the smallest number $n \in \mathbb N$ for which $x \in nU$. Then for any $x \in X$ and any $f \in U^\circ$ one has the estimate $|f(x)| \leqslant n(x)$. This means that $U^\circ \subset \prod _{x \in X}{\mathbb {C}_{n(x)}}$, where $\mathbb {C}_{n(x)}$ denotes the closed disc in $\mathbb {C}$ centered at zero and of radius n(x). By Tikhonov’s theorem on products of compact sets, $\prod _{x \in X}{\mathbb {C}_{n(x)}}$ is compact $\mathbb {C}^X$. Therefore, $U^\circ$ is a closed subset of a compact set, and as such is compact.   $\square$

Corollary 1.

Let U be a neighborhood of zero in the locally convex space X. Then $U^\circ \subset X^*$ is $\sigma (X^*, X)$-compact.

Proof.

Consider first the dual pair $(X,{X}')$. The polar of the set U in ${X}'$ consists of functionals that are bounded on the neighborhood U, i.e., of continuous functionals. Therefore, one can equally well consider the polar $U^\circ$ in the dual pair $(X,{X}')$ or in the dual pair $(X, X^*)$ — we obtain the same set $U^\circ \subset X^*$. By Theorem 2, $U^\circ$ is $\sigma ({X}', X)$-compact. It remains to remark that on $X^*$ the topologies $\sigma ({X}', X)$ and $\sigma (X^*, X)$ coincide.   $\square$

Corollary 2

(L. Alaoglu, 1940).¹ Let X be a Banach space. Then the closed unit ball of the dual Banach space $X^*$ is $\sigma (X^*, X)$-compact.

Proof.

Indeed, $\overline{B}_{X^*} = ({B_X})^\circ$.   $\square$

Recall that in an infinite-dimensional Banach space with the norm topology balls cannot be compact (Riesz’s theorem, Subsection 11.2.1). This restricts considerably the applicability of geometric intuition in the infinite-dimensional case: as it turns out, all arguments relying on the extraction of a convergent subsequence from a bounded sequence are prohibited. However, Alaoglu’s theorem gives rise to the hope that this prohibition can be partially lifted, at least in dual spaces, and not for the convergence in norm, but in the weaker, $\sigma (X^* , X)$-convergence. A difficulty we still need to face is that Alaoglu’s theorem deals with compactness, not sequential compactness; that is, the possibility of extracting convergent subsequences remains under question at this point. This issue will be studied in detail in Sect. 17.2. For the moment, before taking leave from the general theory of duality and passing to the Banach spaces so dear to the author, we will formulate in a series of exercises several additional results, in particular, an important theorem of Mackey and Arens which describes, for a dual pair (X, Y), those topologies on X for which $X^* = Y$. For a detailed exposition the reader is referred to the textbook [37].

Exercises: Topology of Uniform Convergence

Let (X, Y) be a dual pair.

1.

For a set $A \subset Y$ to be $\sigma (Y, X)$-bounded it is necessary and sufficient that its polar $A^\circ \subset X$ is absorbing.

2.

Let $\tau$ be a locally convex topology on X. Then for a functional $y \in Y$ to be $\tau$-continuous it is necessary and sufficient that the set $\{y\}^\circ$ is a neighborhood of zero in the topology $\tau$.

A family $\mathfrak {C}$ of subsets of the space Y is said to be admissible if it obeys the following conditions:

• — $\{y\} \in \mathfrak {C}$ for any $y \in Y$;

• — for any $A \in \mathfrak {C}$ and any scalar $\lambda$, the set $\lambda A$ also belongs to $\mathfrak {C}$;

• — if $A, B \in \mathfrak {C}$, then there exists a $C \in \mathfrak {C}$ such that $A \cup B \subset C$;

• — any element $A \in \mathfrak {C}$ is bounded in the topology $\sigma (Y, X)$.

3.

Show that the following families of subsets of a space Y are admissible:

• — the family $\mathfrak {Fin}(Y)$ of all finite subsets;

• — the family $\mathfrak {Comp}(Y)$ of all absolutely convex $\sigma (Y, X)$-compact subsets;

• — the family $\mathfrak {Bound}(Y)$ of all $\sigma (Y, X)$-bounded subsets.

4.

Any admissible family $\mathfrak {C}$ satisfies the condition $\mathfrak {Fin}(Y)\subset \mathfrak {C}\subset \mathfrak {Bound}(Y)$.

5.

Let $\tau$ be a locally convex topology on X with the property that $X^* \supset Y$. Then the family $\tau ^\circ$ of sets of the form $A^\circ$, where A runs over all neighborhoods of zero in the topology $\tau$, is admissible.

6.

If $\mathfrak {C}$ is an admissible family of subsets of the space Y, then the family of sets $A^\circ$ with $A \in \mathfrak {C}$ constitutes a neighborhood basis of zero in a separated locally convex topology on X.

7.

A sequence $x_n \in X$ converges in the topology of the preceding exercise to a point $x \in X$ if and only if ${\sup }_{y \in A} |\langle {x_n - x, y} \rangle | \rightarrow 0$ as $n \rightarrow \infty$ for all $A \in \mathfrak {C}$.

Let $\mathfrak {C}$ be an admissible family of subsets of the space Y. The topology in which a neighborhood basis of zero is formed by all sets $A^\circ$ with $A \in \mathfrak {C}$ is called the topology of uniform convergence on the sets of the family $\mathfrak {C}$.

A locally convex topology $\tau$ on the space X is said to be compatible with duality, if $(X, \tau )^* = Y$.

8.

Suppose the topology $\tau$ on X is compatible with duality. Then the topology of uniform convergence on the sets of the family $\tau ^\circ$ from Exercise 5 above coincides with $\tau$.

9.

The topology $\sigma (X, Y)$ coincides with the topology of uniform convergence on the sets of the family $\mathfrak {Fin}(Y)$.

The topology of uniform convergence on the sets of family $\mathfrak {Comp}(Y)$ is called the Mackey topology, and is denoted by $\tau (X, Y)$.

10.

Mackey–Arens theorem. The topology $\tau$ on X is compatible with duality if and only if $\sigma (X,Y)\prec \tau \prec \tau (X, Y)$.

11.

Every $\sigma (X, Y)$-bounded set is also $\tau (X, Y)$-bounded, i.e., all topologies compatible with duality have the same supply of bounded sets.

12.

For the dual pair $(c_0 ,\ell _1)$ with the duality pairing $\langle {x, y} \rangle = \sum _{k = 1}^\infty {x_k y_k}$, the Mackey topology $\tau (c_0 ,\ell _1)$ coincides with the norm topology on the space $c_0$.

13.

The identity operator I, regarded as an operator acting from the space $c_0$ endowed with the topology $\sigma (c_0 ,\ell _1)$ into $c_0$ with the norm topology, is an example of a bounded discontinuous linear operator.

The topology of uniform convergence on the sets of the family $\mathfrak {Bound}(Y)$ is called the strong topology, and is denoted by $\beta (X, Y)$ .

14.

$\beta (\ell _1,c_0)$ coincides with the topology of the normed space $\ell _1$. On this example convince yourself that the strong topology is not always compatible with duality.

15.

Describe the topology $\beta (c_0 ,\ell _1)$. Prove that $\beta (c_0 ,\ell _1)$ is compatible with duality.

17.2 Duality in Banach Spaces

In Banach spaces the convergence in norm is called strong convergence. In this section we dwell in detail on two weaker forms of convergence: weak and weak$^*$ (weak star) convergence.

17.2.1 $w^{*}$-Convergence

Throughout this subsection we will consider a dual pair $(X, X^*)$, where X is a Banach space.

Definition 1.

The topology $\sigma (X^*, X)$ is called the $w^*$ -topology or, in words, weak-star topology on the Banach space $X^*$. A sequence of functionals $x_n^* \in X^*$ is said to $w^*$-converge to a functional $x^* \in X^*$ (and one writes $x_n^* \buildrel {w^*}\over \longrightarrow x^*$; the notation $x_n^*{\mathop {\rightharpoonup }\limits ^{*}}x^*$ is also frequently used) if it converges in the $w^*$-topology. In detail: $x_n^* \buildrel {w^*} \over \longrightarrow x^*$ if $x_n^* (x) \rightarrow x^* (x)$ as $n \rightarrow \infty$ for all $x \in X$.

