Jensen's formula

Let $f(z)$ be regular in the disk $\left|z\right| \leqslant r$ and have, inside the disk, roots $a_{1}, a_{2}, \ldots , a_{n}$, distinct from zero. By joining each of the zeros $a_{\nu }$ with a cut to the circle $\left|z\right| = r$ and integrating the corresponding function over the resulting contour, prove Jensen's formula:

Prove that if, under the conditions of the previous problem (problem 5509), the maximum of $\left|f(z)\right|$ equals $M$, $\left|f(0)\right| = m$, and $N$ is the number of zeros of $f(z)$ in the disk $\left|z\right| \leqslant \frac{r}{e}$, then

Let $z_{\nu }$ be the roots of an entire function $f(z)$, and

while for the function $f(z)$ the inequality

holds. Prove that for $\left|z\right| \leqslant 3R$

Let the function $f(z)$ be regular and different from zero for $\left|z\right|\leq r$. Prove that the geometric mean of $\left|f(z)\right|$ on the circle $\left|z\right|=r$

The function $f(z)$ is regular in the disk $\left|z\right|\leq r$ and does not vanish for $z=0$; the zeros of $f(z)$ in the disk are $z_{1},z_{2},\ldots ,z_{n}$, where each multiple zero is represented with its multiplicity. Then the geometric mean of $\left|f(z)\right|$ on the circle $\left|z\right|=r$ is

Under the hypothesis of problem 120 the geometric mean of $\left|f(z)\right|$ on the disk $\left|z\right|\leq r$,

is always smaller than the geometric mean of $\left|f(z)\right|$ on the circle $\left|z\right|=r$,

We have in fact

We assume that the function $f(z)$ is meromorphic in the disk $\left|z\right|\leq 1$, regular and non-zero on the boundary and at the origin. The zeros of $f(z)$ in $\left|z\right|\leq 1$ are $a_{1},a_{2},\ldots ,a_{m}$ and the poles $b_{1},b_{2},\ldots ,b_{n}$ (multiple zeros and poles are listed with correct multiplicity). Then we have Jensen's formula

The function $f(z)$ is meromorphic in the disk $\left|z\right|\leq R$, regular and different from $0$ on the circle $\left|z\right|=R$ and it has inside the circle the zeros $a_{1},a_{2},\ldots ,a_{m}$ and the poles $b_{1},b_{2},\ldots ,b_{n}$, counted with correct multiplicities. If the point $z=re^{i\vartheta }$, $r<R$, is neither a zero nor a pole we have

The function $f(z)$ is meromorphic in the half-plane $\mathrm{Re}\left(z\right)\geq 0$, regular and non-zero on its boundary and it has inside the half-plane the zeros $a_{1},a_{2},\ldots ,a_{m}$ and the poles $b_{1},b_{2},\ldots ,b_{n}$, counted with correct multiplicity. If $f(z)$ is regular at infinity (but also under weaker conditions on its behaviour at infinity) and if $f(z)$ is regular and non-zero at $z=x+iy$, $x>0$, we have

The function $f(z)$ is regular in the domain

different from zero on the boundary of $\mathfrak {D}$ and it has in $\mathfrak {D}$ the zeros $a_{1},a_{2},\ldots ,a_{m}$, $a_{\mu }=r_{\mu }e^{i\vartheta_{\mu }}$, $\mu =1,2,\ldots ,m$. Using the definition $\log \left|f(\varrho e^{i\vartheta })\right|=U(\varrho ,\vartheta )$ we have the formula

where $\chi (R)$ is bounded as $R\to \infty$; $r,f(z)$ are fixed.

(continued). Suppose $f(z)$ does not vanish on $\left|z\right|=r$ and that its zeros in the disk $\left|z\right|<r$ are $c_{1},c_{2},\ldots ,c_{m}$. Then we have for $\left|z\right|<r$

where $\gamma$ is a real constant. [Consequence of problem 231 as problem 120 is a consequence of problem 119.]

Let the function $f(z)$ satisfy the following conditions:

1. $f(z)$ is regular, $\left|f(z)\right|\leq M$ in the disk $\left|z-s\right|\leq r$;

2. $f(z)$ does not vanish in the closed half-disk $\left|z-s\right|\leq r$, $\re (z-s)\geq 0$;

3. $f(z)$ has in the disk $\left|z-s\right|\leq \frac{2}{3}r$ the zeros $c_{1},c_{2},\ldots ,c_{l}$. Then

[We may assume $s=0$; problems 232, 120.]

In contrast to the notation of problem 12, we now denote by $M_{k}(r)$ and $N_{k}(r)$ the maximum modulus and the number of zeros of $(f(z))^{k}$ in the circle $\left|z\right| \leq r$, $k = 1, 2, 3, \ldots$ respectively. The quotient $\dfrac {N_{k}(r)}{\log M_{k}(r)}$ is independent of $k$.

Let $a_{0} \neq 0$. Then

Assuming that $a_{0} \neq 0$, we have

For an arbitrary entire function

Consider an entire function and let $k > 0$. If the infinite integral $\int_{1}^{\infty } r^{-k-1} \log M(r) \, dr$ is convergent, then the infinite series $\sum_{n=q+1}^{\infty } r_{n}^{-k}$ is also convergent (not vice versa!).

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.

For an entire function whose zeros have a finite exponent of convergence $\lambda$

Let $L$ be the maximum of the absolute values of the coefficients of the polynomial

and let $z_1, z_2, \ldots , z_n$ be the zeros of $f(z)$. Then we have

Let $h(z) = c(1+z_1 z)(1+z_2 z) \cdots (1+z_n z)$, $\left|z_\nu \right| \leqq 1$, $\nu = 1, 2, \ldots , n$, $c$ real, $c > 0$. Then, setting $g(\vartheta ) = \left|h(e^{i\vartheta })\right|^2$, we have

Find all complex $z$ for which $\exp (\exp (z)) = 1$. Sketch them as points in the plane. Show that there is no entire function of finite order which has a zero at each of these points (except, of course, $f \equiv 0$).

Suppose $f$ is entire, $f(0) \neq 0,\left|f(z)\right| < \exp \left(\left|z\right|^{p}\right)$ for large $\left|z\right|$, and $\left\{ z_{n}\right\}$ is the sequence of zeros of $f$, counted according to their multiplicities. Prove that $\Sigma \left|z_{n}\right|^{-p-\epsilon } < \infty$ for every $\epsilon > 0$. (Compare with Sec. 15.20).

Suppose $f$ is an entire function, $f(\sqrt{n}) = 0$ for $n = 1,2,3, \ldots$, and there is a positive constant $\alpha$ such that $\left|f(z)\right| < \exp \left(\left|z\right|^{\alpha }\right)$ for all large enough $\left|z\right|$. For which $\alpha$ does it follow that $f(z) = 0$ for all $z$?

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

Put $f(z) = \exp [-(1+z) /(1-z)]$. Using the notation of Theorem 15.19, show that

although $f \in H^{\infty }$. Note the contrast with Theorem 15.24.