Schwarz lemma on general domains via conformal maps

Let $f$ be analytic in the closed unit disc, with $f(-\log 2) = 0$ and $\left|f(z)\right| \leq \left|e^{z}\right|$ for all $z$ with $\left|z\right| = 1$. How large can $\left|f(\log 2)\right|$ be?

Suppose that $f$ is analytic on the open upper half-plane and satisfies $\left|f(z)\right| \leq 1$ for all $z$, $f(i) = 0$. How large can $\left|f(2i)\right|$ be under these conditions?

Let $f(z)$ be analytic on the right half-plane $H = \left\{ z \mid \mathrm{Re}\left(z\right)>0\right\}$ and suppose $\left|f(z)\right| \leq 1$ for $z \in H$. Suppose also that $f(1) = 0$. What is the largest possible value of $\left|f'(1)\right|$?

Suppose that $w = f(z)$ maps a simply connected region $G$ one-to-one and conformally onto a circular disk $D_r$ with center $w = 0$, radius $r$, such that $f(a) = 0$ and $\left|f'(a)\right| = 1$ for some point $a \in G$.

If $f : D(0,1) = \left\{ \left|z\right| < 1\right\} \to \mathbb {C}$ is an analytic function which satisfies $f(0) = 0$, and if

prove that

Let $\Omega = \left\{ z \in \mathbb {C}; -1 < \im z < 1\right\}$, and let $\mathcal F$ be the family of all analytic functions $f : \Omega \to \mathbb {C}$ such that $\left|f\right| < 1$ on $\Omega$ and $f(0) = 0$. Find

Let $f$ be an analytic function on $D = \left\{ z; \left|z\right| < 1\right\}$ such that $f(0) = 1$, and suppose that $\left|1+f(z)\right| \leq 1 + \left|f(z)\right|$ whenever $\left|z\right| < 1$. Prove that $\left|f'(0)\right| \leq 4$.

Let $P$ be the set of holomorphic functions $f$ on the open unit disc so that (i) Both the real and imaginary parts of $f(z)$ are positive for $\left|z\right| < 1$, (ii) $f(0) = 1+i$. Let $E = \left\{ f(\frac12) : f \in P\right\}$. Describe $E$ explicitly.

Let $\Omega = \left\{ z : \left|z-4\right| < 4\right\}$. If $\mathcal F$ is the family of all analytic functions on $\Omega$ such that $\left|f\right| < 1$ in $\Omega$ and $\lim_{w \to \infty } f(w) = 0$, find $\sup_{f \in \mathcal F} \left|f(8)\right|$. Your answer should be an explicit number, and you should prove your assertion.

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 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.)

Suppose $f \in H\left(\mathrm{II}^{+}\right)$, where $\mathrm{II}^{+}$is the upper half plane, and $\left|f\right| \leq 1$. How large can $\left|f^{\prime }(i)\right|$ be? Find the extremal functions. (Compare the discussion in Sec. 12.5).

Let $\Phi$ be the set of all $f \in H(U)$ such that $0 < \left|f(z)\right| < 1$ for $z \in U$, and let $\Phi_{c}$ be the set of all $f \in \Phi$ that have $f(0) = c$. Define

Find $M$, and $M(c)$ for $0 < c < 1$. Find an $f \in \Phi$ with $f^{\prime }(0) = M$, or prove that there is no such $f$. Suggestion: $\log f$ maps $U$ into the left half plane. Compose $\log f$ with a properly chosen map that takes this half plane to $U$. Apply the Schwarz lemma.

Suppose $f$ and $g$ are holomorphic mappings of $U$ into $\Omega , f$ is one-to-one, $f(U) = \Omega$, and $f(0) = g(0)$. Prove that

Let the function $f(z)$ be regular in the strip $\left|\mathrm{Re}\left(z\right)\right| < \pi / 4$, satisfy there the inequality $\left|f(z)\right| < 1$, and vanish at the point $z = 0$. Prove that $\left|f(z)\right| \leqslant \left|\operatorname {tg} z\right|$ in this strip.