As the notation indicates, $w^*$-convergence is a particular case of the pointwise convergence we are familiar with even for operators, not only for functionals. In particular, the following assertions (Theorems 1 and 2 of Subsection 10.4.2) hold true for $w^*$-convergence.

Banach–Steinhaus theorem. If $x_n^* \buildrel {w^*}\over \longrightarrow x^*$, then $\sup _{n \in \mathbb {N}}\Vert x_n^* \Vert < \infty$.   $\square$

$w^*$-convergence criterion. Let $A \subset X$ be a dense subset, and $x_n^*,\, x^* \in X^*$. Then the following conditions are equivalent:

• — $x_n^* \buildrel {w^*} \over \longrightarrow x^*$;

• — $\sup _{n \in \mathbb {N}}\Vert x_n^* \Vert < \infty$ and $x_n^* (x) \rightarrow x^*(x)$ as $n \rightarrow \infty$, for all $x \in A$.    $\square$

Since pointwise convergence on A implies pointwise convergence on $\mathrm{Lin}\, A$, in the last criterion it suffices to require that, instead of the set A itself, its linear span is a dense set.

Based on this general criterion, we next derive $w^*$-convergence criteria in the sequence spaces we are familiar with.

Theorem 1.

Let X be the sequence space $c_0$ or $\ell _p$ with $1 \leqslant p < \infty$, and $X^*$ be the space $\ell _1$ or $\ell _{p'}$ with $1 < p' \leqslant \infty$, respectively. Then for a sequence of elements $x_n = ({x_{n, j}})_{j \in \mathbb {N}} \in X^*$, $n = 1,2,\ldots$, to $w^*$-converge to an element $y = ({y_j})_{j \in \mathbb {N}} \in X^*$, it is necessary and sufficient that this sequence is bounded in norm and converges to y componentwise (coordinatewise): $x_{n, j} \rightarrow y_j$ as $n\rightarrow \infty$, for all $j \in \mathbb {N}$.

Proof.

Consider the standard basis of the sequence space X: $e_1 = (1,0,0,\ldots )$, $e_2 = (0,1,0,0,\ldots ),\ldots \,$. Since $\langle {f, e_j}\rangle = f_j$ for any $f = \left( {f_j} \right) _{j \in \mathbb {N}} \in X^*$, componentwise (coordinatewise) convergence is the convergence on the elements $e_j$ of the standard basis. It remains to use the fact that the set $\mathrm{Lin}\{e_j\}_{j \in \mathbb {N}}$ is dense in X and apply the criterion obtained above.   $\square$

Since componentwise convergence in $\ell _p$ does not imply convergence in norm (the sequence of $e_j$ from the proof of Theorem 1 is a typical example), the $w^*$-convergence in these spaces does not coincide with the norm convergence. This is the case not only in $\ell _p$, but also in all infinite-dimensional normed spaces (Josefson [55], Nissenzweig [72]) .

In Subsection 6.4.3 (Theorem 2 and Exercise 4) it was shown that the norm on the space $X^*$ is lower semi-continuous with respect to weak$^*$ convergence: if $x_n^* \buildrel {w^*} \over \longrightarrow x^*$, then $\Vert x^* \Vert \leqslant {\underline{\lim }}_{n \rightarrow \infty }\Vert x_n^* \Vert$. In fact, the following stronger property holds.

Theorem 2.

The norm on the dual space $X^*$ is lower semi-continuous with respect to the $w^*$-topology.

Proof.

Recall (Subsection 1.2.4) that a function $f:E \rightarrow \mathbb {R}$, defined on a topological space E, is said to be lower semi-continuous if for any $a \in \mathbb {R}$ the set $f^{-1} ((a, + \infty ))$ is open. In our case E is the dual space $X^*$, equipped with the topology $\sigma (X^* , X)$, and the function of interest is $f(x^*) = \Vert x^* \Vert$. Hence, $f^{-1}((a, + \infty ))$ is either the entire space $X^*$ (when $a < 0)$, which is indeed open, or the set $X^* \setminus a\overline{B}_{X^*}$. Since the ball $\overline{B}_{X^*}$ of the space $X^*$ is $\sigma (X^* , X)$-closed (and even $\sigma (X^*, X)$-compact, thanks to Alaoglu’s theorem), $X^* \setminus a\overline{B}_{X^*}$ is a $w^*$-open set.   $\square$

Exercises

1.

Let X be a Banach space with a basis $\{{e_n}\}_1^\infty$. Then, as observed in Subsection 10.5.3, every functional $f \in X^*$ can be identified with the numerical sequence $({f(e_1), f(e_2),\ldots , f(e_n),\ldots })$, and accordingly the space $X^*$ can be identified with the set of all such sequences. Extend Theorem 1 to this case.

2.

Let X be a Banach space with a basis $\{{e_n}\}_1^\infty$. Then the linear span of the set $\{{e_n^*}\}_1^\infty$ of coordinate functionals is $w^*$-dense in $X^*$.

3.

On the example of the space $X = \ell _1$ with the standard basis $e_1 = (1,0,0,\ldots )$, $e_2 = (0,1,0,0,\ldots )$, ..., show that the linear span of the set of coordinate functionals is not necessarily norm-dense in $X^*$.

4.

Using the $w^*$-convergence criterion and the diagonal method, show that from any bounded sequence of functionals $x_n^* \in X^*$ on a separable Banach space X one can extract a $w^*$-convergent subsequence. Below we shall deduce this result from Alaoglu’s theorem and metrizability considerations.

5.

Suppose the Banach space E is the dual of some Banach space F, i.e., $E = F^*$. Suppose any finite linked system of closed balls in E has non-empty intersection. Then E has the linked balls property (see the exercises in Subsection 9.3.3 for the definition).

6.

The real spaces $L_\infty (\Omega , \Sigma , \mu )$ have the linked balls property, and consequently are 1-injective. This extends the result of Exercise 4 in Subsection 9.3.3.

Recall (see Exercises 2–4 in Subsection 8.4.6 and the comments to them) that any function of bounded variation $F:[0,1] \rightarrow \mathbb {R}$ can be regarded as a continuous linear functional on C[0, 1] if one defines its action on elements $f \in C[0,1]$ by the formula $\langle F, f \rangle = \int _0^1 {f\, dF}$. Prove the following theorem of Helly.

7.

Helly’s theorem. Suppose $F_n :[0,1] \rightarrow \mathbb {R}$ is a sequence of functions which converges pointwise to a function F, and the variations of the functions $F_n$ are jointly bounded. Then F also has finite variation, and the functionals on C[0, 1] generated by the functions $F_n$ $w^*$-converge to the functional generated by the function F.

Helly’s theorem admits the following partial converse.

8.

Suppose the functions $F_n , F :[0,1] \rightarrow \mathbb {R}$ of bounded variation are right-continuous everywhere, except possibly at zero, and vanish at zero. Further, suppose that the sequence of functionals on C[0, 1] generated by the functions $F_n$ $w^*$-converge to the functional generated by the function F. Then the variations of the functions $F_n$ are jointly bounded, and $F_n (t) \rightarrow F(t)$ as $n \rightarrow \infty$ in each point of continuity t of the function F.

9.

Helly’s second theorem. From any uniformly bounded sequence of functions $F_n :[0,1] \rightarrow \mathbb {R}$ with jointly bounded variation one can extract a subsequence that converges pointwise on [0, 1].

17.2.2 The Second Dual

Let X be a Banach space. Then $X^*$ is also a Banach space, and so one can in turn consider the dual of $X^*$. This dual space, $( X^*)^*$, is called the second dual of X, and is denoted by $X^{**}$. The elements of the space $X^{**}$ are, by definition, continuous linear functionals on $X^*$. We are already familiar with a wide class of such functionals from the general duality theory: namely, they are the elements of the initial space X, regarded as functionals on $X^*$.

