A normed vector space automatically generates a topology, known as the *norm topology* or *strong topology* on , generated by the open balls . A sequence in such a space *converges strongly* (or *converges in norm*) to a limit if and only if as . This is the topology we have implicitly been using in our previous discussion of normed vector spaces.

However, in some cases it is useful to work in topologies on vector spaces that are weaker than a norm topology. One reason for this is that many important modes of convergence, such as pointwise convergence, convergence in measure, smooth convergence, or convergence on compact subsets, are not captured by a norm topology, and so it is useful to have a more general theory of topological vector spaces that contains these modes. Another reason (of particular importance in PDE) is that the norm topology on infinite-dimensional spaces is so strong that very few sets are compact or pre-compact in these topologies, making it difficult to apply *compactness methods* in these topologies. Instead, one often first works in a weaker topology, in which compactness is easier to establish, and then somehow upgrades any weakly convergent sequences obtained via compactness to stronger modes of convergence (or alternatively, one abandons strong convergence and exploits the weak convergence directly). Two basic weak topologies for this purpose are the weak topology on a normed vector space , and the weak* topology on a dual vector space . Compactness in the latter topology is usually obtained from the Banach-Alaoglu theorem (and its sequential counterpart), which will be a quick consequence of the Tychonoff theorem (and its sequential counterpart) from the previous lecture.

The strong and weak topologies on normed vector spaces also have analogues for the space of bounded linear operators from to , thus supplementing the operator norm topology on that space with two weaker topologies, which (somewhat confusingly) are named the strong operator topology and the weak operator topology.

** — 1. Topological vector spaces — **

We begin with the definition of a *topological vector space*, which is a space with suitably compatible topological and vector space structures on it.

Definition 1Atopological vector spaceis a real or complex vector space , together with a topology such that the addition operation and the scalar multiplication operation or is jointly continuous in both variables (thus, for instance, is continuous from with the product topology to ).

It is an easy consequence of the definitions that the translation maps for and the dilation maps for non-zero scalars are homeomorphisms on ; thus for instance the translation or dilation of an open set (or a closed set, a compact set, etc.) is open (resp. closed, compact, etc.). We also have the usual limit laws: if and in a topological vector space, then , and if in the field of scalars, then . (Note how we need joint continuity here; if we only had continuity in the individual variables, we could only conclude that (for instance) if one of or was constant.)

We now give some basic examples of topological vector spaces.

Exercise 1Show that every normed vector space is a topological vector space, using the balls as the base for the topology. Show that the same statement holds if the vector space is quasi-normed rather than normed.

Exercise 2Every semi-normed vector space is a topological vector space, again using the balls as a base for the topology. This topology is Hausdorff if and only if the semi-norm is a norm.

Example 1Any linear subspace of a topological vector space is again a topological vector space (with the induced topology).

Exercise 3Let be a vector space, and let be a (possibly infinite) family of topologies on , each of which turning into a topological vector space. Let be the topology generated by (i.e. it is the weakest topology that contains all of the . Show that is also a topological vector space. Also show that a sequence converges to a limit in if and only if in for all . (The same statement also holds if sequences are replaced by nets.) In particular, by Exercise 2, we can talk about the topological vector space generated by a family of semi-norms on .

Exercise 4Let be a linear map between vector spaces. Suppose that we give the topology induced by a family of semi-norms , and the topology induced by a family of semi-norms . Show that is continuous if and only if, for each , there exists a finite subset of and a constant such that for all .

Example 2 (Pointwise convergence)Let be a set, and let be the space of complex-valued functions ; this is a complex vector space. Each point gives rise to a seminorm . The topology generated by all of these seminorms is thetopology of pointwise convergenceon (and is also the product topology on this space); a sequence converges to in this topology if and only if it converges pointwise. Note that if has more than one point, then none of the semi-norms individually generate a Hausdorff topology, but when combined together, they do.

Example 3 (Uniform convergence)Let be a topological space, and let be the space of complex-valued continuous functions . If is not compact, then one does not expect functions in to be bounded in general, and so the sup norm does not necessarily make into a normed vector space. Nevertheless, one can still define “balls” in byand verify that these form a base for a topological structure on the vector space, although it is not quite a topological vector space structure because multiplication is no longer continuous. A sequence converges in this topology to a limit if and only if converges uniformly to , thus is finite for sufficiently large and converges to zero as .

