# Variational methods and Weak Topologies: Part 1

One of the central facts that turns functional analysis into a non-trivial generalization of linear algebra is the lack of compactness phenomena that arises in the infinite dimensional analysis. Indeed, in virtue of  Riesz’s lemma the closed balls are never sequentially compact (and therefore never compact) in infinite dimensional NLS’s. On the other hand, since any finite dimensional space is esentially equal to some euclidean space, it can be shown that Heine-Borel’s property holds implying that any closed bounded set is also compact.

This problem suggest to define a weaker topology in the infinite dimensional spaces (i.e. with less open sets) such that the number of compact or precompact sets increases. Some standard topologies used to fulfill this requirement are the weak topology and the weak* topology. However, this approach have a cost, neither of these topologies are metrizable and therefore the topological concepts does not have to be characterized in terms of sequences.

We are not interested in discuss systematically the theory of weak topologies in NLS in this post, for the readers interested in a deep understading in the subject we suggest this Terence Tao’s post or the third chapter of Brezis’ book. Our main interest here is to present a version of a classical result called “the direct method” and to discuss some subtleties in its formulation.

First of all let us recall some important definitions and results required for our analysis.

Definition 1 Given a NLS ${X}$ and its topological dual ${X^*}$ we define:

1. The weak topology of ${X}$ as the coarsest topology that make all the functionals of ${X^*}$ continuous, we denote this topology by ${\sigma(X,X^*)}$.
2. The weak* topology of ${X^*}$ as the coarsest topology that make all the functionals lf ${J(X)}$ continuous, where ${J:X\rightarrow X^{**}}$ is the canonical imbedding of ${X}$ into ${X^{**}}$. We denote this topology by ${\sigma(X^*,X)}$

Remark 1 The weak* topology is useful to work with duals of nonreflexive spaces like ${L^\infty}$. Clearly for ${X^*}$ the weak* topology ${\sigma(X^*,X)}$ is coarser than the weak topology ${\sigma(X^*,X^{**})}$. If ${X}$ is reflexive (i. e. ${J(X)=X^{**}}$) both topologies coincide.

We summarize some required properties about both topologies in the following propositions.

Note: Many results presented does not require the NLS ${X}$ to be complete, however to be coherent with the reference provided (i. e. Brezis’ book), we assume ${X}$ being a Banach space throughout this post.

Proposition 2 Let ${X}$ be a Banach space then:

1. The topologies ${\sigma(X,X^*)}$ and ${\sigma(X^*,X)}$ are Hausdorff.
2. Given ${x_0\in X}$:

$\displaystyle V(f_1,\cdots,f_k;\varepsilon);=\{x\in X|\, \, \, |f_i(x-x_0)|<\varepsilon \quad i=1,\dots ,k \}. \ \ \ \ \ (1)$

with ${\varepsilon>0}$ and ${f_i\in X^*}$ for ${i=1,\dots ,k}$ with any ${k\in {\mathbb N}}$. Constitutes a local basis for ${x_0}$ in ${\sigma(X,X^*)}$.

3. Given ${f_0\in X^*}$:

$\displaystyle V(\varepsilon;x_1,\cdots,x_k);=\{f\in X^*|\, \, \,|(f-f_0)(x_i)|<\varepsilon \quad i=1,\dots ,k \}. \ \ \ \ \ (2)$

with ${\varepsilon>0}$ and ${x_i\in X}$ for ${i=1,\dots ,k}$ with any ${k\in {\mathbb N}}$. Constitutes a local basis for ${f_0}$ in ${\sigma(X^*,X)}$.

Proof: See Brezis’ book chapter 3. $\Box$

Remark 2 We preserve the notation of Brezis for the weak convergence, i.e. if ${\{x_n\}_{n\in {\mathbb N}}}$ converges to ${x}$ in ${\sigma(X,X^*)}$ we write

$\displaystyle x_n\rightharpoonup x \quad \text{in} \quad \sigma(X,X^*). \ \ \ \ \ (3)$

Analoguously, if if ${\{f_n\}_{n\in {\mathbb N}}}$ converges to ${f}$ in ${\sigma(X^*,X)}$ we write

$\displaystyle f_n\rightharpoonup f \quad \text{in} \quad \sigma(X^*,X). \ \ \ \ \ (4)$

Some important properties about the convergence in these topologies are summarized in the following proposition.

