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My question is related to the proof here: https://en.m.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis)

Consider the closed Ball $B$ with radius $d$ around $z_0$ and a holomorphic function $g(z)$ which vanishes at $z_0$. Then the closed ball is mapped to $g(B)$ which should be also closed. Let $e={\rm min}_{z \in \partial B} |g(z)|$ be the minimal value of $|g(z)|$ on the boundary of $B$, so that the open disk $D$ at $0$ with radius $e$ is fully contained in $g(B)$, which is my assumption, since otherwise $|g(z)|$ would have a maximum inside $B$.

Now in the above link it seems to require Rouches theorem to claim that for any $w\in D$ there is a solution to $g(z)=w$ in $B$, but I feel like this statement is somehow trivial which it apparently is not. If $D$ is contained in $g(B)$ shouldn‘t this be automatically the case? What am I overlooking here?


So is the following argument valid?

Using the notation above: The boundary of $D$ - which has modulus $e$ - maps onto the boundary of $g^{-1}(D)$. This is because $g$ is holomorphic and by the maximum modulus principle attains its maximum modulus at the boundary of $g^{-1}(D)$, which however by construction is $e$ and therefore the boundary of $D$. Thus any value inside $g^{-1}(D)$ maps onto a value $g(z)$ with modulus $<e$ since $g$ is not constant.

The same argument is true for $B$ i.e. the maximum modulus is attained at the boundary of $B$. Since $|g(z_0)|=0$ and $|g(z)|$ does not have a local maximum inside $B$, but $|g(\partial B)|\geq e$ we must have $g^{-1}(D) \subseteq B$ or $D\subseteq g(B)$. (For any $\phi \in [0,2\pi]$ and $r_1<r_2<d$ we have $\left|g\left(z_0 + r_1\, {\rm e}^{i\phi}\right)\right|<\left|g\left(z_0 + r_2\, {\rm e}^{i\phi}\right)\right|<\left|g\left(z_0 + d\, {\rm e}^{i\phi}\right)\right|$)

I'd be really thankful if somebody could look over this argument!

map of g(z)

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  • $\begingroup$ You don't automatically have the inclusion $D\subseteq g(B)$, not without proving it. $\endgroup$ – quasi Nov 2 '18 at 1:08
  • $\begingroup$ So would it be also possible to proof this part with maximum modulus principle for $|g(z)|$ which is then attained at the boundary of $g^{-1}(D)$. $\endgroup$ – Diger Nov 2 '18 at 11:41
  • $\begingroup$ I don't see how that would immediately yield $D\subseteq g(B)$. But feel free to post what you think is a proof, either by editing your question, or by posting an answer. That way someone (maybe not me) will be able to review your proposed argument. $\endgroup$ – quasi Nov 2 '18 at 12:19
  • $\begingroup$ I actually have issues to formulate it, but if the maximum is obtained at the boundary (which by definition is $e$), how can any $z$ inside $g^{-1}(D)$ attain a value $g(z)$ larger than $e$ in modulus. $\endgroup$ – Diger Nov 2 '18 at 14:39
  • $\begingroup$ I added a picture above. The point is that I think since B is simply connected so is g(B) and the minimal radius e defines D which then is fully contained in g(B). Also I think a boundary of B becomes a boundary of g(B). $\endgroup$ – Diger Nov 2 '18 at 18:15

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