Example 4 (Uniform convergence on compact sets)Let and be as in the previous example. For every compact subset of , we can define a seminorm on by . The topology generated by all of these seminorms (as ranges over all compact subsets of ) is called thetopology of uniform convergence on compact sets; it is stronger than the topology of poitnwise convergence but weaker than the topology of uniform convergence. Indeed, a sequence converges to in this topology if and only if converges uniformly to on each compact set.

Exercise 5Show that an arbitrary product of topological vector spaces (endowed with the product topology) is again a topological vector space. [I am not sure if the same statement is true for the box topology; I believe it is false.]

Exercise 6Show that a topological vector space is Hausdorff if and only if the origin is closed. (Hint: first use the continuity of addition to prove the lemma that if is an open neighbourhood of , then there exists another open neighbourhood of such that , i.e. for all .)

Example 5 (Smooth convergence)Let be the space of smooth functions . One can define the norm on this space for any non-negative integer by the formulawhere is the derivative of . The topology generated by all the norms for is the

smooth topology: a sequence converges in this topology to a limit if converges uniformly to for each .

Exercise 7 (Convergence in measure)Let be a measure space, and let be the space of measurable functions . Show that the setsfor , , form the base for a topology that turns into a topological vector space, and that a sequence converges to a limit in this topology if and only if it converges in measure.

Exercise 8Let be given the usual Lebesgue measure. Show that the vector space cannot be given a topological vector space structure in which a sequence converges to in this topology if and only if it converges almost everywhere. (Hint: construct a sequence in which does not converge pointwise a.e. to zero, but such that every subsequence has a further subsequence that converges a.e. to zero, and use Exercise 7′ from Notes 8.) Thus almost everywhere convergence is not “topologisable” in general.

Exercise 9 (Algebraic topology)Recall that a subset of a real vector space isalgebraically openif the sets are open for all .

- (i) Show that any set which is open in a topological vector space, is also algebraically open.
- (ii) Give an example of a set in which is algebraically open, but not open in the usual topology. (Hint: a line intersects the unit circle in at most two points.)
- (iii) Show that the collection of algebraically open sets in is a topology.
- (iv) Show that the collection of algebraically open sets in does
notgive the structure of a topological vector space.

Exercise 10 (Quotient topology)Let be a topological vector space, and let be a subspace of . Let be the space of cosets of ; this is a vector space. Let be the coset map . Show that the collection of sets such that is open gives the structure of a topological vector space. If is Hausdorff, show that is Hausdorff if and only if is closed in .

Some (but not all) of the concepts that are definable for normed vector spaces, are also definable for the more general category of topological vector spaces. For instance, even though there is no metric structure, one can still define the notion of a Cauchy sequence in a topological vector space: this is a sequence such that as (or more precisely, for any open neighbourhood of , there exists such that for all ). It is then possible to talk about a topological vector space being *complete* (i.e. every Cauchy sequence converges). (From a more abstract perspective, the reason we can define notions such as completeness is because a topological vector space has something better than a topological structure, namely a uniform structure.)

Remark 1As we have seen in previous lectures, complete normed vector spaces (i.e. Banach spaces) enjoy some very nice properties. Some of these properties (e.g. the uniform boundedness principle and the open mapping theorem extend to a slightly larger class of complete topological vector spaces, namely the Fréchet spaces. AFréchet spaceis a complete Hausdorff topological vector space whose topology is generated by an at most countable family of semi-norms; examples include the space from Exercise 5 or the uniform convergence on compact sets topology from Exercise 4 in the case when is -compact. We will however not study Fréchet spaces systematically here.

One can also extend the notion of a dual space from normed vector spaces to topological vector spaces in the obvious manner: the dual space of a topological space is the space of continuous linear functionals from to the field of scalars (either or , depending on whether is a real or complex vector space). This is clearly a vector space. Unfortunately, in the absence of a norm on , one cannot define the analogue of the norm topology on ; but as we shall see below, there are some weaker topologies that one can still place on this dual space.

** — 2. Compactness in the strong topology — **

We now return to normed vector spaces, and briefly discuss compactness in the strong (or norm) topology on such spaces. In finite dimensions, the Heine-Borel theorem tells us that a set is compact if and only if it is closed and bounded. In infinite dimensions, this is not enough, for two reasons. Firstly, compact sets need to be complete, so we are only likely to find many compact sets when the ambient normed vector space is also complete (i.e. it is a Banach space). Secondly, compact sets need to be totally bounded, rather than merely bounded, and this is quite a stringent condition. Indeed it forces compact sets to be “almost finite-dimensional” in the following sense:

Exercise 11Let be a subset of a Banach space . Show that the following are equivalent:

- (i) is compact.
- (ii) is sequentially compact.
- (iii) is closed and bounded, and for every , lies in the -neighbourhood of a finite-dimensional subspace of .
Suppose furthermore that there is a nested sequence of finite-dimensional subspaces of such that is dense. Show that the following statement is equivalent to the first three:

- (iv) is closed and bounded, and for every there exists an such that lies in the -neighbourhood of .