Recall that an element $x \in X$ acts on an element $x^* \in X^*$ by the rule $x(x^*) = x^* (x)$. Thus, x can be regarded as a linear functional on $X^*$, and in this way X becomes a linear subspace of the second dual $X^{**}$. Furthermore, the formula $\Vert x\Vert _X = \sup _{f \in S_{X^*}} |f(x)|$ (Lemma in Subsection 9.4.1), recast as $\Vert x\Vert _X = \sup _{f \in S_{X^*}} | {x(f)} |$, acquires a new meaning: the norm of the element x in the space X is equal to the norm of x as a linear functional on $X^*$. Accordingly, X can be treated as subspace of the Banach space $X^{**}$. Of individual interest is the question of when this inclusion becomes an equality. Spaces with this property are called reflexive and will be considered in Subsection 17.2.6.

The next theorem, due to Goldstine, is an easy consequence of the bipolar theorem. However, one should note that Goldstine’s theorem appeared earlier and was the original result from which, properly speaking, the bipolar theorem was molded.

Theorem 1

(Goldstine). The closed unit ball of a Banach space X is $w^*$-dense (i.e., dense in the topology $\sigma ({X^{**}, X^*})\mathrm{)}$ in the closed unit ball of the space $X^{**}$. The space X is $w^*$-dense in its second dual $X^{**}$.

Proof.

Consider the dual pair $({X^{**}, X^*})$. The ball ${\overline{B}_X}$ is a convex balanced subset of the space $X^{**}$. By the bipolar theorem, $\overline{B}_X$ is $\sigma ({X^{**}, X^*})$-dense in $({\overline{B}_X })^{\circ \circ }$. But $({\overline{B}_X})^\circ = \overline{B}_{X^*}$ and $({\overline{B}_X})^{\circ \circ } = \overline{B}_{X^{**}}$. Therefore, $\overline{B}_X$ is $w^*$-dense in $\overline{B}_{X^{**}}$.

The fact that X is $w^*$-dense in $X^{**}$ can be alternatively deduced again from the bipolar theorem in the dual pair $({X^{**}, X^*})$: $X^\bot = \{0\}$, $X^{\bot \bot } = X^{**}$, or by using the fact that the space $X^{**}$ is the union of the balls $n\overline{B}_{X^{**}}$, and each ball $n\overline{B}_{X^{**}}$ contains as a $w^*$-dense subset the corresponding ball of the space X.   $\square$

Exercises

1.

A model example of a triple $X, X^*, X^{**}$ is provided by the spaces $c_0$, $\ell _1$, $\ell _\infty$. Based on the $w^*$-convergence criterion in $\ell _\infty$, prove that the unit ball of $c_0$ is dense in the $\sigma (\ell _\infty ,\ell _1)$-topology in the unit ball of $\ell _\infty$. That is, prove Goldstine’s theorem in this particular case.

2.

$c_0$ is a Banach space, and so, being complete, it must be closed in any ambient space. In particular, $c_0$ is closed in $\ell _\infty$. On the other hand, $c_0$ does not coincide with $\ell _\infty$, hence it cannot be dense in $\ell _\infty$. Doesn’t this contradict Goldstine’s theorem?

3.

$X^*$ is also a Banach space, and so one can consider the canonical embedding $X^* \subset X^{***}$. For every element $x^{***} \in X^{***}$, define the element $P(x^{***}) \in X^*$ as the restriction of the functional $x^{** *}$ to the subspace $X\subset X^{**}$. Prove that P is a projector and $\Vert P\Vert = 1$. That is, for any Banach space the first dual is complemented in the third dual.

4.

If $X = X^{**}$, the projector P introduced in the preceding exercise is bijective, i.e., in the present case $X^*=X^{***}$.

5.

The ball $\overline{B}_X$ is a complete set in the topology $\sigma ({X^{**}, X^*})$ if and only if $\overline{B}_X = \overline{B}_{X^{**}}$, i.e., $X = X^{**}$.

6.

The space X is a complete set in the topology $\sigma \left( {X^{**}, X^*} \right)$ if and only if X is finite-dimensional (this fact is connected with the existence of discontinuous linear functionals on $X^*$ in the infinite-dimensional case).

7.

Let $A \in L(X, Y)$. Then one can define the second adjoint operator: $A^{**} = (A^*)^*$, $A^{**} \in L(X^{**}, Y^{**})$. Prove that the restriction of the operator $A^{**}$ to X coincides with the original operator A.

8.

Using the preceding exercise, prove the converse to Schauder’s theorem on the compactness of the adjoint operator (Theorem 3 of Subsection 11.3.2). Namely, show that if the adjoint operator is compact, then so is the original operator.

17.2.3 Weak Convergence in Banach Spaces

Definition 1.

The topology $\sigma (X, X^*)$ is called the weak topology of the Banach space X. A sequence of elements $x_n \in X$ is said to converge weakly to the element $x \in X$ (notation: $x_n \buildrel w \over \longrightarrow x$, also frequently $x_n^*{\mathop {\rightharpoonup }\limits ^{w}}x^*$ or simply $x_n^*\, {\rightharpoonup }\, x^*$) if $f(x_n) \rightarrow f(x)$ as $n\rightarrow \infty$ for any $f \in X^*$ (one says that $(x_n)$ is a weakly convergent sequence) .

Note that $\sigma (X, X^*)$ is the restriction to X of the topology $\sigma \left( {X^{**}, X^*} \right)$, and the weak convergence of a sequence $x_n \in X$ to the elements $x \in X$ is simultaneously the $w^*$-convergence of the same sequence in the space $X^{**}$. Accordingly, the simplest properties of the $w^*$-convergence mentioned in Subsection 17.2.1 carry over to the weak convergence:

• — if $x_n \buildrel w \over \longrightarrow x$, then $\sup _{n \in \mathbb {N}}\Vert {x_n}\Vert < \infty$;

• — if $x_n \buildrel w \over \longrightarrow x$, then $\Vert x{}\Vert \leqslant \underline{\lim }_{n \rightarrow \infty }\Vert x_n\Vert$.

The convergence criterion is also preserved, with the roles of the spaces X and $X^*$ switched. More precisely, let $A \subset X^*$ be a subset whose linear span is dense in $X^*$ in the strong topology; let $x_n , x \in X$. Then the following conditions are equivalent:

• A.   $x_n \buildrel w \over \longrightarrow x$;

• B.   $\mathop {\sup }\limits _{n \in \mathbb {N}} \left\| {x_n} \right\| < \infty$ and $x^* (x_n) \rightarrow x^*$ for all $x^* \in A$.

From this fact one can readily deduce a weak convergence criterion analogous to Theorem 1 of Subsection 17.2.1 for the class of spaces defined below.

Definition 2.

Let X be a Banach space with a basis $\{{e_n}\}_1^\infty$, and $\{{e_n^*}\}_1^\infty$ be the corresponding coordinate (evaluation) functionals. The basis $\{{e_n}\}_1^\infty$ is called a shrinking basis if the linear span of the set $\{{e_n^*}\}_1^\infty$ is dense (in norm) in $X^*$.

Theorem 1.

Let X be a Banach space with a shrinking basis $\left\{ {e_n} \right\} _1^\infty$, and let $x_n , x \in X$, $x_n =\sum _{j = 1}^\infty {a_{n, j} e_j}$, $x = \sum _{j = 1}^\infty {a_j e_j}$. Then for the sequence $(x_n)$ to converge weakly to an element x, it is necessary and sufficient that this sequence is bounded in norm and converges componentwise to x: $a_{n, j} \rightarrow a_j$ as $n\rightarrow \infty$, for all $j \in \mathbb {N}$.

In particular, “boundedness plus coordinatewise convergence” is a weak convergence criterion in such sequence spaces as $c_0$ or $\ell _p$ with $1< p < \infty$, in which the standard basis is shrinking.

