Laplace operator of a smooth function Let $f$ be a smooth function in a domain $\Omega \subset \mathbb{C}$. Prove that for any $a \in \Omega$ we have $$\Delta f \left( a \right ) = \lim_{r \to 0^+} \frac{4}{r^2} \left( \frac{1}{2\pi}\int\limits_{0}^{2\pi} f\left(a+re^{i\theta}\right)d\theta -f\left(a\right)\right),$$
where $\Delta f = f_{xx}+f_{yy}$ is the Laplace operator. 

My idea is using Taylor's expansion: $$f\left(z\right)=f\left(a\right) + f'\left(a\right)\left(z-a\right)+\frac{1}{2}f''\left(a\right)\left(z-a\right)^2+o\left(\left|z-a\right|^3\right).$$
Take $z=a+re^{i\theta}$ we get $$f\left(a+re^{i\theta}\right)- f\left(a\right) = f'\left(a\right)re^{i\theta}+\frac{1}{2}f''\left(a\right)r^2e^{2i\theta}+o\left(r^3\right)$$
However, the problem is that when I integrate both sides of above equation:
$$\frac{2}{r^2}\left(\frac{1}{2\pi}\int\limits_{0}^{2\pi}f\left(a+re^{i\theta}\right)d\theta - f\left(a\right)\right)=\frac{2}{r}f'\left(a\right)\frac{1}{2\pi}\int\limits_{0}^{2\pi}e^{i\theta}d\theta+f''\left(a\right)\frac{1}{2\pi}\int\limits_{0}^{2\pi}e^{2i\theta}d\theta + 2o\left(r\right) = 2o\left(r\right)$$
This is impossible! I don't know where the mistake is. Thank you very much for your help.
 A: Your problem is that you've treated $f$ like it's analytic. If this is the case, the Cauchy–Riemann equations imply that $\Delta f=0$, as your calculation shows. We have to assume that $f$ is only smooth, not analytic, in which case, setting $z=a+x+iy$, $f(a+x+iy) = F(x,y)$, we have the Taylor expansion
$$ F(x,y) = F(0,0) + F_x(0,0)x + F_y(0,0)y + \frac{1}{2}(F_{xx}(0,0)x^2 + 2F_{xy}(0,0)xy +F_{yy}(0,0)y^2) + o(r^2). $$
Then, with $x=r\cos{\theta}$, $y=r\sin{\theta}$,
\begin{align} &\frac{1}{2\pi} \int_0^{2\pi} F(r\cos{\theta},r\sin{\theta}) \, d\theta \\
&= \frac{1}{2\pi} \int_0^{2\pi} F(0,0) + r(F_x(0,0)\cos{\theta} + F_y(0,0)\sin{\theta}) \\
&\qquad + \frac{r^2}{2}(F_{xx}(0,0)\cos^2{\theta} + 2F_{xy}(0,0)\cos{\theta}\sin{\theta} +F_{yy}(0,0)\sin^2{\theta}) + o(r^2) \, d\theta \\
&= F(0,0) + 0r + \frac{r^2}{2} \frac{1}{2\pi} \int_0^{2\pi} F_{xx}(0,0)\cos^2{\theta} + 2F_{xy}(0,0)\cos{\theta}\sin{\theta} +F_{yy}(0,0)\sin^2{\theta} \, d\theta + o(r^2) \\
&= F(0,0) + \frac{r^2}{4} ( F_{xx}(0,0) + F_{yy}(0,0) ) + o(r^2).
\end{align}
by the usual Fourier series–like argument. So I think actually
$$ \Delta F(0) = \lim_{r \to 0} \frac{4}{r^2}  \left( \frac{1}{2\pi} \int_0^{2\pi} F(r\cos{\theta},r\sin{\theta}) \, d\theta - F(0,0) \right), $$
which doesn't quite lead back to your formula: $F(r\cos{\theta},r\sin{\theta}) = f(a+re^{i\theta})$, so the right-hand side is the same up to a factor of 2, but the left-hand side is not, since $\Delta F = f_{xx}-f_{yy}$ if I've got my calculations correct.
