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245B, Notes 11: The strong and weak topologies21 February, 2009 in 245B Real analysis, math.FA, math.GN | Tags: BanachAlaoglu theorem, strong operator topology, strong topology,weak operator topology, weak topology

A normed vector space automatically generates a topology, known as the norm topology or strongtopology on , generated by the open balls . A sequence in such aspace converges strongly (or converges in norm) to a limit if and only if as . This isthe 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 normtopology. 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 normtopology, and so it is useful to have a more general theory of topological vector spaces that contains thesemodes. Another reason (of particular importance in PDE) is that the norm topology on infinitedimensionalspaces is so strong that very few sets are compact or precompact in these topologies, making it difficult toapply compactness methods in these topologies. Instead, one often first works in a weaker topology, in whichcompactness is easier to establish, and then somehow upgrades any weakly convergent sequences obtained viacompactness to stronger modes of convergence (or alternatively, one abandons strong convergence and exploitsthe weak convergence directly). Two basic weak topologies for this purpose are the weak topology on a normedvector space , and the weak* topology on a dual vector space . Compactness in the latter topology isusually obtained from the BanachAlaoglu theorem (and its sequential counterpart), which will be a quickconsequence 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 ofbounded linear operators from to , thus supplementing the operator norm topology on that space with twoweaker topologies, which (somewhat confusingly) are named the strong operator topology and the weakoperator topology.

— 1. Topological vector spaces —

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

Definition 1 A topological vector space is a real or complex vector space, together with a topology such that the addition operation and

the scalar multiplication operation or is jointlycontinuous in both variables (thus, for instance, is continuous from withthe product topology to ).

It is an easy consequence of the definitions that the translation maps for and the dilationmaps for nonzero scalars are homeomorphisms on ; thus for instance the translation or dilationof an open set (or a closed set, a compact set, etc.) is open (resp. closed, compact, etc.). We also have the usuallimit laws: if and in a topological vector space, then , and if in thefield of scalars, then . (Note how we need joint continuity here; if we only had continuity in the

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individual variables, we could only conclude that (for instance) if one of or wasconstant.)

We now give some basic examples of topological vector spaces.

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

Exercise 2 Every seminormed vector space is a topological vector space, againusing the balls as a base for the topology. This topology is Hausdorff if andonly if the seminorm is a norm.

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

Exercise 3 Let be a vector space, and let be a (possibly infinite) family oftopologies on , each of which turning into a topological vector space. Let

be the topology generated by (i.e. it is the weakest topologythat contains all of the . Show that is also a topological vector space. Alsoshow 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.) Inparticular, by Exercise 2, we can talk about the topological vector space generated by a family of seminorms on .

Exercise 4 Let be a linear map between vector spaces. Suppose that wegive the topology induced by a family of seminorms , and thetopology induced by a family of seminorms . Show that is continuous ifand 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 ofcomplexvalued functions ; this is a complex vector space. Each point

gives rise to a seminorm . The topology generated by all of theseseminorms is the topology of pointwise convergence on (and is also the product

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topology on this space); a sequence converges to in this topology if andonly if it converges pointwise. Note that if has more than one point, then none ofthe seminorms individually generate a Hausdorff topology, but when combinedtogether, they do.

Example 3 (Uniform convergence) Let be a topological space, and let bethe space of complexvalued continuous functions . If is not compact,then one does not expect functions in to be bounded in general, and so the supnorm does not necessarily make into a normed vector space. Nevertheless, onecan still define “balls” in by

and 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 isno longer continuous. A sequence converges in this topology to a limit

if and only if converges uniformly to , thus isfinite for sufficiently large and converges to zero as .

Example 4 (Uniform convergence on compact sets) Let and be as in theprevious example. For every compact subset of , we can define a seminorm

on by . The topology generated by all of theseseminorms (as ranges over all compact subsets of ) is called the topology ofuniform convergence on compact sets; it is stronger than the topology of poitnwiseconvergence but weaker than the topology of uniform convergence. Indeed, asequence converges to in this topology if and only if converges uniformly to on each compact set.

