A normed vector space {(X, \| \|_X)} automatically generates a topology, known as the norm topology or strong topology on {X}, generated by the open balls {B(x,r) := \{ y \in X: \|y-x\|_X < r \}}. A sequence {x_n} in such a space converges strongly (or converges in norm) to a limit {x} if and only if {\|x_n-x\|_X \rightarrow 0} as {n \rightarrow \infty}. 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 {X}, and the weak* topology on a dual vector space {X^*}. 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 {B(X \rightarrow Y)} of bounded linear operators from {X} to {Y}, 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 1 A topological vector space {V = (V,{\mathcal F})} is a real or complex vector space {V}, together with a topology {{\mathcal F}} such that the addition operation {+: V \times V \rightarrow V} and the scalar multiplication operation {\cdot: {\bf R} \times V \rightarrow V} or {\cdot: {\bf C} \times V \rightarrow V} is jointly continuous in both variables (thus, for instance, {+} is continuous from {V \times V} with the product topology to {V}).

It is an easy consequence of the definitions that the translation maps {x \mapsto x + x_0} for {x_0 \in V} and the dilation maps {x \mapsto \lambda \cdot x} for non-zero scalars {\lambda} are homeomorphisms on {V}; 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 {x_n \rightarrow x} and {y_n \rightarrow y} in a topological vector space, then {x_n + y_n \rightarrow x+y}, and if {\lambda_n \rightarrow \lambda} in the field of scalars, then {\lambda_n x_n \rightarrow \lambda x}. (Note how we need joint continuity here; if we only had continuity in the individual variables, we could only conclude that {x_n+y_n \rightarrow x+y} (for instance) if one of {x_n} or {y_n} was constant.)

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 {B(x,r)} as the base for the topology. Show that the same statement holds if the vector space is quasi-normed rather than normed.

Exercise 2 Every semi-normed vector space is a topological vector space, again using the balls {B(x,r)} as a base for the topology. This topology is Hausdorff if and only if the semi-norm is a norm.

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

Exercise 3 Let {V} be a vector space, and let {({\mathcal F}_\alpha)_{\alpha \in A}} be a (possibly infinite) family of topologies on {V}, each of which turning {V} into a topological vector space. Let {{\mathcal F} := \bigvee_{\alpha \in A} {\mathcal F}_\alpha} be the topology generated by {\bigcup_{\alpha \in A} {\mathcal F}_\alpha} (i.e. it is the weakest topology that contains all of the {{\mathcal F}_\alpha}. Show that {(V, {\mathcal F})} is also a topological vector space. Also show that a sequence {x_n \in V} converges to a limit {x} in {{\mathcal F}} if and only if {x_n \rightarrow x} in {{\mathcal F}_\alpha} for all {\alpha \in A}. (The same statement also holds if sequences are replaced by nets.) In particular, by Exercise 2, we can talk about the topological vector space {V} generated by a family of semi-norms {(\| \|_\alpha)_{\alpha \in A}} on {V}.

Exercise 4 Let {T: V \rightarrow W} be a linear map between vector spaces. Suppose that we give {V} the topology induced by a family of semi-norms {(\| \|_{V_\alpha})_{\alpha \in A}}, and {W} the topology induced by a family of semi-norms {(\| \|_{W_\beta})_{\beta \in B}}. Show that {T} is continuous if and only if, for each {\beta \in B}, there exists a finite subset {A_\beta} of {A} and a constant {C_\beta} such that {\| Tf\|_{W_\beta} \leq C_\beta \sum_{\alpha \in A_\beta} \| f\|_{V_\alpha}} for all {f \in V}.

Example 2 (Pointwise convergence) Let {X} be a set, and let {{\bf C}^X} be the space of complex-valued functions {f: X \rightarrow {\bf C}}; this is a complex vector space. Each point {x \in X} gives rise to a seminorm {\| f\|_x := |f(x)|}. The topology generated by all of these seminorms is the topology of pointwise convergence on {{\bf C}^X} (and is also the product topology on this space); a sequence {f_n \in {\bf C}^X} converges to {f} in this topology if and only if it converges pointwise. Note that if {X} 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 {X} be a topological space, and let {C(X)} be the space of complex-valued continuous functions {f: X \rightarrow {\bf C}}. If {X} is not compact, then one does not expect functions in {C(X)} to be bounded in general, and so the sup norm does not necessarily make {C(X)} into a normed vector space. Nevertheless, one can still define “balls” {B(f,r)} in {C(X)} by

\displaystyle  B(f,r) := \{ g \in C(X): \sup_{x \in X} |f(x)-g(x)| \leq r \}

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 is no longer continuous. A sequence {f_n \in C(X)} converges in this topology to a limit {f \in C(X)} if and only if {f_n} converges uniformly to {f}, thus {\sup_{x \in X} |f_n(x) -f(x)|} is finite for sufficiently large {n} and converges to zero as {n \rightarrow \infty}.

