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).]

## 101 comments

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22 February, 2009 at 12:31 am

AnonymousExercise 15: I thought in l^1 a sequence converges iff it converges weakly? The first statement is true for 1 < p < oo.

22 February, 2009 at 8:48 am

EricThe answer to the Question after Corollary 5 is yes. More generally, the predual of any (norm-)separable Banach space is separable. Indeed, if V is not separable, then by induction we can construct a well-ordered strictly increasing sequence of closed subspaces of order type . By Hahn-Banach, the orthogonal complements of these spaces are a strictly decreasing sequence of closed subspaces of V*. But this is impossible in a second-countable space, so this implies that V* cannot be norm-separable.

22 February, 2009 at 12:13 pm

Terence TaoThanks for the correction and answer to the question!

30 October, 2020 at 11:18 am

AdamDr. Tao, I have question please, how to prove that the weak zero neighborhood {x \in l_1 : | <= 1 } is unbounded with respect to the norm ||.||_1 ?

Here, w is in l_{infty} \ {0}.

23 February, 2009 at 12:16 pm

anonHow does Banach-Alaoglu not contradict F. Riesz’s Theorem that a TVS is locally compact if and only if it is finite-dimensional?

23 February, 2009 at 12:26 pm

EricRe anon: Banach-Alaoglu says that the

normunit ball in a dual space is weak* compact. This does not say that the space is locally compact in the weak* topology because the unit ball has empty interior in the weak* topology.23 February, 2009 at 1:52 pm

Matt DawsFor Remark 2, consider , which has many, many preduals. Indeed, let be any countable compact space, and consider . The dual is . As $K$ is countable, and as any measure is countably additive, in the canonical way. Then if is a sequence in with say, then in the weak*-topology in , where is the point mass at . So by choosing non-homeomorphic and , we have found preduals for which induce distinct weak*-topologies.

Indeed, preduals of can be very weird indeed. The Argyros-Haydon space is an example!

23 February, 2009 at 2:53 pm

Terence TaoThanks for the example and the comments!

23 February, 2009 at 6:20 pm

less than epsilonIn example 5, is summation index from 0 to k or 0 to infinity?

23 February, 2009 at 6:35 pm

less than epsilonBetween exc 13 and remark 3: ” a seq in V converges in the weak star topology……”

should it be ” a seq in V star …….” ?

23 February, 2009 at 8:51 pm

Terence TaoThanks for the corrections!

28 February, 2009 at 1:20 am

AnonymousThis is in relation to exercise 23, part 2. In class, we showed that pointwise convergence implies by the UBP that the are bounded, and immediately concluded that we get weak convergence. Isn’t a step through exercise 17 needed, to show that is finite so that is in ? Thanks.

28 February, 2009 at 9:17 am

Terence TaoWell, in the exercise is assumed to be in already. But yes, one could use Exercise 17 (which is proven using the UBP in any case); also, from Fatou’s lemma, the pointwise limit of sequences bounded in is automatically in .

28 February, 2009 at 2:15 pm

Yasser TaimaIn the proof of Theorem 3 (Banach-Alaoglu theorem), is it necessary to take the unit ball closed ? To identify with a subset of , I believe if I take the map defined as , then by linearity of , implies , hence in . This seems to work for the closed or open.

Also, the reference to exercise 27 in Notes 10, in the proof of the sequential Banach-Alaoglu theorem might well be to exercise 28 instead. Thanks.

28 February, 2009 at 2:42 pm

Terence TaoDear Yasser: Thanks for the correction! It is true that one can take B to be either closed or open for the proof. (It is essential, though, that be closed.)

2 March, 2009 at 12:48 am

etaleIn exercise 22 :”Show that {W} is closed in the weak topology of {W}. ”

Should “closed” be replaced by some other concepts like “complete”? A space is trivially closed in any topology on it by definition.

2 March, 2009 at 8:17 am

245B, Notes 12: Continuous functions on locally compact Hausdorff spaces « What’s new[…] and -compact. As is equivalent to the dual of the Banach space , it acquires a weak* topology (see Notes 11), known as the vague topology. A sequence of Radon measures then converges vaguely to a limit if […]

2 March, 2009 at 9:09 am

Terence TaoDear Etale: Oops, the exercise is meant to state that W is closed in the weak topology of V. It’s corrected now.

