1
$\begingroup$

Suppose that $f$ is holomorphic on the unit disk $|z|<1$. If $\,\exists$ $r \in (0,1)$ such that $|f(1/n)|\leq r^n$ for $n \in \mathbb{N}$. Then $f=c$ (constant).

I think this problem could be solved by using Cauchy's inequalities formula ( I am not sure) which is: If $f$ is holomorphic in an open set $\Omega$ that contains the closure $C$ of a disc $D$, centered at $z$ and has radius $R$, then

$|f^{(n)}(z)|\leq \dfrac{n! ||f||_C}{R^n} $,

where $||f||_C=\sup_{z \in C}|f(z)|$.

But I don't know how to use it. What is my disc (What are its radius and center?) I have exam in two days, any help is appreciated.

$\endgroup$
2
$\begingroup$

Taking the limit as $n \rightarrow \infty$ you get that $|f(0)| \leq 0$ so that $f(0) =0.$

We can write $f(x) = xg(x)$ where $g$ is holomorphic on the interior unit disk. Notice that $g(0) = f'(0)$. However, $|g(\frac{1}{n})| \leq nr^n$ so repeating what we did for $f$ we have that $g(0)=0$ and in particular that $f'(0)=0$.

You should now generalize this to show that $f^{(n)}(0)=0$.

$\endgroup$
  • $\begingroup$ Is proving $f^{n}(0)=0$, or even $f'(0)=0$, enough to say that $f$ is constant? If $f'(0)=0$, may there is another $z$, say $z_0$, such that $f'(z_0)\neq 0$. $\endgroup$ – Ruzayqat Feb 19 '14 at 9:37
  • $\begingroup$ Also, how to prove that $f''(0)=0$? I don't have a bound for $|f'(z)|$ as that for $|f(z)|$. $\endgroup$ – Ruzayqat Feb 19 '14 at 9:54
  • $\begingroup$ We know that $g(0) = 0$ so we again write $g(x) = xh(x)$ for $h$ analytic on the interior of the disk. $h(0) = f''(0)/2$ and $|h(1/n)| \leq n^2r^n$. Do you see the pattern now? $\endgroup$ – SomeEE Feb 19 '14 at 15:35
  • $\begingroup$ Yes, it is enough just to show $f^{(m)}(0) =0$ for all $m$. Just consider the power series of $f$ centered at the origin. $\endgroup$ – SomeEE Feb 19 '14 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.