Example 4 (Uniform convergence on compact sets) Let {X} and {C(X)} be as in the previous example. For every compact subset {K} of {X}, we can define a seminorm {\| \|_{C(K)}} on {C(X)} by {\|f\|_{C(K)} := \sup_{x \in K} |f(x)|}. The topology generated by all of these seminorms (as {K} ranges over all compact subsets of {X}) is called the topology 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 {f_n \in C(X)} converges to {f \in C(X)} in this topology if and only if {f_n} converges uniformly to {f} on each compact set.

Exercise 5 Show 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 6 Show that a topological vector space is Hausdorff if and only if the origin {\{0\}} is closed. (Hint: first use the continuity of addition to prove the lemma that if {V} is an open neighbourhood of {0}, then there exists another open neighbourhood {U} of {0} such that {U+U \subset V}, i.e. {u+u' \in V} for all {u,u' \in U}.)

Example 5 (Smooth convergence) Let {C^\infty([0,1])} be the space of smooth functions {f: [0,1] \rightarrow {\bf C}}. One can define the {C^k} norm on this space for any non-negative integer {k} by the formula

\displaystyle  \|f\|_{C^k} := \sum_{j=0}^k \sup_{x \in [0,1]} |f^{(j)}(x)|,

where {f^{(j)}} is the {j^{th}} derivative of {f}. The topology generated by all the {C^k} norms for {k=0,1,2,\ldots} is the smooth topology: a sequence {f_n} converges in this topology to a limit {f} if {f_n^{(j)}} converges uniformly to {f^{(j)}} for each {j \geq 0}.

Exercise 7 (Convergence in measure) Let {(X,{\mathcal X},\mu)} be a measure space, and let {L(X)} be the space of measurable functions {f: X \rightarrow {\bf C}}. Show that the sets

\displaystyle  B(f,\epsilon,r) := \{ g \in L(X): \mu( \{ x: |f(x)-g(x)| \geq r \} < \epsilon ) \}

for {f \in L(X)}, {\epsilon > 0}, {r > 0} form the base for a topology that turns {L(X)} into a topological vector space, and that a sequence {f_n \in L(X)} converges to a limit {f} in this topology if and only if it converges in measure.

Exercise 8 Let {[0,1]} be given the usual Lebesgue measure. Show that the vector space {L^\infty([0,1])} cannot be given a topological vector space structure in which a sequence {f_n \in L^\infty([0,1])} converges to {f} in this topology if and only if it converges almost everywhere. (Hint: construct a sequence {f_n} in {L^\infty([0,1])} 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 {U} of a real vector space {V} is algebraically open if the sets {\{ t \in {\bf R}: x+tv \in U \}} are open for all {x,v \in V}.

  • (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 {{\bf R}^2} 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 {V} is a topology.
  • (iv) Show that the collection of algebraically open sets in {{\bf R}^2} does not give {{\bf R}^2} the structure of a topological vector space.

Exercise 10 (Quotient topology) Let {V} be a topological vector space, and let {W} be a subspace of {V}. Let {V/W := \{ v+W: v \in V \}} be the space of cosets of {W}; this is a vector space. Let {\pi: V \rightarrow V/W} be the coset map {\pi(v) := v+W}. Show that the collection of sets {U \subset V/W} such that {\pi^{-1}(U)} is open gives {V/W} the structure of a topological vector space. If {V} is Hausdorff, show that {V/W} is Hausdorff if and only if {W} is closed in {V}.

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 {x_n \in V} in a topological vector space: this is a sequence such that {x_n-x_m \rightarrow 0} as {n,m \rightarrow \infty} (or more precisely, for any open neighbourhood {U} of {0}, there exists {N >0} such that {x_n-x_m \in U} for all {n,m \geq N}). 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 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 a slightly larger class of complete topological vector spaces, namely the Fréchet spaces. A Fréchet space is a complete Hausdorff topological vector space whose topology is generated by an at most countable family of semi-norms; examples include the space {C^\infty([0,1])} from Exercise 5 or the uniform convergence on compact sets topology from Exercise 4 in the case when {X} is {\sigma}-compact. We will however not study Fréchet spaces systematically here.

