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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.

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    $\begingroup$ 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. $\endgroup$ Apr 15, 2014 at 23:32
  • $\begingroup$ 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? $\endgroup$
    – student
    Apr 15, 2014 at 23:47
  • $\begingroup$ 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. $\endgroup$ Apr 16, 2014 at 0:14
  • $\begingroup$ 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 $f$-$z_0$. $\endgroup$
    – student
    Apr 16, 2014 at 0:29
  • $\begingroup$ 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? $\endgroup$
    – Braindead
    Apr 16, 2014 at 1:55

1 Answer 1

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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.

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  • $\begingroup$ Thanks. This is for the map from a metric space to real numbers. I'd like to knwo a general answer. $\endgroup$
    – student
    May 22, 2014 at 13:45
  • $\begingroup$ @QinfengLi The general answer is no. And it w was given in the thread to which you linked. $\endgroup$
    – user147263
    May 23, 2014 at 3:00
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    $\begingroup$ @user147263 But that answer doesn't give a counterexample. The answer just tells me in philosophy it is impossible, but in mathematics a lot of unrelated things end up being closely linked. $\endgroup$
    – student
    May 23, 2014 at 13:18

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