Example 6Let . In order for a set to be compact in the strong topology, it needs to be closed and bounded, and alsouniformly -power integrable at spatial infinityin the sense that for every there exists such thatfor all . Thus, for instance, the “moving bump” example , where is the sequence which equals on and zero elsewhere, is not uniformly power integrable and thus not a compact subset of , despite being closed and bounded.

For “continuous” spaces, such as , uniform integrability at spatial infinity is not sufficient to force compactness in the strong topology; one also needs some uniform integrability at very fine scales, which can be described using harmonic analysis tools such as the Fourier transform. We will not discuss this topic here.

Exercise 12Let be a normed vector space.

- If is a finite-dimensional subspace of , and , show that there exists such that for all . Give an example to show that is not necessarily unique (in contrast to the situation with Hilbert spaces).
- If is a finite-dimensional proper subspace of , show that there exists with such that for all . (cf. the Riesz lemma.)
- Show that the closed unit ball is compact in the strong topology if and only if is finite-dimensional.

** — 3. The weak and weak* topologies — **

Let be a topological vector space. Then, as discussed above, we have the vector space of continuous linear functionals on . We can use this dual space to create two useful topologies, the *weak topology* on and the *weak* topology* on :

Definition 2 (Weak and weak* topologies)Let be a topological vector space, and let be its dual.

- The weak topology on is the topology generated by the seminorms for all .
- The weak* topology on is the topology generated by the seminorms for all .

Remark 2It is possible for two non-isomorphic topological vector spaces to have isomorphic duals, but with non-isomorphic weak* topologies. (For instance, has a very large number of preduals, which can generate a number of different weak* topologies on .) So, technically, one cannot talk abouttheweak* topology on a dual space , without specifying exactly what the predual space is. However, in practice, the predual space is usually clear from context.

Exercise 13Show that the weak topology on is a topological vector space structure on that is weaker than the strong topology on . Also, if (and hence and ) are normed vector spaces, show that the weak* topology on is a topological vector space structure on that is weaker than the weak topology on (which is defined using the double dual . When is a reflexive normed vector space, show that the weak and weak* topologies on are equivalent.

From the definition, we see that a sequence converges in the weak topology, or *converges weakly* for short, to a limit if and only if for all . This weak convergence is often denoted , to distinguish it from strong convergence . Similarly, a sequence converges in the weak* topology to if for all (thus , viewed as a function on , converges pointwise to ).

Remark 3If is a Hilbert space, then from the Riesz representation theorem for Hilbert spaces we see that a sequence converges weakly (or in the weak* sense) to a limit if and only if for all .

Exercise 14Show that if is a normed vector space, then the weak topology on and the weak* topology on are both Hausdorff. (Hint:You will need the Hahn-Banach theorem.) In particular, we conclude the important fact that weak and weak* limits, when they exist, are unique.

The following exercise shows that the strong, weak, and weak* topologies can all differ from each other.

Exercise 15Let , thus and . Let be the standard basis of either , , or .

- Show that the sequence converges weakly in to zero, but does not converge strongly in .
- Show that the sequence converges in the weak* sense in to zero, but does not converge in the weak or strong senses in .
- Show that the sequence for converges in the weak* topology of to zero, but does not converge in the weak or strong senses. (Hint: use a generalised limit functional).

Remark 4Recall from Exercise 11 of Notes 9 that sequences in which converge in the weak topology, also converge in the strong topology. We caution however that the two topologies are not quite equivalent; for instance, the open unit ball in is open in the strong topology, but not in the weak.

Exercise 16Let be a normed vector space, and let be a subset of . Show that the following are equivalent:

- is strongly bounded (i.e. is contained in a ball).
- is weakly bounded (i.e. is bounded for all ).
(Hint: use the Hahn-Banach theorem and the uniform boundedness principle.) Similarly, if is a subset of , and is a Banach space, show that is strongly bounded if and only if is weak* bounded (i.e. is bounded for each ).) Conclude in particular that any sequence which is weakly convergent in or weak* convergent in is necessarily bounded.