Caution! The standard basis is not a basis in the space $\ell _\infty$, while in the space $\ell _1$ it is a basis, but it is not shrinking (Exercise 3 of Subsection 17.2.1). Consequently, in $\ell _1$ and $\ell _\infty$ boundedness plus componentwise convergence are not sufficient for weak convergence. In $\ell _1$ the weak convergence criterion is rather unusual: in this space, according to a theorem of Schur,² the weak and strong convergence coincide. In $\ell _\infty$ there is no conveniently verifiable criterion for weak convergence.

Theorem 2

(weak convergence criterion in C(K)). Let K be a compact topological space. For functions $x_n , x \in C(K)$ the following conditions are equivalent:

• (i) $x_n \buildrel w \over \longrightarrow x$;

• (ii) $\sup _{n \in \mathbb {N}}\Vert {x_n}\Vert < \infty$ and $x_n (t)\rightarrow x(t)$ as $n\rightarrow \infty$ in all points $t \in K$.

Proof.

(i) $\Longrightarrow$ (ii). The boundendness of a weakly convergent sequence is a general result. The condition $x_n(t) \rightarrow x(t)$ as $n \rightarrow \infty$ is simply convergence on the evaluation functional $\delta _t \in C(K)^*$, acting by the rule $\delta _t (f) = f(t)$.

(ii) $\Longrightarrow$ (i). We have to show that $F(x_n) \rightarrow F(x)$ as $n\rightarrow \infty$ for any $F \in C(K)^*$. Taking into account the general form of a linear functional on C(K), we need to show that $\int _K {x_n d\nu } \rightarrow \int _K {x\, d\nu }$ for any regular Borel charge $\nu$ on K. By hypothesis, we are given a sequence of functions $x_n$ that are uniformly bounded and converge pointwise to the function x. It remains to apply the Lebesgue dominated convergence theorem.    $\square$

The results stated below point to a closer connection between weak and strong convergence than the one observed between $w^*$-convergence and strong convergence.

Theorem 3.

Let A be a convex subset of the Banach space X. Then the following conditions are equivalent:

• (a) A is weakly closed;

• (b) A is weakly sequentially closed, i.e., if $x_n \buildrel w \over \longrightarrow x$ and $x_n \in A$, then $x \in A$;

• (c) A is closed in the strong topology.

Proof.

The implication (a) $\Longrightarrow$ (b) is obvious (closedness implies sequential closedness in any topology, not only in the weak one).

(b) $\Longrightarrow$ (c). Let $x_n \in A$ and $\Vert {x_n - x}\Vert \rightarrow 0$. Then $x_n \buildrel w \over \longrightarrow x$ and, by condition (b), $x \in A$.

(c) $\Longrightarrow$ (a). This implication was proved in Theorem 1 of Subsection 17.1.1 not only for Banach spaces, but also for arbitrary locally convex ones.   $\square$

Theorem 4

(Mazur). Suppose the sequence $(x_n)$ of elements of the Banach space X converges weakly to an element $x \in X$. Then x lies in the strong closure of the convex hull of the sequence $(x_n)$. Moreover, there exists a sequence $(y_n)$ of convex combinations of the elements $x_n$ that converges strongly to x, such that $y_n \in \mathrm{conv}\{x_k\}_{k = n}^\infty$, $n = 1,\, 2,\ldots$.

Proof.

Let us show that for each $n \in \mathbb {N}$ there exists an element $y_n \in \mathrm{conv}\{x_k\}_{k = n}^\infty$ such that $\Vert {y_n - x}\Vert < 1/n$. Then $(y_n)$ will be the sought-for sequence. Indeed, let $A_n$ denote the strong closure of the set $\mathrm{conv}\{x_k\}_{k = n}^\infty$. By the preceding theorem, $A_n$ is also a weakly closed set. Since $A_n$ contains all $x_k$ with $k \geqslant n$, $A_n$ also contains the weak limit x. Therefore, x lies in the strong closure of the set $\mathrm{conv}\{x_k\}_{k = n}^\infty$, and x can be arbitrarily well approximated by elements of this convex hull.   $\square$

Theorem 5.

Let X, Y be Banach spaces. Then for a linear operator $T:X \rightarrow Y$ the following conditions are equivalent:

• (A) T is continuous in the weak topologies of the spaces X and Y;

• (B) T maps any sequence that converges weakly to zero again into one that converges weakly to zero;

• (C) T is continuous in the strong topologies of the spaces X and Y.

Proof.

The implication (A) $\Longrightarrow$ (B) is obvious.

(B) $\Longrightarrow$ (C). We apply assertion (3) of Theorem 1 of Subsection 6.4.1: T is continuous if and only if it maps sequences that converge to zero into bounded sequences. Suppose $\Vert {x_n}\Vert \rightarrow 0$ as $n \rightarrow \infty$. Then $x_n \buildrel w \over \longrightarrow 0$. By condition (B), this means that $Tx_n \buildrel w \over \longrightarrow 0$. But then $\sup _{n \in \mathbb { N}} \Vert {Tx_n}\Vert < \infty$.

(C) $\Longrightarrow$ (A). Here it suffices to use Corollary 1 in Subsection 17.1.3.

Exercises

1.

Let Y be a subspace of the Banach space X. Then the restriction of the topology $\sigma (X, X^*)$ to Y coincides with the topology $\sigma (Y, Y^*)$. In particular, any sequence that converges weakly in Y also converges weakly in X, and any weakly compact subset of Y is also weakly compact in X. Henceforth these facts will be used without further clarifications.

2.

On the example of the standard basis verify that the weak convergence criterion established in $\ell _p$, $1< p < \infty$, (boundedness plus componentwise convergence) fails in $\ell _1$.

3.

Consider the sequence $x_1 = (1,0,0,\ldots )$, $x_2 = (1,1,0,0,\ldots )$, $x_3 = (1,1,1, 0,\ldots )$, $\ldots$ in $\ell _\infty$. Verify that this sequence is bounded and converges componentwise to $x = (1,\, 1,\, 1,\ldots , 1,\ldots )$. However, Mazur’s theorem does not hold for the sequence $(x_n)$ , i.e., $(x_n)$ does not converge weakly to x. Thus, the weak convergence criterion “bounded plus componentwise convergence” is not valid in $\ell _\infty$.

4.

Prove the following weak convergence criterion in the space $L_p [0,1]$, $1< p < \infty$: $f_n \buildrel w \over \longrightarrow f$ if and only if $\sup _{n \in \mathbb {N}}\Vert {f_n}\Vert < \infty$ and $\int _a^b {f_n (t)\,dt} \rightarrow \int _a^b {f(t)\, dt}$ for any subinterval $[a, b] \subset [0,1]$.

5.

Using the example of the sequence $f_n = 2^n\big ({\mathbbm {1}_{[0,2^{ - (n + 1)}]} - \mathbbm {1}_{[2^{ - (n + 1)}, 2^{ - n}]}} \big )$ convince yourself that in $L_1 [0,1]$ the criterion established in Exercise 4 fails.

6.

Prove the following weak convergence criterion in the space $L_1 [0,1]$: $f_n \buildrel w \over \longrightarrow f$ if and only if $\sup _{n \in \mathbb {N}}\Vert {f_n}\Vert < \infty$ and $\int _A {f_n(t)\,dt} \rightarrow \int _A {f(t)\, dt}$ for any measurable subset $A\subset [0,1]$.

7.

Let K be a convex weakly compact set in a Banach space that possesses a normal structure (see the exercises in Subsection 15.3.1 for the definition). Show that the Kakutani fixed point theorem remains valid for mappings of the set K.

Let D be a metric space. A mapping $f:D \rightarrow D$ is said to be non-expansive if for any $x_1, x_2 \in D$ one has $\rho (f(x_1), f(x_2))\leqslant \rho (x_1, x_2)$.

8.

In the setting of Exercise 7, show that every non-expansive mapping $f:K \rightarrow K$ has a fixed point.

9.

Does there exists a common fixed point for all non-expansive mappings $f:[0,1] \rightarrow [0,1]$?

17.2.4 Total and Norming Sets. Metrizability Conditions

Definition 1.

