Let $G\subseteq \mathbb{C}$ be a bounded domain with $0 \in G$ and $f:\bar{G}\rightarrow\mathbb{C}$ is continous. Furthermore let $f$ be holomorphic in $G$ and let $|f(z)| \geq e^{Re (z)}$ for all $z\in\partial G$ and $|f(0)|<1$. I've got to show that $f$ has a zero.

My attempt:

Suppose $f(z)$ has no root. So that $|f(z)|>0$ is always fullfilled. Since the conditions for the maximum modulus theorem are fullfilled we apply it to $f(z)$ and $\frac{1}{f(z)}$ (which is also holomorphic on $G$ as $f(z) \neq 0$)

That is:

$0 \leq |e^{Re (z)}|\leq |f(z)| \leq f(C_1)$ for some $C_1 \in \partial G$, and

$0 \leq | \frac{1}{f(z)}|\leq |\frac{1}{e^{Re (z)}}|\leq \frac{1}{f(C_2)}$ for some $C_2 \in \partial G$

Now in the second row we get for $z=0$:

$\frac{1}{f(0)}\leq 1$ which is contradiction to the assumption that $|f(0)|<1$.

Is this proof correct? I mean can $0 \in \partial G$ be true when only $0\in G$ is assumed? Thank you for your help.

  • $\begingroup$ What do you mean by "$f$ has a root"? $f$ is a function not an equation. $\endgroup$ – Empty Aug 14 at 16:30
  • $\begingroup$ Of which inequality? Well, I've got to show that f has a zero (or root). $\endgroup$ – Thesinus Aug 14 at 16:40
  • $\begingroup$ I forget the exact question but I will look for it if needed but someone said that for a 0 solution in these cases you can use SOLUTION(X) - SOLUTION(X) ... $\endgroup$ – Jay Aug 14 at 16:41

I don't think that your proof is correct. The inequality $|f(z)| \geq e^{\operatorname{Re} (z)}$ holds only on the boundary of $G$, but not inside $G$. (If it did then $f$ could not have a zero in the domain.)

If one “sees” that $ e^{\operatorname{Re} (z)} = |e^z|$, so that the condition $|f(z)| \geq e^{\operatorname{Re} (z)}$ is equivalent to $|f(z)| \ge |e^z|$ or $\left|\frac{e^z}{f(z)} \right| \le 1$ then the proof becomes apparent:

Assume that $f$ has no zeros in $G$. Then $h(z) = \frac{e^z}{f(z)}$ is holomorphic in $G$. On the boundary we have $$ |h(z)| = \frac{e^{\operatorname{Re} (z)}}{|f(z)|} \le 1 \, , $$ so that $|h(z)| \le 1$ for all $z \in G$. On the other hand, $|h(0)| = \frac{1}{|f(0)|} >1$. This is a contradiction, therefore $f$ must have a zero in $G$.

  • $\begingroup$ Good reasoning! (Deduction :P) $\endgroup$ – Jay Aug 14 at 16:44
  • $\begingroup$ Thank you for your genial post! To my attempt: Doesn't a function always 'send' an argument to just 'one' value. I mean if $f(0)<1$ for $0 \in G$ then it always has to be true no matter where $0$ is. Of course by hypothesis, we can assume $0\in \partial G$or not?, $\endgroup$ – Thesinus Aug 14 at 16:56
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    $\begingroup$ @Thesinus: I do not understand what you are saying, to be honest. But $0\in \partial G$ is not possible, because then $1 > |f(0)| \ge e^{\operatorname{Re} 0} = 1$. $\endgroup$ – Martin R Aug 14 at 16:59
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    $\begingroup$ @Thesinus: It is also given that $0 \in G$, that excludes the possibility that $0$ is on the boundary of $G$: For an open set, $G$ and $\partial G$ are disjoint. $\endgroup$ – Martin R Aug 14 at 17:08

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