## Agregation 2013 – Analysis – Part 4

**Part IV: To be or not to be with separable dual**

1. Consider a segment () of and a closed linear subspace of such that each function in is of class on . For such that we denote where .

(a) Prove that .

(b) Prove that for every we have

(c) Deduce that there exists which verifies for each and

(d) Let be a finite sequence of points in such that (for ) and . Prove that

2. Let be a closed linear subspace of such that every function in is of class on . Prove that has finite dimension.

3. Consider a closed linear subspace of such that every function of is of class on .

(a) Give an example of such a subspace with infinite dimension. (hint: use part III).

Fix a sequence of strictly increasing towards ,a sequence of of strictly increasing towards and a sequence of integers strictly increasing towards such that and

(b) Let be an application which maps to . Prove that this application is well defined and satisfies for every .

(c) Deduce that is isomorphic to a subspace of and therefore is isomorphic to a subspace of .

(d) Justify that is isomorphic to and therefore is separable. (hint: use Hahn-Banach)

4. We wish to prove by contradiction that is not separable. Suppose that there exists a sequence which is dense in . For and we denote the evaluation on .

(a) Justify that and that for every with we have

Denote by the application which associates to an integer such that .

(b) Prove that we define like this a one to one map from to and conclude.

**Hints:** 1. (a) This is obvious, since is just a linear combination of the evaluations of at and .

(b) By the mean value theorem we know that there exists between and such that and since is continuous on , for fixed is bounded above (for ).

(c) Use Banach Steinhaus theorem for the family of linear maps defined on with values in and deduce that

which translates to

exactly what we needed to prove.

(d) Pick . By (c) we can see that . Denote the point of maximum for on (which exists since is compact). Then is in an interval and it is closer to one side, WLOG suppose that it is closer to . Then by (c) we have , which implies the desired result.

2. Choose and pick a sequence as in 1. (d). Next, suppose that are two functions in such that . Note that is also in (because this is a vector subspace of , and the difference of two functions is ). Then the estimate in 1. (d) is valid for . But that means

and therefore , which means that .

This means that each function is uniquely determined by its values . Therefore the application

is an isomorphism and the conclusion follows.

3. (a) From part III one such space is for such that .

(b) Since we can see that converges to so the application is well defined. Furthermore, by 1. (d) we know that from where we deduce that . The last inequality is trivial from the definitions of the sup norms.

(c) Using the same argument as in 2. we find that is injective, so is isomorphic to which is a subspace of . Use the application defined in Part I 3. to see that is isomorphic to which is a subspace of .

(d) In general, if are two isomorphic normed vector spaces (with an isomorphism with continuous) then to each functional we can associate the functional by

This application defined on with values in is linear, continuous and injective (by its definition and by the surjectivity of ).

Since is invertible, we can do the same thing the other way around: to we associate by

We have thus constructed the inverse of the above linear continuous application, which means that is isomorphic to .

In order to deduce separability of we prove that is separable. To do that, a good candidate for a countable dense set is . Pick and . Then there exists such that for every we have . For all other first terms, we can find an -close number in so that there exists an element in at distance smaller than of . This means that .

Any subset of a separable space is also separable, so we are done. 4. (a) The evaluation is obviously a linear functional. If then we have

and we can construct a continuous function on such that and (for example a piecewise linear interpolation). (b) If then if and we cannot have . If then by (a) we have

a contradiction. Therefore when and the application which associates to a positive integer is injective. This would mean that is countable, which is certainly not the case. As a consequence, our initial assumption was false and is not separable.