6 March, 2009 at 2:35 am

Sebastian ScholtesDear Prof. Tao,

I have a question regarding Exercise 8, if the statement was true the following argument would be valid (if I’m not mistaken):

Let in , every subsequence also converges in , such that there is a subsubsequence which converges pointwise almost everywhere. With a subsubsequence convergence principle (sometimes called Urysohn’s principle) for topological spaces one concludes for the whole sequence almost everywhere, which we know is false, i.e. [Counterexamples in Analysis, Gelbaum, Olmsted, Example 40, p.111 (iii) (i)].

I would be very grateful if you could clarify whatever is wrong with my reasoning or the exercise, since I don’t need another sleepless night (-; .

Best regards

Sebastian Scholtes

6 March, 2009 at 7:28 am

Terence TaoDear Sebastian,

Hmm, you’re right. I was trying to glue together the various notions of uniform convergence outside of small sets as given by Egoroff’s theorem, but I didn’t get the quantifiers right and indeed this does not generate a topology, by your Urysohn principle argument. So I’ve replaced the exercise with its negation. :)

28 March, 2009 at 10:13 pm

liuxiaochuanDear Professor Tao:

The last bracket of exercise 16, it should be , instead of

[Corrected, thanks – T]23 April, 2009 at 7:44 am

PDEbeginnerDear Prof. Tao,

I guess in example 2 ‘‘should be ‘‘.

[Corrected, thanks – T]29 May, 2010 at 12:12 am

AnonymousDear Prof. Tao

I have a question about the compactness of the closed unit ball of a dual space.I want to know wheather or not it’s compact in the strong operator topology.what’s the reason?

Best regards

Jafar soltani farsani

16 September, 2010 at 8:51 am

Daniel MckenzieHey prof. Tao

Not that it makes much difference, but shouldn’t the statement ‘the weak-$latex*$ topology on the dual of a separable space is metrisable’ read ‘the weak-$latex*$ topology on the dual of a separable space is metrisable on norm-bounded subsets of ‘?

[Corrected, thanks – T.]regards

Daniel

26 October, 2010 at 5:14 pm

245A, Notes 4: Modes of convergence « What’s new[…] not attempt to exhaustively enumerate these modes here (but see this Wikipedia page, and see also these 245B notes on strong and weak convergence). We will, however, discuss some of the modes of convergence that arise from measure theory, when […]

2 November, 2010 at 3:34 am

JeanDoes the weak* convergence in {L^\infty} implies the convergence almost everywhere (up to a subsequence if necessary)?

2 November, 2010 at 7:25 am

Terence TaoNo. For instance, on the torus , the exponential functions converge in the weak-* sense to zero (the Riemann-Lebesgue lemma), but clearly do not converge almost everywhere to zero, even after taking subsequences. The point is that weak-* convergence allows for an “escape to frequency infinity” that is not convergent under any convergence method based on the absolute value rather than on the phase.

9 November, 2010 at 9:01 am

245A, Notes 3: Integration on abstract measure spaces, and the convergence theorems « What’s new[…] The claim then follows by another appeal to the definition of lim inf. Remark 12 Informally, Fatou’s lemma tells us that when taking the pointwise limit of unsigned functions , that mass can be destroyed in the limit (as was the case in the three key moving bump examples), but it cannot be created in the limit. Of course the unsigned hypothesis is necessary here (consider for instance multiplying any of the moving bump examples by ). While this lemma was stated only for pointwise limits, the same general principle (that mass can be destroyed, but not created, by the process of taking limits) tends to hold for other “weak” notions of convergence. We will see some instances of this in 245B. […]

13 December, 2010 at 12:51 pm

AnonymousDear Prof. Tao,

I’m greatly confused about Example 3, in particular about continuity of multiplication with respect to such topology. If and , it seems to me that for arbitrary , for any r. Doesn’t it imply than multiplication is not continuous, since has no neighbourhood which is mapped into ? It’s probably just my misunderstanding, but I would really appreciate pointing out my mistake.

Best regards,

Marcin Łoś

13 December, 2010 at 7:51 pm

Terence TaoAck, you’re right. I’ve rewritten the example accordingly.

