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 1Given a NLS and its topological dual we define:

- The weak topology of as the coarsest topology that make all the functionals of continuous, we denote this topology by .
- The weak* topology of as the coarsest topology that make all the functionals lf continuous, where is the canonical imbedding of into . We denote this topology by

Remark 1The weak* topology is useful to work with duals of nonreflexive spaces like . Clearly for the weak* topology is coarser than the weak topology . If is reflexive (i. e. ) both topologies coincide.

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

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

Proposition 2Let be a Banach space then:

- The topologies and are Hausdorff.
- Given :
with and for with any . Constitutes a local basis for in .

- Given :
with and for with any . Constitutes a local basis for in .

*Proof:* See Brezis’ book chapter 3.

Remark 2We preserve the notation of Brezis for the weak convergence, i.e. if converges to in we write

Analoguously, if if converges to in we write

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

Proposition 3Let be a Banach space, let be a sequence in and let be a sequence in , then:

- if and only if for all .

- if and only if for all .

- Strong convegence (convergence in norm) in ( ) implies convergence in ().

- If , then is bounded and
- If , then is bounded and

*Proof:* See Brezis’ book chapter 3.

The following proposition shows the relationships and differences between the strong closed sets and the closed sets in the weak topology .

Proposition 4Let be a NLS then

- If is finite dimensional then the weak topology coincides with the norm topology.
- If is infnite dimensional then
- Let be a convex subset of then is strong closed if and only if is closed in .

*Proof:* See Brezis’ book chapter 3.

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

Definition 5Let be a topological space, we say that is sequentially closed if

Remark 3It is well known that in metric space a set is closed if and only if is sequentally closed. However, in a topological space the equivalence is not true.

Proposition 6Let be a topological space, then any closed subset of is sequentally closed. If is first countable then the converse is true.

*Proof:* Let be a closed subset of . Let such that there exist such that . If then there exist an open neighbourhood of given by , that not contains any element of the sequence, contradicting the definition of convergence.

Conversely, suppose that is sequentally closed and that is first countable. Let us show that is open in . If not, there exist a point such that any open neighbourhood of intersects . Since is first countable we can find a countable local basis for given by such that . Recursively, we can define a sequence with , it is not difficult to prove that , implying that .

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

Theorem 7Let be a infinite dimensional Banach space then endowed with the weak topology is never first countable. Particularly, the weak topology is never metrizable.

*Proof:* To prove that is not first countable let us reason by contradiction. Suppose that has a countable local basis given by such that . By **Proposition 3** any open neighbourhood of contains a closed subspace of , indeed, this last proposition guarantees that there exist and functionals such that

Considering the linear functional defined by it is not dificult to see that the kernel of is a closed infinite dimensional subspace of that fulfills our requirement.

Then for any we can pick with . **Proposition 5** implies that has not weakly convergent subsequence.

Given any and any the set is an open neighbourhood of , therefore there exist such that for any

This implies that in , which is absurd.

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

Remark 4The spaces where the sequentially closure and the ordiary closure coincide are known asFréchet-Uryhsonspaces. 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 be a Banach space, then closed unit ball

is compact in .

*Proof:* The proof follows from noticing that is a closed subset of the set , 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.

Remark 5It is temptative to claim that Banach-Alaoglu’s theorem implies that endowed with the topology is locally compact, howeverProposition 3shows that any open set of must be unbounded, implying that the balls have empty interior in . 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 into the former space .

Theorem 9 (Kakutani)Let be a Banach space. Then is reflexive if and only if the closed unit ball in is weakly compact.

*Proof:* See Brezis’ book Theorem 3.17.

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

Theorem 10 (Eberlein–Šmulian)Let be a reflexive Banach space. Then is weakly compact if and only if is sequentally weakly compact if and only if 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.

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

*Proof:* Let be a bounded sequence in . Therefore there exist a closed ball that contains this sequence, and by **Eberlein–Šmulian**‘s theorem it follows that has a weakly convergent subsequence.

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 12Let be a Banach space. Then the closed ball is metrizable in if and only if is separable with the norm topology.Analogously, the closed ball is metrizable in if and only if is separable with the norm topology.

*Proof:* See Brezis’ book Theorems 3.28 and 3.29.

Corollary 13 (Sequential Banach-Alaoglu)Let be a separable Banach space then any bounded and sequentially closed set in the is sequentially compact in .

*Proof:* Let be a bounded and sequentially closed set in the . Let us consider a sequence . We can assume (normalizing the sequence if necessary) that belongs to the closed unit ball of which is compact by Banach-Alaoglu’s theorem and metrizable by the last theorem, therefore there exist such that in and since is sequentally closed in it follows that .

Remark 6Contrary 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 is essential for the proof, it can be shown (Terence Tao’s notes Remark 5) that the closed unit ball of 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 14Let be a Banach space then:

- If is sequentially closed in then is closed in the strong topology.
- If is a convex subset of then is closed in if and only if is sequentially closed in .
- If is reflexive and is a bounded subset of then is closed in if and only if is sequentially closed in .
- If is separable and is a bounded subset of then is closed in if and only if is sequentially closed in .
- If , a separable Hilbert space, then there exist sequentially closed subsets in that are not closed in .

*Proof:*

- This first part follows from the fact that strong convergence implies weak convergence, i.e.
**Proposition 5**. - IF is convex
**Proposition 6**implies that is strong closed if and only if weak closed. Hence if is weakly sequentially closed, by the first part of this proposition we have that is closed and therefore is weakly closed. - If is reflexive and bounded in ,
**Corollary 16**implies that any sequence in has a convergent subsequence, and since is weakly sequentially closed the limit of this subsequence belongs to , implying that is weakly sequentially compact.**Eberlein–Šmulian**‘s theorem implies that is also weakly compact, and since the weak topology is Hausdorff (**Proposition 3**) it follows that is weakly closed too. - If is separable then the unit closed ball in , is metrizable. Since is bounded there exist a positive constant , such that . Clearly is weakly sequentially closed in . Hence, since is metrizable (
**Theorem 17**) it follows that is a closed subspace of with the subspace topology inherited from the weak topology .

On the other hand, noticing that is a strong closed convex set of ,**Proposition 6**implies that is weakly closed. It is not difficult to show that if is weakly closed then 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 , then without loss of generality we can prove this part only for .

Let be the canonical Hilbert basis of , let us define the sequence for any . Since has no bounded subsequence**Proposition 5**implies that has no weakly convergent subsequence, therefore is weakly sequentially closed in .On the other hand, combining**Proposition 3**and Riesz representation theorem for HIlbert spaces the sets of the form:with and for , constitutes a local basis for .

Reasoning by contradiction let us suppose that does not belong to the weak clousure of , then there exist and for such that .

Let us define . Clearly and satisfies

But this contradicts the fact that .

Finally, we conclude that is weakly sequentially closed but not closed.

Remark 7It is well known that a sequence in 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 . However, byProposition 6we know that the strong and weak topologies does not coincide in , therefore there must exist a weak closed subset of 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.