Intereting Posts

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I don’t understand the final step of an argument I read.

Suppose $f$ is holomorphic in a neighborhood containing the closed unit disk, nonconstant, and $|f(z)|=1$ when $|z|=1$. There is some point $z_0$ in the unit disk such that $f(z_0)=0$.

By the maximum modulus principle, it follows that $|f(z)|<1$ in the open unit disk. Since the closed disk is compact, $f$ obtains a minimum on the closed disk, necessarily on the interior in this situation.

- If $f^2$ and $f^3$ are analytic prove that $f$ is analytic at every point of $\mathbb{C}$.
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- Laplace transform of the Bessel function of the first kind

But why does that imply that $f(z_0)=0$ for some $z_0$? I’m aware of the minimum modulus principle, that the modulus of a holomorphic, nonconstant, nonzero function on a domain does not obtain a minimum in the domain. But I’m not sure if that applies here.

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If not, consider $g(z)=\frac 1{f(z)}$ on the closure of the unit disc. We have $|g(z)|=1$ if $|z|=1$ and $|g(z)|>1$ if $|z|<1$. Since $g$ is holomorphic on the unit disk, the maximum modulus principle yields a contradiction.

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