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

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

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Example 5 (Smooth convergence) Let be the space of smooth functions . One can define the norm on this space for any nonnegative integer

by the formula

where 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 sets

for , , form the base for a topology that turns into atopological vector space, and that a sequence converges to a limit in thistopology if and only if it converges in measure.

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

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

(i) Show that any set which is open in a topological vector space, is alsoalgebraically open.(ii) Give an example of a set in which is algebraically open, but not open in theusual topology. (Hint: a line intersects the unit circle in at most two points.)

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(iii) Show that the collection of algebraically open sets in is a topology.(iv) Show that the collection of algebraically open sets in does not give thestructure of a topological vector space.

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

Some (but not all) of the concepts that are definable for normed vector spaces, are also definable for the moregeneral category of topological vector spaces. For instance, even though there is no metric structure, one canstill 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 suchthat 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 notionssuch as completeness is because a topological vector space has something better than a topological structure,namely a uniform structure.)

Remark 1 As 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 aslightly larger class of complete topological vector spaces, namely the Fréchetspaces. A Fréchet space is a complete Hausdorff topological vector space whosetopology is generated by an at most countable family of seminorms; examplesinclude the space from Exercise 5 or the uniform convergence on compactsets topology from Exercise 4 in the case when is compact. We will howevernot study Fréchet spaces systematically here.

One can also extend the notion of a dual space from normed vector spaces to topological vector spaces inthe 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 thenorm topology on ; but as we shall see below, there are some weaker topologies that one can still place onthis 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

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such spaces. In finite dimensions, the HeineBorel theorem tells us that a set is compact if and only if it isclosed and bounded. In infinite dimensions, this is not enough, for two reasons. Firstly, compact sets need to becomplete, so we are only likely to find many compact sets when the ambient normed vector space is alsocomplete (i.e. it is a Banach space). Secondly, compact sets need to be totally bounded, rather than merelybounded, and this is quite a stringent condition. Indeed it forces compact sets to be “almost finitedimensional”in the following sense:

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

(i) is compact.(ii) is sequentially compact.(iii) is closed and bounded, and for every , lies in the neighbourhood

of a finitedimensional subspace of .

Suppose furthermore that there is a nested sequence of finitedimensional subspaces of such that is dense. Show that the followingstatement is equivalent to the first three:

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

Example 6 Let . In order for a set to be compact in the strongtopology, it needs to be closed and bounded, and also uniformly powerintegrable at spatial infinity in the sense that for every there exists suchthat

for 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 andbounded.

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

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Exercise 12 Let be a normed vector space.

If is a finitedimensional subspace of , and , show that there exists such that for all . Give an example to show that is notnecessarily unique (in contrast to the situation with Hilbert spaces).If is a finitedimensional 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 ifand only if is finitedimensional.

— 3. The weak and weak* topologies —

Let be a topological vector space. Then, as discussed above, we have the vector space of continuouslinear functionals on . We can use this dual space to create two useful topologies, the weak topology on andthe 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 2 It is possible for two nonisomorphic topological vector spaces to haveisomorphic duals, but with nonisomorphic weak* topologies. (For instance, has a very large number of preduals, which can generate a number of differentweak* topologies on .) So, technically, one cannot talk about the weak*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 13 Show that the weak topology on is a topological vector spacestructure 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.

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From the definition, we see that a sequence converges in the weak topology, or converges weakly forshort, 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 theweak* topology to if for all (thus , viewed as a function on , convergespointwise to ).

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

Exercise 14 Show 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 HahnBanach theorem.) In particular, we conclude the important fact that weak andweak* limits, when they exist, are unique.

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

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

Show that the sequence converges weakly in to zero, but does notconverge strongly in .Show that the sequence converges in the weak* sense in to zero, butdoes 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 ageneralised limit functional).

