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Let $\{a_n \}$ be a sequence of real numbers such that for any $z \in \mathbb{C}$ with $\operatorname{Im}z>0$, the series $$ h(z)=\sum_{n=1}^\infty a_n \sin(nz) $$ is convergent. Show that $h$ is holomorphic on $\{\operatorname{Im} z>0\}$.

I tried to prove this by considering the partial sum of $h$, which is holomorphic for all finite sum. While it seems that I need the sequence to converge uniformly on each compact subset of the upper half space. If this is true then the limit $h$ could be holomorphic, but I have trouble proving it. Is this a viable path?

Thanks for any help.

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  • $\begingroup$ I wonder if Morera's theorem can do it? $\endgroup$ Commented Jan 5, 2015 at 0:49
  • $\begingroup$ I tried it before, but I had problem proving the integral over a closed path to be zero, since it would need to move the limit into the integral. This is also why I think maybe we need the uniform convergence. $\endgroup$
    – Liu
    Commented Jan 5, 2015 at 0:52
  • $\begingroup$ . . . and how is one to make use of the fact that the domain is the upper half-plane? ${}\qquad{}$ $\endgroup$ Commented Jan 5, 2015 at 0:57
  • $\begingroup$ ...I'm not really sure, but if this works then since it would be real on the real line, we could use Schwarz reflection theorem to say $h$ is indeed entire. Is it right? $\endgroup$
    – Liu
    Commented Jan 5, 2015 at 1:07
  • $\begingroup$ . . . provided it's holomorphic on the real line. $\endgroup$ Commented Jan 5, 2015 at 3:48

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Hint: using $\sin w = \frac1{2i}(e^{iw}-e^{-iw})$ and thus $|\sin iny| = \frac 12(e^{ny}-e^{-ny})$ for $y>0$, show that convergence of $h(z)$ at $z=iy$ implies $|a_n|e^{ny} \to 0$ as $n\to 0$, which in turn implies uniform absolute convergence of $h(z)$ for all $z$ with $|\Im z|\le y-1$ say. That shows that $h$ is entire even.

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  • $\begingroup$ Could you explain why $|a_n|e^{y} \rightarrow 0$ as $n \rightarrow 0$ implies the uniform convergence of $h$? (For each fixed $y$, we only get $|a_n| \rightarrow 0$.) $\endgroup$
    – Liu
    Commented Jan 5, 2015 at 2:12
  • $\begingroup$ Bah, I left out a bunch of $n$s - that's why my answer didn't make sense. Hopefully it does now.... $\endgroup$ Commented Jan 5, 2015 at 5:38

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