24 December, 2010 at 2:29 am

AnonymousDear Prof. Tao,

Thank you for taking time to reply. I believe there might be a problem with ex. 19, 3rd dot. If we take any unit vector , the sequence is Césaro-convergent to 0 (since norm of a partial sum is bounded by 1), but surely there is no subsequence weakly convergent to 0. Sorry if it’s just some stupid misunderstanding. If it’s not, is it still possible to prove the 4th dot?

Merry Christmas,

Marcin Łoś

24 December, 2010 at 9:55 am

Terence TaoOops, you’re right, the third item is not correct and should be deleted. The fourth item still seems correct to me though (the main trick in proving the “only if” direction being to normalise x to zero and then to use weak convergence choose a subsequence where the vectors are approximately orthogonal to each other).

13 March, 2011 at 1:02 pm

AnonymousI’m confused by the concepts here. It seems that the strong closure of each set is included in its weak closure. So one can say “weakly closed sets are strongly closed” or “strongly closed sets are weakly closed”?

13 March, 2011 at 1:11 pm

Terence TaoYes, and you should be able to work out which way the inclusion goes. (Hint:

a good example to settle one’s intuition here is the unit sphere in, say, .)

13 March, 2011 at 1:52 pm

AnonymousHmm, maybe the standard orthonormal basis in is enough? So is weakly closed and also strongly closed, but is only strongly closed not weakly closed?

13 March, 2011 at 3:11 pm

AnonymousYes :-), I think so.

4 May, 2011 at 11:53 am

AnonymousDear Prof. Tao,

I think that the dual space of is not complete under the weak topology (since we can approximate the elements in by under this topology), and that we can apply the Banach-Alaoglu Thm to .

But you mentioned that one should have complete topology if he wants to apply Banach-Alaoglu Thm. For this reason, I am a little bit doubted if I am right.

Thanks a lot!

4 May, 2011 at 12:07 pm

Terence TaoCompleteness is a notion that only makes sense for metric spaces, not for topological spaces. The weak-* topology is not metrisable, and so there is no meaningful way to discuss completeness (unless one uses the notion of completeness for uniform spaces or something, but this does not seem to be your intention here).

Note also that the weak-* topology on is distinct from the weak-* topology on , because convergence in the former requires convergence when evaluated at all members of , whereas convergence in the latter only requires convergence when evaluated at all members of the smaller space .

4 May, 2011 at 12:48 pm

AnonymousDear Prof. Tao,

Many thanks for your prompt reply! I got some point!

I am still a little confused on Exercise 21, the weak* topology of the dual of Banach space is complete.

Could you give some explanation for this? It seems the completeness of the functional space plays an important role. Thanks!

28 September, 2015 at 3:45 pm

AnonymousDoes the article in wikipedia (https://en.wikipedia.org/wiki/Complete_topological_space) suggest that people can still define the notion of completeness in a topological space using Cauchy nets?

[If the topological space comes with the structure of a Cauchy space, then yes. -T.]8 September, 2011 at 2:11 pm

254A, Notes 2: Building Lie structure from representations and metrics « What’s new[…] norm on the vector space. However, not every topological vector space is generated by a norm. See these notes for some further […]

17 February, 2012 at 1:29 pm

RexDoes the “algebraic topology” of Exercise 9 actually arise in some natural context (if so, where?), or is it something cooked up solely to serve as a counterexample?

17 February, 2012 at 2:39 pm

Terence TaoThe notion of algebraic openness is natural when considering geometric versions of the Hahn-Banach theorem (see Theorem 4 of Notes 6). But outside of that, I don’t know of any other major application of this topology.

17 February, 2012 at 4:25 pm

RexIncidentally, it might be worthwhile to mention the Zariski topology, which is also another topology (the simplest and most natural I can think of) on or that does not make it into a topological vector space.

10 July, 2012 at 12:19 pm

Anonymousin Remark 1, on line 3 from bottom, “compacta” should be “compact”.

[Corrected, thanks. -T]12 July, 2012 at 7:59 am

anonI do not understand the last part of Exercise 13: “Show that the weak* topology on the dual is weaker than the weak topology on the dual (which is defined using the double dual)”.

But how do we define the double dual (V*)* if the dual V* does not have a topology yet?

