# Does maximal principle imply open mapping theorem for any continuous function?

At first I spent a lot of time looking for counterexamples because I had never seen such a claim that M.P. implies O.M.T.. But later I realized the claim might be true, so I just had a try and proved it! I checked my steps and I think the proof is correct. But I still feel unsafe, so I asked this question.

There is a same question in math stackexchange. Here is the link: Prove the open mapping theorem by using maximum modulus principle. I found the answer unsatisfactory, at least it didn't provide a counterexample. I posted my proof right behind the original answer.

The following is the proof. Just for convenience for the reader, I copied it again here.

Suppose $$f$$ is the non constant continuous function that satisfies the maximal principle property. 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 using translation $$g$$ to compose with $$f$$, to make sure $$|g(f(x)|$$ has the maximal moduli over $$D$$. We can do that because $$f(D)$$ is compact, so there exits $$z_0$$ in $$f(D)$$ such that dist($$f(x), z_0$$) is equal to the maximum of the distance from $$f(x)$$ to all points in $$f(D)$$, then consider $$g(z)=z-z_0$$. Now $$g(f(x))$$ achieves its local maximal point over $$D$$ at $$x$$, by hypothesis, $$g(f(x))$$ is constant function on $$D$$ since maximal principle is preserved under translation, i.e., a non constant function still gets its local maximal modulus on the boundary of the local neighborhood after translation.

Therefore, $$f$$ is a constant function on $$D$$. Using typical method (for example, Lebesgue lemma) for path-connectness of a domain, we say that $$f$$ is constant on its domain.

So, it seems that maximal modulus principle implies open mapping theorem for any continuous functions, but I've never seen such a claim before and nobody mentioned it. Any comments would be appreciated.

• Real-valued harmonic functions have the maximum modulus property, but aren't open mappings. You'd need to put more properties of holomorphic functions into a proof. Apr 15, 2014 at 23:32
• But let's just assume u to be a harmonic function defined on an open subset of plane. So u is locally the real part of a holomorphic function f. f satisfies open mapping theorem, thus the image of f is open, and image of u is the intersection of the real line and image of f, so image of u is open. Then u is open. Is it correct? Apr 15, 2014 at 23:47
• As a mapping to $\mathbb{R}$, it is indeed open. But not as a mapping to $\mathbb{C}$. Anyway, if an interior point of $D$ is mapped to a boundary point of $f(D)$, you need not be able to apply a rigid motion to get that to be a point of maximal modulus. Consider the possibility that $f(D)$ is an annulus, and $z\in D$ is mapped to a point on the inner boundary circle. Apr 16, 2014 at 0:14
• Yeah, I mean u mapping to $\mathbb{R}$. I've edited the proof, replacing rigid motion with translation, since "rigid motion" is not proper here. I mean, since $f(D)$ is compact, there exits $z_0$ in $f(D)$ such that dist($f(x), z_0$) is equal to the maximum of the distance from $f(x)$ to all points in $f(D)$. Now consider the function $f$-$z_0$. Apr 16, 2014 at 0:29
• I don't understand the essence of your argument. What is the point of the translation, and how does this give you a local maximum in the interior? Apr 16, 2014 at 1:55

The function $f(x)=x^2$ on the line (or its analog $f(x)=|x|^2$ in higher dimensions) satisfies the maximum principle, but is not open as a map into $\mathbb R$.
If you require both maximum and minimum principles to hold for a continuous function $f$, in their strict form, then yes, it follows that $f$ is open as a map into $\mathbb R$. Proof by contrapositive: if the image of an open set is not open, then it's either a closed or a half-closed interval. In both cases we have an interior extremum point, contrary to the maximum/minimum principles.