Herglotz and Caratheodory functions

The function $f(z)$ is regular, its real part is positive in the open disk $\left|z\right|<R$ and continuous in the closed disk $\left|z\right|\leq R$. If the real part becomes identically zero on an arc of the circle the imaginary part of $f(z)$ changes on this arc always in the same sense: it decreases as $\arg z$ increases.

The function

is assumed to be regular and to have positive real part in the open disk $\left|z\right|<1$. Then $\left|a_{n}\right|\leq 1$, $n=1,2,\ldots$ In none of these inequalities can we replace $1$ by a smaller number.

We assume that the function $f(z) = a_{0}+a_{1}z+\cdots +a_{n}z^{n}+\cdots$ is regular in the disk $\left|z\right|<R$ and we denote by $\Delta (f)$ the largest oscillation of the real part of $f(z)$ in $\left|z\right|<R$, i.e.$\Delta (f)$ is the least upper bound of $\left|\re f(z_{1})-\re f(z_{2})\right|$ for $\left|z_{1}\right|<R$, $\left|z_{2}\right|<R$. Then we have

The factor $\dfrac {2}{\pi }$ cannot be replaced by a smaller one.

Interpret this proposition geometrically.

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

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

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