12 July, 2012 at 9:46 am

Terence TaoSorry, in the last two parts of that question V should also be assumed to be a normed vector space; the question has been updated accordingly.

13 July, 2012 at 1:11 pm

anonOk, thanks for the reply!

3 December, 2013 at 6:03 pm

Linear elliptic equations with measure data I | Partial Differential Australians[…] and we say that a sequence in converges to in weak- if for all . Further details can be found in Terry Tao’s notes on the subject. […]

12 May, 2014 at 7:43 am

André CaldasDear professor,

In Exercise 12, item 1, if you consider the space spawned by and , then the problem is reduced to the finite-dimensional normed space , right? It seems to me that the conclusion is wrong. In special, because of your comment: “in contrast to the situation with Hilbert spaces”. But is a Hilbert space, right?

12 May, 2014 at 10:03 am

Terence TaoUniqueness can fail for some normed spaces V (including some finite dimensional spaces), but can hold for other spaces (such as Hilbert spaces). Uniqueness is

necessarilytrue if V is a Hilbert space, but isnot necessarilytrue if one only assumes that V is a normed vector spaces.12 May, 2014 at 10:24 am

André CaldasSorry! I fell for the “all norms are equivalent” cavet! Very silly of me… the example is very easy to construct in dimension two and the supremum norm.

Thank you very much for your time.

21 November, 2014 at 10:29 am

AnonymousSome people define the weak topology on with respect to as the weakest topology on so that all the linear functional remain continuous in this new topology. Is this definition equivalent to the one in Definition 2?

21 November, 2014 at 4:21 pm

Terence TaoYes; I recommend verifying this as an exercise.

4 March, 2015 at 8:12 pm

AnonymousIn the proof of Theorem 3 (Alaoglu):

one can identify with a subset of , the space of functions from to.For “identify”, do you mean a “homeomorphism” (topological isomorphism) or just a “bijection” (set isomorphism) ?

If one defines by . Then is bijective.

One easily verifies that the weak* topology on is nothing more than the product topology of restricted to .Let be the weak* topology on and be the product topology of restricted to . One can show that a net converges to a in if and only if converges to in . Does it suggest that these two topologies are the “same”? (By Exercise 13 in Note 8, this only suggest that is continous.) As I’ve seen from Note 8, “nets” are in the “optional” section. Do you suggest that there is an alternative way to show that and are the same?

5 March, 2015 at 8:17 am

Terence TaoInitially, the identification is only on the level of sets (i.e. as a bijection), but once one verifies that the topologies are compatible (in your notation) then the identification is at the level of topological spaces (i.e. as a homeomorphism).

Exercise 13 (combined with the fact that one has an “if and only if” rather than just an “only if” in the relation between convergence of nets) shows that f and its inverse are both continuous, i.e. that f is a homeomorphism. One can also demonstrate the homeomorphism property by considering the inverse image or forward image of sub-basic sets.

5 March, 2015 at 1:16 pm

AnonymousFor “sub-basic” sets, do you mean sets in a subbasis of a topological space? (People use different spelling such as “subbase”, “sub-base”, etc. I’m usually confused about which one is grammatically correct.)

If we use and consider instead of , I guess we could avoid working back and forth between and since is a subset (the elements of which have the linear property) of . Still need to check the topology though, it might be slightly faster.

12 October, 2015 at 7:24 am

AnonymousSub-basic sets for can be with open in . Sub-basic sets for are with open in and . Would you elaborate how these two things can be used to show the homeomorphism?

12 October, 2015 at 8:34 am

Terence TaoActually, the subbasic sets for are not of the form (which lives in the wrong space, namely B), but rather of the form for and open in .

12 October, 2015 at 10:08 am

AnonymousAh, so the subbasic sets for are not be of the form , but rather of the form .

12 October, 2015 at 10:32 am

AnonymousFor , the subbasic sets are of the form

for and open in . For , the subbasic sets are of the form for and open in . Denotes these two subbases as and .

To show is an open map, it suffices to show . To show is continuous it suffices to show that .

Let be open in and . How can one show that the image of this subbasic set is in ? (Since is not necessarily in , I don’t see how one can make the argument here…)

12 October, 2015 at 11:20 am

Terence TaoEven if is not in , some positive scalar multiple of will lie in . One can then proceed using the homogeneity property of linear functionals.

