# Application of Maximum modulus theorem

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?

• Your answer is wrong because $f(\frac 1 {n\pi})=0$ for all $n$. – Kavi Rama Murthy Nov 26 '18 at 8:27
• Can I put z=2/pi? – user227158 Nov 26 '18 at 8:36
• $\frac 2 {\pi}$ is inside the domain. But I don't understand how you are applying MMT. – Kavi Rama Murthy Nov 26 '18 at 8:39
• How you deduce it is equal to zero for all z? – user227158 Nov 26 '18 at 8:40
• I just try to plug in different value to find the maximum value of |f|. – user227158 Nov 26 '18 at 8:41

## 1 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 $$0$$.

• So the problem cannot be solved using maximum modulus theorem? – user227158 Nov 26 '18 at 8:55
• As far as I can see MMT theorem is of no use here. @kenkenb – Kavi Rama Murthy Nov 26 '18 at 8:58
• Could you explain the last sentence? – user227158 Nov 26 '18 at 9:26
• What happens to 2/pi? – user227158 Nov 26 '18 at 9:27
• At $\frac 2 {\pi}$ we only know that $|f(\frac 2 {\pi})|\leq \sqrt {\frac {\pi} 2}$. This is not very useful. – Kavi Rama Murthy Nov 26 '18 at 9:29