Let X be a linear space and $Y \subset X$ be a subspace. A set $F \subset {X}'$ is called a total set over Y, if it separates the points of the subspace Y. In detail, F is total over Y if for any $y\in Y \setminus \{0\}$ there exists an $f \in F$ such that $f(y) \ne 0$.

Definition 2.

Let X be a Banach space, $Y \subset X$ a linear subspace, and $\theta \in (0,1]$. A set $F \subset X^*$ is said to be $\theta$-norming over Y if

$$\begin{aligned} \sup _{f \in F\backslash \{0\}} \dfrac{|f(y)|}{\Vert f\Vert } \geqslant \theta \Vert y\Vert \end{aligned}$$

(1)

for all $y \in Y$. The set $F \subset X^*$ is said to be norming over Y if there exists a $\theta \in (0,1)$ such that F is $\theta$-norming over Y.

Obviously, every norming set is total: if $y \ne 0$, then condition (1) means that, in any case, there exists an $f \in F$ such that $f(y) \ne 0$.

In the case when F lies on the unit sphere of the space $X^*$, condition (1) reads $\sup _{f \in F} |f(y)| \geqslant \theta \Vert y\Vert$. In view of the homogeneity in y of this inequality, it suffices to verify it for $y \in S_Y$. Therefore, for $F \subset S_{X^*}$ Definition 2 can be restated as follows: the set F is $\theta$-norming over Y if the inequality

$$\begin{aligned} \sup _{f \in F} |f(y)| \geqslant \theta \end{aligned}$$

(2)

holds for all $y \in S_Y$.

Theorem 1.

Let X be a Banach space, $Y \subset X$ a linear subspace, and $G \subset S_{X^*}$ a set that is 1-norming over Y. Further, let $\varepsilon \in (0,1)$ and suppose that on the unit sphere of the subspace Y there is given an $\varepsilon$-net D. For each element $x\in D$ fix a functional $f_x \in G$ such that $f_x (x)> 1 - \varepsilon$. Then the set $F = \{{f_x}\}_{x \in D}$ is $\theta$-norming over Y, with $\theta = 1 - 2 \varepsilon$.

Proof.

Fix a $y \in S_Y$ and let us show that estimate (2) holds for y. By the definition of an $\varepsilon$-net, there exists an element $x_0 \in D$ such that $\Vert {y - x_0}\Vert < \varepsilon$. We have:

$$ \sup _{f \in F} |f(y)| = \sup _{x \in D} | {f_x (y)} | \geqslant | {f_{x_{_0}} (y)}| = |{f_{x_{_0}} (x_0) - f_{x_{_0}} (y - x_0)}| $$

$$ \geqslant 1 - \varepsilon - \Vert {y - x_0}\Vert > 1 - 2\varepsilon = \theta . \qquad {\square } $$

From this theorem, recalling that in a finite-dimensional space the unit sphere contains a finite $\varepsilon$-net, one derives the following

Corollary 1.

Let X be a Banach space, and let $Y \subset X$ be a finite-dimensional subspace. Then for any $\theta \in (0,1)$ there exists a finite $\theta$-norming set over Y.

Here is a slightly more complex variant that will prove suitable in the next subsection.

Corollary 1$'$. Let E be a Banach space, and let $Y\subset E^{**}$ be a finite-dimensional subspace. Then for any $\theta \in (0,1)$ there exists a finite $\theta$-norming set over Y consisting of elements of the unit sphere of the space $E^*$.

Proof.

Apply Theorem 1 with $X = E^{**}$ and $G = S_{E^*}$.    $\square$

For infinite-dimensional separable spaces we obtain the following result.

Corollary 2.

Let X be a Banach space, and $Y\subset X$ a separable subspace. Then there exists a countable 1-norming set over Y.

Proof.

Take as the set D figuring in Theorem 1 a countable dense subset of the unit sphere in the subspace Y and put $G = S_{X^*}$. The set D forms an $\varepsilon$-net in $S_Y$ for all $\varepsilon >0$. Therefore, the set $F = \{{f_x}\}_{x \in D}$ consisting of the support functionals ($\Vert {f_x}\Vert = 1 = f_x(X)$) is $\theta$-norming over Y for all $\theta < 1$. But if condition (2) is satisfied for all $\theta < 1$, then it is also satisfied for $\theta = 1$.   $\square$

Recall (Exercises 15–18 in Subsection 16.2.1) that a Hausdorff topological vector space is metrizable if and only if it has a countable neighborhood basis of zero. Further, according to Exercise 3 in Subsection 16.3.2, the topology $\sigma (X, X^*)$ has a countable neighborhood basis of zero if and only if the space $X^*$ has an at most countable Hamel basis. Since a Hamel basis of an infinite-dimensional Banach space is necessarily uncountable (Exercise 4 of Subsection 6.3.2), the weak topology of an infinite-dimensional space, if regarded on the whole space, is not metrizable. The situation changes drastically when instead we consider the weak topology on subsets.

Theorem 2.

Let (X, E) a dual pair, let $B \subset X$ be compact in the topology $\sigma (X, E)$, and let $Y = \mathrm{Lin}\, B$. Denote by $\sigma _B(X, E)$ the topology induced on B by the weak topology $\sigma (X, E)$. Further, suppose there exists a countable total set over Y, $F = \{{f_1, f_2, \ldots }\} \subset E$. Then there exists a norm p on Y such that the norm topology generated on B by p coincides with $\sigma _B (X, E)$. In particular, on B the weak topology is metrizable.

Proof.

Since B is a weakly compact set and the functionals $f_n$ are continuous in the weak topology $\sigma (X, E)$, each of the $f_n$’s is bounded on B in modulus by some number $a_n$. With no loss of generality, we may assume that $a_n \leqslant 1$ (otherwise, we multiply $f_n$ by the factor $1/{a_n}$, without affecting the totalness of the sequence of functionals on Y). For each $y \in Y$ we put $p(y) = \sum _{n = 1}^\infty 2^{-n}|f_n(y)|$. Let us show that p is the requisite norm. Each of the terms $2^{-n}|f_n(y)|$ is non-negative and satisfies the triangle inequality and the positive homogeneity condition. Therefore, p inherits the same properties. The non-degeneracy ($p(y) = 0 \Longrightarrow y = 0)$ follows from the fact that the set F is total.

Now let us compare the topologies involved. Let $x \in B$, $r> 0$. Consider the set $U({x, r}) = \{{y \in B: p (x - y) <r}\}$, i.e., the ball in B of radius r and center x generated by the norm p. Pick an $N\in \mathbb {N}$ such that ${2^{-N}} < r/2$, and consider the following weak neighborhood of the point x in the set B:

$$ V = \left\{ y \in B: \max _{1 \leqslant k \leqslant N} | {f_k (x - y)} | <{r}/{4}\right\} . $$

If $y \in V$, then

$$\begin{aligned} p (x - y) = \sum \limits _{n = 1}^\infty {\dfrac{\left| {f_n (x - y)}\right| }{2^n}}&\leqslant \sum \limits _{n = 1}^N{\dfrac{\left| {f_n (x - y)}\right| }{2^n}}+\dfrac{1}{2^N}\\&<2\mathop {\max }\limits _{1\leqslant k\leqslant N}\left| {f_k (x - y)} \right| + \dfrac{r}{2} <r \end{aligned}$$

and $y \in U ({x, r})$. That is, $V \subset U ({x, r})$. This establishes that on B the topology generated by the norm p is weaker than the topology $\sigma _B (X, E)$. Since it is not possible to strictly weaken the topology of a compact set so that it remains separated (see Subsection 1.2.3, second paragraph), we conclude that $\sigma (X, E)$ and p induce on B one and the same topology.   $\square$

Corollary 3.

Let X be a Banach space. Then on any separable weakly compact subset B of X the weak topology $\sigma (X, X^*)$ is metrizable.

Proof.

It suffices to apply Theorem 2 to the dual pair $(X, X^*)$. Here the existence of a countable total set follows from Corollary 2.   $\square$

Corollary 4.

Let X be a separable Banach space. Then the $w^*$-topology $\sigma (X^*, X)$ is metrizable on the bounded subsets of the space $X^*$.