Proposition 3 Let ${X}$ be a Banach space, let ${\{x_n\}}$ be a sequence in ${X}$ and let ${\{f_n\}}$ be a sequence in ${X^*}$ , then:

1. ${x_n\rightharpoonup x \quad \text{in} \quad \sigma(X,X^*)}$ if and only if ${g(x_n)\rightarrow g(x)}$ for all ${g\in X^*}$.
1. ${f_n\rightharpoonup f \quad \text{in} \quad \sigma(X^*,X)}$ if and only if ${f_n(y)\rightarrow f(y)}$ for all ${y\in X}$.
1. Strong convegence (convergence in norm) in ${X}$ ( ${X^*}$) implies convergence in ${\sigma(X,X^*)}$ (${\sigma(X^*,X)}$).
1. If ${x_n\rightharpoonup x \quad \text{in} \quad \sigma(X,X^*)}$, then ${\{||x_n||\}_{n\in {\mathbb N}}}$ is bounded and

$\displaystyle ||x||\leq \liminf_{n\rightarrow \infty} ||x_n||. \ \ \ \ \ (5)$

2. If ${f_n\rightharpoonup f \quad \text{in} \quad \sigma(X^*,X)}$, then ${\{||f_n||\}_{n\in {\mathbb N}}}$ is bounded and

$\displaystyle ||f||\leq \liminf_{n\rightarrow \infty} ||f_n||. \ \ \ \ \ (6)$

Proof: See Brezis’ book chapter 3. $\Box$

The following proposition shows the relationships and differences between the strong closed sets and the closed sets in the weak topology ${\sigma(X,X^*)}$.

Proposition 4 Let ${X}$ be a NLS then

1. If ${X}$ is finite dimensional then the weak topology ${\sigma(X,X^*)}$ coincides with the norm topology.
2. If ${X}$ is infnite dimensional then

$\displaystyle \overline{\mathbb{S}_X}^{\sigma(X,X^*)}=\overline{B_X}:=\{x\in X| ||x||\leq 1\}. \ \ \ \ \ (7)$

3. Let ${C}$ be a convex subset of ${X}$ then ${C}$ is strong closed if and only if ${C}$ is closed in ${\sigma(X,X^*)}$.

Proof: See Brezis’ book chapter 3. $\Box$

We are in position now to discuss one of the main topics of this post.

Definition 5 Let ${(A,\tau)}$ be a topological space, we say that ${F\subset A}$ is sequentially closed if

$\displaystyle F=\{x\in A| \exists \{x_n: n\in {\mathbb N}\}\subset F\quad \text{such that}\quad x_n\rightarrow x\} \ \ \ \ \ (8)$

Remark 3 It is well known that in metric space ${(M,d)}$ a set ${F}$ is closed if and only if ${F}$ is sequentally closed. However, in a topological space the equivalence is not true.

Proposition 6 Let ${(A,\tau)}$ be a topological space, then any closed subset of ${A}$ is sequentally closed. If ${A}$ is first countable then the converse is true.

Proof: Let ${F}$ be a closed subset of ${A}$. Let ${x\in A}$ such that there exist ${\{x_n: n\in {\mathbb N}\}\subset F}$ such that ${ x_n\rightarrow x}$. If ${x\notin F}$ then there exist an open neighbourhood of ${x}$ given by ${F^c}$, that not contains any element of the sequence, contradicting the definition of convergence.

Conversely, suppose that ${F}$ is sequentally closed and that ${A}$ is first countable. Let us show that ${F^c}$ is open in ${A}$. If not, there exist a point ${x\in F^c}$ such that any open neighbourhood of ${x}$ intersects ${F}$. Since ${A}$ is first countable we can find a countable local basis for ${x}$ given by ${\{V_n\}_{n\in {\mathbb N}}}$ such that ${V_{n+1}\subset V_n}$. Recursively, we can define a sequence ${\{x_n: n\in {\mathbb N}\}\subset F}$ with ${x_i\in V_i\cap F}$, it is not difficult to prove that ${x_n\rightarrow x}$, implying that ${x\in F}$. $\Box$

However, as stated in the following theorem for infinite dimensional Banach spaces the weak topology is never first countable.

Theorem 7 Let ${X}$ be a infinite dimensional Banach space then ${X}$ endowed with the weak topology ${\sigma(X,X^*)}$ is never first countable. Particularly, the weak topology is never metrizable.