Exercise 17Let be a Banach space, and let converge weakly to a limit . Show that the sequence is bounded, andObserve from Exercise 15 that strict inequality can hold (cf. Fatou’s lemma). Similarly, if converges in the weak* topology to a limit , show that the sequence is bounded and that

Again, construct an example to show that strict inequality can hold. Thus we see that weak or weak* limits can lose mass in the limit, as opposed to strong limits (note from the triangle inequality that if converges strongly to , then converges to ).

Exercise 18Let be a Hilbert space, and let converge weakly to a limit . Show that the following statements are equivalent:

- converges strongly to .
- converges to .

Exercise 19Let be a separable Hilbert space. We say that a sequenceconverges in the Césaro senseto a limit if converges strongly to as .

- Show that if converges strongly to , then it also converges in the Césaro sense to .
- Give examples to show that weak convergence does not imply Césaro convergence, and vice versa. On the other hand, if a sequence converges both weakly and in the Césaro sense, show that the weak limit is necessarily equal to the Césaro limit.
- Show that a sequence converges weakly to if and only if every subsequence has a further subsequence that converges in the Césaro sense to .

Exercise 20Let be a Banach space. Show that the closed unit ball in is also closed in the weak topology, and the closed unit ball in is closed in the weak* topology.

Exercise 21Let be a Banach space. Show that the weak* topology on is complete.

Exercise 22Let be a normed vector space, let be a subspace of which is closed in the strong topology of .

- Show that is closed in the weak topology of .
- If is a sequence and , show that converges to in the weak topology of if and only if it converges to in the weak topology of . (Because of this fact, we can often refer to “the weak topology” without specifying the ambient space precisely.)

Exercise 23Let with the uniform (i.e. ) norm, and identify the dual space with in the usual manner.

- Show that a sequence converges weakly to a limit if and only if the are bounded in and converge pointwise to .
- Show that a sequence converges in the weak* topology to a limit if and only if the are bounded in and converge pointwise to .
- Show that the weak topology in is not complete.
(More generally, it may help to think of the weak and weak* topologies as being analogous to pointwise convergence topologies.)

One of the main reasons why we use the weak and weak* topologies in the first place is that they have much better compactness properties than the strong topology, thanks to the Banach-Alaoglu theorem:

Theorem 3 (Banach-Alaoglu theorem)Let be a normed vector space. Then the closed unit ball of is compact in the weak* topology.

This result should be contrasted with Exercise 12.

*Proof:* Let’s say is a complex vector space (the case of real vector spaces is of course analogous). Let be the closed unit ball of , then any linear functional maps the closed unit ball of into the disk . Thus one can identify with a subset of , the space of functions from to . One easily verifies that the weak* topology on is nothing more than the product topology of restricted to . Also, one easily shows that is closed in . But by Tychonoff’s theorem, is compact, and so is compact also.

One should caution that the Banach-Alaoglu theorem does *not* imply that the space is locally compact in the weak* topology, because the norm ball in has empty interior in the weak* topology unless is finite dimensional. In fact, we have the following result of Riesz:

Exercise 24Let be a locally compact Hausdorff topological vector space. Show that is finite dimensional. (Hint: If is locally compact, then there exists an open neighbourhood of the origin whose closure is compact. Show that for some finite-dimensional subspace , where . Iterate this to conclude that for any . On the other hand, use the compactness of to show that for any point there exists such that is disjoint from . Conclude that and thence that .)

The sequential version of the Banach-Alaoglu theorem is also of importance (particularly in PDE):

Theorem 4 (Sequential Banach-Alaoglu theorem)Let be a separable normed vector space. Then the closed unit ball of is sequentially compact in the weak* topology.

*Proof:* The functionals in are uniformly bounded and uniformly equicontinuous on , which by hypothesis has a countable dense subset . By the sequential Tychonoff theorem, any sequence in then has a subsequence which converges pointwise on , and thus converges pointwise on by Exercise 28 of Notes 10, and thus converges in the weak* topology. But as is closed in this topology, we conclude that is sequentially compact as required.

Remark 5One can also deduce the sequential Banach-Alaoglu theorem from the general Banach-Alaoglu theorem by observing that the weak* topology on (bounded subsets of) the dual of a separable space is metrisable. The sequential Banach-Alaoglu theorem can break down for non-separable spaces. For instance, the closed unit ball in is not sequentially compact in the weak* topology, basically because the space of ultrafilters is not sequentially compact (see Exercise 12 of these lecture notes).

If is reflexive, then the weak topology on is identical to the weak* topology on . We thus have

Corollary 5If is a reflexive normed vector space, then the closed unit ball in is weakly compact, and (if is separable) is also sequentially weakly compact.