Proof.

Consider the dual pair $(X^*, X)$ and take $B=\overline{B}_{X^*}$. By Alaoglu’s theorem, B is compact in the $\sigma (X^*, X)$-topology. By assumption, X contains a countable dense set F. This set F is total over $X^*$. Theorem 2 shows that the $w^*$-topology on the unit ball of the space $X^*$ is metrizable. To complete the proof, it remains to observe that any bounded subset of $X^*$ lies in some ball of the form $r\overline{B}_{X^*}$.   $\square$

Corollary 5.

Let X be a separable Banach space. Then from any bounded sequence of functionals $x_n^* \in X^*$ one can extract a convergent subsequence.

Proof.

With no loss of generality we can assume that $x_n^* \in \overline{B}_{X^*}$ (otherwise we multiply all $x_n^*$ by the factor $(\sup _n \Vert x_n^*\Vert )^{-1})$. By Alaoglu’s theorem, $\overline{B}_{X^*}$ is $w^*$-compact, and by Corollary 4, this compact set is metrizable. It remains to recall that from any sequence of elements of a compact metric space one can extract a convergent subsequence.   $\square$

Exercises

1.

Let X be a linear space, and $Y \subset X$ a subspace. Then there exists a finite total set over Y if and only Y is finite-dimensional.

2.

Let X be a Banach space, and $\theta \in (0,1)$. A set $F \subset X^*$ is $\theta$-norming for X if and only if the $w^*$-closure of the absolute convex hull of F contains $\theta \overline{B}_{X^*}$.

3.

For a Banach space X the following conditions are equivalent:

• — there exists a countable total set over X;

• — there exists an injective continuous linear operator mapping X into $\ell _2$.

4.

For a Banach space X the following conditions are equivalent:

• — there exists a countable norming set over X;

• — there exists a bounded below continuous linear operator mapping X into $\ell _\infty$.

5.

For a Banach space X the following conditions are equivalent:

• — there exists a countable 1-norming set over X;

• — there exists a linear isometric embedding of the space X into $\ell _\infty$.

6.

In particular, every separable Banach space admits an isometric embedding into the space $\ell _\infty$.

A separable Banach space E is called universal if among its subspaces are isometric copies of all separable Banach spaces. The following two exercises give a sketch of the proof of the famous Banach–Mazur theorem on the universality of the space C[0, 1] .

7.

Let X be a separable Banach space and $\mathcal {K}$ be the Cantor set. Equip the closed ball $\overline{B}_{X^*}$ with the $w^*$-topology. By Exercise 6 of Subsection 1.4.4 and Alaoglu’s theorem, there exists a surjective continuous mapping $F:\mathcal {K} \rightarrow \overline{B}_{X^*}$. Define the operator $T:X \rightarrow C(\mathcal {K})$ by the formula $(Tx)(t) = \langle {F(t), x} \rangle$. Prove that the operator T effects a linear isometric embedding of the space X into $C(\mathcal {K})$. This will establish that the space $C(\mathcal {K})$ of continuous functions on the Cantor set is universal.

8.

Prove that $C(\mathcal {K})$, where $\mathcal {K}$ is the Cantor set, admits an isometric embedding into C[0, 1]. Deduce from this the universality of the space C[0, 1] .

Since $X \subset X^{**}$, one can talk about subsets of the Banach space X that are total or norming over $X^*$.

9.

For a linear subspace Y of a Banach space X the following conditions are equivalent:

• — Y is total over $X^*$;

• — Y is a norming set over $X^*$;

• — Y is dense in X.

10.

Denote by $e_n^*$ the functional on $\ell _\infty$ which assigns to each element $x = (x_1,x_2 ,\ldots )$ of the space $\ell _\infty$ its n-th coordinate: $e_n^* (x) = x_n$. On the example of the sequence $(e_n^*)_1^\infty$ convince yourself that on non-separable spaces there exists bounded sequences of functionals which contain no $w^*$-convergent subsequences. In particular, this example shows that, despite Alaoglu’s theorem, the unit ball of the dual space is not necessarily $w^*$-sequentially compact.

17.2.5 The Eberlein–Smulian Theorem

The last exercise in the preceding subsection reminds us that in non-metrizable topological spaces compactness and sequential compactness are, generally speaking, distinct properties. The weak topology of an infinite-dimensional Banach space is not metrizable. It is therefore even more surprising that the weak compactness of a set in a Banach space is equivalent to its weak sequential compactness. This theorem of Eberlein and Smulian consists of two parts, the first of which was proved by V.L. Smulian (in the literature one encounters also the spelling Šmulian or Shmulian) in 1940, and the second by W.F. Eberlein in 1947 .

Theorem 1

(Smulian). Let K be a weakly compact set in a Banach space X. Then from every sequence $x_n \in K$, $n=1,2,\ldots$, one can extract a weakly convergent subsequence.

Proof.

Consider the closed linear span Y of the sequence $(x_n)_{n=1}^\infty$ and let $\widetilde{K}=K\cap Y$. Since Y is separable, $\widetilde{K}$ is a separable set. Further, any closed linear subspace is a weakly closed set (Theorem 3 of Subsection 17.2.3), hence $\widetilde{K}$, being the intersection of a weakly compact set with a weakly closed set, is a weakly compact set. By Corollary 3 of Subsection 17.2.4, the weak topology $\sigma (X, X^*)$ is metrizable on $\widetilde{K}$. Finally, $x_n \in \widetilde{K}$ by construction, and we know that from any sequence of elements of a metrizable compact set one can extract a convergent (in the present case, weakly convergent, since we are dealing with the weak topology) subsequence.    $\square$

Theorem 2

(Eberlein). Let K be a weakly sequentially compact subset of a Banach space, i.e., such that from every sequence $x_n \in K$, $n = 1,2, \ldots$, one can extract a weakly convergent subsequence, and the limit of the subsequence again lies in K. Then K is weakly compact.

Proof.

To begin with, note that K is a bounded set. Indeed, assuming that K is not bounded, there would exist a sequence $x_n \in K$ such that $\Vert {x_n}\Vert \rightarrow \infty$. But such a sequence cannot contain bounded subsequences, and hence it cannot contain weakly convergent subsequences.

Since any bounded set can be transformed into a subset of the unit ball by multiplying it by a small positive number, we can assume for simplicity that $K \subset \overline{B}_X$. Further, using the embedding $\overline{B}_X \subset \overline{B}_{X^{**}}$, K can be regarded as a subset of the ball $\overline{B}_{X^{**}}$. Since on the space X, and hence also on the set K, the topologies $\sigma (X, X^*)$ and $\sigma (X^{**}, X^*)$ coincide, it suffices to show that K is $\sigma (X^{**}, X^*)$-compact. By Alaoglu’s theorem, $\overline{B}_{X^{**}}$ is $\sigma (X^{**}, X^*)$-compact. Hence, to prove that K is $\sigma (X^{**}, X^*)$-compact, it in turn suffices to show that K is $\sigma (X^{**}, X^*)$-closed in $\overline{B}_{X^{**}}$.

So, let $x^{**} \in \overline{B}_{X^{**}}$ be an arbitrary $\sigma (X^{**}, X^*)$-limit point of the set K. We need to verify that $x^{**} \in K$. Recalling the form of the neighborhoods in the topology $\sigma (X^{**}, X^*)$, we see that the condition that $x^{**}$ is a $\sigma (X^{**}, X^*)$-limit point can be expressed as follows:

(A) for any finite set of functionals $D \subset X^*$ and any $\varepsilon > 0$, there exists an element ${(A) }$ $x \in K$ such that $\mathop {\max }_{y^* \in D} | {y^* (x^{**} - x)} | < \varepsilon$.

The main idea of the proof is to construct a sequence $x_n \in K$ such that none of its subsequences can converge weakly to a point different from $x^{**}$. Since by assumption any sequence $x_n \in K$ contains a subsequence that converges to some point of K, this will establish that $x^{**} \in K$.