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

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

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is strongly bounded (i.e. is contained in a ball). is weakly bounded (i.e. is bounded for all ).

(Hint: use the HahnBanach theorem and the uniform boundedness principle.)Similarly, if is a subset of , and is a Banach space, show that is stronglybounded 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 17 Let be a Banach space, and let converge weakly to a limit . Show that the sequence is bounded, and

Observe from Exercise 15 that strict inequality can hold (cf. Fatou’s lemma).Similarly, if converges in the weak* topology to a limit , show thatthe sequence is bounded and that

Again, construct an example to show that strict inequality can hold. Thus we seethat 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 18 Let be a Hilbert space, and let converge weakly to a limit . Show that the following statements are equivalent:

converges strongly to . converges to .

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

Show that if converges strongly to , then it also converges in the Césaro sense to.

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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 inthe 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 hasa further subsequence that converges in the Césaro sense to .

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

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

Exercise 22 Let be a normed vector space, let be a subspace of which isclosed 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 weaktopology of if and only if it converges to in the weak topology of . (Becauseof this fact, we can often refer to “the weak topology” without specifying theambient space precisely.)

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

Show that a sequence converges weakly to a limit if and only ifthe 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 beinganalogous 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

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better compactness properties than the strong topology, thanks to the BanachAlaoglu theorem:

Theorem 3 (BanachAlaoglu theorem) Let be a normed vector space. Then theclosed 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 bethe 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 iscompact also.

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

Exercise 24 Let be a locally compact Hausdorff topological vector space. Showthat is finite dimensional. (Hint: If is locally compact, then there exists an openneighbourhood of the origin whose closure is compact. Show that forsome finitedimensional subspace , where .Iterate this to conclude that for any . On the other hand, use thecompactness of to show that for any point there exists such that

is disjoint from . Conclude that and thence that .)

The sequential version of the BanachAlaoglu theorem is also of importance (particularly in PDE):

Theorem 4 (Sequential BanachAlaoglu theorem) Let be a separable normedvector 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 hypothesishas a countable dense subset . By the sequential Tychonoff theorem, any sequence in then has asubsequence 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 issequentially compact as required.

Remark 5 One can also deduce the sequential BanachAlaoglu theorem from thegeneral BanachAlaoglu theorem by observing that the weak* topology on the dualof a separable space is metrisable. The sequential BanachAlaoglu theorem can

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break down for nonseparable spaces. For instance, the closed unit ball in isnot sequentially compact in the weak* topology, basically because the space ofultrafilters 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 5 If is a reflexive normed vector space, then the closed unit ball in isweakly compact, and (if is separable) is also sequentially weakly compact.

Remark 6 If is a normed vector space that is not separable, then one can showthat is not separable either. Indeed, using transfinite induction on firstuncountable ordinal, one can construct an uncountable proper wellordered chain ofclosed separable subspaces of the inseparable space , which by the HahnBanachtheorem induces an uncountable proper wellordered 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 spacescan have nonseparable duals; consider , for instance.]

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

In later notes we will see that the (sequential) BanachAlaoglu theorem will combine very nicely with the Rieszrepresentation 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 normedvector spaces, then is the space of bounded linear transformations from to . This is a normedvector space with the operator norm

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

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

Note that a sequence converges in the strong operator topology to a limit ifand only if strongly in for all , and converges in the weak operator topology. (Incontrast, converges to in the operator norm topology if and only if converges to uniformly onbounded 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 7 When 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 theweak topology on . Meanwhile, coincides with , and the operatornorm topology coincides with the strong topology on , while the strong and weakoperator 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 asequence of bounded linear operators from a Banach space to a normed vectorspace , let be another bounded linear operator, and let be a densesubspace 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 25 Let the hypotheses be as in Proposition 7, but now assume that isalso a Banach space. Show that the conclusion of Proposition 7 continues to hold if“strong operator topology” is replaced by “weak operator topology”.

