Question about holomorphic function and Cauchy-Riemann equation Let $f(x+iy)=u(x,y)+iv(x,y)$ where $u,v$ are real. I have a theorem that say that $f$ is holomorphic at $z_0=x_0+iy_0$ then $u,v$ are $\mathcal C^\infty$ at $z_0$ and harmonic at $z_0$. Let
$$f(z)=\begin{cases}z^2\sin\frac{1}{|z|}&z\neq 0\\ 0&otherwise.\end{cases}$$
We have that $f$ is holomorphic at $z=0$, but $\Delta u(x_0,y_0)\neq 0$. How is it possible ? Is my theorem wrong ?
For information, we have that $$u(x,y)=\begin{cases}(x^2-y^2)\sin \frac{1}{\sqrt{x^2+y^2}}&(x,y)\neq (0,0)\\ 0&otherwise\end{cases}$$
$$v(x,y)=\begin{cases}2xy\sin \frac{1}{\sqrt{x^2+y^2}}&(x,y)\neq (0,0)\\ 0&otherwise\end{cases}.$$
 A: The problem is that your theorem is slightly off. It has been shown that if $f: \mathbf{C} \to \mathbf{C}$ is pointwise complex differentiable at each point in its domain , in the sense that the limit
$$ \lim_{h \to 0} \frac{f(z + h) - f(z)}{h} $$
exists for each $z \in \mathbf{C}$, then $f'$ is also complex differentiable at every point, and by induction this shows $f$ is infinitely differentiable, and therefore that $u$ and $v$ are $C^\infty$ and harmonic everywhere ($n$'th partial derivatives only exist if the function is in $C^{n-1}$). It makes sense that differentiability at a single point does not imply $C^\infty$ differentiability of the component functions $u$ and $v$, because we can reexpress the condition that $f$ is differentiable at $z$ as the existence of a complex number $w$ such that $f(z + h) = f(z) + wh + o(h)$, where $o(h)$ is the class of functions $g$ such that
$$ \lim_{h \to 0} \frac{g(h)}{h} = 0 $$
If $u$ and $v$ were automatically $C^\infty$ around $z$ if $f'(z)$ existed, this would imply that every function in $o(h)$ is in $C^\infty$, and unfortunately this isn't true (there are some really pathological nasties in this set).
