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.


1 Answer 1


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 $f^{(n)}(0)=0$.

  • $\begingroup$ 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
    Commented Feb 19, 2014 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
    Commented Feb 19, 2014 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
    Commented Feb 19, 2014 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
    Commented Feb 19, 2014 at 15:38

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