One can also extend the notion of a dual space {V^*} from normed vector spaces to topological vector spaces in the obvious manner: the dual space {V^*} of a topological space is the space of continuous linear functionals from {V} to the field of scalars (either {{\bf R}} or {{\bf C}}, depending on whether {V} is a real or complex vector space). This is clearly a vector space. Unfortunately, in the absence of a norm on {V}, one cannot define the analogue of the norm topology on {V^*}; 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 11 Let {K} be a subset of a Banach space {V}. Show that the following are equivalent:

  • (i) {K} is compact.
  • (ii) {K} is sequentially compact.
  • (iii) {K} is closed and bounded, and for every {\epsilon > 0}, {K} lies in the {\epsilon}-neighbourhood {\{ x \in V: \|x-y\| < \epsilon \hbox{ for some } y \in W \}} of a finite-dimensional subspace {W} of {V}.

Suppose furthermore that there is a nested sequence {V_1 \subset V_2 \subset \ldots} of finite-dimensional subspaces of {V} such that {\bigcup_{n=1}^\infty V_n} is dense. Show that the following statement is equivalent to the first three:

  • (iv) {K} is closed and bounded, and for every {\epsilon > 0} there exists an {n} such that {K} lies in the {\epsilon}-neighbourhood of {V_n}.

Example 6 Let {1 \leq p < \infty}. In order for a set {K \subset \ell^p({\bf N})} to be compact in the strong topology, it needs to be closed and bounded, and also uniformly {p^{th}}-power integrable at spatial infinity in the sense that for every {\epsilon > 0} there exists {n > 0} such that

\displaystyle  (\sum_{m > n} |f(m)|^p)^{1/p} \leq \epsilon

for all {f \in K}. Thus, for instance, the “moving bump” example {\{ e_1, e_2, e_3, \ldots \}}, where {e_n} is the sequence which equals {1} on {n} and zero elsewhere, is not uniformly {p^{th}} power integrable and thus not a compact subset of {\ell^p({\bf N})}, despite being closed and bounded.

For “continuous” {L^p} spaces, such as {L^p({\bf R})}, 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 12 Let {V} be a normed vector space.

  • If {W} is a finite-dimensional subspace of {V}, and {x \in V}, show that there exists {y \in W} such that {\|x-y\| \leq \|x-y'\|} for all {y' \in W}. Give an example to show that {y} is not necessarily unique (in contrast to the situation with Hilbert spaces).
  • If {W} is a finite-dimensional proper subspace of {V}, show that there exists {x \in V} with {\|x\|=1} such that {\|x-y\| \geq 1} for all {y \in W}. (cf. the Riesz lemma.)
  • Show that the closed unit ball {\{ x \in V: \|x\| \leq 1 \}} is compact in the strong topology if and only if {V} is finite-dimensional.

— 3. The weak and weak* topologies —

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

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

  • The weak topology on {V} is the topology generated by the seminorms {\| x\|_\lambda := |\lambda(x)|} for all {\lambda \in V^*}.
  • The weak* topology on {V^*} is the topology generated by the seminorms {\| \lambda\|_x := |\lambda(x)|} for all {x \in V}.

Remark 2 It is possible for two non-isomorphic topological vector spaces to have isomorphic duals, but with non-isomorphic weak* topologies. (For instance, {\ell^1({\bf N})} has a very large number of preduals, which can generate a number of different weak* topologies on {\ell^1({\bf N})}.) So, technically, one cannot talk about the weak* topology on a dual space {V^*}, without specifying exactly what the predual space {V} is. However, in practice, the predual space is usually clear from context.

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

From the definition, we see that a sequence {x_n \in V} converges in the weak topology, or converges weakly for short, to a limit {x \in V} if and only if {\lambda(x_n) \rightarrow \lambda(x)} for all {\lambda \in V^*}. This weak convergence is often denoted {x_n \rightharpoonup x}, to distinguish it from strong convergence {x_n \rightarrow x}. Similarly, a sequence {\lambda_n \in V^*} converges in the weak* topology to {\lambda \in V^*} if {\lambda_n(x) \rightarrow \lambda(x)} for all {x \in V} (thus {\lambda_n}, viewed as a function on {V}, converges pointwise to {\lambda}).