Proof: To prove that ${X}$ is not first countable let us reason by contradiction. Suppose that ${0\in X}$ has a countable local basis given by ${\{V_n\}_{n\in {\mathbb N}}}$ such that ${V_{n+1}\subset V_n}$. By Proposition 3 any open neighbourhood ${V_n}$ of ${0}$ contains a closed subspace of ${X}$, indeed, this last proposition guarantees that there exist ${\varepsilon>0}$ and functionals ${f_1,\cdots, f_k\in X^*}$ such that

$\displaystyle V(f_1,\cdots, f_k;\varepsilon)\subset V_n. \ \ \ \ \ (9)$

Considering the linear functional ${F:X\rightarrow {\mathbb R}^k}$ defined by ${F(x)=(f_1(x),\cdots, f_k)}$ it is not dificult to see that the kernel of ${F}$ is a closed infinite dimensional subspace of ${X}$ that fulfills our requirement.

Then for any ${n\in {\mathbb N}}$ we can pick ${y_n\in V_n}$ with ${||y_n||=n}$. Proposition 5 implies that ${\{y_n\}_{n\in {\mathbb N}}}$ has not weakly convergent subsequence.

Given any ${f\in X^*}$ and any ${\varepsilon>0}$ the set ${U_{f,\varepsilon}:=\{x\in X| |f(x)|<\varepsilon\}}$ is an open neighbourhood of ${0}$, therefore there exist ${N\in {\mathbb N}}$ such that for any ${n\geq N}$

$\displaystyle y_n\in U_{f,\varepsilon}. \ \ \ \ \ (10)$

This implies that ${x_n\rightharpoonup 0}$ in ${\sigma(X,X^*)}$, which is absurd.

Since every metrizable space is first countable it follows that the weak topology is never metrizable (for infinite dimensional Banach spaces). $\Box$

Remark 4 The spaces where the sequentially closure and the ordiary closure coincide are known as Fréchet-Uryhson spaces. There are examples of Fréchet-Uryhson spaces that are not first countable, therefore the last theorem is not enough to show that sequential closedness does not coincide with closedness in the weak topology. Before giving some striking examples of this fact it is useful to illustrate the importance of reflexive and separable spaces in the weak and weak* topologies.

The following is maybe the most important theorem related with the weak* topology,  because it shows that actually this construction increases considerably the number of compact sets in the dual of a Banach space.

Theorem 8 (Banach-Alaoglu) Let ${X}$ be a Banach space, then closed unit ball

$\displaystyle \overline{B_{X^*}}:=\{f\in X^*| ||f||\leq 1\}, \ \ \ \ \ (11)$

is compact in ${\sigma(X^*,X)}$.

Proof: The proof follows from noticing that ${\overline{B_{X^*}}}$ is a closed subset of the set ${[-1,1]^{\overline{B_X}}:=\{f:\overline{B_X}\rightarrow [-1,1]\}}$  , endowed with the product topology, which is compact by Tychonoff’s theorem.

See Brezis’ book, theorem 3.16 for details or Terence Tao’s post Theorem 3 for the case of complex Banach spaces. $\Box$

Remark 5 It is temptative to claim that Banach-Alaoglu’s theorem implies that ${X^*}$ endowed with the topology ${\sigma(X^*,X)}$ is locally compact, however Proposition 3 shows that any open set of ${X^*}$ must be unbounded, implying that the balls have empty interior in ${\sigma(X^*,X)}$. Moreover, it can be shown that any locally compact Hausdorff topological vector space must be finite dimensional (see Exercise 24 of Terence Tao’s post).

The following theorem gives us a way to translate this compactness condition in the dual ${X^*}$ into the former space ${X}$.

Theorem 9 (Kakutani) Let ${X}$ be a Banach space. Then ${X}$ is reflexive if and only if ${\overline{B_X}}$ the closed unit ball in ${X}$ is weakly compact.

Proof: See Brezis’ book Theorem 3.17. $\Box$

Surprisingly, in the weak topology compactness, sequential compactness and Heine-Borel’s property are equivalent in reflexive spaces.

Theorem 10 (Eberlein–Šmulian) Let ${X}$ be a reflexive Banach space. Then ${K\subset X}$ is weakly compact if and only if ${K}$ is sequentally weakly compact if and only if ${K}$ is weakly closed and bounded.

Proof: Some of these implications follows from the theory developed in the chapter 3 of Brezis’ book, for the complete proof see Yosida’s book. $\Box$

Corollary 11 Let ${X}$ be a reflexive Banach space. Then any bounded sequence has a weakly convergent subsecuence in ${X}$.

