Advanced Analysis, Notes 11: Banach spaces (weak topologies, Alaoglu’s theorem)
by Orr Shalit
Let be the Banach space of continuous functions on the interval with the sup norm. Consider the following sequence of functions defind as follows. and for all , is equal to zero on the interval between and , and is linear in the intervals where we haven’t defined it yet (visualize!). The sequence is tending to zero pointwise, but the norm of does not detect this. The sequence tends to in the norm, but the norm is not in the game. Can the Banach space structure of detect the convergence of to ?
The answer is yes. Recall the Riesz–Markov representation theorem from a course in measure theory:
Theorem 1: Let be a locally compact Hausdorff space, and let denote the space of continuous functions that vanish at infinity. Then , where is the space of regular complex Borel measures on .
Let . Since everywhere it converges almost everywhere with respect to any measure. Since for all , the dominated convergence theorem tells us that
Our little discussion motivates the following definition.
Definition 2: Let be a normed space. A sequence is said to converge weakly to if for all , .
We see that the above sequence converges weakly to . It is clear that if in the norm then weakly (sometimes convergence in norm is called strong convergence). The converse is false.
Example: Consider an orthonormal sequence in a Hilbert space . Then for all , so the sequence does not converge in the norm. However, for any , Bessel’s inquality implies that , so 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 be a Banach space.
Definition 1: The weak topology on is the weakest topology in which all the functionals in are continuous. In other words, it is the topology generated by the sets
where and .
Recall that given a family of subsets of a set , one can define the topology generated by simply as the smallest topology that contains , and that this topology turns out to be the topology with the base where is any integer and . We see that the weak topology has a base
Exercise A: Prove that the weak topology has a base given by the sets
One may also replace be .
Proposition 2: The weak topology on 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 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 is infinite dimensional. For this it suffices to show that the weak topology is not metrizable. By the Hahn-Banach theorem, is an infinite dimensional space, and we already observed that it is Banach. Suppose that the weak topology of was metrizable. Then the point 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 be a sequence which contains any that arises in this countable local base at . For every , the set is a neighborhood of , hence it must contain some set of the form . By Lemma 4 below, implies that is in the span of . We have just proved that every is in the algebraic span of , and therefore 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 be a subspace of that separates the points of . The –weak topology on , denoted , is the weakest topology on in which all of the functional in are continuous. In other words, it is the topology generated by the sets
where and . Let be the image of in . The is called the weak* topology on .
Lemma 4: Let be a linear functional on , , and suppose that
Then there are scalars such that .
Proof: Denote . We will first show that . Indeed, if , then for all , hence for all , whence .
Now it is an exercise in linear algebra to show that implies that for some scalars . This is proved by induction. We may assume that all functionals in the discussion are not zero. If , let such that . Then for all , so is one dimensional and the result follows. If we proved the result for , then consider . Then , so . Thus vanishes on and is therefore a multiple of .
Proposition 5: Let be a subspace. A linear functional on is continuous in the –topology if and only if .
Proof:One direction is the definition, so nothing to prove. Assume that is continuous in the topology. Then is open, so there are such that
The lemma now shows that .
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 in a topological space is closed if and only if for every net one has . A net in converges to in the –topology if and only if for all .
2. Weakly convergent sequences
Theorem 6: Every weakly convergent sequence is bounded. Every weak* convergent sequence is bounded.
Proof: Suppose that is weak* convergent. Then for every , the sequence is convergent, and in particular it is bounded. By the principle of uniform boundedness, . If is weakly convergent, then the isometrically embedded sequence in , given by , is weak* convergent, hence bounded, by the previous argument.
Exercise B: Construct a weakly convergent net that is not bounded.
Example: Let be an orthonormal basis in a Hilbert space , and let . The sequence has no bounded sub-sequence, hence it has no weakly convergent sub-sequence. However, , and it particular, the sequence has a subnet that weakly converges to .
3. Alaoglu’s theorem
Proposition 5 shows that if is not reflexive, then the weak* topology on is strictly weaker then the weak topology on (the –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 of a normed space is compact if and only if is finite dimensional.
Theorem 7 (Alaoglu’s theorem): The closed unit ball of 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 , let . Define . By Tychonoff’s theorem, is compact in the product topology. Now is a space of complex valued functions on , so can be naturally identified with a (topological) subspace of . Note that the topology on induced by the product topology is precisely the weak* topology on in (use nets). If we show that is closed, we will be done, because a closed subset of a compact Hausdorff space is compact.
Suppose that in the product topology. By definition of , if we show that is linear we will be done, because then will be bounded of norm . But if , then
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 be a subspace of that separates the points of . Let be a convex subset of closed in the topology and let . Then there exists a functional which strictly separates and .
Corollary 9: Let be a subspace in , and let be a subspace of that separates the points of . A point is in the closure of if and only if for every which annihilates .
Exercise D: Prove the above theorem and corllary.
Recall (from the homework exercises) that a set is said to be a closed half space if , where and .
Corollary 10: A set is convex and closed in the weak topology if an only if 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 be convex and closed in the weak topology, and denote for each by a functional that separates from as guaranteed by Theorem 7. The separation provides us with a closed half space such that and . Then we obtain that
That completes the proof.
Corollary 11: A convex set in is weakly closed if and only if it is strongly closed.
Proof: In the homework exercises you proved that a set is convex and closed in the norm if and only if is the intersection of a family of closed half spaces. But we have just seen that the precise same condition is equivalent to being convex and closed in the weak topology.
Let us return to the sequence of functions considered in the introduction to this post. We saw that this series converges weakly to , so that the Banach space structure “sees” that this sequence goes to . 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 which go to 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 obtained from the by shifting to the right so the peak is at . Every convex combination of the s is of norm .
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 be a sequence in a Hilbert space that converges weakly to . Then there exists a subsequence such that the arithmetic averages
converge strongly to .
Exercise E: Prove Theorem 11.
5. The orthogonality relations revisited
At the end of Notes 7 we obtained the relations and . We also noted that in general .
Proposition 13: Let be a subspace of and be a subspace of . Then and .
Here, denotes the norm closure of , and denotes the weak* closure of .
Proof: We already observed the first relation as a consequences of Hahn–Banach, so we prove the second. by definition, and since the left hand side is weak* closed, we have . Now let be in the complement of . By Corollary 9, there is some such that and . Thus . But the latter is equal to , we we are done.
Corollary 14: Let . Then .
6. Continuity properties
Exercise F: Let and be Banach spaces. A linear operator is norm–norm continuous is and only if it is weak–weak continuous.
Exercise G: Let and be Banach spaces. A linear operator is weak*–weak* continuous is and only if there exists such that .
The operator is sometimes denoted and is referred to as the pre-adjoint of .