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Is there an analytic function $f$ in the open unit disc such that $|f(z)|=e^{|z|}$ therein?

My try:

Suppose such a function exists .Then $|f(0)|=e^0=1$ Also $|f|\ge 0$.

Now $e^{|z|}$ attains its minimum value i.e. $1$ at $0$. Thus $|f(z)|\ge 1$

I was trying to use the minimum modulus theorem which states that if $f$ is a non- constant analytic function such that $|f|$ attains its minimum value in an interior point of the unit disc then $f$ is constant.

Here $|f|$ attains its minimum value at $0$ which is an interior point of the unit disc and thus $f$ is constant which is false as $|f|=e^{|z|}$.

Thus no such analytic function exists.

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    $\begingroup$ You are right. You should add that $f$ is nonzero on the open disc, so that you can take its reciprocal and apply the maximum modulus theorem, but this is immediate, since $|f(z)|=e^{|z|}>0$. $\endgroup$ Apr 5, 2016 at 7:47
  • $\begingroup$ Thank you very much @Jean-ClaudeArbaut $\endgroup$
    – Learnmore
    Apr 5, 2016 at 10:36
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    $\begingroup$ "if f is a non- constant analytic function such that |f| attains its minimum value in an interior point of the unit disc then f is constant." You need to say f is never 0 for this $\endgroup$
    – zhw.
    Apr 5, 2016 at 18:29

1 Answer 1

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This is a community wiki answer to notice that this question was answered in comments and remove the question from the unanswered list -- the answer is yes, the OP's proof is correct.

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  • $\begingroup$ The OP's statement "If |$f|$ attains its minimum in the interior of the unit disc then $f$ is constant" is incorrect. It is only correct if $f$ is never $0$ in the open unit disc, as stated in the comment by zhw on Apr.5/16. $\endgroup$ Aug 6, 2017 at 18:13
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    $\begingroup$ Which follows immediately from the condition $|f| = e^{|z|}$. Whether or not to call the OP's argument "correct" with this detail omitted comes down to a question of how rigorously his or her course is graded; there is no substantive issue. $\endgroup$ Aug 6, 2017 at 18:15

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