## Agregation 2013 – Analysis – Part 3

**Part III: Muntz spaces and the Clarkson-Edros Theorem**

Recall that for every we define and that is a strictly increasing sequence of positive integers.

1. Suppose that and . Let . Define and by recurrence for define by

(a) Calculate and prove that

(b) Prove that for every is a linear combination of .

(c) Prove that for every we have .

(d) Deduce that .

(e) Conclude that .

From here on suppose that is arbitrary and the series converges. For denote . For denote the cardinal of the set .

2. (a) Justify that .

(b) Compute .

3. Using the results in the second part prove that there exists a function continuous and bounded by on such that for every we have

4. (a) Prove that for every we have .

(b) Deduce that for every we have

5. For fixed prove that:

(a) .

Define for and (by continuity).

(b) Prove that for we have and deduce that

where .

For we pose and we admit that we have in the same way the minoration

6. (a) Prove that there exists a sequence such that for every we have

(b) Deduce from there that for we have .

7. Take and prove that there exists (depending only on and ) such that for every and for every we have

where (with all zero from a point on). (hint: apply 6. (a))

Consider the limit of a sequence where for every the polynomial is written (where the coefficients of every polynomial are zero from a point on).

(b) Prove that for fixed, the sequence converges to a complex number .

(c) Deduce that for every we have .

8. **Clarkson-Erdos Theorem**

(a) Let . For we define by . Prove that .

(b) Prove that is exactly the space of functions which are continuous on for which there exists a sequence of complex numbers such that

**Hints:** (a) Simple integration yields . The estimate comes from

(we have used .

(b) Simple induction.

(c) Use induction: (a) is the first step. To prove the induction step, use the recurrence relation and majorize directly with the RHS just like we did in (a).

(d) For great enough we have and moreover

Therefore is bounded by which tends to zero, as the series is divergent.

As a consequence there exists a sequence given by (b) and the above estimate says that converges uniformly to so .

(e) From (d) and we conclude that all monomials are in . (note that here we have used , and it is not possible to prove that using (c),(d), because in (c) the estimate doesn’t work for ) Hence, all polynomials are in and since the space of polynomials is dense in for the uniform convergence, we are done.

2. (a) is the cardinal of the set . From Part I.2 we deduce the desired result.

(b)

and since is increasing, as the sequence above has the limit equal to .

3. As indicated in the official hint, we can use the second part, namely, the definition of and the final point II.6. We choose so that the series converges. Choose for which , and we are done.

4. (a) If then using the above formula we get

this extends to every finite linear combination of monomials, and therefore to all .

(b) One definition of is the following:

and since it is clear that

for every . (here we have used (a))

5. (a) Estimate term by term:

where . This means that

where the last inequality comes from the fact that . Summing up all these relations for we get the desired result.

(b) Use the inequality and induction to prove that .

Since is strictly increasing, if we denote by all the terms in the interval then we have so the product of all the denominators is greater than .

All the denominators are smaller than so the first part of the inequality is proved.

In order to prove the second part, let’s estimate

and then

Now it should be clear what to do… Use the first inequality provided to see that

Now increase a bit the denominator and we are done. (do some similar reasoning for the part )

6. (a) Use 3,4 and 5 (formula for the infimum, formula for the integral, split product in 3 parts , and the corresponding estimates)to prove that a choice of gamma is

(b) a clear consequence of (a).

7. (a) If we use 6. (a) as indicated for and (note that if we have nothing to prove) we get

and all that we have to do in order to prove the desired estimate is to prove that there exists such that

which is equivalent to proving that

which is true since the LHS tends to zero as . Note that depends only on and .

(b) Note that the uniform convergence of to implies that is Cauchy in the norm , and using (a) we have

As a consequence is Cauchy in for fixed and therefore it is convergent to some complex number .

(c) If we denote , then using the above relation for we know that

From the above inequality we can see that we have problems near , but on the other hand, for every we can choose small enough so that the last fraction is positive (and well defined on ) to see that is uniformly bounded by a constant times so converges uniformly to which was the desired result.

8. (a) If we assume that does not converge uniformly to then there exists such that . As a consequence there exists a sequence of maximum values for on which (up to a subsequence) can be taken convergent to . Therefore we have

but the last term of the inequality goes to as by the continuity of . This is a contradiction, and therefore the assumption made was false and

(b) Until here we have proven only that if then has the given form. We still have to prove the converse. Take

for . The key is (of course) to use point (a). Take the functions which have the representation

The functions are defined on but behave as on which is a compact subset of the domain of convergence. Therefore converges uniformly and so . Apply (a) and we are done.