Exercise 26 Show that the operator norm topology, strong operator topology, andweak operator topology, are all Hausdorff. As these topologies are nested, we thusconclude that it is not possible for a sequence of operators to converge to one limit

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in one of these topologies and to converge to a different limit in another.

Example 8 Let , and for each , let be the translationoperator by : . If is continuous and compactly supported, then(e.g. from dominated convergence) we see that in as . Since thespace of continuous and compactly supported functions is dense in , thisimplies (from the above proposition, with some obvious modifications to deal withthe continuous parameter instead of the discrete parameter ) that converges inthe strong operator topology (and hence weak operator topology) to the identity. Onthe 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 easilyverifies that as for any compactly supported , andhence for all by the usual limiting argument, and hence converges inthe weak operator topology to zero.

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

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

Show that in the operator norm topology if and only if for anybounded sequences .Show that in the strong operator topology if and only if for anyconvergent sequence and any bounded sequence .Show that in the weak operator topology if and only if for anyconvergent 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 toshow that the corresponding claim for the strong operator topology is false.

There is a counterpart of the BanachAlaoglu theorem (and its sequential analogue), at least in the case ofHilbert spaces:

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

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

Exercise 29 Let be a Hilbert space, let be sequences ofoperators, 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 weakoperator) topology.If in the operator norm topology, and is bounded in the operator normtopology, show that and in the operator norm topology.If in the strong operator topology, and is bounded in the operator normtopology, show that in the strong operator norm topology.Give an example where in the strong operator topology, and in theweak operator topology, but does not converge to zero even in the weakoperator topology.

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

Show that is compact if and only if it is the limit of finite rankoperators in the operator norm topology. Conclude in particular that is aclosed 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 operatorsin the strong operator topology.If , show that maps weakly convergent sequences to stronglyconvergent sequences. (This property is known as complete continuity.)Show that is a subspace of , which is closed with respect toleft and right multiplication by elements of . (In other words, the space of

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56 comments Comments feed for this article

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22 February, 2009 at 12:31 amAnonymous

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22 February, 2009 at 8:48 amEric

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22 February, 2009 at 12:13 pmTerence Tao

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

Reply

The answer to the Question after Corollary 5 is yes. More generally, the predual ofany (norm)separable Banach space is separable. Indeed, if V is not separable, thenby induction we can construct a wellordered strictly increasing sequence of closed

subspaces of order type . By HahnBanach, the orthogonal complements of these spaces are a strictlydecreasing sequence of closed subspaces of V*. But this is impossible in a secondcountable space, so thisimplies that V* cannot be normseparable.

Reply

Thanks for the correction and answer to the question!

Reply

compact operators is an twoideal in the algebra of bounded operators.)

The weak operator topology plays a particularly important role on the theory of vonNeumann algebras, which we will not discuss here. We will return to the study ofcompact operators next quarter, when we discuss the spectral theorem.

[Update, Feb 23: Corrections, another exercise and remark added (noterenumbering).]

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23 February, 2009 at 12:16 pmanon

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23 February, 2009 at 12:26 pmEric

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23 February, 2009 at 1:52 pmMatt Daws

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23 February, 2009 at 2:53 pmTerence Tao

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23 February, 2009 at 6:20 pmless than epsilon

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23 February, 2009 at 6:35 pmless than epsilon

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23 February, 2009 at 8:51 pmTerence Tao

28 February, 2009 at 1:20 amAnonymous

How does BanachAlaoglu not contradict F. Riesz’s Theorem that a TVS islocally compact if and only if it is finitedimensional?

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Re anon: BanachAlaoglu says that the norm unit 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.

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For Remark 2, consider , which has many, many preduals. Indeed, let be anycountable compact space, and consider . The dual is . As $K$ iscountable, 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 nonhomeomorphic and , we have found preduals for

which induce distinct weak*topologies.

