### Advanced Analysis, Notes 11: Banach spaces (weak topologies, Alaoglu’s theorem)

Let $X$ be the Banach space $C([0,1])$ of continuous functions on the interval $[0,1]$ with the sup norm. Consider the following sequence of functions $\{f_n\}$ defind as follows. $f(0) = 0$ and $f_n(1/(n+1)) = 1$ for all $n = 1, 2, \ldots$,  $f_n$ is equal to zero on the interval between $2/(n+1)$ and $1$, and $f_n$ is linear in the intervals where we haven’t defined it yet (visualize!). The sequence is tending to zero pointwise, but the norm of $X$ does not detect this. The sequence tends to $0$ in the $L^1$ norm, but the $L^1$ norm is not in the game. Can the Banach space structure of $X$ detect the convergence of $f_n$ to $0$?

The answer is yes. Recall the Riesz–Markov representation theorem from a course in measure theory:

Theorem 1: Let $X$ be a locally compact Hausdorff space, and let $C_0(X)$ denote the space of continuous functions that vanish at infinity. Then $C_0(X)^* = M(X)$, where $M(X)$ is the space  of regular complex Borel measures on $X$

Let $\mu \in M([0,1])$. Since $f_n \rightarrow 0$ everywhere it converges almost everywhere with respect to any measure. Since $|f_n|\leq 1$ for all $n$, the dominated convergence theorem tells us that

$\langle f_n, \mu \rangle = \int f_n d\mu \rightarrow 0 .$

Our little discussion motivates the following definition.

Definition 2: Let $X$ be a normed space. A sequence $\{x_n\}$ is said to converge weakly to $x \in X$ if for all $f \in X^*$, $f(x_n) \rightarrow f(x)$

We see that the above sequence converges weakly to $0$. It is clear that if $x_n \rightarrow x$ in the norm then $x_n \rightarrow x$ weakly (sometimes convergence in norm is called strong convergence). The converse is false.

Example: Consider an orthonormal sequence $\{e_n\}$ in a Hilbert space $H$. Then $\|e_n - e_m \|^2 = 2$ for all $m,n$, so the sequence does not converge in the norm. However, for any $x \in H$, Bessel’s inquality implies that $(x,e_n) \rightarrow 0$, so $e_n \rightarrow 0$ weakly.

Weak convergence turns out to be the notion of convergent sequence with respect to a certain topology, the weak topology. As you know from a course in topology, the convergent sequences do not, in general, give us all the information on a topological space.

#### 1. The weak topology

For the rest of this post, let $X$ be a Banach space.

Definition 1: The weak topology on $X$ is the weakest topology in which all the functionals in $X^*$ are continuous. In other words, it is the topology generated by the sets

$V(x; f; \epsilon) = \{y\in X : |f(x-y)|< \epsilon \}$

where $x \in X$ and $f \in X^*$

Recall that given a family of subsets $\mathcal{F}$ of a set $A$, one can define the topology generated by $\mathcal{F}$ simply as the smallest topology that contains $\mathcal{F}$, and that this topology turns out to be the topology with the base $\cap_{i=1}^n F_i$ where $n$ is any integer and $F_i \in \mathcal{F}$. We see that the weak topology has a base

$V(x_1, \ldots, x_k,; f_1, \ldots, f_k; \epsilon_1, \ldots, e_k) = \{y \in X : \forall i . |f_i(x_i-y)|<\epsilon_i\} .$

Exercise A: Prove that the weak topology has a base given by the sets

$V(x; f_1, \ldots, f_k; \epsilon) = \{y \in X : \forall i . |f_i(x-y)|<\epsilon\} .$

One may also replace $\epsilon$ be $1$.

Proposition 2: The weak topology on $X$ is a Hausdorff topology in which the vector space operations are continuous. In the weak topology every point has a local base consisting of convex sets. The weak topology is weaker than the norm topology, and if $X$ is infinite dimensional then it is strictly weaker and non-metrizable.

Proof: The Hahn-Banach theorem implies Hausdorff-ness. Continuity of the vector space operations is more instructive as an exercise, so please prove it. The basic open sets that we wrote down above are convex.

