Does there exists an entire function $f$ such that $f(0)=1$ and $|f(z)|\leq \frac{1}{|z|} $ for all $|z|\geq 5$?

Question

Does there exists an entire function $f$ such that $f(0)=1$ and $|f(z)|\leq \frac{1}{|z|} $ for all $|z|\geq 5$?

Answer 1

If $\bigl|f(z)\bigr|\leqslant\frac1{|z|}$, then $\bigl|zf(z)\bigr|\leqslant1$. So, you know that $zf(z)$ is bounded in the region $\{z\in\mathbb{C}\,|\,|z|\geqslant5\}$. But it is also bounded in the closed disk centered at $0$ with radius $5$. Therefore it is bounded and then Liouville's theorem implies that $zf(z)$ is constant. But it cannot be the null function, because then we would not have $f(0)=1$. And we also cannot have $f(z)=\frac kz$ for some $k\in\mathbb C$, since $f$ is entire.

Therefore, there is no such function.

Answer 2

Let $g(z)=zf(z)$. Then $|g(z)| \le 1$ for $|z| \ge 5$.

$g$ is bounded on $|z| \le 5$. Hence $g$ is bounded on $ \mathbb C$.

Liouville: there is $c$ such that $zf(z)=c$ for all $z$. With $z=0$ we get $c=0$. Hence $f(z)=0$ for all $z \ne 0$. Therefore $f(z)=0$ for all $z$. This contradicts $f(0)=1$.