Proof: Let ${\{x_n\}_{n\in {\mathbb N}}}$ be a bounded sequence in ${X}$. Therefore there exist a closed ball that contains this sequence, and by Eberlein–Šmulian‘s theorem it follows that ${\{x_n\}_{n\in {\mathbb N}}}$ has a weakly convergent subsequence. $\Box$

Even though Theorem 10 shows that the weak topology is never metrizable in infinite dimensional Banach spaces it is still possible to endow the closed unit ball with a metric structure.

Theorem 12 Let ${X}$ be a Banach space. Then the closed ball ${B_{X^*}}$ is metrizable in ${\sigma(X^*,X)}$ if and only if ${X}$ is separable with the norm topology.

Analogously, the closed ball ${B_{X}}$ is metrizable in ${\sigma(X,X^*)}$ if and only if ${X^*}$ is separable with the norm topology.

Proof: See Brezis’ book Theorems 3.28 and 3.29. $\Box$

Corollary 13 (Sequential Banach-Alaoglu) Let ${X}$ be a separable Banach space then any bounded and sequentially closed set in the ${\sigma(X^*,X)}$ is sequentially compact in ${\sigma(X^*,X)}$.

Proof: Let ${F}$ be a bounded and sequentially closed set in the ${\sigma(X^*,X)}$. Let us consider a sequence ${\{f_n\}_{n\in {\mathbb N} }}$. We can assume (normalizing the sequence if necessary) that ${\{f_n\}_{n\in {\mathbb N} }}$ belongs to the closed unit ball of ${X^*}$ which is compact by Banach-Alaoglu’s theorem and metrizable by the last theorem, therefore there exist ${f\in X^*}$ such that ${f_n\rightharpoonup f}$ in ${\sigma(X^*,X)}$ and since ${F}$ is sequentally closed in ${\sigma(X^*,X)}$ it follows that ${f\in F}$. $\Box$

Remark 6 Contrary to the case of closed sets, the compactness and sequantial compactness are properties that cannot be compared in arbitrary topological spaces. However, in this case we overcome this difficulty using the fact that these two properties are the same in metric spaces.

The separability of ${X}$ is essential for the proof, it can be shown (Terence Tao’s notes Remark 5) that the closed unit ball of ${(l^\infty({\mathbb N}))^*}$ is not sequentially compact in the weak* topology.

Furnished with this theory we can provide a useful (though not general) result that shows whether closed and sequentially closed sets coincide in the weak topology.

Proposition 14 Let ${X}$ be a Banach space then:

• If ${F}$ is sequentially closed in ${\sigma(X,X^*)}$ then ${F}$ is closed in the strong topology.
• If ${F}$ is a convex subset of ${X}$ then ${F}$ is closed in ${\sigma(X,X^*)}$ if and only if ${F}$ is sequentially closed in ${\sigma(X,X^*)}$.
• If ${X}$ is reflexive and ${F}$ is a bounded subset of ${X}$ then ${F}$ is closed in ${\sigma(X,X^*)}$ if and only if ${F}$ is sequentially closed in ${\sigma(X,X^*)}$.
• If ${X^*}$ is separable and ${F}$ is a bounded subset of ${X}$ then ${F}$ is closed in ${\sigma(X,X^*)}$ if and only if ${F}$ is sequentially closed in ${\sigma(X,X^*)}$.
• If ${X=H}$ , a separable Hilbert space, then there exist sequentially closed subsets in ${\sigma(H,H^*)}$ that are not closed in ${\sigma(H,H^*)}$.

Proof:

