## Agregation 2013 – Analysis – Part 2

**Part II: A Blaschke product**

1. Let be a sequence of complex numbers.

(a) Prove that for every we have

(b) Prove that for every we have

2. Let be a sequence of holomorphic functions on an open set such that the series of general terms converges normally on every compact subset of . Prove that the sequence of functions defined for all by

converges uniformly on every compact subset of to a function which is holomorphic on .

(a series of functions converges normally on a set if for every we have that on and the series is convergent.)

Until the end of this second part is a sequence of reals in such that the series converges. We denote . For we define

3. (a) Prove that for every and every such that we have .

(b) Prove that the sequence of functions converges uniformly on every compact of to a function which is holomorphic on .

We consider the function defined by .

(c) Prove that the function is integrable on .

Denote .

(d) For fixed prove that . (*Indication:* we could consider a half circle with radius with endpoints glued with the segment )

5. (a) Prove that for every the integral is well defined and that the function is continuous on .

(b) Prove that we can extend by continuity on with .

6. Deduce that for every we have .

**Hints:** 1. (a) Expand all the products and note that the only difference between the LHS and the RHS is that in the left we have a absolute value of a sum and in the right we have the sum of absolute values, hence it is a simple triangle inequality application.

(b) For every real we have . Apply this for and take the product.

2. Use 1. (a),(b) to prove that the product is uniformly convergent on every compact subset of , and then use the fact that the limit of a uniformly convergent sequence of holomorphic function is again holomorphic.

3. (a) We have the following equivalences:

(b) Choose a compact and note that where

We would like to show that there exists a constant (independent of ) such that . Then we could apply 2. and we are done.

Denote . Then we have

Denote

suppose that (at least for a subsequence which we relabel) and denote the values in which is attained. Since the series is convergent, we have . The sequence is contained in the compact so we it contains a convergent subsequence which converges to . Therefore we have

which is a contradiction. As a consequence, is bounded, and there exists a constant as desired. (c) Since for we have , we have and as a consequence

and is integrable on . 4. Consider the closed contour defined by the segment and the semicircle which passes through . Integrating on this contour and using Cauchy’s integral formula we have

But using 3. (a) the last integral is bounded in absolute value by

and taking we get the desired formula.

5. (a) Take . Then exists, because the absolute value of the integrand is the same as the absolute value of , which is an integrable function. The function is obviously continuous if we use the properties of the exponential.

(b)

6. Use the results obtained in 4,5 and Fubini’s Theorem.

Hi Beni,

is there an official website for aggregate? also, are there problems from previous years?

Yes, there is an official site, and you can find all sorts of informations and the problems from previous years. Here it is: agreg.org.