Remark 3 If {V} is a Hilbert space, then from the Riesz representation theorem for Hilbert spaces we see that a sequence {x_n \in V} converges weakly (or in the weak* sense) to a limit {x \in V} if and only if {\langle x_n, y\rangle \rightarrow \langle x, y \rangle} for all {y \in V}.

Exercise 14 Show that if {V} is a normed vector space, then the weak topology on {V} and the weak* topology on {V^*} 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 15 Let {V := c_0({\bf N})}, thus {V^* \equiv \ell^1({\bf N})} and {V^{**} \equiv \ell^\infty({\bf N})}. Let {e_1, e_2, \ldots} be the standard basis of either {V}, {V^*}, or {V^{**}}.

  • Show that the sequence {e_1, e_2, \ldots} converges weakly in {V} to zero, but does not converge strongly in {V}.
  • Show that the sequence {e_1, e_2, \ldots} converges in the weak* sense in {V^*} to zero, but does not converge in the weak or strong senses in {V^*}.
  • Show that the sequence {\sum_{m=n}^\infty e_m} for {n=1,2,\ldots} converges in the weak* topology of {V^{**}} to zero, but does not converge in the weak or strong senses. (Hint: use a generalised limit functional).

Remark 4 Recall from Exercise 11 of Notes 9 that sequences in {V^* \equiv \ell^1({\bf N})} 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 {\ell^1({\bf N})} is open in the strong topology, but not in the weak.

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

  • {E} is strongly bounded (i.e. {E} is contained in a ball).
  • {E} is weakly bounded (i.e. {\lambda(E)} is bounded for all {\lambda \in V^*}).

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

Exercise 17 Let {V} be a Banach space, and let {x_n \in V} converge weakly to a limit {x \in V}. Show that the sequence {x_n} is bounded, and

\displaystyle  \|x\|_V \leq \liminf_{n \rightarrow \infty} \|x_n\|_V.

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

\displaystyle  \|\lambda\|_{V^*} \leq \liminf_{n \rightarrow \infty} \|\lambda_n\|_{V^*}.

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 {x_n} converges strongly to {x}, then {\|x_n\|_V} converges to {\|x\|_V}).

Exercise 18 Let {H} be a Hilbert space, and let {x_n \in H} converge weakly to a limit {x \in H}. Show that the following statements are equivalent:

  • {x_n} converges strongly to {x}.
  • {\|x_n\|} converges to {\|x\|}.

Exercise 19 Let {H} be a separable Hilbert space. We say that a sequence {x_n \in H} converges in the Césaro sense to a limit {x \in H} if {\frac{1}{N} \sum_{n=1}^N x_n} converges strongly to {x} as {n \rightarrow \infty}.

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

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

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

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

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

Exercise 23 Let {V := c_0({\bf N})} with the uniform (i.e. {\ell^\infty}) norm, and identify the dual space {V^*} with {\ell^1({\bf N})} in the usual manner.

  • Show that a sequence {x_n \in c_0({\bf N})} converges weakly to a limit {x \in c_0({\bf N})} if and only if the {x_n} are bounded in {c_0({\bf N})} and converge pointwise to {x}.
  • Show that a sequence {\lambda_n \in \ell^1({\bf N})} converges in the weak* topology to a limit {\lambda \in \ell^1({\bf N})} if and only if the {\lambda_n} are bounded in {\ell^1({\bf N})} and converge pointwise to {\lambda}.
  • Show that the weak topology in {c_0({\bf N})} 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 {V} be a normed vector space. Then the closed unit ball of {V^*} is compact in the weak* topology.

This result should be contrasted with Exercise 12.

Proof: Let’s say {V} is a complex vector space (the case of real vector spaces is of course analogous). Let {B^*} be the closed unit ball of {V^*}, then any linear functional {\lambda \in B^*} maps the closed unit ball {B} of {V} into the disk {D := \{ z \in {\bf C}: |z| \leq 1 \}}. Thus one can identify {B^*} with a subset of {D^B}, the space of functions from {B} to {D}. One easily verifies that the weak* topology on {B^*} is nothing more than the product topology of {D^B} restricted to {B^*}. Also, one easily shows that {B^*} is closed in {D^B}. But by Tychonoff’s theorem, {D^B} is compact, and so {B^*} is compact also. \Box

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

Exercise 24 Let {V} be a locally compact Hausdorff topological vector space. Show that {V} is finite dimensional. (Hint: If {V} is locally compact, then there exists an open neighbourhood {U} of the origin whose closure is compact. Show that {U \subset W + \frac{1}{2} U} for some finite-dimensional subspace {W}, where {W+\frac{1}{2} U := \{ w + \frac{1}{2} u: w \in W, u \in U \}}. Iterate this to conclude that {U \subset W + \varepsilon U} for any {\varepsilon > 0}. On the other hand, use the compactness of {\overline{U}} to show that for any point {x \in V \backslash W} there exists {\varepsilon > 0} such that {x - \varepsilon U} is disjoint from {W}. Conclude that {U \subset W} and thence that {V = W}.)

