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These two questions I didn't even find the way to solve So please if you can help

  1. Suppose $f (z)$ is entire with $|f(z)| \le |\exp(z)|$ for every $z$ I want to prove that $f(z) = k\exp(z)$ for some $|k| \le 1$

  2. Can a non constant entire function be bounded in half a plane? Prove if yes , example if not.

Ayman Hourieh
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someone
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3 Answers3

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(1) can actually be generalized to show that no entire function $f$ can dominate another entire function $g$ unless $g = \lambda f$ for some constant $\lambda$. Let $Z(f)$ be the set of zeros of $f$. If $|g| \le |f|$, then: $$ \left|\frac{g(z)}{f(z)}\right| \le 1 \quad \forall z \in \Bbb C - Z(f) $$

But $g/f$ is bounded in a deleted neighborhood of each $a \in Z(f)$. Thus each $a \in Z(f)$ is a removable singularity of $g/f$. Therefore $g/f$ is entire and bounded, hence constant by Liouville's theorem. It follows that $g = \lambda f$.

In your particular question, $f(z) = \exp(z)$ and this function doesn't have any zeros, so it isn't necessary to discuss the zeros of $f$.

Ayman Hourieh
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  • i have some troubles understanding the simple meaning of the deleted neighborhood so if you can tell it to me simple to get the point the rest i understand thank you very much – someone Mar 06 '13 at 00:58
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    @someone Deleted neighborhood of a point means that you take a neighborhood of that point and then you remove the point itself. The open disk $D(p, \epsilon)$ is a neighborhood of $p$. The set $D(p, \epsilon) - {p}$ is a deleted neighborhood of $p$. Does it make sense now? – Ayman Hourieh Mar 06 '13 at 08:02
  • wow you really helped it seems so easy to me now thank you :) – someone Mar 07 '13 at 03:25
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    Using the boundedness of $|g/f|$ is much more convenient than going through a case by case analysis of the potential types singularities of the function. This is an excellent theorem to keep in mind whenever working with singularities in general, and one that I often forget about in the heat of battle – cnolte Aug 15 '16 at 18:37
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Hint for 2) you have written such a function in your post.

mrf
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Hint:

1) Liouville + $\,\displaystyle{\frac{f(z)}{e^z}}\,$ is analytic and bounded...

2) Develop the function in power series and extend analytically into the "other" halpf plane.

DonAntonio
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