Suppose the function is an analytic function on $\{\omega={z:0<|z|\le 1}$}.Moreover $f$ satisfies $ |f(z)| \le \frac{1}{|z|^{1/2}}|\sin(\frac{1}{|z|})|$ for all $z$ in $\omega$. $a_n$ is the coefficient of Laurent series. I want to show that $f$ is a constant function and find the value of $f$ in the domain $\omega$. I tries to use maximum modulus theorem.$|f(z)|\le \sqrt{\frac{\pi}{2}}$ for all $z$ in the domain and hence $f=\sqrt{\frac{\pi}{2}}$ for all $z$ in the domain.Is this valid?
Second attempt:
the function is bounded by 0 in the open disc of radius $\frac{1}{\pi}$. And then by maximum modulus theorem it is zero throughout the disk and since it is analytic after redefining then by unique of coefficient of power series, the result follows. Is this valid?
Best Answer
Let $g(z)=zf(z)$. Then $|g(z)| \leq \sqrt {|z|}$ so $g$ has a removable singularity at $0$ and it becomes analytic in the unit disk if define $g(0)$ to be $0$. Hence $\frac {g(z)} z$ has a removable singularity at $0$ which means $f$ has a removable singularity at $0$. Now, since $f (\frac 1 {n\pi}) =0$ for all $n$ the zeros of $f$ have a limit point so $f$ must be identically $0$.