Home > Analysis, Complex analysis, Real Analysis > Agregation 2013 – Analysis – Part 2

Agregation 2013 – Analysis – Part 2


Part II: A Blaschke product

1. Let {(z_n)_{n \in \Bbb{N}}} be a sequence of complex numbers.

(a) Prove that for every {N \in \Bbb{N}} we have

\displaystyle \left| \left[ \prod_{j=0}^N (1+z_j) -1 \right] \right|\leq \left[\prod_{j=0}^N (1+|z_j|) \right] -1.

(b) Prove that for every {N \in \Bbb{N}} we have

\displaystyle \prod_{j=0}^N (1+|z_j|) \leq \exp \left(\sum_{j=0}^n |z_j|\right).

2. Let {(g_j)} be a sequence of holomorphic functions on an open set {U \subset \Bbb{C}} such that the series of general terms {g_j} converges normally on every compact subset of {U}. Prove that the sequence of functions {(G_N)} defined for all {z \in U} by

\displaystyle G_N(z)= \prod_{j=0}^N (1+g_j(z))

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

(a series of functions {\sum_{j \geq 0}g_j} converges normally on a set {K} if for every {j} we have that {|g_j|\leq u_j} on {K} and the series {\sum_{j \geq 0}u_j|} is convergent.)

Until the end of this second part {(\alpha_n)_{n \in \Bbb{N}}} is a sequence of reals in {(0,1]} such that the series {\sum_{n \geq 0}\alpha_n} converges. We denote {\Omega_1= \{z \in \Bbb{C} : \text{re}(z)>-1\}}. For {z \in \Omega_1} we define

\displaystyle f_n(z)=\frac{1-\alpha_n z}{1+\alpha_n z}.

3. (a) Prove that for every {n \in \Bbb{N}} and every {z \in \Bbb{C}} such that {\text{re}(z) \geq 0} we have {|f(z)|\leq 1}.

(b) Prove that the sequence of functions {z \mapsto \frac{2}{(1+z)^2}\prod_{n=0}^N f_n(z)} converges uniformly on every compact of {\Omega_1} to a function {F} which is holomorphic on {\Omega_1}.
We consider the function {\varphi: \Bbb{R} \rightarrow \Bbb{C}} defined by {\varphi(s)=F(is)}.

(c) Prove that the function {\varphi} is integrable on {\Bbb{R}}.
Denote {\Omega_0=\{z \in \Bbb{C} : \text{re}(z)>0\}}.

(d) For {z \in \Omega_0} fixed prove that {\displaystyle F(z) =\frac{1}{2\pi}\int_{-\infty}^\infty \frac{\varphi(s)}{z-is}}. (Indication: we could consider a half circle with radius {R>|z|} with endpoints glued with the segment {[-iR,iR]})

5. (a) Prove that for every {x \in (0,1]} the integral {\mu(x)=\frac{1}{2\pi} \int_{-\infty}^\infty \varphi(s) e^{-is \ln(x)}ds} is well defined and that the function {\mu: x \mapsto mu(x)} is continuous on {(0,1]}.

(b) Prove that we can extend {\mu} by continuity on {[0,1]} with {\|\mu\|_\infty=1}.

6. Deduce that for every {z \in \Omega_0} we have {\displaystyle F(z)=\int_0^1 x^{(z-1)}\mu(x)dx}.

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 {x} we have {e^{x}\geq x+1}. Apply this for {|z_i|,\ i=1..n} and take the product.

2. Use 1. (a),(b) to prove that the product is uniformly convergent on every compact subset of {U}, 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:

\displaystyle |f_n(z)|\leq 1 \Leftrightarrow |1-\alpha_n z|^2 \leq |1+\alpha_n z|^2 \Leftrightarrow -\alpha_n(z+\overline z)\leq \alpha_n(z+\overline z) \Leftrightarrow \text{Re}(z) \geq 0.

(b) Choose a compact {K \subset \Omega_1} and note that {f_n=1+g_n} where

\displaystyle g_n(z)=-\frac{2\alpha_nz}{1+\alpha_n z}.

We would like to show that there exists a constant (independent of {n}) such that {|g_n(z)| \leq C(K)\alpha_n}. Then we could apply 2. and we are done.

Denote {z=x+yi}. Then we have

\displaystyle |g_n(z)|=2\alpha_n \sqrt{\frac{x^2+y^2}{(1+\alpha_n x)^2+\alpha_n^2y^2}}.

Denote

\displaystyle M_n=\max_{x+yi\in K}\sqrt{\frac{x^2+y^2}{(1+\alpha_n x)^2+\alpha_n^2y^2}}

suppose that {M_n \rightarrow \infty} (at least for a subsequence which we relabel) and denote {(x_n,y_n)\in K} the values in which {M_n} is attained. Since the series {\sum\alpha_n} is convergent, we have {\alpha_n \rightarrow 0}. The sequence {(x_n,y_n)} is contained in the compact {K} so we it contains a convergent subsequence which converges to {(x_0,y_0)}. Therefore we have

\displaystyle \sqrt{x_0^2+y_0^2} =\lim_{n \rightarrow \infty}M_n =\infty,

which is a contradiction. As a consequence, {M_n} is bounded, and there exists a constant {C(K)} as desired. (c) Since for {z=is} we have {|f_n(z)|=1}, we have {|\varphi(s)|=|F(is)|=2/|1+is|^2=2/(1+s^2)} and as a consequence

\displaystyle \left|\int_\Bbb{R} \varphi(s)ds\right| \leq \int_\Bbb{R} |\varphi(s)|ds =\int_\Bbb{R} \frac{2}{1+s^2}=2\pi.

and {\varphi} is integrable on {\Bbb{R}}. 4. Consider the closed contour {\gamma} defined by the segment {[iR,-iR]} and the semicircle {S_r} which passes through {iR,-iR,R}. Integrating on this contour and using Cauchy’s integral formula we have

\displaystyle F(z) =\frac{1}{2\pi i}\int_\gamma \frac{F(w)}{w-z}dw=\frac{1}{2\pi}\int_{-R}^R \frac{\varphi(s)}{z-is}+\frac{1}{2\pi i}\int_{S_r} \frac{F(w)}{w-z}dw.

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

\displaystyle \frac{1}{2\pi} \int_{S_r} \frac{2}{|w+1|^2\max_{S_r}|w-z|}dw=O(1/R)

and taking {R \rightarrow \infty} we get the desired formula.

5. (a) Take {x \in (0,1]}. Then {\mu(x)} exists, because the absolute value of the integrand is the same as the absolute value of {\varphi(x)}, 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.

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  1. June 21, 2013 at 7:34 am

    Hi Beni,

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

    • June 21, 2013 at 8:29 am

      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.

  1. June 19, 2013 at 9:07 pm

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