The Maximum Principle

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.

The function $f(z)$ is assumed to be regular and $\left|f(z)\right|<1$ in the disk $\left|z\right|<1$. If $f(0)=0$ either the stricter inequality $\left|f(z)\right|<\left|z\right|$ holds for $z\neq 0$ or $f(z)=e^{i\alpha }z$, $\alpha =$ real.

We denote by $z=\varphi (\zeta )$ and $w=\psi (\zeta )$ two schlicht maps of the unit disk $\left|\zeta \right|<1$ into the regions $\mathfrak {R}$ and $\mathfrak {S}$ of the $z$- and $w$-plane resp.; the images of the origin $\zeta =0$ are the points $z=z_{0}$ and $w=w_{0}$ resp. In addition let $0<\varrho <1$ and $\mathfrak {r}$ and $\mathfrak {s}$ be the images of the disk $\left|\zeta \right|\leq \varrho$ under the above mentioned maps. We assume that $w=f(z)$ is a regular analytic function in $\mathfrak {R}$ the range of which belongs to $\mathfrak {S}$ and that $f(z_{0})=w_{0}$. Then $f(z)$ assumes in the subdomain $\mathfrak {r}$ of $\mathfrak {R}$ values that belong to the subdomain $\mathfrak {s}$ of $\mathfrak {S}$. These values are in the interior of $\mathfrak {s}$ unless $f(z)$ is a schlicht function that maps the region $\mathfrak {R}$ onto the region $\mathfrak {S}$.

Let $f(z)$ be regular and $\left|f(z)\right|<1$ for $\left|z\right|<1$. Then

The relation will be an equality only for the linear function

Suppose that $f(z)$ is regular for $\left|z\right|<R$ and that $A(r)$ denotes the maximum of the real part of $f(z)$ for $\left|z\right|\leq r$, $0\leq r<R$. Then we have the inequality

where $\lim \limits_{r\to R-0} A(r)=A(R)$ [$A(r)$ increases monotonically with $r$, problem 313]. There is equality only for the linear function

(continued). The maximum $M(r)$ of the absolute value of $f(z)$ in the disk $\left|z\right|\leq r$ is restricted by

We assume that $f(z)$ is regular, non-zero and bounded for $\left|z\right|<R$. Then

where $\lim \limits_{r\to R-0} M(r) = M(R)$.

The functions $f_{1}(z),f_{2}(z),f_{3}(z),\ldots ,f_{n}(z),\ldots$ are supposed to be regular, non-zero, and of absolute value smaller than $1$, for $\left|z\right|<1$. If the series

is absolutely convergent, the series

is absolutely convergent for $\left|z\right|\leq \dfrac {1}{3}$.

We assume that the function $f(z)$ is regular, has positive real part for $\left|z\right|<1$ and that $f(0)$ is real. Then we have

There is equality only if

Let $f(z)$ be a regular function with $\left|\re f(z)\right|<1$ for $\left|z\right|<1$. If $f(0)=0$ the stronger inequality

holds; in addition

Equality occurs if and only if

Suppose that the function $f(z)$ is regular for $\left|z\right|<R$ and that $\Delta$ denotes the oscillation of its real part in the disk, i.e.

for $\left|z_{1}\right|<R$, $\left|z_{2}\right|<R$. Then the oscillation of its real part in the smaller disk $\left|z\right|\leq r$, $r<R$, is

The oscillation of the imaginary part is restricted by the inequality

We denote by $\mathfrak {T}$ the infinite region of the $z$-plane that is symmetric with respect to the $x$-axis and whose points are characterized by the inequalities

where $k(x)$ is a positive and continuous function defined for $x\geq 0$. Then there exists a positive function $h(x)$ that depends only on $\mathfrak {T}$ and has the following properties: If $F(z)$ is regular in $\mathfrak {T}$ and bounded from below, $\left|F(z)\right|>c$, then

is bounded from above as $x$ increases to infinity. (The theorem is particularly interesting when $k(x)$ converges monotonically to $0$; i.e.when the region $\mathfrak {T}$ is ``tapering''; whereas an analytic function can increase arbitrarily fast along a ray (Part IV, problem 180) the rate of increase is limited if the function grows to infinity in some properly chosen neighbourhood of the ray. The result can be formulated with reference to the points of the modular graph somewhat vaguely so: If none should fall below a certain minimum standard none may rise arbitrarily high.)

Let $f(z)$ be a regular function and $\left|f(z)\right|<1$ in the unit disk $\left|z\right|<1$; in addition let $f(z)$ be regular at $z=1$ and $f(0)=0$, $f(1)=1$. Then $f'(1)$ must be real and $f'(1)\geq 1$.

