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).
Show that for any real inner product space
and for a complex inner product space
- and ;
Moreover, let us recall that the norm induced by the inner product is given by for all . With this in mind, let us proceed in demonstrating the desired identities:
Let be a real inner product space. Recall that for all , and observe that the following holds for all :
Now, let be a complex inner product space. Then, observe that the following holds for all :
Let with supremum norm and define to be the set of functions such that
Let with supremum norm and define . Let us consider the linear continuous function defined by . As the preimage of a closed set under a continuous function is closed, the preimage of the closed set must also be closed under the continuous function . Following from the fact that by definition, we may see that is closed.
Now, let be arbitrary and fix . Then, we may observe:
It follows that, for all , it must be the case that whenever . Thus, is convex.
Further, for all , let us observe that the following holds:
That is, for all .
Observe that if we desire to show that is not a Hilbert space, it suffices to show that there exists no unique element in such that .
To do so, we construct a sequence in which approaches the minimum of the norm, but which does not converge to an element of . Setting for convenience, let us then consider the sequence defined by:
While rather undesirably messy when expressed as such, this is simply the following function:
Observe by inspection (for convenience, as this is already cumbersome enough) that for all $t > 0$, and whenever .
Moreover, note that and as . But, observe as well that point-wise as , which is not a continuous function. Thus, we have constructed a sequence of functions in which grow arbitrarily close to the minimum of the norm, but which do not converge point-wise to an element of . Correspondingly, there cannot exist an element of minimal norm in with regard to , and thus is not a Hilbert space.
We seek to compute .
Let us begin by observing that the set is a closed, linear subspace of the Hilbert space with inner product . Further, note that it may be easily checked via routine computation that is an orthonormal basis for .
Observing that , let be the unique element of with minimal norm in regard to . That is, let be such that
Then, we may compute by taking the projection of onto . As is an orthonormal basis for , it follows that
As a Hilbert space is necessarily complete, the following then holds
Thus, the given integral is minimized when , , and .
Suppose is a Hilbert space, and suppose the closed unit ball of , denoted by , is compact. For some index set , let be an orthonormal basis for . Then, it follows that for all . Moreover, as for any two with , observe that we may deduce:
Let us denote the open ball of radius about by and construct the set:
Observe that each open ball in contains precisely one element of . Further, by the polar identity, let us note that, for all , the following holds:
From this, we may see that each lies within at least one open ball in the set . Thus, , and it follows that is indeed an open cover of .
But, as is compact, every open cover admits a finite subcover. As no open ball in may be omitted without also omitting an element of the basis , it must then be the case that is finite. Correspondingly, as the cardinality of is the same as the cardinality of , it must be the case that our orthonormal basis is finite as well. Thus, dim.