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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.)

mr eyeglasses
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1 Answers1

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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 $a \in \partial G$. $\blacksquare$

Alonso Delfín
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  • If $f$ has no zeroes, does that automatically make $f$ a constant? – mr eyeglasses May 21 '15 at 03:23
  • 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 ? – Alonso Delfín May 21 '15 at 03:27
  • I don't understand why $[1/f]$ attains a maximum – mr eyeglasses May 21 '15 at 03:27
  • 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 $\partial G$ – Alonso Delfín May 21 '15 at 03:28
  • 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 $\partial$ – Alonso Delfín May 21 '15 at 03:31
  • The maximum principle says if $1/f$ is a non-constant holomorphic function in $G$ then $1/f$ has no maximum in $G$. So $1/f$ is continuous in $G$, but how do you know it's a non-constant or constant function? And how do you know if it's holomorphic in $G$, it's continuous in $\partial G$? – mr eyeglasses May 21 '15 at 03:32
  • Let us continue this discussion in chat. – Alonso Delfín May 21 '15 at 03:32
  • We only need part II) actually, what is the point to post part I)? – Bach Mar 23 '19 at 12:39