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Minimum modulus principle: If $f$ is a non-constant holomorphic function a bounded region $G$ and continuous on $\bar{G}$, then either $f$ has a zero in $G$ or $|f|$ assumes its minimum value on $\partial G$.

1) Why does $f$ has to be continuous on $\bar{G}$ and not just $G$?

Proof: Let $F(z) = e^{f(z)}$. Then $|F(z)| = |e^{f(z)}| = e^{Re(f(z))} \geq e^{Re(f(a))}$ on $B(a, r)$.

2) Why is $e^{Re(f(z))} \geq e^{Re(f(a))}$ on $B(a, r)$? How do we know that the real part of $f(z)$ has a minimum in $B(a, r)$?

Proof (cont'd): Then $F(a) = 0$ or $F(z)$ is constant on $B(a,r)$.

3) How do we know that $F(a) = 0$ or $F(z)$ is constant on $B(a,r)$?

Proof (cont'd): $F(z) \neq 0$ so $F(z)$ is constant on $B(a,r)$, thus $f(z)$ is constant on $B(a,r)$ and by the Identity theorem, $f(z)$ is constant on $G$.

(I understand this last part.)

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  • $\begingroup$ Really ? It is not complete at all. You want a complete proof? I can post it as an answer $\endgroup$ – Alonso Delfín May 21 '15 at 2:41
  • $\begingroup$ @AlonsoDelfín Yes, then maybe I will understand it $\endgroup$ – mr eyeglasses May 21 '15 at 2:41
  • $\begingroup$ Ok, I assume you do know and understand the maximun modulus principle right? It is essential to the proof I'll be giving $\endgroup$ – Alonso Delfín May 21 '15 at 2:43
  • $\begingroup$ @AlonsoDelfín I know it, but understanding it is different I guess...I'll try to understand it $\endgroup$ – mr eyeglasses May 21 '15 at 2:44
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    $\begingroup$ Ok I have post an answer, it took me a little bit long than I expected but I hope is clear enough, let me know if have trouble with anything. $\endgroup$ – Alonso Delfín May 21 '15 at 3:15
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I am giving a different proof of the one you have since I believe is not complete and that is why you don´t get it. You must already know the Maximum Principle (not modulus), in case you don´t here it is:

Maximum principle If $f: G \to \mathbb{C}$ is a non-constant holomorphic function in a region $G$, then $|f|$ has no maximum in $G$.

A proof of this can be easily obtained as a corollary of the Open mapping theorem.

Minimum Modulus principle If $f$ is a non-constant holomorphic function a bounded region $G$ and continuous on $\overline{G}$, then either $f$ has a zero in $G$ or $|f|$ assumes its minimum value on $\partial G$.

Proof:

I) Lets assume first that $f$ has no zeros in $\overline{G}$, then of course $1/f$ is holomorphic in $G$ and continuous in $\overline{G}$. Thus, by the Maximum Principle above $|1/f|$ attains its maximum in $\partial G$, that is there exist $a \in \partial G$ such that $$ \left| \frac{1}{f(z)} \right| \leq \left| \frac{1}{f(a)} \right| \ \forall \ z \in \overline{G} $$ So indeed for all $z \in \overline{G}$, $|f(a)|\leq |f(z)|$, thus $|f|$ assumes its minimum value in $a \in \partial G$.

II) On the other side, if $|f|$ does not assumes its minimum value on $ \in \partial G$, then there exist $b \in G$ such that $$ |f(b)| \leq |f(z)| \ \forall \ z \in G $$ Lets prove that $f(b)$ must be $0$. Assume not, then $$ \left| \frac{1}{f(z)} \right| \leq \left| \frac{1}{f(b)} \right| \ \forall \ z \in G $$ but since $b \not\in \partial G$, the Maximum Principle gives that $f$ must be constant (since the maximum is not attained at the boundary) a contradiction. Thus since supposing that $f(b)\neq 0$ implies that $f$ must be constant, then since $f$ is non-constant by hypothesis we can conclude that $f(b)=0$.

It follows now from I) and II) that either $f$ has a zero $b \in G$ or $|f|$ assumes its minimum value on $a \in \partial G$. $\blacksquare$

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  • $\begingroup$ If $f$ has no zeroes, does that automatically make $f$ a constant? $\endgroup$ – mr eyeglasses May 21 '15 at 3:23
  • $\begingroup$ Not quite. In part I) I have proved that if $f$ has no zeros then $f$ assumes its minimum value at the boundary. While in part II) I have proved that if $f(b)\neq 0$ for that specific $b$, then $f$ is constant, which is a contradiction and then $f(b)$ must be $0$. Is it clear @ᴇʏᴇs ? $\endgroup$ – Alonso Delfín May 21 '15 at 3:27
  • $\begingroup$ I don't understand why $[1/f]$ attains a maximum $\endgroup$ – mr eyeglasses May 21 '15 at 3:27
  • $\begingroup$ Ok, since $1/f$ is holomorphic in $G$ and continuous in $\partial G$ then by the Maximum Principle, $1/f$ can not assume a maximum in $G$ right ? Thus the since $\overline{G}$ is compact the maximum must be at $\partial G$ $\endgroup$ – Alonso Delfín May 21 '15 at 3:28
  • $\begingroup$ The relation is required in the maximum modulus principle. For $1/f$ to have a maximum on $\overline{G}$ is must be continuos in $\overline{G}$, since it is holomorphic by hypothesis on $G$ we must ask $1/f$ to be continuos only in $\partial$ $\endgroup$ – Alonso Delfín May 21 '15 at 3:31

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