Suppose $f(z)$ is an entire function and $|f(z)| \leq e^{Re(z)}$ for all $z$. Show that $f(z) = ce^{z}$ for some constant $c$.

Attempt: $|e^z|=e^{Re(z)}$, so $|f(z)| \leq |e^z|$ then $|e^{-z}f(z)| \leq 1$. Now how do I know $e^{-z}$ is an entire function? If it is, then $|e^{-z}f(z)| \leq 1$ is a bounded entire function, so by Liouville's Theorem, $e^{-z}f(z)=c$ then the result follows for all $z$ and some constant $c$

  • $\begingroup$ $e^{-z}$ is clearly entire and the product of two entire functions is always entire. Note that if $h(z)$ is entire, then $1/h(z)$ is entire iff $h(z)$ has no zeros. This is due to the holomorphic = analytic theorem (checking holomorphy is much easier than checking analyticity, since it involves only the 1st derivative). You also have that $f(z),g(z)$ entire $\implies f(g(z))$ entire (and of course $f(z)+g(z)$ entire). $\endgroup$ – reuns Sep 21 '16 at 12:44
  • $\begingroup$ I see, thanks for the comment. $\endgroup$ – James.Welch Sep 21 '16 at 13:50
  • $\begingroup$ I'm not convinced you see it. Do you know the holomorphic = analytic theorem ? Can you check that $e^{-z}$ is holomorphic ? $\endgroup$ – reuns Sep 21 '16 at 13:57
  • $\begingroup$ Yes, you start with a holomorphic function, then you show by way of Cauchy's Integral Formula, that you can write the Taylor Expansion of the holomorphic function. $\endgroup$ – James.Welch Sep 22 '16 at 15:57

Your attempt seems correct to me. The function $e^{-z}$ is entire, since $e^z$ is an entire function and $p(z) = -z$ is a polynomial, hence also entire; Finally, the composition of two entire functions will again yield an entire function $\Longrightarrow e^{-z}$ is entire. This closes the "gap" in your argument. Nice work!

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  • $\begingroup$ Ok I see, thanks for your comment. $\endgroup$ – James.Welch Sep 21 '16 at 12:02
  • $\begingroup$ You're welcome! $\endgroup$ – ComplexFlo Sep 21 '16 at 12:03

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