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Let $f$ be an entire function.

Let $m:[0,\infty)\to\mathbb{R}$ be a function defined by $m(r)=\min\limits_{|z|=r}|f(z)|$.

Suppose that $\lim\limits_{r\to +\infty}m(r)$ exists and equals to a positive real number.

Prove that $f$ is constant.

From the given by setting $\lim\limits_{r\to +\infty}m(r)=M>0$ we get that there exists $R>0$ such that $\forall r>R$ we have $m(r)>\frac{M}{2}$.

Therefore, we have that $f$ is lower bounded by $\frac{M}{2}$ in $D=\{z\in\mathbb{C}:|z|>R\}$.

We can look at $\frac{1}{f}$ and say it is upper bounded by $\frac{2}{M}$ in $D$. The problem is that it may have poles in $\mathbb{C}\setminus D$.

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The function $f$ cannot be polynomial, because then $\lim_{r\to+\infty}m(r)=+\infty$.

So, if $g(z)=f\left(\frac1z\right)$, then $g$ has an essential singularity at $0$. So, the Casoratti-Weierstrass theorem tells us that if $V$ is a neighbourhood of $0$, $g(V\setminus\{0\})$ is dense. In other words, if $R>0$, $f(\{z\in\mathbb{C}\,|\,|z|>R\})$ is dense and, in particular, it contains numbers as close to $0$ as you want. Therefore, $\lim_{r\to+\infty}m(r)=0$.

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