I'm quite confused by my course notes for Complex Analysis. On the one hand, it says that a complex function $f:\mathbb{C} \to \mathbb{C}$ is differentiable at a point $z_0$ iff the corresponding vector-function $F:\mathbb{R^2} \to \mathbb{R^2}$ is differentiable and the Cauchy-Riemann equations are satisfied. Now differentiability of a multivariate real function is equivalent to the continuity of partial derivatives, so this would imply for $f(z) = u(x,y)+iv(x,y)$ that $u(x,y)$ and $v(x,y)$ have continuous partial derivatives.
On the other hand, the proof that is given for the Cauchy-Goursat theorem is an extremely long, convoluted one, and at the end of it, it mentions that a much more elegant and short proof is possible if one assumes that $u(x,y)$ and $v(x,y)$ have continuous partial derivatives (it makes use of Green's theorem). Why is this an 'assumption'? Doesn't it necessarily follow from the assumption of $f$ differentiable in the theorem statement?
Best Answer
No, differentiability of a multivariate real function is not equivalent to the continuity of partial derivatives. For instance, if$$f(x,y)=\begin{cases}x^2\sin\left(\frac1x\right)&\text{ if }x\ne 0\\0&\text{ if }x=0,\end{cases}$$then $f$ is differentiable at $(0,0)$, but $\frac{\partial f}{\partial x}$ is discontinuous there.
However, it is true that the continuity of partial derivatives implies differentiability.