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

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

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

Remark 5 One can also deduce the sequential Banach-Alaoglu theorem from the general Banach-Alaoglu theorem by observing that the weak* topology on 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 {\ell^\infty({\bf N})} is not sequentially compact in the weak* topology, basically because the space {\beta {\bf N}} of ultrafilters is not sequentially compact (see Exercise 12 of these lecture notes).

If {V} is reflexive, then the weak topology on {V} is identical to the weak* topology on {(V^*)^*}. We thus have

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

Remark 6 If {V} is a normed vector space that is not separable, then one can show that {V^*} is not separable either. Indeed, using transfinite induction on first uncountable ordinal, one can construct an uncountable proper chain of closed separable subspaces of the inseparable space {V}, which by the Hahn-Banach theorem induces an uncountable proper chain of closed subspaces on {V^*}, 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 {\ell^1({\bf N})}, 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 {V} 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 {L^1} or {L^\infty}. (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 {X} and {Y} are normed vector spaces, then {B(X \rightarrow Y)} is the space of bounded linear transformations from {X} to {Y}. This is a normed vector space with the operator norm

\displaystyle  \|T\|_{op} := \sup \{ \|Tx\|_Y: \|x\|_X \leq 1 \}.

This norm induces the operator norm topology on {B(X \rightarrow Y)}. Unfortunately, this topology is so strong that it is difficult for a sequence of operators {T_n \in B(X \rightarrow Y)} to converge to a limit; for this reason, we introduce two weaker topologies.

Definition 6 (Strong and weak operator topologies) Let {X, Y} be normed vector spaces. The strong operator topology on {B(X \rightarrow Y)} is the topology induced by the seminorms {T \mapsto \| T x \|_Y} for all {x \in X}. The weak operator topology on {B(X \rightarrow Y)} is the topology induced by the seminorms {T \mapsto |\lambda(Tx)|} for all {x \in X} and {\lambda \in Y^*}.

Note that a sequence {T_n \in B(X \rightarrow Y)} converges in the strong operator topology to a limit {T \in B(X \rightarrow Y)} if and only if {T_n x \rightarrow Tx} strongly in {Y} for all {x \in X}, and {T_n} converges in the weak operator topology. (In contrast, {T_n} converges to {T} in the operator norm topology if and only if {T_n x} converges to {Tx} 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 7 When {X} is the scalar field, then {B(X \rightarrow Y)} is canonically isomorphic to {Y}. In this case, the operator norm and strong operator topology coincide with the strong topology on {Y}, and the weak operator norm topology coincides with the weak topology on {Y}. Meanwhile, {B(Y \rightarrow X)} coincides with {Y^*}, and the operator norm topology coincides with the strong topology on {Y^*}, while the strong and weak operator topologies correspond with the weak* topology on {Y^*}.

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

Proposition 7 (Uniform boundedness principle) Let {T_n \in B(X \rightarrow Y)} be a sequence of bounded linear operators from a Banach space {X} to a normed vector space {Y}, let {T \in B(X \rightarrow Y)} be another bounded linear operator, and let {D} be a dense subspace of {X}. Then the following are equivalent:

  • {T_n} converges in the strong operator topology of {B(X \rightarrow Y)} to {T}.
  • {T_n} is bounded in the operator norm (i.e. {\|T_n\|_{op}} is bounded), and the restriction of {T_n} to {D} converges in the strong operator topology of {B(D \rightarrow Y)} to the restriction of {T} to {D}.

