Prove the open mapping theorem by using maximum modulus principle

Question

The open mapping theorem says a non constant analytic function maps open sets to open sets.

The maximum modulus principle says if $f$ a non constant analytic function on an open connected set $D\subset\mathbb{C}$, then $|f|$ does not attain a local maximum on $D$.

It it known that one application of the open mapping theorem is to prove the maximum modulus principle. But what about the other way around? Can we use the maximum modulus principle(possibly plus some other results) to prove the open mapping theorem?

The reason I am interested in this question is because after I see the proof in wiki-pedia, personally I found the idea in this proof somewhat "hidden", it is not that intuitive (at least to me).

Answer 1

Sorry for the late response. Here's a proof of the open mapping theorem assuming the maximum modulus principle.

First, we need the "minimum modulus principle". That is, if $f$ is a non-constant analytic function on an open connected set $D \subset \mathbb{C}$, and $f$ has no zeroes in $D$, then $| f |$ cannot attain a minimum in $D$. The proof follows trivially by applying the maximum modulus principal to the function $1/f$ which is analytic on $D$.

Now suppose $D \subset \mathbb{C}$ is open and connected, and $f$ is a non-constant analytic function on $D$. Let $U \subset D$ be open, and let $w_0 \in f(U)$, say $w_0=f(z_0)$ with $z_0 \in U$. We must show that there is a disc centered at $w_0$ which is contained in $f(U)$.

Choose $t>0$ so that $\overline {D_t(z_0)} \subset U$ and $f(z) \neq w_0$ for any $z \in \overline {D_t(z_0)}$ other than $z_0$. Let $m=\inf \{|f(z)-w_0| : |z-z_0|=t \} > 0 $. Suppose $|w-w_0| < m/3$, and that there is no $z \in U$ such that $f(z)=w$. Then the function $g(z)=f(z)-w$ is analytic, non-constant, and has no zeroes in the open connected set $D_t(z_0)$, so the minimum modulus principle shows that $g$ cannot attain a minimum modulus in $D_t(z_0)$. However, $g$ does attain a minimum modulus in the compact set $ \overline {D_t(z_0)} $, so this minimum modulus must occur on the boundary circle defined by $|z-z_0|=t$. But if $|z-z_0|=t$, then

$|g(z)|=|f(z)-w| \geq |f(z)-w_0| - |w_0-w| \geq 2m/3 $, and

$|g(z_0)| = |w_0-w| < m/3 < 2m/3$.

This gives a contradiction since $z_0$ is obviously in the interior of the disc in question. Therefore $f(z)=w$ for some $z \in D_t(z_0)$, and $D_{m/3}(w) \subset f(U)$, showing that $f(U)$ is open and proving the theorem.

Answer 2

I think maximal principle implies open mapping theorem.

Suppose $f$ is a non-constant analytic function. If open mapping theorem is not true, then $f$ maps an interior point $x$ of a small closed neighborhood $D$ to the point $f(x)$ which is on the boundary of $f(D)$.

Then there exists a $b \in \mathbb{C}$ such that $|f(z) - b|$ has a local maximum modulus over $D$. We can do that because $f(D)$ is compact, so imagine a vector starting at $f(z)$ and pointing out of $f(D)$. Choose $b$ accordingly so that the magnitude of that vector is $|f(z) - b|$.

Now $f(z) - b$ is analytic and hence has the maximum modulus principle (MMP). $|f(z) - b|$ achieves a local maximum over $D$ at $z$ in the interior. Hence the MMP guarantees that $f(z) - b$ is constant on $D$. Therefore $f$ is constant on $D$.

Finally, Using typical method (like Lebesgue lemma) for path-connectedness of a domain, we say that $f$ is constant on its domain.

Answer 3

You can't prove the open mapping theorem (OMT) with the maximum modulus principle. Because the maximum modulus principle is not a tool that is suitable for proving the open mapping theorem. The maximum modulus principle is insufficiently sophisticated to understand the topology of the complex plane. It is merely a statement about $|f|$, not about $f$ itself. The maximum modulus principle holds for lots of maps, e.g., the product of $f$ with any unimodular function, which have no reason to be open. Etc., etc.

I will try to explain the idea of the OMT proof instead. Pick a point $z_0$ in the domain of $f$. Consider the series $f(z)=f(z_0)+a_k(z-z_0)^k+\dots$ where by $k$ is the first index after $0$ such that $a_k\ne 0$. Let $F(z)=f(z_0)+a_k(z-z_0)^k$. The map $F$ is open, because it's just a power of $z$ composed with some shifts. (The power of $z$ can be written out explicitly in polar coordinates, to check that it's open.)

When $z$ is sufficiently close to $z_0$, $|f(z)-F(z)|$ is very small, because it consists of powers of $z-z_0$ higher than $k$. This allows us to use Rouché's theorem to conclude that the values attained by $F$ will also be attained by $f$ (I am omitting the details of estimates here). Since the image of $F$ contains a neighborhood of $f(z_0)$, so does the image of $f$.