The requisite sequence $x_n \in K$ will be constructed recursively, using at each step property (A) and Corollary 1$'$ of Subsection 17.2.4. Fix some $\theta \in (0,1)$ and a sequence $\varepsilon _n \rightarrow 0$. Consider $Y_0 = \mathrm{Lin}\,\{x^{**}\}$. By Corollary 1$'$ of Subsection 17.2.4, there exists a finite $\theta$-norming set $D_0 \subset S_{X^*}$ over $Y_0$. Using property (A), choose a point $x_1 \in K$ such that $\mathop {\max }_{y^* \in D_0}|y^* (x^{**} - x_1)| < \varepsilon _1$. Next, consider $Y_1 =\mathrm{Lin}\,\{x^{**}, x_1 \}$. Again, by the same Corollary 1$'$ of Subsection 17.2.4, there exists a $\theta$-norming set $D_1 \subset S_{X^*}$ over $Y _1$. With no loss of generality we may assume that $D_0 \subset D_1$: otherwise, we can replace $D_1$ by the union $D_0 \cup D_1$. Using again property (A), we choose $x_2 \in K$ such that $\mathop {\max }_{y^* \in D_1} | {y^* (x^{**} - x_2)} | < \varepsilon _2$. Continuing this construction, we obtain a sequence of elements $x_n \in K$, a sequence of subspaces $Y_n = \mathrm{Lin}\,\{x^{**}, x_1,x_2 ,\ldots , x_n\}$, and a sequence of finite subsets $D_0 \subset D_1 \subset D_2 \subset \cdots$ of the unit sphere of the space $X^*$, such that $D_n$ is a $\theta$-norming set over $Y_n$ and

$$\begin{aligned} \mathop {\max }\limits _{y^* \in D_{n - 1}} | {y^* (x^{** } - x_n)} | < \varepsilon _n . \end{aligned}$$

(3)

Denote $\bigcup _{n = 0}^\infty {D_n}$ by D, and the norm-closure of the subspace $\mathrm{Lin} \{x^{**}, x_1, x_2 , \ldots \}$ by Y. Suppose that some subsequence $(x_{n_j})$ of the sequence $(x_n)$ converges weakly to a point $x \in K$. Let us prove that $x = x^{**}$: as we explained above, this will complete the proof of the entire theorem. First we note that, by Mazur’s theorem (Theorem 4 in Subsection 17.2.3), $x \in Y$. Hence, $x - x^{**} \in Y$, too. By construction, the set D is $\theta$-norming over all subspaces $Y_n$. Consequently, D is $\theta$-norming over $\bigcup _{n = 1}^\infty {Y_n}$, and hence also over Y, the strong closure of this union. It follows that

$$ \Vert {x - x^{**}}\Vert \leqslant \dfrac{1}{\theta }\mathop {\sup }\limits _{y^* \in D} | {y^* (x^{**} - x)} |. $$

We claim that the right-hand side of the last inequality is equal to zero. Indeed, for every $y^* \in D$ there exists a number $N \in \mathbb {N}$ such that $y^* \in D_m$ for all $m \geqslant N$. By condition (3), this means that $| {y^* (x^{**} - x_m )} | < \varepsilon _m$ for all $m > N$. Since x is a weak limit point of the sequence $(x_n)$, this allows us to conclude that $| {y^* (x^{**} - x)} | =0$.

   $\square$

Exercises

Prove the following two assertions, treated as obvious above:

1.

Let $Y_0 \subset Y_1 \subset Y_2 \subset \cdots$ be an increasing chain of subspaces, and D be a $\theta$-norming set over all $Y_n$’s. Then D is also $\theta$-norming over $\bigcup _{n = 1}^\infty {Y_n}$.

2.

Let D be a $\theta$-norming set over the linear subspace E. Then D is also $\theta$-norming over the strong closure of the subspace E.

The Eberlein–Smulian theorem may create the illusion that in the weak topology of a Banach space all topological properties may be adequately expressed in the language of sequences. The exercises below will help to dispel this illusion.

3.

In the space $\ell _2$ consider the standard basis $\{ e_n\}$. Set $x_n = n^{1/4}e_n$. Prove that:

• — 0 is a limit point of the sequence $(x_n)$ in the weak topology;

• — the sequence $(x_n)$ contains no bounded subsequence, and hence no weakly convergent subsequence.

The weak sequential closure of the set A in the Banach space X is defined to be the set of weak limits of all weakly convergent sequences of elements of A.

4.

von Neumann’s example. In the space $\ell _2$ consider the set A of the vectors $x_{n, m} = e_n + ne_m$, $n,\, m \in \mathbb {N}$, $m > n$ (where $e_n$ is, as above, the standard basis). Verify that

• — for $m \rightarrow \infty$ and fixed n, the sequence $(x_{n, m})$ converges weakly to $e_n$. That is to say, all the vectors $e_n$ lie in the weak sequential closure of the set A.

• — 0 does not belong to the weak sequential closure of A.

• — The weak sequential closure of the set A is not a weakly sequentially closed set.

17.2.6 Reflexive Spaces

A Banach space X is said to be reflexive if $X = X^{**}$. In a reflexive space, thanks to the equality $X = X^{**}$, the weak topology $\sigma (X, X^*)$ coincides with $\sigma (X^{**}, X^*)$, and together with the usual properties of a weak topology (equivalence of closedness and weak closedness of convex sets, equivalence of continuity and weak continuity of linear operators), it also enjoys the main nice feature of the weak$^*$ topology, namely, the compactness of the unit ball. This combination of properties makes reflexive spaces far more convenient in applications.

Theorem 1.

For a Banach space X the following conditions are equivalent:

• (i) X is reflexive;

• (ii) the closed ball $\overline{B}_X$ is weakly compact;

• (iii) from any bounded sequence $x_n \in X$ one can extract a weakly convergent subsequence.

Proof.

(i) $\Longrightarrow$ (ii). If $X = X^{**}$, then $\overline{B}_X = \overline{B}_{X^{**}}$; and, by Alaoglu’s theorem, $\overline{B}_{X^{**}}$ is $\sigma (X^{**}, X^*)$-compact.

(ii) $\Longrightarrow$ (i). Since the restriction of the topology $\sigma (X^{**}, X^*)$ to X coincides with the weak topology $\sigma (X, X^*)$, (ii) implies that $\overline{B}_X$ is a $\sigma (X^{**}, X^*)$-compact subset of the space $X^{**}$. In particular, $\overline{B}_X \subset \overline{B}_{X^{**}}$ is a $w^*$-closed subset. By Goldstine’s theorem (Theorem 1 of Subsection 17.2.2), $\overline{B}_X$ is $w^*$-dense in $\overline{B}_{X^{**}}$, and hence $\overline{B}_X = \overline{B}_{X^{**}}$. Passing to linear spans, we obtain the needed equality $X = X^{**}$.

Finally, the equivalence (ii) $\Longleftrightarrow$ (iii) follows from the Eberlein–Smulian theorem.

   $\square$

Theorem 2.

If the Banach space X is reflexive, then its subspaces and quotient spaces are reflexive.

Proof.

Let Y be a subspace of the Banach space X. By our convention, Y is a closed linear subspace. Therefore, Y is weakly closed in X. Since $\overline{B}_X$ is weakly compact, the set $\overline{B}_Y = Y \cap \overline{B}_X$ will also be weakly compact. This establishes the reflexivity of the space Y.

Now let us consider the quotient space X / Y. Recall that the quotient mapping $q:X \rightarrow X / Y$, which sends each element of the space X into its equivalence class, is a continuous linear operator. Hence, q is a weakly continuous operator. In particular, $q({\overline{B}_X})$, being the image of a weakly compact set under a weakly continuous map, is a weakly compact subset of the space X / Y. Further, for any $[x] \in \overline{B}_{X/Y}$ there exists a representative $\widetilde{x} \in [x]$ such that $\Vert \widetilde{x}\Vert \leqslant \Vert {[x]}\Vert + 1 \leqslant 2$. That is, $\overline{B}_{X / Y} \subset 2q({\overline{B}_X})$. Therefore, the set $\overline{B}_{X / Y}$ is weakly compact, as a weakly closed subset of a weakly compact set.    $\square$

Theorem 3.