We assume that the function $f(z)$ is regular and $\left|f(z)\right|<1$ in the unit disk $\left|z\right|<1$, and that, besides, $f(z)$ is regular for $z=1$ and $f(1)=1$. Then the derivative $f'(1)$ is real and

Suppose that $f(z)$ is regular and $\im f(z)>0$ in the upper half-plane $\mathrm{Im}\left(z\right)>0$ and that, besides, $f(z)$ is regular at the point $z=a$ of the real axis and $f(a)=b$, $b$ real. Then $f'(a)$ is real and positive and we have the inequality

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 $f(z)\equiv 0$.

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.

In three dimensional space, the $n$ points $P_{1},P_{2},\ldots ,P_{n}$ are given and $P$ denotes a variable point. The function

($\overline{PP_{\nu }}$ is the distance between $P$ and $P_{\nu }$) of the point $P$ assumes its maximum in any domain on the boundary. (Generalization of problem 137.)

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.

Let the function $f(z)$ be regular in the disk $\left|z\right|<R$. Suppose

Then

This means that in an orthogonal system of coordinates the graph of $\log M(r)$ as a function of $\log r$ appears as a convex curve. (Hadamard's three circle theorem.) [Examine $z^{\alpha }f(z)$ with suitably chosen $\alpha$.]

(continued). The function $\log M(r)$ is a strictly convex function of $\log r$ unless $f(z)$ is of the form $az^{\alpha }$, $a,\alpha$ constants, $\alpha$ real: this is the only type of function for which the inequality in problem 304 becomes an equality.

Suppose that the function $f(z)$ is regular for $\left|z\right|<R$ but not of the form $cz^{n}$, $c$ constant, $n$ integer; let

denote the arithmetic mean of $\left|f(z)\right|^{2}$ on the circle $\left|z\right|=r$, $r<R$. The function $I_{2}(r)$ is monotone increasing with $r$ and $\log I_{2}(r)$ is a strictly convex function of $\log r$.

Let $f(z)$ be regular for $\left|z\right|<R$;

denotes the geometric mean of $\left|f(z)\right|$ on the circle $\left|z\right|=r$, $r<R$. The function $\mathfrak {G}(r)$ is monotone increasing with $r$ and a convex function of $\log r$ (in the wide sense).

The function $f(z)$ is supposed to be regular for $\left|z\right|<R$ and not a constant. We put

The function $I(r)$ is monotone increasing with $r$ and $\log I(r)$ is a convex function of $\log r$. [Problems 299, 304.]

Let $f(z)$ be regular and not a constant in the disk $\left|z\right|<R$. The function $w=f(z)$ maps the circle $\left|z\right|=r$, $r<R$, in the $z$-plane onto a curve in the $w$-plane with length $l(r)$. The ratio $\dfrac {l(r)}{2\pi r}$ is monotone increasing with $r$.

We suppose that $f(z)$ is regular and not constant for $\left|z\right|<R$ and that $p$ is a positive number. We define

The function $I_{p}(r)$ is monotone increasing with $r$ and $\log I_{p}(r)$ is a convex function of $\log r$. (Cf.problems 306, 308, 307, 267 and 304, for an analogous case see Part IV, problem 19.)

An analytic function that is regular in the closed disk $\mathfrak {R}$ cannot assume real values at all the boundary points of $\mathfrak {R}$ except if it is a real constant.

We denote by $u(x,y)$, $z=x+iy$, a harmonic function that is regular in the disk $(x-x_{0})^{2}+(y-y_{0})^{2}\leq r^{2}$. The value of $u(x,y)$ at the center is

consequently

If a harmonic function is regular in a closed disk its absolute value at the center does not exceed the arithmetic mean of its absolute values on the boundary. Under what conditions is there equality?

A harmonic function is supposed to be regular and single-valued in a domain $\mathfrak {D}$. Then it attains its maximum and its minimum on the boundary and only on the boundary unless it is a constant.

A harmonic function that is regular in the domain $\mathfrak {D}$ and vanishes at all the boundary points of $\mathfrak {D}$ is identically zero.

The equilibrium described in solution 31 is not stable.

A harmonic function is assumed to be single-valued in the domain $\mathfrak {D}$ and regular with the exception of finitely many points at which it becomes $-\infty$ (i.e.the function converges to $-\infty$ as $z$ approaches such a point). Then it assumes its maximum on the boundary of $\mathfrak {D}$.

