Schwarz's Lemma

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 $f(z)\equiv 0$.