The Banach space X is reflexive if and only if its dual $X^*$ is reflexive.

Proof.

Suppose $X = X^{**}$. Passing to dual spaces, we have that $X^* = X^{***}$ (see also Exercises 3 and 4 of Subsection 17.2.2). Thus, the reflexivity of the original space implies the reflexivity of its dual. Conversely, suppose that the space $X^*$ is reflexive. Then, as we have just seen, its dual $X^{**}$ is also reflexive. But X is a subspace of $X^{**}$, and as such X must be reflexive.   $\square$

Let us list several properties of reflexive spaces which find application in problems of approximation theory and variational calculus.

Theorem 4.

Let $A \ne \emptyset$ be a convex closed subset of a reflexive Banach space X. Then for any $x \in X$, the set A contains a closest point to x.

Proof.

Denote $\rho (x, A)$ by r and consider the sets

$$ A_n = \{a \in A: \Vert a - x \Vert \leqslant r + (1/n) \} = A \cap \left( x + (r + ( 1 / n ))\overline{B}_X \right) . $$

Since A is a weakly closed and $\overline{B}_X$ is weakly compact, each of the sets $A_n$ is weakly compact. Any decreasing chain of compact sets (in fact, even any centered family of compact sets) has a non-empty intersection. Every element $y \in \bigcap _n {A_n}$ lies in A and lies at distance r from x. Thus, y is the sought-for point of A that is closest to x.    $\square$

We suggest to the reader to compare Theorem 4 with the theorem on best approximation in a Hilbert space and Exercises 4–6 in Subsection 12.2.1. In particular, under the assumptions of Theorem 4, the uniqueness of the closest point is not guaranteed even in the finite-dimensional case (see Exercise 6 below).

Theorem 5.

For any continuous linear functional f on a reflexive Banach space X there exists an element $x \in S_X$ such that $f(x) = \Vert f\Vert$. Therefore, in a reflexive Banach space every linear functional attains the maximum of its modulus on the unit sphere.

Proof.

By Theorem 1 of Subsection 9.2.1, for any point $f \in X^*$ there exists a supporting functional $x \in S_{X^{**}}$ at the point f. For this element one has $f(x) = \Vert f\Vert$. It remains to recall that $X = X^{**}$, i.e., x does not just lie somewhere in the second dual space, but, as required, it lies on the unit sphere of the original space X.    $\square$

Let us note that the converse to Theorem 5 also holds: if the Banach space X is not reflexive, then there exists a functional $f \in X^*$ which does not attain on $S_X$ its supremum. The proof of this by far not simple theorem of R.C. James can be found in the first chapter of J. Diestel’s monograph [12] .

To conclude, let us record which of the spaces known to us are reflexive, and which are not.

• — All finite-dimensional spaces are reflexive.

• — All spaces $L_p$ and $\ell _p$ with $1< p < \infty$ are reflexive (this follows from the theorem on the general form of linear functionals on $L_p$).

• — The space $c_0$ is not reflexive, since $({c_0})^{**} = \ell _\infty \ne c_0$.

• — The space $\ell _1$ is not reflexive, since it is the dual of the non-reflexive space $c_0$.

• — The space $\ell _\infty$ is not reflexive, since it is the dual of the non-reflexive space $\ell _1$.

• — The space C(K), where K is an infinite compact space, is not reflexive, since it contains a non-reflexive subspace isomorphic to $c_0$. The reader is invited to verify that for any sequence of functions $f_n \in S_{C(K)}$ with disjoint supports, $\overline{\mathrm{Lin}}\{f_n\}$ is a subspace of C(K) isometric to $c_0$. In particular, the space C[0, 1] is not reflexive.

• — The space $L_1({\Omega , \Sigma , \mu } )$, where $\Omega$ cannot be decomposed into a finite union of atoms of the measure $\mu$, is not reflexive; indeed, it contains a non-reflexive subspace isomorphic to $\ell _1$. (For any sequence of functions $f_n \in S_{L_1(\Omega , \Sigma , \mu )}$ with disjoint supports, $\overline{\mathrm{Lin}}\{f_n\}$ is a subspace of $L_1(\Omega , \Sigma , \mu )$ isometric to $\ell _1$). In particular, the space $L_1 [0,1]$ is not reflexive.

Remark 1.

The definitions of the spaces $L_p$ and $\ell _p$ look more complicated and less natural than those of the spaces C[0, 1], $L_1 [0,1]$, or $c_0$. The list given above sheds light on the reason for the wide utilization of the spaces $L_p$: their relatively complicated definition is more than compensated by their pleasant features, first and foremost, by their reflexivity.

Exercises

1.

In each of the examples of non-reflexive spaces listed above construct explicitly a bounded sequence that contains no weakly convergent subsequence. This will lead to another way of establishing that the spaces in question are not reflexive.

Using the weak compactness of the unit ball, prove the following generalization of Theorem 5.

2.

Let $T\in L (X, Y)$, with X reflexive and Y finite-dimensional. Then there exists an element $x\in S_X$ such that $\Vert {T(x)}\Vert = \Vert T\Vert$.

3.

Provide an example of a continuous linear functional on $c_0$ which does not attain its supremum on the unit sphere. Give a complete description of such functionals.

4.

Solve the analogues of the preceding exercise for the spaces $\ell _1$, $L_1 [0,1]$, and C[0, 1].

5.

Let A be a weakly compact subset of a Banach space X. Then for any $x \in X$ in A there exists a point closest to x.

6.

In the space $\mathbb {R}^2$ equipped with the norm $\Vert (x_1, x_2) \Vert = \max \{|x_1|, |x_2| \}$, consider the set $A = \{(1, a): a \in [- 1,1] \}$. Verify that for $x = 0$ the point in A closest to x is not unique.

7.

An operator $A\in L (X, Y)$ is called a Dunford–Pettis operator if it maps weakly convergent sequences into norm-convergent sequences. Show that any compact operator is a Dunford–Pettis operator. If the space X is reflexive, then every Dunford–Pettis operator is compact. In non-reflexive spaces (say, in C[0, 1]) there exist non-compact Dunford–Pettis operators.

Comments on the Exercises

Subsection 17.2.1

Exercise 7. Denote by A the set of all points where at least one of the functions $F_n$, F is discontinuous. Consider the space X of all bounded functions on [0, 1] that have a left and a right limit at every point, are continuous at all the points of the set A, and have at most finitely many discontinuity points. Equip X with the norm $\Vert f\Vert = \mathop {\sup } _{[0,1]}|f (t)|$. The required relation $\int _0^1 {f\, dF_n} \rightarrow \int _0^1 {f\, dF}$ is simpler to prove not for $f \in C [0,1]$, but for the wider class of $f \in X$. To this end we need to show the relation $\int _0^1 {f\, dF_n} \rightarrow \int _0^1{f\, dF}$ for $f = \mathbbm {1}_{[a, b]}$, then extend it by linearity to the set of all piecewise-constant functions, and use the pointwise convergence criterion for operators (convergence on a dense subset + boundedness in norm).

Exercise 8. Pass to the Borel charges $\nu _n$, for which $F_n (t)= \nu _n ( [0,t] )$ for $t \in (0,1]$. First deal with the case where $\nu _n$ are measures.

Exercise 9. Use the fact that the set of discontinuity points of a function of bounded variation is at most countable, and Exercises 4 and 8. For another argument, see the textbook by A. Kolomogorov and S. Fomin, Chap. VI, § 6.

Subsection 17.2.5

Exercise 3. The example is taken from the paper [51]. The indicated sequence $(x_n)$ will converge to zero with respect to the statistical filter $\mathfrak {F}_\mathrm{s}$ (Exercise 7 in Subsection 16.1.1). The following general result holds true [59]: for a sequence of numbers $a_n > 0$ the following conditions are equivalent: (1) there exists a sequence $x_n \in \ell _2$ with $\Vert {x_n}\Vert =a_n$, for which 0 is a weak limit point, and (2) $\sum {a_n^{-2}} = \infty$.