4
$\begingroup$

Which of the following are not true?

$(a)$ There exists an analytic function $f:\mathbb{C}\to\mathbb{C}$ such that for all $z\in \mathbb{C}$ , $Re(f(z))=e^x$.

$(b)$ There exists an analytic function $f:\mathbb{C}\to\mathbb{C}$ such that $f(0)=1$ and for all $z\in \mathbb{C}$ such that $|z|\geq1$ such that $|f(z)|\leq e^{-|z|}$.

could you please give me some hints?

I have applied liouville thorem for $(b)$ and got $f(z)$ is constant but not necessarily $f(z)=1$. Am I correct?

$\endgroup$
  • $\begingroup$ Well, if $f$ is constant and $f(0)=1$, then ... But does that comply with $|f(z)|\le e^{-|z|}$? Or more directly: Use Cauchy integral to compute $f(0)$ and compare with the given $f(0)=1$. $\endgroup$ – Hagen von Eitzen May 8 '17 at 6:18
  • $\begingroup$ Yes. Could you please give me some hints? $\endgroup$ – user444042 May 8 '17 at 6:19
3
$\begingroup$

(a): If $u=Re(f)$, then $u$ is harmonic, this means: $u_{xx}+u_{yy}=0$. Is this possible if $u(x,y)=e^x$ ?

(b) For $|z| \ge 1$ we have $e^{-|z|} \le 1/e$, hence $f$ is bounded for $|z| \ge 1$.

The set $K=\{z :|z| \le 1\}$ is compact, hence $f$ is bounded on $K$.

Conclusion: $f$ is bounded on $ \mathbb C$, hence $f$ is constant, thus, for some $c$:

$f(z)=c$ for all $z$. From $f(0)=1$ we get $c=1$. It follows:

$1=f(1) \le 1/e$, a contradiction.

$\endgroup$
  • $\begingroup$ Awesome. Thank you very much. $\endgroup$ – user444042 May 8 '17 at 12:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.