A Variety of Functional Analysis Exercises:
As a helpful review, here are a variety of problem and solutions to exercises from an introductory functional analysis class. This is the second installment in a series of functional analysis exercises (Part 1, Part 2, Part 3, Part 4)
Let us first recall the general statement of the Banach-Alaoglu theorem:
However, note that this weak-* topology is on instead of the usual , and is correspondingly generated by the seminorms for all . But, observe that is reflexive, which is to say that all are such that , where is the canonical embedding of some element of into its second dual. Thus, our previous statement then gives that is weak-* compact in the topology generated by the seminorms for all . By reflexivity, this then gives that is weakly compact as well.
Thus, the weak-* closure of the canonical embedding of the weakly compact set into is simply . However, observe that a weakly compact set is weak-* closed, and as the weak closure coincides with norm closure for convex sets, it then follows that the canonical embedding of the ball of is precisely the ball of , which is to say that is reflexive.
Thus, we need only recall that a bounded operator on a Banach space is correspondingly continuous to conclude our proof, as the image of a compact set under a continuous function is, itself, compact. Thus, whenever is a reflexive Banach space, it follows that every is weakly compact.
We first seek to show that is compact. To do so, observe first that the sequence \newline , as . Following from this, it then holds that every subsequence of also converges to zero. Thus, as sequential compactness is equivalent to compactness in a metric space, is indeed compact.
We now seek to show that is bounded. That is, we seek to show that for any neighborhood of zero, there exists such that . Observe that, by the linearity of our space, it suffices to show that is contained within a dilation of the open unit ball, denoted here by . Observe
Clearly, this supremum is attained by . Thus, for any , and is correspondingly bounded.
Next, we seek to show that is not closed. To do so, let us set , , and so on, reaching the general form . Observe that for all . However, note that and , as is not a finite convex combination. Thus, we have constructed a sequence in which does not converge to an element of , and is correspondingly not closed.
Finally, we seek to calculate . To do so, let us first recall Milman’s theorem:
To use this result, we first need a lemma:
Observe that the operator defined by
is bounded (and thus continuous) by part (b). By problem 5 in A variety of Functional Analysis Exercises: Part 2, this operator is also compact, as . Moreover, observe . As is a closed subset of a compact set, it is correspondingly compact as well.
Now, note that is a locally convex space, is a compact, convex set in , and . Clearly, , so it follows that .
For the reverse inclusion, observe that the nonzero elements of are linearly independent, and as convex combinations are a subset of linear combinations, the nonzero elements of must be in . Moreover, observe that convex combinations are indeed linear combinations with strictly positive coefficients, and as consists of positive real scalings of orthogonal vectors, zero must be in as well. Thus, , and we may conclude that .
If is a -finite measure space, show that the set of extreme points of is .
Let be a -finite measure space. We seek to show:
To this end, let us first suppose there exists such that . Let be a real constant such that . Then, observe that and . But then,
So, we may see that all elements of must have norm one.
Now, let us suppose with , but . Let . Then, observe that . Let be a real constant such that . Then, observe that and . But then,
So, we may see that
To conclude our proof, let us demonstrate the reverse inclusion. To this end, let \newline . Then, suppose there exist such that . But, this then yields that or, equivalently, that . But then, as , it must then be the case that and . Thus, , and we may conclude
Show that subspaces of reflexive Banach spaces are reflexive. Also, if is reflexive and is a subspace of , then is reflexive.
Let be a reflexive Banach space and be a subspace of . We seek to show first that is reflexive as well.
To do so, let us first observe the theorem proved in problem :
Observe then that is a closed subset of (as the closure and weak closure coincide for convex sets in locally convex spaces), which is correspondingly weakly compact by the above result. As a closed subset of a compact set is compact as well, our desired result then follows.
We now seek to show that the quotient is reflexive. To do so, recall that is indeed a Banach space, and note that the image of a compact set under a continuous function is compact. Recalling that the quotient map is a linear operator with norm (and thus continuous), note that the image of the unit ball of under the quotient map remains weakly compact whenever is reflexive (via the previously cited result). Correspondingly, the unit ball of is weakly compact and thus reflexive.
Theorems characterizing the linear isometries of a space onto itself rely heavily on the Krein-Milman theorem. Here is the statement of the famous Banach-Stone theorem:
Give a proof of the Banach-Stone theorem for .
Recall that is compact Hausdorﬀ space, and let be a complex linear subalgebra of . Assume that contains the constant function , and suppose that is a linear map from onto which is isometric, i.e. . We seek to show that has the form = where , , and is an algebra automorphism.
First, let us note that is an abelian -algebra when equipped with conjugation as an involution. By a proof given in Conway, it then follows that the Gelfand transform is an isometric isomorphism. Similarly, recall that is compact and Hausdorff, and note that by theorem III.5.7 of Conway, there exists an isometric isomorphism of onto , where is the space of all -valued regular Borel measures on with the total variation norm.
As such, let us consider the map . Note that is an isometric isomorphism, and by the properties of the adjoint, recall that . Moreover, as is invertible, observe that must also be invertible, and correspondingly an isometric isomorphism as well. It then follows that is a weak-* homeomorphism of onto . Moreover, as is linear, observe that distributes over convex combinations. Hence, it is clear that:
Now, recall that, by Conway theorem V.8.4, . So, for each , there exists unique and scalar such that and .
We now show that is continuous. Let be a net in and . Then, in the weak-* topology of . Hence, in the weka-* topology of . In particular, . Thus, is continuous.
Continuing on, we now demonstrate that is a homoeomorphism. To do so, observe that we need only show that is continuous, as a continuous bijection between compact sets is a homeomorphism. But, as and are continuous and is invertible, the continuity of follows immediately. Thus, is a homeomorphism.
Now, observe that if and , then
Thus, as the Gelfand transformation is an isometric isomorphism, our desired result then follows.