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

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Thanks for the example and the comments!

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In example 5, is summation index from 0 to k or 0 to infinity?

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Between exc 13 and remark 3: ” a seq in V converges in the weak startopology……”

should it be ” a seq in V star …….” ?

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Thanks for the corrections!

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This is in relation to exercise 23, part 2. In class, we showed that pointwiseconvergence implies by the UBP that the are bounded, and immediately

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28 February, 2009 at 9:17 amTerence Tao

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28 February, 2009 at 2:15 pmYasser Taima

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28 February, 2009 at 2:42 pmTerence Tao

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2 March, 2009 at 12:48 ametale

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2 March, 2009 at 8:17 am245B, Notes 12: Continuous functions on locally compact Hausdorff spaces « What’s new

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6 March, 2009 at 2:35 amSebastian Scholtes

concluded that we get weak convergence. Isn’t a step through exercise 17 needed, to show that is finiteso that is in ? Thanks.

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Well, in the exercise is assumed to be in already. But yes, one could useExercise 17 (which is proven using the UBP in any case); also, from Fatou’slemma, the pointwise limit of sequences bounded in is automatically in .

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In the proof of Theorem 3 (BanachAlaoglu theorem), is it necessary to take theunit 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 BanachAlaoglu theorem mightwell be to exercise 28 instead. Thanks.

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

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In 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 istrivially closed in any topology on it by definition.

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[…] and compact. As is equivalent to the dual of the Banach space , it acquires a weak* topology (see Notes 11), known as the vaguetopology. A sequence of Radon measures then converges vaguely to a limit if […]

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Dear Etale: Oops, the exercise is meant to state that W is closed in the weak topologyof V. It’s corrected now.

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Dear Prof. Tao,

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

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6 March, 2009 at 7:28 amTerence Tao

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28 March, 2009 at 10:13 pmliuxiaochuan

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23 April, 2009 at 7:44 amPDEbeginner

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29 May, 2010 at 12:12 amAnonymous

16 September, 2010 at 8:51 amDaniel Mckenzie

Let in , every subsequence also converges in , such that there is a subsubsequence whichconverges pointwise almost everywhere. With a subsubsequence convergence principle (sometimes calledUrysohn’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’tneed another sleepless night (; .

Best regardsSebastian Scholtes

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Dear Sebastian,

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

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Dear Professor Tao:

The last bracket of exercise 16, it should be , instead of [Corrected, thanks –T]

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Dear Prof. Tao,

I guess in example 2 ‘ ‘should be ‘ ‘.

[Corrected, thanks – T]

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

Best regardsJafar soltani farsani

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Hey prof. Tao

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26 October, 2010 at 5:14 pm245A, Notes 4: Modes of convergence « What’s new

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9 November, 2010 at 9:01 am245A, Notes 3: Integration on abstract measure spaces, and the convergence theorems « What’s new

13 December, 2010 at 12:51 pmAnonymous

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 aseparable space is metrisable on normbounded subsets of ‘?

[Corrected, thanks – T.]

regardsDaniel

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[…] not attempt to exhaustively enumerate these modes here (but see this Wikipedia page, and see also these 245B notes on strong and weakconvergence). We will, however, discuss some of the modes of convergence that arise from measure theory, when […]

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Does the weak* convergence in L^\infty implies the convergence almosteverywhere (up to a subsequence if necessary)?

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No. For instance, on the torus , the exponential functions converge inthe weak* sense to zero (the RiemannLebesgue lemma), but clearly do notconverge 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 convergencemethod based on the absolute value rather than on the phase.

