**A Variety of Functional Analysis Exercises: **

**Part 1**

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).

**Problem 1**

**Show that for any real inner product space**

**and for a complex inner product space**

**Solution**

Recall that if is an inner product on a vector space , the following hold:

- ;
- ;
- 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 :

**Problem 2**

**Let with supremum norm and define to be the set of functions such that**

**Prove that is a closed, convex subset of , but that has no element of minimal norm. This gives an alternative proof of the fact that is not a Hilbert space.**

**Solution**

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.

**Problem 3**

**Compute**

**Solution**

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 .

**Problem 4**

**If the closed unit ball of is compact, show that .**

**Solution**

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.