I am supposed to prove the following: $$4\frac{\partial}{\partial z}\frac{\partial}{\partial\bar{z}}=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\,\,\,$$
Using the definitons $$\frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x} + \frac{1}{i}\frac{\partial}{\partial y}\right)$$ and $$\frac{\partial}{\partial \bar{z}}=\frac{1}{2}\left(\frac{\partial}{\partial x} - \frac{1}{i}\frac{\partial}{\partial y}\right),$$ I get that $$4\frac{\partial}{\partial z}\frac{\partial}{\partial\bar{z}}=\frac{\partial^2}{\partial x^2}-\frac{1}{i}\frac{\partial^2}{\partial x\partial y}+\frac{1}{i}\frac{\partial^2}{\partial y\partial x}+\frac{\partial^2}{\partial y^2}.$$
This leaves me with a few questions.
Assume $f:\Omega\to\mathbb{C}$, and let $u(x)=\textrm{Re}(f(x))$ and $v(x)=\textrm{Im}(f(x))$. Then if $\frac{\partial^2 u}{\partial x\partial y}$, $\frac{\partial^2 u}{\partial y\partial x}$, $\frac{\partial^2 v}{\partial x\partial y}$ and $\frac{\partial^2 v}{\partial y\partial x}$ are continuous, then $\frac{\partial^2 u}{\partial x\partial y} = \frac{\partial^2 u}{\partial y\partial x}$ and $\frac{\partial^2 v}{\partial x\partial y} = \frac{\partial^2 v}{\partial y\partial x}$, so $\frac{\partial^2 f}{\partial x\partial y} = \frac{\partial^2 f}{\partial y\partial x}$. Accordingly the equality in the title holds for $f$.
So I wonder: Are these assumptions about $u$ and $v$ needed? Or can the equality be proven in another manner without them?
In fact a holomorphic function can be defined only to differentiable function with values on $\mathbb{C}$, and you can prove that if its is differentiable on an open set it is in fact analitic on this open set. Then, if you weaken this hypotesis won't improve the result, it will just give you a different proof, and maybe harder.