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[…] The claim then follows by another appeal to the definition of lim inf. Remark 12 Informally, Fatou’s lemma tells us that when takingthe 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 movingbump examples by ). While this lemma was stated only for pointwise limits, the same general principle (that mass can be destroyed, but notcreated, by the process of taking limits) tends to hold for other “weak” notions of convergence. We will see some instances of this in 245B.[…]

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Dear Prof. Tao,I’m greatly confused about Example 3, in particular about continuity ofmultiplication 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

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13 December, 2010 at 7:51 pmTerence Tao

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24 December, 2010 at 9:55 amTerence Tao

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13 March, 2011 at 1:11 pmTerence Tao

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13 March, 2011 at 3:11 pmAnonymous

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

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Dear 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 Césaroconvergent to 0

(since norm of a partial sum is bounded by 1), but surely there is no subsequence weakly convergent to 0. Sorryif it’s just some stupid misunderstanding. If it’s not, is it still possible to prove the 4th dot?Merry Christmas,Marcin Łoś

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Oops, you’re right, the third item is not correct and should be deleted. The fourthitem 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).

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

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Yes, 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, .)

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

weakly closed?

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Yes :), I think so.

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4 May, 2011 at 11:53 amAnonymous

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4 May, 2011 at 12:07 pmTerence Tao

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8 September, 2011 at 2:11 pm254A, Notes 2: Building Lie structure from representations and metrics « What’s new

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17 February, 2012 at 1:29 pmRex

17 February, 2012 at 2:39 pm

Dear Prof. Tao,

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

But you mentioned that one should have complete topology if he wants to apply BanachAlaoglu Thm. For thisreason, I am a little bit doubted if I am right.

Thanks a lot!

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Completeness is a notion that only makes sense for metric spaces, not for topologicalspaces. The weak* topology is not metrisable, and so there is no meaningful way todiscuss 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 , becauseconvergence in the former requires convergence when evaluated at all members of , whereas convergencein the latter only requires convergence when evaluated at all members of the smaller space .

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Dear 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 importantrole. Thanks!

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[…] norm on the vector space. However, not every topological vector space is generated by a norm. See these notes for some further […]

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Does 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?

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The notion of algebraic openness is natural when considering geometric versions of the

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3 December, 2013 at 6:03 pmLinear elliptic equations with measure data I | Partial Differential Australians

12 May, 2014 at 7:43 amAndré Caldas

HahnBanach theorem (see Theorem 4 of Notes 6). But outside of that, I don’tknow of any other major application of this topology.

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

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

[Corrected, thanks. T]

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

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

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Sorry, in the last two parts of that question V should also be assumed to be a normedvector space; the question has been updated accordingly.

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Ok, thanks for the reply!

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[…] and we say that a sequence in converges to in weakif for all . Further details can be found in Terry Tao’s

notes on the subject. […]

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Dear professor,

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

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4 March, 2015 at 8:12 pmAnonymous

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Uniqueness can fail for some normed spaces V (including some finite dimensionalspaces), but can hold for other spaces (such as Hilbert spaces). Uniqueness isnecessarily true if V is a Hilbert space, but is not necessarily true if one only assumes that

V is a normed vector spaces.

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

Thank you very much for your time.

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

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Yes; I recommend verifying this as an exercise.

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In 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” (setisomorphism) ?

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 showthat 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 iscontinous.) As I’ve seen from Note 8, “nets” are in the “optional” section. Do you suggest that there is analternative way to show that and are the same?

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5 March, 2015 at 8:17 amTerence Tao

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31 March, 2015 at 2:49 pmSina

Initially, the identification is only on the level of sets (i.e. as a bijection), but once oneverifies that the topologies are compatible (in your notation) then the identification isat 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 relationbetween 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 ofsubbasic sets.

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For “subbasic” sets, do you mean sets in a subbasis of a topological space? (Peopleuse different spelling such as “subbase”, “subbase”, etc. I’m usually confused aboutwhich one is grammatically correct.)

If we use and consider instead of , I guess we could avoid working backand 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.

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nice and systematic explanation of topological vector spaces and weak norms. Wasvery useful as all these concepts were kind of blurry in my head. It is much moreneater in my head now, thanks.

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