problem on existence of entire function Does there exist an entire function $f$ such that $f(0)=1$ and $\lvert f(z)\rvert\leq1/\lvert z\rvert$ for all $\lvert z\rvert\geq5$?
My Attempt:
As $\lvert z\rvert\to\infty, \lvert f(z)\rvert\to0$.
So, $f(z)$ has a removable singularity at $z=\infty$, so $f$ must be identically $1$, but the answer says that there does not exist any such entire function.
What is wrong in the above?
Thanks in advance...
 A: From your given condition , $|zf(z)|\le 1$ fort all $z$ such that $|z|\ge 5$. So $zf(z)$ is bounded in $|z| \ge 5$.
Again $zf(z)$ is analytic and as $|z| \le 5$ is compact domain so , $zf(z)$ is bounded in $|z| \le 5$. Therefore , $zf(z)$ is bounded in $\Bbb C$ and so by Liouville's theorem , $zf(z)$ is constant.
Let , $zf(z)=C.$ Now $f(0)=1$ givs $C=0$, then $f(z)=0$ , which contradicts $f(0)=1$.
So such function does not exists.
A: Suppose there exists an entire function $f$with $f(0) = 1$ and $|f(z)| \leq \frac{1}{|z|} \leq \frac{1}{5} \forall |z| \geq 5$,
$\underline{\textit{Wrongly Applied Liouville's Theorem}}:$
Applying Liouville's theorem as $f$ is entire and bounded so $f$ must be constant .
Also as given $f(0) = 1$, so $f = 1$ for all $z$ such that $|z| \geq 5$, but then it fails to satisfy the second inequality gven $|f(z)| \leq \frac{1}{|z|} \leq \frac{1}{5}\forall |z| \geq 5$.
In this method we have not proved that $f$ is bounded over the whole complex plane ,we did so only for the region $|z| \geq 5$.Hence we cannot apply Liouvilles theorem until we prove that $f(z)$ is bounded in $|f(z)|\leq 5$
$\textbf{EDIT}$
As corrected , $|zf(z)| \leq 1 \forall |z | \geq 5$ and also $|zf(z)| \leq 1 \forall |z| \leq 5$ thus $|zf(z)|$ is bounded in the Whole of the Complex Plane.
Implying that $zf(z) = c$ (a constant function) where c is any real number.
By using the given condition $f(0) = 1$ we get $c = 0$ so $zf(z) = 0$ possible only when $f(z) = 0$, which violates thegiven condition that $f(0) = 1$.
Ths such an entire function does not exist.
