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}}.


\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.


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

  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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