We saw above an example in a Hilbert space of a weakly convergent sequence that is not strongly convergent, but we will now show that the weak topology is always weaker when $X$ is infinite dimensional. For this it suffices to show that the weak topology is not metrizable. By the Hahn-Banach theorem, $X^*$ is an infinite dimensional space, and we already observed that it is Banach. Suppose that the weak topology of $X$ was metrizable. Then the point $0$ would have a countable local base. It is then an exercise in topology to show that we may assume that this local base consists of basic open sets of the form given in Exercise A. Let $\{f_n\}_{n=1}^\infty$ be a sequence which contains any $f \in X^*$ that arises in this countable local base at $0$. For every $g \in X^*$, the set $V(0;g;1)$ is a neighborhood of $0$, hence it must contain some set of the form $V(0;f_{n_1}, \ldots, f_{n_k}; \epsilon)$. By Lemma 4 below, $V(0;f_{n_1}, \ldots, f_{n_k}; \epsilon) \subseteq V(0;g;1)$ implies that $g$ is in the span of $f_{n_1}, \ldots, f_{n_k}$. We have just proved that every $g \in X^*$ is in the algebraic span of $\{f_n\}_{n=1}^\infty$, and therefore $X^*$ has a countable Hamel basis. This contradicts Exercise H in Notes 2 (Exercise H said that an infinite dimensional Hilbert space cannot have a countable Hamel basis. Although there is a proof that works only for Hilbert spaces (using orthonormal bases and the Riesz–Fischer theorem) the usual proof — which we discussed in class — works just as well for banach spaces). The contradiction completes the proof.

A vector space with a topology on it that satisfies the conditions in the first sentence of the proposition is called a topological vector space. A topological vector space that satisfies the second sentence in the proposition is called a locally convex (topological vector) space. There are many locally convex spaces that arise in analysis, and in operator theory in particular. Many of them (though not all) arise in the following way.

Definition 3: Let $Y$ be a subspace of $X^*$ that separates the points of $X$. The $Y$–weak topology on $X$, denoted $\sigma(X,Y)$, is the weakest topology on $X$ in which all of the functional in $Y$ are continuous. In other words, it is the topology generated by the sets

$V(x; f; \epsilon) = \{y\in X : |f(x-y)|< \epsilon \}$

where $x \in X$ and $f \in Y$. Let $\hat{X}$ be the image of $X$ in $X^{**}$. The $\sigma(X^*, \hat{X})$ is called the weak* topology on $X^*$

Lemma 4: Let $f$ be a linear functional on $X$, $f_1, \ldots, f_k \in X^*$, and suppose that

$V(0;f_1, \ldots, f_n;\epsilon) \subseteq V(0;f;1)$

Then there are scalars $a_1, \ldots, a_n$ such that $f = a_1 f_1 + \ldots + a_n f_n$

Proof: Denote $M = \cap_{i=1}^n N(f_i)$. We will first show that $M \subseteq N(f)$. Indeed, if $x \in M$, then $nx \in M \subseteq V(0;f_1, \ldots, f_n;\epsilon)$ for all $n$, hence $n|f(x)|<1$ for all $n$, whence $f(x) = 0$.

Now it is an exercise in linear algebra to show that $M \subseteq N(f)$ implies that $f = a_1 f_1 + \ldots + a_n f_n$ for some scalars $a_i$. This is proved by induction. We may assume that all functionals in the discussion are not zero. If $n=1$, let $y$ such that $f_1(y) = 1$. Then for all $x \in X$, $x - f_1(x)y \in N(f_1)$ so $X/N(f_1)$ is one dimensional and the result follows. If we proved the result for $n-1$, then consider $f\big|_{N(f_n)}, f_1 \big|_{N(f_n)}, \ldots, f_{n-1} \big|_{N(f_n)}$. Then $\cap_{i=1}^{n-1} N(f_{i}\big|_{N(f_n)}) \subseteq N(f\big|_{N(f_n)})$, so $f\big|_{N(f_n)} = \sum_{i=1}^{n-1} b_i f_{i}\big|_{N(f_n)}$. Thus $f - \sum_{i=1}^{n-1} b_i f_{i}$ vanishes on $N(f_n)$ and is therefore a multiple of $f_n$.

Proposition 5: Let $Y \subseteq X^*$ be a subspace. A linear functional $f$ on $X$ is continuous in the $\sigma(X,Y)$–topology if and only if $f \in Y$

Proof:One direction is the definition, so nothing to prove. Assume that $f$ is continuous in the $\sigma(X,Y)$ topology. Then $\{x:|f(x)|<1\} = V(o;f;1)$ is $\sigma(X,Y)$ open, so there are $f_1, \ldots, f_k \in Y$ such that

$V(0;f_1, \ldots, f_n;\epsilon) \subseteq V(0;f;1)$.

