If $f$ satisfies $|f(z)|\le |f(z^2)|$ then $f$ is constant

Asked Nov 30 '15 at 15:59

Active Nov 30 '15 at 15:59

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If $f:\{|z|<1\}\to \mathbb C$ be analytic such that $|f(z)|\le |f(z^2)|$ for all $z$ in $|z|<1$ then prove that $f$ is constant.

We have , $|f(z)|\le |f(z^2)|\le|f(z^4)|\le|f(z^8)|\le\cdots \le |f(z^n)|\to |f(0)|$ , as $n\to \infty$. As , $|z|<1$ so , $z^n\to 0$. From here I can show that $f$ is bounded. But $f$ is NOT entire so that we can say that $f$ is constant. From here how I can proceed ?

asked Nov 30 '15 at 15:59

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Did you already bump into this Wiki page: https://en.wikipedia.org/wiki/Maximum_modulus_principle ? – Giovanni De Gaetano Nov 30 '15 at 16:05
Ah, and obviously if you show that $f$ is constant in any neighborhood of zero, then by analyticity... – Giovanni De Gaetano Nov 30 '15 at 16:07
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Hint: Maximum Modulus Principle. – bartgol Nov 30 '15 at 16:07
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Compare http://math.stackexchange.com/questions/1374959/analytic-function-on-the-open-unit-disc. – Martin R Nov 30 '15 at 16:10

If $f$ satisfies $|f(z)|\le |f(z^2)|$ then $f$ is constant

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