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.