The lemma now shows that $f \in Y$.

Since the weak topologies are not metrizable, convergent sequences do not give a full description of the topologies. We will be working with nets. These completely determine the topology: a set $F$ in a topological space is closed if and only if for every net $F \ni x_\alpha \rightarrow x$ one has $x \in F$. A net $\{x_\alpha\}$ in $X$ converges to $x$ in the $\sigma(X,Y)$–topology if and only if $\langle x_\alpha, y \rangle \rightarrow \langle x, y \rangle$ for all $y \in Y$.

#### 2. Weakly convergent sequences

Theorem 6: Every weakly convergent sequence is bounded. Every weak* convergent sequence is bounded.

Proof: Suppose that $\{x_n^*\}$ is weak* convergent. Then for every $x \in X$, the sequence $\langle x^*_n, x \rangle$ is convergent, and in particular it is bounded. By the principle of uniform boundedness, $\sup_n \|x^*_n\| < \infty$. If $\{x_n\}$ is weakly convergent, then the isometrically embedded sequence in $X^{**}$, given by $\{\hat{x}_n\}$, is weak* convergent, hence bounded, by the previous argument.

Exercise B: Construct a weakly convergent net that is not bounded.

Example: Let $\{e_n\}_{n=1}^\infty$ be an orthonormal basis in a Hilbert space $H$, and let $x_n = \sqrt{n} e_n$. The sequence $\{x_n\}$ has no bounded sub-sequence, hence it has no weakly convergent sub-sequence. However, $0 \in \overline{\{x_n\}}^w$, and it particular, the sequence has a subnet that weakly converges to $0$.

#### 3. Alaoglu’s theorem

Proposition 5 shows that if $X$ is not reflexive, then the weak* topology on $X^*$ is strictly weaker then the weak topology on $X^*$ (the $\sigma(X^*,X^{**})$–topology). One reason to prefer a weaker topology is that a weaker topology has a better chance of being compact. One reason why the weak* topology is useful are the following exercise and  theorem.

Exercise C: Prove that in the closed unit ball $X_1$ of a normed space $X$ is compact if and only if $X$ is finite dimensional.

Theorem 7 (Alaoglu’s theorem): The closed unit ball $(X^*)_1$ of $X^*$ is compact in the weak* topology.

Remark: From the theorem it follows that the closed unit ball in a reflexive Banach space is weakly compact. It is true that the closed unit ball of a Banach space is weakly compact if and only if the space is reflexive.

Proof: For every $x \in X$, let $D_x = \{z \in \mathbb{C} : |z| \leq \|x\|\}$. Define $Z = \Pi_{x\in X}D_x$. By Tychonoff’s theorem, $Z$ is compact in the product topology. Now $Z$ is a space of complex valued functions on $X$, so $(X^*)_1$ can be naturally identified with a (topological) subspace of $Z$. Note that the topology on $(X^*)_1$ induced by the product topology is precisely the weak* topology on in (use nets). If we show that $(X^*)_1$ is closed, we will be done, because a closed subset of a compact Hausdorff space is compact.

Suppose that $x^*_\alpha \rightarrow z$ in the product topology. By definition of $Z$, if we show that $z$ is linear we will be done, because then $z$ will be bounded of norm $1$. But if $x, y \in X, a \in \mathbb{C}$, then

$z(ax + y) = \lim_\alpha x^*_\alpha(ax + y) = \lim_\alpha a x^*_\alpha (x) + x^*_\alpha (y) = az(x) + z(y).$

#### 4. The Hahn–Banach theorem for weak topologies

We proved many Banach space theoretic corollaries to the Hahn–Banach theorem. One may also prove Hahn–Banach type theorems in weak topologoes. For example, we have the following useful theorem.

Theorem 8: Let $Y$ be a subspace of $X^*$ that separates the points of $X$. Let $C$ be a convex subset of $X$ closed in the $\sigma(X,Y)$ topology and let $x \notin C$. Then there exists a functional $f \in Y$ which strictly separates $C$ and $x$

Corollary 9: Let $M$ be a subspace in $X$, and let $Y$ be a subspace of $X^*$ that separates the points of $X$. A point $x$ is in the $\sigma(X,Y)$ closure of $M$ if and only if $f(x) = 0$ for every $f \in Y$ which annihilates $M$

Exercise D: Prove the above theorem and corllary.

