Let $\Omega$ be a simply connected open set in the complex plane and $\gamma$ be a simple path inside $\Omega$. Suppose $f_n$ is a sequence of holomorphic functions converging pointwise to 0 on $\gamma$. Does it imply that $f_n$ converges pointwise on the region enclosed by $\gamma$?
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4$\begingroup$ I don't see why the vote to close, since the question does not assume the sequence $(f_n)$ is bounded on $\gamma$ $\endgroup$– Yemon ChoiCommented Feb 8, 2015 at 20:42
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2$\begingroup$ If you assume that $\int_\gamma|f_n|$ has an upper bound independent of $n$, I can give a proof. Do you make any uniform assumptions of this spirit? $\endgroup$– Joonas IlmavirtaCommented Feb 8, 2015 at 20:46
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1 Answer
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For a counterexample, let $\gamma$ be the unit circle. Let $$A_n = \{z \in \gamma:\; \text{Im}(z) \in [-1,0] \cup [1/n, 1]\}$$ By Runge's theorem there is a polynomial $f_n$ such that $|f_n| < 1/n$ on $A_n$ but $f_n(0) = (-1)^n$. We then have $f_n \to 0$ pointwise on $\gamma$ but $f_n(0)$ does not converge.
answered Feb 8, 2015 at 22:07
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$\begingroup$ I understand that Runge's theorem gives you the first condition, but how do you control $f_n(0)$? $\endgroup$– PabloCommented Feb 9, 2015 at 1:13
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$\begingroup$ Use Runge on $A_n \cup \{0\}$, and add a small constant term. $\endgroup$ Commented Feb 9, 2015 at 20:49
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2$\begingroup$ So, to which function (or sequence of functions) do you want to converge uniformly in Runge's theorem? Can you please make your answer more detailed? $\endgroup$– PabloCommented Feb 9, 2015 at 21:26
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2$\begingroup$ Take $g_n(z) = 0$ on $A_n$, $g_n(0)=(-1)^n$, polynomial $h_n$ so $|h_n - g_n| < 1/(2n)$ on $A_n \cup \{0\}$, and $f_n = h_n + ((-1)^n - h_n(0))$. $\endgroup$ Commented Feb 9, 2015 at 21:55
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$\begingroup$ Fantastic! By the way, we can do without the correction $((-1)^n-h_n(0))$, which goes to 0 uniformly anyway. $\endgroup$ Commented Feb 13, 2015 at 2:39