# If a continuous function is bounded outside a bounded set then show that it is entire.

If $f:\Bbb C \to \Bbb C$ is continuous such that $f$ is analytic outside the bounded set $S$ then prove that $f$ is entire.

To prove this we have to prove that $f$ is analytic in $S$. For this I I think it can be done by using Morera's theorem , which states that " If $f$ is continuous in a simply connected domain $D$ and $\int_Cf(z)\,dz=0$ , for every closed contour $C$ in $D$ then $f$ is analytic in $D$."

But here how I can show that $\int_C f(z)\,dz=0$ for every closed contour $C$ in the bounded domain $S$?

I've seen this similar type question , but there are no enough proof which I want. So , please don't make this question as a duplicate question , and give me some hints to show the integral value is $0$.

• What are the conditions on $S$? Is it a line segment? This is definitely not true for arbitrary bounded set $S$. – Owen Sizemore Mar 31 '17 at 20:09
• @OwenSizemore Actually it is given just a Bounded Set. I don't know whether it is correct or not. If you think there re something wrong then clarify it , please. – Empty Mar 31 '17 at 20:11

$$f(x) = \begin{cases} z \textrm{ if } |z| \geq 1 \\ z\cdot e^{i\cdot(|z|-1)} \textrm{ if } \frac{1}{2} \leq |z| < 1 \\ z\cdot e^{i\cdot|z|} \textrm{ if } |z| < \frac{1}{2} \end{cases}$$
Basically the idea is that if the set $S$ contains an open set then you have enormous freedom to make the function because being continuous inside this set is a fairly weak condition while being analytic is a $\textit{very}$ strong condition.