• This first part follows from the fact that strong convergence implies weak convergence, i.e. Proposition 5.
• IF ${F}$ is convex Proposition 6 implies that ${F}$ is strong closed if and only if ${F}$ weak closed. Hence if ${F}$ is weakly sequentially closed, by the first part of this proposition we have that ${F}$ is closed and therefore is weakly closed.
• If ${X}$ is reflexive and ${F}$ bounded in ${X}$, Corollary 16 implies that any sequence in ${F}$ has a convergent subsequence, and since ${F}$ is weakly sequentially closed the limit of this subsequence belongs to ${F}$, implying that ${F}$ is weakly sequentially compact. Eberlein–Šmulian‘s theorem implies that ${F}$ is also weakly compact, and since the weak topology is Hausdorff (Proposition 3) it follows that ${F}$ is weakly closed too.
• If ${X^*}$ is separable then the unit closed ball in ${X}$, ${\overline{B_X}}$ is metrizable. Since ${F}$ is bounded there exist a positive constant ${M>0}$, such that ${\frac{1}{M} F \subset \overline{B_X}}$. Clearly ${\frac{1}{M} F }$ is weakly sequentially closed in ${\overline{B_X}}$. Hence, since ${\overline{B_X}}$ is metrizable (Theorem 17) it follows that ${\frac{1}{M} F}$ is a closed subspace of ${\overline{B_X}}$ with the subspace topology inherited from the weak topology ${\sigma(X,X^*)}$.
On the other hand, noticing that ${\overline{B_X}}$ is a strong closed convex set of ${X}$, Proposition 6 implies that ${\frac{1}{M} F}$ is weakly closed. It is not difficult to show that if ${\frac{1}{M} F}$ is weakly closed then ${F}$ is weakly closed too, finishing this part of the proof.
• As it is well known any infinite dimensional separable Hilbert space can be indentified with ${l^2({\mathbb N})}$, then without loss of generality we can prove this part only for ${l^2({\mathbb N})}$.
Let ${\{e_n\}_{n\in {\mathbb N}}}$ be the canonical Hilbert basis of ${l^2({\mathbb N})}$, let us define the sequence ${x_n=\sqrt{n}e_n}$ for any ${n\in {\mathbb N}}$. Since ${\{x_n\}_{n\in {\mathbb N}}}$ has no bounded subsequence Proposition 5 implies that ${\{x_n\}_{n\in {\mathbb N}}}$ has no weakly convergent subsequence, therefore ${\{x_n| \, n\in {\mathbb N}\}}$ is weakly sequentially closed in ${l^2({\mathbb N})}$.On the other hand, combining Proposition 3 and Riesz representation theorem for HIlbert spaces the sets of the form:

$\displaystyle V(y_1,\cdots,y_k;\varepsilon):=\{x\in l^2({\mathbb N})|\, \, |\langle x,y_i\rangle|<\varepsilon, \quad i=1,\cdots k \} \ \ \ \ \ (12)$

with ${\varepsilon>0}$ and ${y_i \in l^2({\mathbb N})}$ for ${i=1,\cdots k}$, constitutes a local basis for ${0}$.

Reasoning by contradiction let us suppose that ${0}$ does not belong to the weak clousure of ${\{x_n| \, n\in {\mathbb N}\}}$, then there exist ${\varepsilon_0>0}$ and ${y_i=\{y_i^n\}_{n\in {\mathbb N}} \in l^2({\mathbb N})}$ for ${i=1,\cdots k}$ such that ${\{x_n| \, n\in {\mathbb N}\}\nsubseteq V(y_1,\cdots,y_k;\varepsilon_0)}$.

Let us define ${a_n:=\sum_{i=1}^{k}|y_i^n|}$. Clearly ${\{a_n\}_{n\in {\mathbb N}}\in l^2({\mathbb N})}$ and satisfies

$\displaystyle |a_n|\geq \frac{\varepsilon_0}{\sqrt{n}} \quad \forall n\in {\mathbb N}. \ \ \ \ \ (13)$

But this contradicts the fact that ${\{a_n\}_{n\in {\mathbb N}}\in l^2({\mathbb N})}$.

Finally, we conclude that ${\{x_n| \, n\in {\mathbb N}\}}$ is weakly sequentially closed but not closed.

$\Box$

Remark 7 It is well known that a sequence in ${l^1({\mathbb N})}$ converges in norm if and only if converges in the weak topology (see for instance, Terence Tao’s notes Remark 4). Then the weakly sequentially closed sets and the closed sets are the same in ${l^1({\mathbb N})}$. However, by Proposition 6 we know that the strong and weak topologies does not coincide in ${l^1({\mathbb N})}$, therefore there must exist a weak closed subset of ${l^1({\mathbb N})}$ that is not strong closed and therefore that is not weakly sequentially closed.

This astonishing example also delivers an emphatic message “the topologies are not characterized by its convergent sequences”.  This conclusion also gives us a intiutive idea of the reason that makes weak sequentially closed and weak closed sets differ in general.

In the second part of these notes we will discuss what kind of conditions must be imposed on functionals defined in Banach spaces in order to guarantee the existence of relative extrema.