Remark 6If is a normed vector space that is not separable, then one can show that is not separable either. Indeed, using transfinite induction on first uncountable ordinal, one can construct an uncountable proper well-ordered chain of closed separable subspaces of the inseparable space , which by the Hahn-Banach theorem induces an uncountable proper well-ordered chain of closed subspaces on , which is not compatible with separability. As a consequence, a reflexive space is separable if and only if its dual is separable. [On the other hand, separable spaces can have non-separable duals; consider , for instance.]

In particular, any bounded sequence in a reflexive separable normed vector space has a weakly convergent subsequence. This fact leads to the very useful *weak compactness* method in PDE and calculus of variations, in which a solution to a PDE or variational problem is constructed by first constructing a bounded sequence of “near-solutions” or “near-extremisers” to the PDE or variational problem, and then extracting a weak limit. However, it is important to caution that weak compactness can fail for non-reflexive spaces; indeed, for such spaces the closed unit ball in may not even be weakly complete, let alone weakly compact, as already seen in Exercise 23. Thus, one should be cautious when applying the weak compactness method to a non-reflexive space such as or . (On the other hand, weak* compactness does not need reflexivity, and is thus safer to use in such cases.)

In later notes we will see that the (sequential) Banach-Alaoglu theorem will combine very nicely with the Riesz representation theorem for measures, leading in particular to Prokhorov’s theorem.

** — 4. The strong and weak operator topologies — **

Now we turn our attention from function spaces to spaces of operators. Recall that if and are normed vector spaces, then is the space of bounded linear transformations from to . This is a normed vector space with the operator norm

This norm induces the *operator norm topology* on . Unfortunately, this topology is so strong that it is difficult for a sequence of operators to converge to a limit; for this reason, we introduce two weaker topologies.

Definition 6 (Strong and weak operator topologies)Let be normed vector spaces. The strong operator topology on is the topology induced by the seminorms for all . The weak operator topology on is the topology induced by the seminorms for all and .

Note that a sequence converges in the strong operator topology to a limit if and only if strongly in for all , and converges in the weak operator topology. (In contrast, converges to in the operator norm topology if and only if converges to *uniformly* on bounded sets.) One easily sees that the weak operator topology is weaker than the strong operator topology, which in turn is (somewhat confusingly) weaker than the operator norm topology.

Example 7When is the scalar field, then is canonically isomorphic to . In this case, the operator norm and strong operator topology coincide with the strong topology on , and the weak operator norm topology coincides with the weak topology on . Meanwhile, coincides with , and the operator norm topology coincides with the strong topology on , while the strong and weak operator topologies correspond with the weak* topology on .

We can rephrase the uniform boundedness principle for convergence (Corollary 1 from Notes 9) as follows:

Proposition 7 (Uniform boundedness principle)Let be a sequence of bounded linear operators from a Banach space to a normed vector space , let be another bounded linear operator, and let be a dense subspace of . Then the following are equivalent:

- converges in the strong operator topology of to .
- is bounded in the operator norm (i.e. is bounded), and the restriction of to converges in the strong operator topology of to the restriction of to .

Exercise 25Show that the conclusion of Proposition 7 continues to hold if “strong operator topology” is replaced by “weak operator topology” throughout.

Exercise 26Show that the operator norm topology, strong operator topology, and weak operator topology, are all Hausdorff. As these topologies are nested, we thus conclude that it is not possible for a sequence of operators to converge to one limit in one of these topologies and to converge to a different limit in another.

Example 8Let , and for each , let be the translation operator by : . If is continuous and compactly supported, then (e.g. from dominated convergence) we see that in as . Since the space of continuous and compactly supported functions is dense in , this implies (from the above proposition, with some obvious modifications to deal with the continuous parameter instead of the discrete parameter ) that converges in the strong operator topology (and hence weak operator topology) to the identity. On the other hand, does not converge to the identity in the operator norm topology. Indeed, observe for any that , and thus .In a similar vein, does not converge to anything in the strong operator topology (and hence does not converge in the operator norm topology either) in the limit , since (say) does not converge strongly in . However, one easily verifies that as for any compactly supported , and hence for all by the usual limiting argument, and hence converges in the weak operator topology to zero.

The following exercise may help clarify the relationship between the operator norm, strong operator, and weak operator topologies.

Exercise 27Let be a Hilbert space, and let be a sequence of bounded linear operators.

