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Let $f$ be an holomrphic function in $\mathbb C$. Is the functions $\sup_{|z| = r} |f(z)|$ continuous as a function of $r$?

I think that yes but I'm not sure.

(I edited the question, adding the modulus for the function)

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First, I guess that you mean $\sup_{|z|=r} |f(z)|$, because if your function has complex values, there is no $\sup$ defined unless you take the absolute value.

By the maximum principle, if $f$ is holomorphic then the maximum principle holds, and $f$ is either constant, either the supremum of the values of the function $f$ on a disk is attained on the boundary. Therefore $\sup_{|z|=r} |f(z)|=\sup_{|z|\leq r} |f(z)|$, and the latter function is continuous.

Denote $M(r)=\sup_{|z|\leq r}|f(z)|$, and because $f$ is also continuous we have $M(r)=f(z_r)$ with $|z_r|=r$. Take now $r \to R$. Then $(z_r)$ contains a convergent subsequence, denoted for simplicity with $(z_r)$ and $z_r \to z_0$. If $|f(z_0)|$ is not the maximum for $|f(z)|$ on $|z|=R$ then there exists $z_1$ with $|z_1|=R$ and $|f(z_1)|>|f(z_0)|$. Because of the continuity of $f$, we can find a neighborhood of $z_1$ on which $|f(z)| \geq |f(z_0)|+\varepsilon$. Since this neighborhood intersects circles of radius close to $R$, we can contradict the fact that $|f(z_r)|$ is the maximum on $|z|=r$ for $r$ close to $R$.

I think we don't even need the fact that $f$ is holomorphic. $f$ continuous suffices.

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Yes, I mean't the modulus of the function. Thanks for the answer. But why the latter function is continuous? – Nestor Apr 8 '12 at 11:29
The fact that f is holomorphic is needed in the first step to apply the maximum principle... – revers Apr 16 '12 at 9:52
@revers: Yes, but in the initial problem the supremum is taken only on the circle, not on the entire disk. – Beni Bogosel Apr 16 '12 at 11:04

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