Blaschke products and the Blaschke condition

Let the function $f$ be analytic in the unit disc, with $\left|f(z)\right| \leq 1$ and $f(0) = 0$. Assume that there is a number $r$ in $(0,1)$ such that $f(r) = f(-r) = 0$. Prove that

Assume $f$ is analytic in a neighborhood of $\closure D$, $f$ maps $D$ into $D$, and $f$ maps $\partial D$ into $\partial D$, where $D = \left\{ z : \left|z\right| < 1\right\}$.

Consider the unit disk $D$ and an analytic function $f : D \to D$. If $f(0) = a \neq 0$, show that $f$ does not have a zero in the disk $\left\{ z : \left|z\right| < \left|a\right|\right\}$.

Let $f$ be an analytic function on $D = \left\{ z : \left|z\right| < 1\right\}$, $f(D) \subset D$ and $f(0) = 0$.

Prove the theorem: if for $\left|z\right| < 1$ the function $f(z)$ is regular and $\left|f(z)\right| < M$, while $f(\alpha ) = 0$, where $\left|\alpha \right| < 1$, then for $\left|z\right| < 1$ the inequality

holds.

Prove the theorem: if $f(z)$ is regular for $\left|z\right| < 1$; $f(0) \neq 0$; $\left|f(z)\right| < M$, and $f(z) = 0$ at $z = z_{1}, z_{2}, z_{3}, \ldots$, then the infinite product $\prod_{\nu =1}^{\infty } \left|z_{\nu }\right|$ converges.

Let the function $f(z)$ be regular, have the zeros $z_{1},z_{2},\ldots ,z_{n}$ and let $f(z)$ be bounded, $\left|f(z)\right|\leq M$, in the unit disk $\left|z\right|<1$. Then the stronger inequality

holds for $\left|z\right|<1$. We have equality either at every point or at no point of the open disk $\left|z\right|<1$. (Proposition 280 is a special case, $n=1$, $z_{1}=0$.)

Let the function $f(z)$ be regular, have the zeros $z_{1},z_{2},\ldots ,z_{n}$ and let $f(z)$ be bounded, $\left|f(z)\right|\leq M$, in the open half-plane $\mathrm{Re}\left(z\right)>0$. Then the stronger inequality

holds for all $z$ with $\mathrm{Re}\left(z\right)>0$. We have equality either at every point or at no point of the right half-plane $\mathrm{Re}\left(z\right)>0$.

A function that is meromorphic in a closed disk and of constant absolute value on the boundary circle is a rational function; in fact it is, up to a constant factor, the product of linear fractional functions that map the disk in question either onto the interior or the exterior of the unit circle.

[W. Blaschke: Sber. Naturf. Ges. Lpz. Vol. 67, p.194 (1915).] We assume that the function $f(z)$ is regular and bounded in the disk $\left|z\right|<1$ and vanishes at the points $z_{1},z_{2},z_{3},\ldots$ Then

(the sum of the distances of the zeros from the unit circle) is finite or else $f(z)\equiv 0$.

We assume that the function $f(z)$ is regular and bounded in the half-plane $\mathrm{Re}\left(z\right)>0$ and vanishes at the points $z_{1},z_{2},z_{3},\ldots$ outside the unit disk in the half-plane, i.e.$\left|z_{n}\right|>1$, $\mathrm{Re}\left(z_{n}\right)>0$, $n=1,2,3,\ldots$ Then the sum of the series

is finite or else $f(z)\equiv 0$.

Consider a function which is regular in the interior of the unit circle, and let $k > 0$. If the integral $\int_{0}^{1} (1-t)^{k-1} \log M(t) \, dt$ is convergent then the series $\sum_{n=1}^{\infty } (1 - r_{n})^{k+1}$ is also convergent.

Suppose $R$ is a rational function such that $\left|R(z)\right| = 1$ if $\left|z\right| = 1$. Prove that

where $c$ is a constant, $m$ is an integer, and $\alpha_{1}, \ldots , \alpha_{k}$ are complex numbers such that $\alpha_{n} \neq 0$ and $\left|\alpha_{n}\right| \neq 1$. Note that each of the above factors has absolute value 1 if $\left|z\right| = 1$.

Obtain an analogous description of those rational functions which are positive on $T$. Hint: Such a function must have the same number of zeros as poles in $U$. Consider products of factors of the form

where $\left|\alpha \right| < 1$ and $\left|\beta \right| < 1$.