- Show that in the operator norm topology if and only if for any bounded sequences .
- Show that in the strong operator topology if and only if for any convergent sequence and any bounded sequence .
- Show that in the weak operator topology if and only if for any convergent sequences .
- Show that in the operator norm (resp. weak operator) topology if and only if in the operator norm (resp. weak operator) topology. Give an example to show that the corresponding claim for the strong operator topology is false.

There is a counterpart of the Banach-Alaoglu theorem (and its sequential analogue), at least in the case of Hilbert spaces:

Exercise 28Let be Hilbert spaces. Show that the closed unit ball (in the operator norm) in is compact in the weak operator topology. If and are separable, show that is sequentially compact in the weak operator topology.

The behaviour of convergence in various topologies with respect to composition is somewhat complicated, as the following exercise shows.

Exercise 29Let be a Hilbert space, let be sequences of operators, and let be another operator.

- If in the operator norm (resp. strong operator or weak operator) topology, show that and in the operator norm (resp. strong operator or weak operator) topology.
- If in the operator norm topology, and is bounded in the operator norm topology, show that and in the operator norm topology.
- If in the strong operator topology, and is bounded in the operator norm topology, show that in the strong operator norm topology.
- Give an example where in the strong operator topology, and in the weak operator topology, but does not converge to zero even in the weak operator topology.

Exercise 30Let be a Hilbert space. An operator is said to be finite rank if its image is finite dimensional. is said to be compact if the image of the unit ball is precompact. Let denote the space of compact operators on .

- Show that is compact if and only if it is the limit of finite rank operators in the operator norm topology. Conclude in particular that is a closed subset of in the operator norm topology.
- Show that an operator is compact if and only if is compact.
- If is separable, show that every is the limit of finite rank operators in the strong operator topology.
- If , show that maps weakly convergent sequences to strongly convergent sequences. (This property is known as
complete continuity.)- Show that is a subspace of , which is closed with respect to left and right multiplication by elements of . (In other words, the space of compact operators is an two-ideal in the algebra of bounded operators.)
The weak operator topology plays a particularly important role on the theory of von Neumann algebras, which we will not discuss here. We will return to the study of compact operators next quarter, when we discuss the spectral theorem.

[

Update, Feb 23: Corrections, another exercise and remark added (note renumbering).]

## 127 comments

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26 February, 2017 at 6:58 am

Variational methods and Weak Topologies: Part 1 | Knowing the odds[…] 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 […]

17 April, 2017 at 3:32 pm

keejRegarding this comment in Remark 5: “…the weak* topology on the dual of a separable space is metrisable.” This is not quite true, right? Wouldn’t this imply that the weak topology on every separable Hilbert space is metrizable?

[Yes, this is a special case of Remark 5. -T.]19 April, 2017 at 11:11 pm

keejDear Professor Tao, thanks for your reply, but I am still confused… I should have mentioned this in the original comment, but I think a well-known example is the set in a Hilbert space with countably infinite basis . The weak closure of contains , but no sequence in converges to . So the weak topology cannot be metrizable. Am I still making a mistake somewhere?

20 April, 2017 at 2:08 pm

Terence TaoAh, I see the issue now. It is the bounded subsets of a normed vector space with the weak* topology that are metrisable; as you point out, unbounded sets need not be metrisable. (This was fixed in the published version of these notes, but not in this blog post, but I will correct this now.)

13 June, 2018 at 12:32 am

Mike NeelyIn Exercise 16, should “completeness” be added? Else we can

define V as the set of all real-valued infinite sequences with a finite

number of nonzero entries, define ||x||^2=sum_{i=1}^{\infty} x_i^2,

and E = {(1,0,0,…), (0,1,0,0…), (0,0,1,0,…), …} which is strongly

bounded but not weakly.

13 June, 2018 at 9:51 am

Terence TaoAll strongly bounded sets are weakly bounded (note that linear functionals in the dual space are, by definition, bounded – we are not referring here to the algebraic dual).

13 June, 2018 at 10:36 am

Mike NeelyThanks for your reply! I likely am using “algebraic dual” unwittingly, since I don’t know another definition of V*. If we use algebraic dual, it seems E is not weakly bounded since we can consider the linear functional f:V—>R defined by mapping the basis vectors

{e_1, e_2, e_3, …} as follows: f(e_k)=k for all k in {1, 2, 3, …}.

