Is there a holomorphic function on the unit disk that satisfies $f(1/n) = 1/\sqrt{n}$?

Thougthts so far: I know that $f(z) = \sqrt{z}$ won't work, as it is not analytic at $0$. My intuition says that this is impossible. Just a hint at this point would be most helpful, as I am preparing for a qual.

  • 2
    $\begingroup$ If such an $f$ existed, it would in particular be complex differentiable at $0$. $\endgroup$ – Daniel Fischer Aug 26 '15 at 19:00
  • $\begingroup$ Way too complicated. What about difference quotients? $\endgroup$ – Daniel Fischer Aug 26 '15 at 19:16

If such a function existed, by continuity, we'd have $f(0) = 0$. Thus we have the difference quotients

$$\frac{f\bigl(\frac{1}{n}\bigr) - f(0)}{\frac{1}{n} - 0} = \frac{1/\sqrt{n}}{1/n} = \sqrt{n},$$

so such a function cannot be differentiable at $0$.

  • $\begingroup$ I see how this can be used to show that $f'(0) = 0$, but why does that mean it can't be differentiable at $0$? $\endgroup$ – user19817 Aug 26 '15 at 19:27
  • $\begingroup$ How do you get $f'(0) = 0$ from that? The sequence of difference quotients doesn't converge to $0$. $\endgroup$ – Daniel Fischer Aug 26 '15 at 19:29
  • $\begingroup$ Ah I see, the $1/n$ tripped me up, I was thinking that $n \to 0$. You're right, this is simpler. $\endgroup$ – user19817 Aug 26 '15 at 19:30

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.