Existence of an analytic function with certain properties Let $S = \{z \in \mathbb{C} : 0 \leq Re(z) \leq 1\}$, and let $K > 0$ and $z_0$ in the interior of $S$ be given. I want to know whether there is a function $f : S \rightarrow \mathbb{C}$ satisfying the following:


*

*$f$ is analytic in the interior of $S$

*$f$ is continuous on $S$

*$f(z_0) = 1$

*$f$ is never $0$

*For every $z$ in the boundary of $S$, $\left|f(z)\right| \geq K$


Thanks.
 A: Yes, such a function exists for every $K > 0$.
If $K \leqslant 1$, we can simply take the constant function $f(z) \equiv 1$, but we can also find a non-constant function with the following construction.
The $0^{\text{th}}$ step is to translate the strip by some purely imaginary constant, $\varphi_0 \colon z \mapsto z + iu$ for some $u\in \mathbb{R}$ that depends on $K$ and $z_0$. We will later explain what values of $u$ are suitable.
The first step is then to map the strip to the left half-plane via
$$\varphi_1 \colon z \mapsto \exp \Bigl(\pi i\bigl(z + \tfrac{1}{2}\bigr)\Bigr).$$
The boundary line $\operatorname{Re} z = 0$ of $S$ is mapped (bijectively) to the half of the imaginary axis in the upper half-plane, $\{ z : \operatorname{Re} Z = 0, \operatorname{Im} z > 0\}$, and the boundary line $\operatorname{Re} z = 1$ is mapped bijectively to the half of the imaginary axis in the lower half-plane.
Then we map the left half-plane to the punctured unit disk via $\exp$. Then $\psi = \exp \circ\, \varphi_1 \circ \varphi_0$ is an entire function, $\psi$ doesn't attain the value $0$, it maps the boundary of the strip $S$ (on)to the unit circle, and the strip $S$ to the punctured unit disk.
To satisfy the condition $f(z_0) = 1$, we take
$$f(z) = \frac{\psi(z)}{\psi(z_0)}.$$
Form the construction so far, it is clear that $f$ satisfies the first four bullet points (whichever $u$ was chosen in the $0^{\text{th}}$ step), and that
$$\lvert f(z)\rvert = \frac{1}{\lvert \psi(z_0)\rvert}\quad\text{for}\quad z \in \partial S.$$
So all that remains is to see that we can choose $u$ in the $0^{\text{th}}$ step so that $\lvert \psi(z_0)\rvert \leqslant K^{-1}$. For $K \leqslant 1$, that is no restriction, all $u\in \mathbb{R}$ work.
We know $\lvert e^w\rvert = e^{\operatorname{Re} w}$, so
$$\lvert \psi(z_0) \rvert \leqslant K^{-1} \iff \operatorname{Re}\bigl(\varphi_1(z_0+iu)\bigr) \leqslant -\log K.\tag{$\ast$}$$
With
$$\varphi_1(z+iu) = \exp \Bigl(\pi i\bigl(z + iu + \tfrac{1}{2}\bigr)\Bigr) = i e^{-\pi (u+\operatorname{Im} z)}\cdot e^{\pi i \operatorname{Re} z},$$
we find
$$\operatorname{Re} \varphi_1 (z_0 + iu) = - e^{-\pi(u+\operatorname{Im} z_0)}\cdot \sin \bigl(\pi\cdot \operatorname{Re} z_0\bigr).$$
Since $0 < \operatorname{Re} z_0 < 1$, we have $\sin \bigl(\pi\cdot \operatorname{Re} z_0\bigr) > 0$, and we see that
$$\lim_{u\to -\infty} \operatorname{Re} \varphi_1(z_0+iu) = -\infty,$$
so for all sufficiently small $u$ (where "sufficiently" depends on $z_0$) the condition $(\ast)$ is satisfied.