This maps every sequence in V to a real number (since sequences

in V have at most a finite number of nonzeros). And so

f(E) = {f(e_1), f(e_2), f(e_3), …} = {1,2, 3, …}

is not a bounded set. Is this correct reasoning if we use “algebraic dual space”? The wiki page here defines “dual space” without the boundedness condition you give: https://en.wikipedia.org/wiki/Dual_space

14 June, 2018 at 7:43 am

Terence TaoAs stated in the second paragraph of the wikipedia page you link, in the category of topological vector spaces, one usually works with the continuous dual space instead of the algebraic dual space, in which the linear functional is required to be continuous (or equivalently, bounded). See also Remark 1 in the previous notes https://terrytao.wordpress.com/2009/01/26/245b-notes-6-duality-and-the-hahn-banach-theorem/ . Certainly many results in functional analysis involving the continuous dual space will break down if the algebraic dual space is used instead, and as such the latter is rarely used in this subject.

22 August, 2020 at 1:31 am

Kenneth HarrisFor proposition 7, do we need Y to be complete also? Otherwise doesn’t the following counterexample hold?

https://math.stackexchange.com/questions/690326/uniform-boundedness-principle-for-norm-convergence

23 August, 2020 at 5:19 am

Kenneth HarrisPlease ignore the last question! The fact that the existence of T is stated in the hypothesis rules out this as a counterexample, and means we don’t need to require Y complete.

18 October, 2020 at 6:57 am

AnonymousIn definition 1, if instead of joint continuity, one uses separate continuity, does one still has ?

18 October, 2020 at 7:58 am

Terence TaoNo. For instance if one gives the cofinite topology then the sequences and both converge to 0 (as well as to every other real number) but their sum does not. Granted, in this particular case scalar multiplication is also discontinuous, but one can fix this by e.g., working with an infinite-dimensional vector space endowed with the topology where the non-empty open sets are the complement of subsets of finite-dimensional subspaces that are closed in the natural topology of those subspaces, and then working with and rather than and .

31 March, 2021 at 1:15 am

SK RayamajhiRegarding example 8: What about operator $T_{B_t} f(x) = f(B_t(x))$ where $B_t$ is some mapping in domain of function.

6 April, 2022 at 8:08 am

Anonymousdear prof. Tao, in exercise 21, what does it mean to say that the weak* topology on is complete? similarly, in exercise 23, i don’t understand the statement that the weak topology in is not complete. thank you.

[See https://en.wikipedia.org/wiki/Complete_topological_vector_space -T]7 April, 2022 at 8:35 pm

Anonymousthank you for your help! i was confused partially because the paragraph before remark 1 defined completeness in TVS using Cauchy sequences, which seems to usually be called sequential completeness in the literature. the claim in exercise 21 seems to hold for sequential completeness but not for completeness (as defined using Cauchy nets as in the Wikipedia article)

some useful links for anyone else who gets stuck here:

https://math.stackexchange.com/questions/1863811/is-x-complete-with-weak-topology

https://math.stackexchange.com/questions/3161333/on-weak-sequentially-completeness

Click to access incompleteness.pdf

31 December, 2022 at 5:45 am

AnonymousHas a topology of been defined somewhere in this set of notes? Maybe another good example after Example 4.

2 January, 2023 at 8:21 am

AnonymousIn Exercise 3, if for every , in , then since is _part_ of , it is trivial to see that . Why is the inverse also true? is generated by the union instead of the intersections of . Why it can’t be true that in but in for only one ?

7 January, 2023 at 7:54 am

Terence TaoYour implications are the wrong way around: in implies in , but not conversely. (I recommend writing out these convergence statements explicitly using quantifiers.)

7 January, 2023 at 2:11 pm

AnonymousAh, I somehow messed up with the implications. The trivial one should be implies for every , which follows from the fact that each is part of . The other direction follows from the fact that the collections of all the open sets in all form a sub-basis of .

1. Suppose and . If is any open set in that contains , then since is also open in , it must contain for all sufficiently large . This completes the trivial direction.

2. Now suppose in for every . Suppose for an open set in . Since is generated by the collection (), can be written as a union of finite intersections of open sets from and thus particularly, where is a finite subset of and is an open set in . By the assumption, contains for all sufficiently large . The proof is done.

2 January, 2023 at 9:15 am

AnonymousIn Exercise 4, suppose for each , there exists a finite subset of and a constant such that for all .

To show that is continuous, it suffices to show that $latext T^{-1}(B)$ is open in $V$ for every ball since the inverse commutes with union and intersections and the balls form a subbasis.

Let be a ball. One can write for some r,

How can one write this as a union of finite intersections of balls in ? I expect that one can use the finite set in the assumption. But I don't see how to construct the desired balls.

7 January, 2023 at 2:24 pm

AnonymousIn Folland’s book, the author uses nets. I am wondering if one can work Exercise 4 out from basic definitions of continuity of maps between topological spaces.

