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Let $f$ be a holomorphic function on $[0,1]\times \mathbb{R}$. If for each $x\in [0,1]$ fixed, $\lim_{y\to\infty}f(x,y)=0$, prove that $f$ is bounded.

My idea: I do not know how to prove and I also do not know whether the holomorphic condition is essential or not.

Let $f(x,y)$ be a continuous (complex valued) function on $[0,1]\times \mathbb{R}$. Suppose for each $x\in [0,1]$ fixed, $\lim_{y\to\infty}f(x,y)=0$. Then can we prove that $f(x,y)$ is bounded over $[0,1]\times \mathbb{R}$?

(I want to show that for any $\epsilon>0$, there exists $N>0$ such that for any $(x,y)\in [0,1]\times \mathbb{R}$ satisfying $|y|>N$, we have $|f(x,y)|<\epsilon$. Is this statement true? Any counterexamples? How about smooth but not holomorphic functions?)

Thank you a lot.

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  • $\begingroup$ Caution: supremum of continuous functions is not necessarily continuous. $\endgroup$ – Robert Israel Apr 13 '14 at 8:33
  • $\begingroup$ Prof. Arthur's comment does not prove that $g(x)$ is continuous. $\endgroup$ – Shiquan Apr 13 '14 at 10:13
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Try $f(x,y) = xy^2 e^{-xy}$. What is $f(x,1/x)$ for $x > 0$?

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  • $\begingroup$ Yeah the smooth case is not true. How to prove for holomorphic functions? $\endgroup$ – Shiquan Apr 13 '14 at 10:17
  • $\begingroup$ If you had a bound of the form $|f(z)| < \exp(A \exp(B |z|)$ the Phragmen-Lindelof theorem would apply. I don't see how to prove it without such a bound. $\endgroup$ – Robert Israel Apr 14 '14 at 1:18

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