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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.

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    $\begingroup$ 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
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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 $0$.

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  • $\begingroup$ 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

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