Exercise 25 Let the hypotheses be as in Proposition 7, but now assume that {Y} is also 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, 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 8 Let {X = L^2({\bf R})}, and for each {t \in {\bf R}}, let {T_t: X \rightarrow X} be the translation operator by {t}: {T_t f(x) := f(x-t)}. If {f} is continuous and compactly supported, then (e.g. from dominated convergence) we see that {T_t f \rightarrow f} in {L^2} as {t \rightarrow 0}. Since the space of continuous and compactly supported functions is dense in {L^2({\bf R})}, this implies (from the above proposition, with some obvious modifications to deal with the continuous parameter {t} instead of the discrete parameter {n}) that {T_t} converges in the strong operator topology (and hence weak operator topology) to the identity. On the other hand, {T_t} does not converge to the identity in the operator norm topology. Indeed, observe for any {t > 0} that {\| (T_t - I) 1_{[0,t]} \|_{L^2({\bf R})} = \sqrt{2} \| 1_{[0,t]} \|_{L^2({\bf R})}}, and thus {\| T_t - I \|_{op} \geq \sqrt{2}}.

In a similar vein, {T_t} does not converge to anything in the strong operator topology (and hence does not converge in the operator norm topology either) in the limit {t \rightarrow \infty}, since {T_t 1_{[0,1]}} (say) does not converge strongly in {L^2}. However, one easily verifies that {\langle T_t f, g \rangle \rightarrow 0} as {t \rightarrow \infty} for any compactly supported {f, g \in L^2({\bf R})}, and hence for all {f, g \in L^2({\bf R})} by the usual limiting argument, and hence {T_t} 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 27 Let {H} be a Hilbert space, and let {T_n \in B(H \rightarrow H)} be a sequence of bounded linear operators.

  • Show that {T_n \rightarrow 0} in the operator norm topology if and only if {\langle T_n x_n, y_n \rangle \rightarrow 0} for any bounded sequences {x_n, y_n \in H}.
  • Show that {T_n \rightarrow 0} in the strong operator topology if and only if {\langle T_n x_n, y_n \rangle \rightarrow 0} for any convergent sequence {x_n \in H} and any bounded sequence {y_n \in H}.
  • Show that {T_n \rightarrow 0} in the weak operator topology if and only if {\langle T_n x_n, y_n \rangle \rightarrow 0} for any convergent sequences {x_n, y_n \in H}.
  • Show that {T_n \rightarrow 0} in the operator norm (resp. weak operator) topology if and only if {T_n^\dagger \rightarrow 0} 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 28 Let {H, H'} be Hilbert spaces. Show that the closed unit ball (in the operator norm) in {B(H \rightarrow H')} is compact in the weak operator topology. If {H} and {H'} are separable, show that {B(H \rightarrow H')} 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 29 Let {H} be a Hilbert space, let {S_n, T_n \in B(H \rightarrow H)} be sequences of operators, and let {S \in B(H \rightarrow H)} be another operator.

  • If {T_n \rightarrow 0} in the operator norm (resp. strong operator or weak operator) topology, show that {ST_n \rightarrow 0} and {T_n S \rightarrow 0} in the operator norm (resp. strong operator or weak operator) topology.
  • If {T_n \rightarrow 0} in the operator norm topology, and {S_n} is bounded in the operator norm topology, show that {S_n T_n \rightarrow 0} and {T_n S_n \rightarrow 0} in the operator norm topology.
  • If {T_n \rightarrow 0} in the strong operator topology, and {S_n} is bounded in the operator norm topology, show that {S_n T_n \rightarrow 0} in the strong operator norm topology.
  • Give an example where {T_n \rightarrow 0} in the strong operator topology, and {S_n \rightarrow 0} in the weak operator topology, but {T_n S_n} does not converge to zero even in the weak operator topology.

Exercise 30 Let {H} be a Hilbert space. An operator {T \in B( H \rightarrow H )} is said to be finite rank if its image {T(H)} is finite dimensional. {T} is said to be compact if the image of the unit ball is precompact. Let {K( H \rightarrow H )} denote the space of compact operators on {H}.

  • Show that {T \in B(H \rightarrow H)} is compact if and only if it is the limit of finite rank operators in the operator norm topology. Conclude in particular that {K(H \rightarrow H)} is a closed subset of {B(H \rightarrow H)} in the operator norm topology.
  • Show that an operator {T \in B(H \rightarrow H)} is compact if and only if {T^\dagger} is compact.
  • If {H} is separable, show that every {T \in B(H \rightarrow H)} is the limit of finite rank operators in the strong operator topology.
  • If {T \in K(H \rightarrow H)}, show that {T} maps weakly convergent sequences to strongly convergent sequences. (This property is known as complete continuity.)
  • Show that {K(H \rightarrow H)} is a subspace of {B(H \rightarrow H)}, which is closed with respect to left and right multiplication by elements of {B(H \rightarrow H)}. (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).]