Suppose $0 < \alpha < 1,0 < \beta < 1, f \in H(U), f(U) \subset U$, and $f(0) = \alpha$. How many zeros can $f$ have in the $\operatorname {disc} \bar{D}(0 ; \beta )$? What is the answer if

Under what conditions on a sequence of real numbers $y_{n}$ does there exist a bounded holomorphic function in the open right half plane which is not identically zero but which has a zero at each point $1+i y_{n}$? In particular, can this happen if

Suppose $0 < \left|\alpha_{n}\right| < 1, \Sigma \left(1-\left|\alpha_{n}\right|\right) < \infty$, and $B$ is the Blaschke product with zeros at the points $\alpha_{n}$. Let $E$ be the set of all points $1 / \bar{\alpha }_{n}$ and let $\Omega$ be the complement of the closure of $E$. Prove that the product actually converges uniformly on every compact subset of $\Omega$, so that $B \in H(\Omega )$, and that $B$ has a pole at each point of $E$. (This is of particular interest in those cases in which $\Omega$ is connected).

Put $\alpha_{n} = 1-n^{-2}$, for $n = 1,2,3, \ldots$, let $B$ be the Blaschke product with zeros at these points $\alpha$, and prove that $\lim_{r \rightarrow 1} B(r) = 0$. (It is understood that $0 < r < 1$).

More precisely, show that the estimate

is valid if $\alpha_{N-1} < r < \alpha_{N}$.

Prove that there is a sequence $\left\{ \alpha_{n}\right\}$ with $0 < \alpha_{n} < 1$, which tends to 1 so rapidly that the Blaschke product with zeros at the points $\alpha_{n}$ satisfies the condition

Hence this $B$ has no radial limit at $z = 1$.

Let $\varphi$ be a linear fractional transformation which maps $U$ onto $U$. For any $z \in U$ define the $\varphi$-orbit of $z$ to be the set $\left\{ \varphi_{n}(z)\right\}$, where $\varphi_{0}(z) = z, \varphi_{n}(z) = \varphi \left(\varphi_{n-1}(z)\right), n = 1,2,3, \ldots$ Ignore the case $\varphi (z) = z$.

Suppose $\left|\alpha_{1}\right| \leq \left|\alpha_{2}\right| \leq \left|\alpha_{3}\right| \leq \cdots < 1$, and let $n(r)$ be the number of terms in the sequence $\left\{ \alpha_{j}\right\}$ such that $\left|\alpha_{j}\right| \leq r$. Prove that

If $B(z) = \Sigma c_{k} z^{k}$ is a Blaschke product with at least one zero off the origin, is it possible to have $c_{k} \geq 0$ for $k = 0,1,2, \ldots$?

Suppose $B$ is a Blaschke product all of whose zeros lie on the segment $(0,1)$ and

Prove that the derivative of $f$ is bounded in $U$.

Suppose $\Omega \supset \bar{U}, f \in H(\Omega ),\left|f\left(e^{i \theta }\right)\right| \geq 3$ for all real $\theta , f(0) = 0$, and $\lambda_{1}, \lambda_{2}, \ldots , \lambda_{N}$ are the zeros of $1-f$ in $U$, counted according to their multiplicities. Prove that

Suggestion: Look at $B /(1-f)$, where $B$ is a certain Blaschke product.

Prove the following converse of Theorem 15.24: If $f \in H(U)$ and if

then $f$ is a Blaschke product. Hint: $\left({ }^{*}\right)$ implies

Since $\log^{+}\left|f\right| \geq 0$, it follows from Theorems 17.3 and 17.5 that $\log^{+}\left|f\right| = 0$, so $\left|f\right| \leq 1$. Now $f = B g, g$ has no zeros, $\left|g\right| \leq 1$, and $\left({ }^{*}\right)$ holds with $1 / g$ in place of $f$. By the first argument, $\left|1 / g\right| \leq 1$. Hence $\left|g\right| = 1$.

Let the function $f(z)$ be regular for $\mathrm{Re}\left(z\right) > 0$, satisfy there the inequality $\left|f(z)\right| < 1$, and vanish at the points $z_{1}, z_{2}, \ldots , z_{m}$. Prove that

Let the function $f(z)$ be regular and bounded in the half-plane $\mathrm{Re}\left(z\right) > 0$, and vanish at a sequence of points $\left\{ z_{n}\right\} , z_{n} \rightarrow \infty$, of this half-plane. Prove that either $f(z) \equiv 0$, or the series $\sum_{n = 1}^{\infty } \mathrm{Re}\left(\frac{1}{z_{n}}\right)$ converges.

Let the function $f(z)$ be regular and bounded in the disk $\left|z\right| < R$, and vanish at a sequence $\left\{ z_{n}\right\}$ of points of this disk. Prove that either $f(z) \equiv 0$, or the series $\sum_{n = 1}^{\infty }\left(R-\left|z_{n}\right|\right)$ converges.