Can one drop the “linear” assumption?

[Exercise 4 fails for nonlinear maps, even from to . You should be able to find a counterexample without difficulty – T.]7 January, 2023 at 2:29 pm

AnonymousIs it by default assumed that and are topological vector spaces with the mentioned topologies?

[This follows from the remark at the end of Exercise 3 -T.]8 January, 2023 at 7:05 am

AnonymousThe question regarding Exercise 4 earlier was not summarized well, and I’d like to rephrase it here.

Exercise 4 is a generalization of “Lemma 1 (Continuity is equivalent to boundedness for linear operators)” in Notes 3. One can try to adapt the argument to prove Exercise 4.

The implication that being continuous implies the inequality for seminorms seems easy using homogeneity as in “Lemma 1”:

If is continuous, then it is continuous at particularly. If is a neighborhood of , then contains an open neighborhood of and is a finite intersection of balls by the seminorms in . So there exists and finitely many seminorms in such that

whenever

One can then work with homogeneity as in "Lemma 1" to get .

The other implication seems more tricky. Suppose the seminorm inequalit(ies) are true. Then by linearity, as in "Lemma 1", to show that is continuous, it suffices to show that is continuous at . In order to do that, here is the question: how could one show that

is an open set in ? (When the finite set contains only

oneelement, this is rather trivial though.)12 January, 2023 at 9:44 am

Terence TaoEach seminorm is a continuous function, so any finite sum of these seminorms is also continuous.

12 January, 2023 at 5:22 pm

AnonymousSorry, I may misunderstand your comment, but I seem to get something circular then: in order to show that any finite sum of these seminorms is also continuous in the first place, I need to show that

that

is an open set in , which what is to be proved.

13 January, 2023 at 6:17 am

Terence TaoThe sum of any finite number of real-valued continuous functions is automatically continuous. (This is ultimately because the addition map defined by is continuous.)

13 January, 2023 at 3:41 pm

AnonymousIf one takes two seminorms, and , then in order to show the continuity of using the continuity of the addition map of real numbers, I think I need to show first that is continuous. But then I end up needing to show that the sets of the form

are open in , which is quite the same as what is to be proved… What do I miss?

13 January, 2023 at 7:37 pm

AnonymousAh, never mind. I was lost. That should follow from properties of the product topology on .

14 January, 2023 at 7:08 pm

AnonymousDr. Tao: How do you have so much patience answering such simple questions? People shouldn’t abuse this blog.

2 January, 2023 at 1:39 pm

AnonymousExercise 13 and Theorem 3 in Section 3 in this set of notes use the norm on . But the notion of operator norms, of which the norm on is a special case, is not defined until Section 4. Do I miss anything? Was it defined or mentioned already before this set of notes?

[See Notes 6. -T]15 January, 2023 at 5:12 am

AnonymousInstead, one often first works in a weaker topology, in which compactness is easier to establish, and then somehow upgrades any weakly convergent sequences obtained via compactness to stronger modes of convergence …

In this situation, could one have worked (in principle maybe) at the very beginning directly with the stronger topologies that give the “stronger modes of convergence” by rearranging the arguments without referring to any “weak” topologies?

15 January, 2023 at 8:21 pm

AnonymousIt is usually the case that stronger topologies don’t work, then we go to weak topologies.

30 January, 2023 at 6:01 am

AnonymousIn Exercise 26, how does one know that the three topologies are nested? The chain of inequalities:

shows that sub-basic sets in these topologies are nested. For instance,

where . Does this imply that the operator norm topology is finer than the strong operator topology?

In general, if one has two topologies and with subbases and respectively, and if for every there exists such that , can one conclude that is finer than ?

30 January, 2023 at 10:57 am

AnonymousYes. What you wrote is correct.

1 February, 2023 at 4:46 am

AnonymousIn Exercise 10, is it less useful to define the open sets in using the forward image of open sets (in ) under the map ?

1 February, 2023 at 9:42 am

AnonymousWe want it to be a topology, not sure what will one achieve by using the forward image of open sets.

3 February, 2023 at 7:24 pm

JTo show the continuity of addition in V/W for Exercise 10, suppose let and is an open set that contains . Then is in the open set . By continuity of addition in one has open sets in such that and .

To show the continuity of the addition of V/W, one needs an open set in that contains and the image of under addition is contained in . How can one get the desired ?

3 February, 2023 at 7:29 pm

JBy chasing the maps, one can see that and it contains . But unfortunately, it is not necessarily open.