The Maximum Principle of Analytic Functions

The maximum of $\left|f(z)\right|$ on the disk $\left|z\right|\leq r$ is $M(r)$.

The maximum $M(r)$ is monotone increasing with $r$ unless $f(z)$ is a constant.

We assume that the function $f(z)$ is regular in the simply connected region $\left|z\right|>R$. $M(r)$ denotes the maximum of $\left|f(z)\right|$ on the circle $\left|z\right|=r$, $r>R$. Then $M(r)$ is also the maximum of $\left|f(z)\right|$ for $\left|z\right|\geq r$ and $M(r)$ is monotone decreasing unless $f(z)$ is a constant.

Let $f(z)$ denote a polynomial of degree $n$; then

Equality is attained only if the polynomial is of the form $cz^{n}$.

Suppose that $f(z)$ is a polynomial of degree $n$ and that

on the real interval $-1\leq z\leq 1$. Then we have

for any $z$ outside this interval; $a$ and $b$ are the semi-axes of the ellipse through $z$ and with foci $-1$ and $1$. What does the proposition imply for $z\to \infty$?

We assume that $f(z)$ is a polynomial of degree $n$, that $E_{1}$ and $E_{2}$ denote two homofocal ellipses with semi-axes $a_{1},b_{1}$ and $a_{2},b_{2}$, $a_{1}<a_{2}$, $b_{1}<b_{2}$. The maximum of $\left|f(z)\right|$ on $E_{1}$ and $E_{2}$ is denoted by $M_{1}$ and $M_{2}$ resp.; then

Derive problems 269 and 270 from this proposition.

If an analytic function is regular in a closed disk and not a constant its absolute value at the center of the disk is smaller than the arithmetic mean of its absolute value on the boundary of the disk.

If the absolute value of an analytic function $f(z)$ is constant in an open set of the $z$-plane (e.g.in a disk) $f(z)$ must be a constant.

We suppose that the functions $\varphi (z)$ and $\psi (z)$ are regular in the closed disk $\left|z\right|\leq 1$ and non-zero in the open disk $\left|z\right|<1$ and that, besides, $\varphi (0)$ and $\psi (0)$ are real and positive. If $\varphi (z)$ and $\psi (z)$ have the same modulus on the circle $\left|z\right|=1$ then $\varphi (z)=\psi (z)$ identically.

The function $f(z)$ is regular and single-valued in the interior of the closed domain $\mathfrak {D}$ and continuous in $\mathfrak {D}$, boundary included; the maximum of $\left|f(z)\right|$ on the boundary of $\mathfrak {D}$ is called $M$. Under these conditions we have the inequality

in the interior of $\mathfrak {D}$ unless $f(z)$ is a constant. [This statement is stronger than problem 135.]

Let $\mathfrak {D}$ denote a domain, $\zeta$ be an inner point of $\mathfrak {D}$ and $\mathfrak {B}$ be the set of those boundary points of $\mathfrak {D}$ whose distance to $\zeta$ does not exceed $\varrho$. The circle of radius $\varrho$ and center $\zeta$ is assumed to have an arc that does not belong to $\mathfrak {D}$ and the length of which is not smaller than $\dfrac {2\pi \varrho }{n}$, $n$ integer.

We suppose that the function $f(z)$ is regular and single-valued in the interior of $\mathfrak {D}$ and continuous on the boundary and that, in particular, $\left|f(z)\right|\leq a$ at the points of $\mathfrak {B}$ and $\left|f(z)\right|\leq A$ at the remaining boundary points of $\mathfrak {D}$; $a<A$. Then

[Examine the product $\displaystyle \prod_{\nu =0}^{n-1} f[\zeta +(z-\zeta )\omega^{-\nu }]$ in a suitable domain, $\omega =e^{2\pi i/n}$.]

We assume that the function $f(z)$ is regular and bounded in the sector $0<\arg z<\alpha$, continuous on the real axis and that $\lim \limits_{x\to \infty } f(x)=0$, $x$ real, $x>0$. Then the limit relation

holds uniformly in any sector $0\leq \arg z\leq \alpha -\varepsilon <\alpha$.

We denote by $M$ a positive constant and by $\mathfrak {R}$ a connected open region. The analytic function $f(z)$ is assumed to have the following properties:

1. $f(z)$ is regular at every point of $\mathfrak {R}$;

2. $f(z)$ is single-valued in $\mathfrak {R}$;

3. each boundary point of $\mathfrak {R}$ has for every positive number $\varepsilon$ a neighbourhood such that at every point $z$ of $\mathfrak {R}$ in this neighbourhood the inequality

   $$\left|f(z)\right| < M+\varepsilon$$

   is satisfied. These conditions imply

and even $\left|f(z)\right|<M$ unless $f(z)$ is a constant. [Stronger than problem 275.]

Let $f(z)$ be regular and bounded on the disk $\left|z\right|<1$ and let

hold uniformly in a sector $\alpha \leq \vartheta \leq \beta$, $\alpha <\beta$. Then $f(z)$ vanishes identically.