I'm trying to do this problem from Gamelin's book:

Let $f_n(z)$ be a sequence of analytic functions on a domain (= open connected set) $D$ such that $f_n(D) \subset D,$ and suppose that $f_n$ converges to $f$ uniformly on each compact subset of $D$. Show that either $f(D)\subset D$, or $f(D)$ consists of a single point in $\partial D$.

Now, I can see that $f(D) \subset D\cup \partial D$. If $f$ is not constant, then by open mapping theorem, $f(D)$ is open, so no point in the boundary can be in the image of $f$, hence in this case $f(D) \subset D$. But if $f$ is constant, I can't figure out why the mage of $f$ must be a point in the boundary. Why can't the image be in $D$?

Thanks in advance.


The image can be in $D$. I think you are reading the statement of the problem incorrectly:

"show that either $f(D) \subset D$, or $f(D)$ consists of a single point in $\partial D$."

That isn't saying that if $f$ is a constant, then the constant must be attained on the boundary. Instead it's saying, if the image of $f$ contains a point on the boundary, then it is a constant.

  • $\begingroup$ I think you are right. It says that if the image of the limiting function is not in $D$, i.e. contains some point of the boundary, then the limiting function is constant. $\endgroup$ – A.Γ. Jul 14 '15 at 17:15

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