12 October, 2015 at 1:31 pm

AnonymousIf , then the image is

Now I have a problem that is not necessarily open in and is not necessarily in …

12 October, 2015 at 3:36 pm

Terence TaoA set that is open in D is the intersection with D of a set which is open in . So one should really just work with open sets in , which behave well with respect to dilations by or . (The restriction to D is already implicit in the definition of B and B^* anyway.)

26 October, 2015 at 11:47 am

AnonymousI eventually come up with a proof, which looks very ugly. I think that might be a reason why people prefer an argument via net convergence.

Consider in the open set for some fixed and open in . Note that

which is the same(!) as

since and . Note that this is a subbasic set in . It follows that is continuous (exercise).

Now we show that is an open map, which gives the continuity of . Consider in the open set

for and open in . Note that for some and . Then

with , which is also open in . Since , for each , we have

It follows that

and thus

which is a subbasic set in . Since is a bijection, implies that is an open map.

21 October, 2015 at 7:56 am

AnonymousIs the form of the subbasic sets for also wrong? Should it be (with open in ) instead?

[Yes – T.]21 October, 2015 at 10:46 am

AnonymousMaybe I’m still looking at things in a stupid wrong way. It makes sense now for me that for , the subbasic sets are of the form

for and open in . For , the subbasic sets are of the form for and open in .

Now consider in the open set for some fixed and open in . Why is open in (which shows that is continuous)? The operator is rather confusing.

21 October, 2015 at 11:53 am

Terence Taois the set , which is a subbasic set in the topology. (Note that a linear functional on V is uniquely determined by its restriction to B.)

21 October, 2015 at 12:52 pm

AnonymousAh, one should really write as

for and open in . Then is

which is the same(!) as

since and .

(I don’t know what is the comment “Note that a linear functional on V is uniquely determined by its restriction to B.” used for though.)

Thanks a lot!

31 March, 2015 at 2:49 pm

Sinanice and systematic explanation of topological vector spaces and weak norms. Was very useful as all these concepts were kind of blurry in my head. It is much more neater in my head now, thanks.

13 September, 2015 at 5:32 pm

AnonymousCan boundedness be defined in a topological vector space?

9 October, 2015 at 8:38 pm

AnonymousAre the strong and weak topologies in this post essentially the final topology (https://en.wikipedia.org/wiki/Final_topology) and initial topology (https://en.wikipedia.org/wiki/Initial_topology) defined in Wikipedia?

10 October, 2015 at 5:32 am

AnonymousThe weak one is quite obvious. I’m sure about the

strongtopology though.10 October, 2015 at 12:00 pm

Terence TaoWell, at a tautological level,

anytopology on a space X is an initial topology, namely the initial topoology corresponding to the identity map from to . So the strong topology is an initial topology for trivial reasons. But I don’t think there is any especially interesting way to think of the strong topology as an initial topology.There is however the notion of an algebraically open set (see Notes 6) which is openness in the initial topology coming from the collection of affine maps from the real line into a vector space. This topology is far stronger than either the strong or weak topologies.

10 October, 2015 at 7:20 am

AnonymousI have checked throughout the 245B notes, but I don’t see the definition of “the topology generated by the seminorms”. One can use

anorm (seminorm) generate a topology by considering the metric (semimetric) defined by this norm and then the metric topology. What is going on with a family of seminorms? Would you elaborate?10 October, 2015 at 11:57 am

Terence TaoThe definition of the topology generated by a family of seminorms is given in Exercise 3 of this post (i.e. it is the weakest topology that contains all the topologies of each single seminorm).

19 October, 2015 at 7:30 am

AnonymousI saw some text defines the weak topology on as the topology generated by the seminorms for all finite . Is this definition equivalent to Definition 2?

[Yes, basically because open sets are closed under the operation of taking finite intersections; it’s a good exercise to work this out in detail. -T.]18 December, 2015 at 8:04 am

AnonymousGive the topology induced by a family of semi-norms . Would you elaborate why the map

(for fixed and )

is continuous from to ?

Denote the map above as . Consider an open set in for instance. Why is open in ?

[Use the triangle inequality to show that every point in contains a basic open neighbourhood that is contained in . -T.]18 December, 2015 at 1:19 pm

AnonymousLet for some fixed .

