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I'm trying to understand Liouville's theorem, and I don't see why $f(z)=e^{-|z|^2}$ isn't a counterexample. It's bounded ($0 < f(z) \leq 1$), so it must be somehow that it's not holomorphic. Isn't it differentiable everywhere?

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No, it's nowhere differentiable (except at the origin). First off, the modulus function $f(z)=|z|$ is nowhere differentiable in $\mathbb{C}$, and since any point (other than the origin) is contained in some open set with a holomorphic branch of logarithm and square root defined, if $g(z) = e^{-|z|^2}$ were differentiable then so too would $(-\log(g(z)))^{1/2}=|z|$, which is a contradiction.

At the origin, $g(0)=1$ so: $$\frac{g(re^{i\theta})-g(0)}{re^i\theta}=\frac{e^{-r^2}-1}{re^{i\theta}} $$

So taking the limit as $r\rightarrow 0$ we get: $$e^{-i\theta}\frac{d}{dx}\bigg(e^{-x^2}\bigg)\biggr\rvert_0$$ Where we take the real function $e^{-x^2}$, which is (real) differentiable at the origin. As pointed out in the comments below, this derivative is equal to zero so the function is differentiable at the origin with derivative $0$.

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  • $\begingroup$ Nitpick: I think it's differentiable at $z=0$. $\endgroup$ – Nate Eldredge May 28 '13 at 20:01
  • $\begingroup$ @NateEldredge Thanks for pointing that out, I was just thinking about it. I have just edited in something that I think shows that it is not differentiable at the origin, please let me know if I'm wrong! $\endgroup$ – Tom Oldfield May 28 '13 at 20:09
  • $\begingroup$ For real $x$ near $x=0$, $e^{-x^2} = 1 - x^2 + o(x^4)$. So $$\frac{e^{-|z|^2}-1}{z} = \frac{-|z|^2 + o(|z|^4)}{z} = -\bar{z} + o(|z|^3)$$ which goes to 0 as $z \to 0$. Your computation above is correct, but since $\frac{d}{dx} e^{-x^2} \biggr\rvert_0 = 0$, the value of the derivative is 0 regardless of the value of $\theta$. $\endgroup$ – Nate Eldredge May 28 '13 at 20:12
  • $\begingroup$ I see... complex differentiability is much more complicated than real differentiability on $R^2$. (I'm completely new to complex analysis.) $\endgroup$ – Adam Crume May 28 '13 at 20:14
  • $\begingroup$ @Nate very good point, thank you! $\endgroup$ – Tom Oldfield May 28 '13 at 20:14
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Hint: $|z|^{2}=z\bar{z}$ but a complex function of complex variable $z\to\bar{z}$ is not differentiable (although its real and imaginary part are differentiable when considered as real functions).

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