After reading the text on analyticity and viewing different theorems, somehow I made the following conclusions. Can please someone validate and throw some light on them?
- If a function is differentiable at a point $z_0$, then the partial derivatives may or may not be continuous at that point.
So, if the partial derivatives are continuous at that point, then this would imply analyticity at that point.
And if the partial derivatives are not continuous at that point, then it will not be analytic at that point.
- However, if a function is analytic at a point $z_0$, then this assures the existence of continuous partial derivatives at that point.
Well it seems true to me, as differentiability at a point and differentiability at a point along with its some neighborhood are two different things.
At last, Can I have an example of a function f(z) which is differentiable at a point $z_0$ but does not have continuous partial derivatives at that point? I know this can happen.
The second statement is false. Analyticity always applies to an open neighborhood, so analyticity requires continuous partial derivatives in an open neighborhood (since it implies continuous differentiability). In the complex case, it is equivalent to all partial derivatives existing, being continuous and satisfying the Cauchy-Riemann equations in an open neighborhood. The fourth statement can also be improved: analyticity implies continuous partial derivatives in an open neighborhood, not just at the point.
And to your last question, yes, a differentiable function with discontinuous partial derivatives is possible. Here's an example:
$$f(x+\mathrm iy):=\begin{cases} x^2\sin(1/x) & x\neq0\\0&x=0\end{cases}$$
It has partial derivatives $u_x=2x\sin(1/x)-\cos(1/x)$ if $x\neq0$, and all other partial derivatives (including $u_x$ at $x=0$) are $0$. You can probably see that $u_x$ is discontinuous on the imaginary axis ($x=0$). But $f$ is still complex differentiable on the axis: it's real differentiable (proof left to the reader), and since all partials are $0$, the Cauchy-Riemann equations hold, so it's complex differentiable. Or you can calculate the limit of the difference quotient, which will come out as $0$, again making it complex differentiable.