Let the function $f(z)$ be regular in the disk $\left|z\right|\leq R$. The function $w=f(z)$ maps the disk onto a certain domain of the $w$-plane that is assumed to be star-shaped with respect to $w=0$. Let $f(0)=0$. Then the images of the concentric circles $\left|z\right|=r$, $r<R$, are also star-shaped with respect to the origin.

Suppose that the function $f(z)$ is regular in the disk $\left|z\right|\leq R$ and that it maps the disk onto a convex domain of the $w$-plane. Then the image of an arbitrary circle in the open disk $\left|z\right|<R$ is also convex.

Let $u_{1}(x,y),u_{2}(x,y),\ldots ,u_{n}(x,y)$, $z=x+iy$, be regular harmonic functions in a domain $\mathfrak {D}$. The continuous function

assumes its maximum on the boundary of $\mathfrak {D}$.

Consider a regular harmonic function in the disk $\left|z\right|<R$. We denote by $A(r)$ its maximum on the circle $\left|z\right|=r$, $r<R$. When $0<r_{1}<r_{2}<r_{3}<R$ we have

i.e.$A(r)$ is a convex function of $\log r$.

Deduce Hadamard's three circle theorem 304 from problem 320 and vice versa problem 320 from Hadamard's three circle theorem.

Let $\alpha$ be given, $0<\alpha <\dfrac {\pi }{2}$. The function $f(z)$ is assumed to be regular in the sector $-\alpha \leq \vartheta \leq \alpha$ ($z=re^{i\vartheta }$). In addition $f(z)$ has the following properties:

1. there exist two positive constants $A$ and $B$ such that

   $$\left|f(z)\right| < Ae^{B\left|z\right|} \quad \text{for} \quad -\alpha \leq \vartheta \leq \alpha , \; z=re^{i\vartheta };$$

2. $\left|f(z)\right|\leq 1$ for $\vartheta =-\alpha$ and $\vartheta =\alpha$ (i.e.on the boundary of the sector considered). Then the inequality $\left|f(z)\right|\leq 1$ holds in the entire sector.

Condition (1) of problem 322 can be generalized insofar as the inequality

need not be required to hold in the entire sector but only on the arcs of the circles $\left|z\right|=r_{1},\left|z\right|=r_{2},\ldots ,\left|z\right|=r_{n},\ldots$ intercepted by the rays $\vartheta =-\alpha$, $\vartheta =\alpha$, $\lim \limits_{n\to \infty } r_{n}=\infty$. The conclusion, namely $\left|f(z)\right|\leq 1$ in the entire sector, remains the same. What more general curves can replace the circular arcs?

We modify condition (2) of problem 322 in the following way: There exist in the sector $-\alpha \leq \vartheta \leq \alpha$ two curves $\Gamma_{1}$ and $\Gamma_{2}$ connecting the points $z=0$ and $z=\infty$ that do not intersect and along which $\left|f(z)\right|\leq 1$. This modified condition together with condition (1) as stated in problem 322 implies the inequality $\left|f(z)\right|\leq 1$ in the domain bounded by $\Gamma_{1}$ and $\Gamma_{2}$.

The function $f(z)$ is assumed to be regular in the half-plane $\mathrm{Re}\left(z\right)\geq 0$ and to satisfy the following conditions:

1. there exist two constants $A$ and $B$, $A>0$, $B>0$ such that in the entire half-plane

   $$\left|f(z)\right| \leq Ae^{B\left|z\right|};$$

2. we have for $r\geq 0$

   $$\left|f(ir)\right| \leq 1, \quad \left|f(-ir)\right| \leq 1;$$
3. $$$$

\limsup _{r\to +\infty } \frac{\log \left|f(r)\right|}{r} \leq 0.

\left|f(z)\right| \leq 1 \quad \text{for} \quad \mathrm{Re}\left(z\right)\geq 0.

Let the function $f(z)$ be regular in the half-plane $\mathrm{Re}\left(z\right)\geq 0$ and satisfy the following conditions:

1. there exist two constants $A$ and $B$, $A>0$, $B>0$, such that in the entire half-plane

   $$\left|f(z)\right| < Ae^{B\left|z\right|};$$

2. $f(z)$ is bounded on the imaginary axis,

   $$\left|f(ir)\right| \leq 1, \quad \left|f(-ir)\right| \leq 1, \quad r\geq 0;$$

3. there exists an angle $\alpha$, $-\dfrac {\pi }{2}<\alpha <\dfrac {\pi }{2}$ such that

   $$\lim _{r\to +\infty } \frac{\log \left|f(re^{i\alpha })\right|}{r} = -\infty .$$