Maximum modulus principle (theorem statement)

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.

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.]

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.]

The sum of the absolute values of several analytic functions attains its maximum on the boundary. Here is a more detailed statement: The functions $f_{1}(z),f_{2}(z),f_{3}(z),\ldots ,f_{n}(z)$ are supposed to be regular and single-valued in the domain $\mathfrak {D}$. Then the function

which is continuous in $\mathfrak {D}$, assumes its maximum on the boundary of $\mathfrak {D}$.

(continued). The function $\varphi (z)$ assumes its maximum only on the boundary of $\mathfrak {D}$ unless all the functions $f_{1}(z),f_{2}(z),\ldots ,f_{n}(z)$ are constants.

We assume that the functions $f_{1}(z),f_{2}(z),\ldots ,f_{n}(z)$ are regular and single-valued in the domain $\mathfrak {D}$. Let $p_{1},p_{2},\ldots ,p_{n}$ denote positive numbers. The function

is continuous in $\mathfrak {D}$. It reaches its maximum only on the boundary of $\mathfrak {D}$ unless all the functions $f_{1}(z),f_{2}(z),\ldots ,f_{n}(z)$ are constants.

The function $f(z)$ is supposed to be regular in the multiply connected closed domain $\mathfrak {D}$ and $\left|f(z)\right|$ single-valued in $\mathfrak {D}$. [$f(z)$ is not necessarily single-valued.] The absolute value $\left|f(z)\right|$ attains its maximum at a boundary point of $\mathfrak {D}$. The maximum cannot be attained at an inner point of $\mathfrak {D}$ unless $f(z)$ is a constant.

The assumptions of problem 278 are weakened insofar as (3) is satisfied in all but possibly finitely many boundary points $z_{1},z_{2},\ldots ,z_{n}$ of $\mathfrak {R}$. An other assumption, however, is added, namely that there exists a positive number $M'$ for which the inequality

holds everywhere in $\mathfrak {R}$. (Only the case $M'>M$ is interesting.) This modification of the hypothesis does not change the conclusion of problem 278 that under those conditions $\left|f(z)\right|\leq M$, $\left|f(z)\right|<M$ respectively. [In the case where the point at infinity belongs to $\mathfrak {R}$ and all the boundary points of $\mathfrak {R}$ lie in the disk $\left|z\right|<r$ we examine the comparison function $(2r)^{n}\displaystyle \prod_{\nu =1}^{n} (z-z_{\nu })^{-1}$.]

Prove that a function, regular in some domain and not identically constant, cannot attain its greatest absolute value at an interior point of this domain. (The maximum modulus principle.)

Let the function $f(z)$ be regular in a domain $G$. Prove that if, for every sequence of points $z_{n} \in G$ converging to some boundary point of the domain $G$, the inequality

holds, then either the inequality $\left|f(z)\right| < M$ holds at every interior point of the domain $G$, or $f(z) \equiv M e^{i \varphi }$.

Let $P(z)$ be a polynomial of degree $n$, and let $M(r) = \max_{\left|z\right| = r}\left|P(z)\right|$. Prove that for $0 < r_{1} < r_{2}$ the inequality

holds, and equality, for even one pair of values $r_{1}$ and $r_{2}$, is possible only for a polynomial of the form $P(z) = a z^{n}$.

Let $P(z) = z^{n}+a_{1} z^{n-1}+\ldots +a_{n}$. Prove that either the inequality $\left|P(z)\right| > 1$ holds at some point of the circle $\left|z\right| = 1$, or $P(z) \equiv z^{n}$.

Let $P(z)$ be a polynomial of degree $n$ satisfying, on the interval $(-1,1)$, the inequality $\left|P(z)\right| \leqslant M$. Prove that at every point $z_{0}$ lying outside this interval the inequality $\left|P\left(z_{0}\right)\right| \leqslant M(a+b)^{n}$ holds, where $a$ and $b$ are the semi-axes of the ellipse with foci $-1$ and $1$ passing through the point $z_{0}$.

Let the functions $f_{1}(z), \ldots , f_{m}(z)$ be regular in a domain $G$, and let $M = \varlimsup_{z \rightarrow \partial G}\left\{ \left|f_{1}(z)\right|+\ldots +\left|f_{m}(z)\right|\right\}$. Prove that if at least one of the $f_{k}(z)$ is not identically constant, then at every point of $G$ the inequality $\left|f_{1}(z)\right|+\ldots +\left|f_{m}(z)\right| < M$ holds.