Explicitly, we have

.

Suppose is a point in , then . Want to find such that

implies

By the triangle inequality,

.

I don’t see how to go on…

[Choose small enough that . -T.]18 December, 2015 at 9:23 am

AnonymousLet be a vector space with a norm . Define for each with

.

Let be the initial topology with respect to the family . I’m trying to show that this is the same topology as the one in Exercise 1.

The continuity of shows that the ball (in exercise 1)

where is open. To show forms a base for , let be an open nbhd at . How can I find contained in ?

18 December, 2015 at 10:32 am

AnonymousLet be a vector space and be an arbitrary function. One define a “ball” as

.

What makes such topology less interesting/useful than the case when is a seminorm?

[Use < and > instead of < and > to avoid portions of text being interpreted incorrectly as HTML and thus lost. -T.]13 January, 2016 at 4:15 am

James RobinsonIn Remark 5, should that be “no sequential weak-* compactness when , i.e. you consider the closed unit ball in ? The closed unit ball in is sequentially weak-* compact since is separable. (I was very worried for a while about this!)

[Corrected, thanks – T.]1 February, 2016 at 12:47 pm

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 (or alternatively, one abandons strong convergence and exploits the weak convergence directly).Why could one “somehow upgrades any weakly convergent sequences obtained via compactness to stronger modes of convergence”? Is this similar to upgrading the regurality of weak solutions in PDE? Do you have a simple example to illustrate this philosophy?

1 February, 2016 at 2:22 pm

Terence TaoYes, elliptic regularity is the type of upgrading that I had in mind, in which the sequence (or its limit) is known to obey some additional properties, such as being a solution of an elliptic PDE. For instance, one way to solve the Dirichlet problem on a domain is to take a sequence of approximate minimisers to the Dirichlet energy functional, use weak compactness to extract a weak limit, show that this limit is weakly harmonic, and then use elliptic regularity to show that it is smooth. Using this and the strong convexity of the Dirichlet functional, one can then show that the minimising sequence converged strongly in H^1 rather than just weakly along a subsequence.

8 March, 2016 at 6:40 pm

AnonymousWhen is a normed space, one has two topologies for , the norm topology and the weak-* topology. Which one is used when one defines ?

[By default, the double dual is defined using the norm topology on the dual. One could potentially define an alternate double dual using the weak-* topology, but this space is almost never used. -T.]22 April, 2016 at 8:47 am

A quick application of the closed graph theorem | What's new[…] In functional analysis, it is common to endow various (infinite-dimensional) vector spaces with a variety of topologies. For instance, a normed vector space can be given the strong topology as well as the weak topology; if the vector space has a predual, it also has a weak-* topology. Similarly, spaces of operators have a number of useful topologies on them, including the operator norm topology, strong operator topology, and the weak operator topology. For function spaces, one can use topologies associated to various modes of convergence, such as uniform convergence, pointwise convergence, locally uniform convergence, or convergence in the sense of distributions. (A small minority of such modes are not topologisable, though, the most common of which is pointwise almost everywhere convergence; see Exercise 8 of this previous post). […]

19 June, 2016 at 2:11 pm

NickWith respect to Ex. 5, is it not straightforward that the statement is true even in the box topology? Say is your infinite product and you have a box neighborhood of , then for each simply take neighborhoods of respectively such that , and then take products. This shows continuity of the addition, and scalar multiplication is clear. Am I doing something wrong here?

26 June, 2016 at 4:15 am

NickIn Exercise 26 I think it is not necessary to assume is a Banach space. The second statement applies the first by a direct limiting argument. Conversely, in order to deduce boundedness of the operator norms from weak operator convergence, it is enough to apply Ex. 16 together with the uniform boundedness principle, neither of which assumes that is complete.

[Fair enough; I’ve edited the exercise accordingly. -T.]13 November, 2016 at 2:30 pm

IanProfessor Tao:

In exercise 16, is the suggested use of the Hahn-Banach Theorem just to verify that the norm of an element $x \in V$ is the same as the norm of its corresponding functional $\iota (x) \in V^{**}$, or is this Theorem used in another substantial way?

[This is basically where HB is used, yes -T.]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 .