Recall (from the homework exercises) that a set $H \subset X$ is said to be a closed half space if $H = \{x\in X : Re f(x) \leq c\}$, where $f \in X^*$ and $c \in \mathbb{R}$.

Corollary 10: A set $C \subsetneq X$ is convex and closed in the weak topology if an only if $C$ is the intersection of a a family of closed half spaces.

Proof: A closed half space is closed in the weak topology and convex, thus the intersection of a family of closed half spaces is convex and closed in the weak topology. Conversely, let $C$ be convex and closed in the weak topology, and denote for each $x \notin C$ by $f_x$ a functional that separates $C$ from $x$ as guaranteed by Theorem 7. The separation provides us with a closed half space $H_x$ such that $C \subseteq H_x$ and $x \notin H_x$. Then we obtain that

$C = \cap_{x \notin C}H_x .$

That completes the proof.

Corollary 11: A convex set in $X$ is weakly closed if and only if it is strongly closed.

Proof: In the homework exercises you proved that a set $C \subsetneq X$ is convex and closed in the norm if and only if $C$ is the intersection of a family of closed half spaces. But we have just seen that the precise same condition is equivalent to $C$ being convex and closed in the weak topology.

Let us return to the sequence of functions $\{f_n\} \subset C([0,1])$ considered in the introduction to this post. We saw that this series converges weakly to $0$, so that the Banach space structure “sees” that this sequence goes to $0$. However, we used something a little high tech. Can we only use the norm and the vector space structure to see the convergence? According to Corollary 11, there exist convex combination of the elements of the sequence $\{f_n\}$ which go to $0$ in the norm. It is easy to construct such a sequence of convex combinations by hand. It is interesting to contrast the behavior of this sequence with that of the sequence $g_n$ obtained from the $f_n$ by shifting to the right so the peak is at $1/2$. Every convex combination of the $g_n$s is of norm $1$.

The following theorem is closely related to our discussion. Note that it is not a trivial consequence of Corollary 11.

Theorem 12 (Banach–Saks Theorem): Let $\{x_n\}$ be a sequence in a Hilbert space $H$ that converges weakly to $x$. Then there exists a subsequence $\{x_{n_k}\}$ such that the arithmetic averages

$\frac{x_{n_1} + \ldots + x_{n_k}}{k}$

converge strongly to $x$

Exercise E: Prove Theorem 11.

#### 5. The orthogonality relations revisited

At the end of Notes 7 we obtained the relations $N(T^*) = R(T)^\perp, N(T) = ^\perp R(T^*)$ and $^\perp N(T^*) = \overline{R(T)}$. We also noted that in general $\overline{R(T^*)} \neq N(T)^\perp$.

Proposition 13: Let $M$ be a subspace of $X$ and $N$ be a subspace of $X^*$. Then $^\perp (M^\perp) = \overline{M}^{\| \cdot \|}$ and $(^\perp N)^\perp = \overline{N}^{w*}$

Here, $\overline{M}^{\|\cdot \|}$ denotes the norm closure of $M$, and $\overline{N}^{w*}$ denotes the weak* closure of $N$.

Proof: We already observed the first relation as a consequences of Hahn–Banach, so we prove the second. $(^\perp N)^\perp \supseteq N$ by definition, and since the left hand side is weak* closed, we have $(^\perp N)^\perp \supseteq \overline{N}^{w*}$. Now let $x^*$ be in the complement of $\overline{N}^{w*}$. By Corollary 9, there is some $x \in X$ such that $x \in ^\perp \overline{N}^{w*}$ and $\langle x, x^*\rangle \neq 0$. Thus $x^* \notin (^\perp \overline{N}^{w*})^\perp$. But the latter is equal to $(^\perp N)^\perp$, we we are done.

Corollary 14: Let $T \in B(X,Y)$. Then $N(T)^\perp = \overline{R(T^*)}_{w*}$

#### 6. Continuity properties

Exercise F: Let $X$ and $Y$ be Banach spaces. A linear operator $T: X \rightarrow Y$ is norm–norm continuous is and only if it is weak–weak continuous.

Exercise G: Let $X$ and $Y$ be Banach spaces. A linear operator $T: Y^* \rightarrow X^*$ is weak*–weak* continuous is and only if there exists $S \in B(X,Y)$ such that $S^* = T$.

The operator $S$ is sometimes denoted $T_*$ and is referred to as the pre-adjoint of $T$.