# formal partial complex derivatives and symmetry

This is probably a silly question but let me ask.

As it is well known for a general function $f:\mathbb R^2\to \mathbb R$ which posesses partial derivatives of second order it is not necessarily true that

$$\frac{\partial}{\partial x}(\frac{\partial f}{\partial y})=\frac{\partial}{\partial y}(\frac{\partial f}{\partial x})$$ and one concrete example is $$f(x,y)=\frac{xy^3}{x^2+y^2},$$ for $(x,y)\neq (0,0)$ nad $0$ at the origin. My question is about formal complex partial derivatives

$$\frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)$$ $$\frac{\partial}{\partial \bar z}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)$$ in the complex plane $\mathbb C\cong \mathbb R^2 (x+iy\cong z)$.

Is it always true that

$$\frac{\partial}{\partial z}(\frac{\partial f}{\partial \bar z})=\frac{\partial}{\partial \bar z}(\frac{\partial f}{\partial z}),$$ when both sides make sense or there exists an example of the above type (of course it can not be too regular)?

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What do you mean by "formal complex derivative making sense"? Do you mean that $\partial_x f$ makes sense and $\partial_y f$ makes sense and you just define $\partial_z f$ to be the complex linear combination of the two? Then wouldn't the example you gave already work as a counterexample? –  Willie Wong Oct 26 '12 at 14:33
Yes that is exactly what "formal complex derivative making sense" means. –  ahmed sulejmani Oct 26 '12 at 15:06

$$\frac{\partial }{\partial z}\frac{\partial f}{\partial \bar z} = \frac14 \Delta f +\frac{i}{4}\left(\frac{\partial }{\partial x}\frac{\partial f}{\partial y} -\frac{\partial }{\partial y}\frac{\partial f}{\partial x}\right)$$ and $$\frac{\partial }{\partial \bar z}\frac{\partial f}{\partial z} = \frac14 \Delta f - \frac{i}{4}\left(\frac{\partial }{\partial x}\frac{\partial f}{\partial y} -\frac{\partial }{\partial y}\frac{\partial f}{\partial x}\right)$$ This means that the $z$- and $\bar z$- derivatives commute if and only if $x$- and $y$- derivatives commute. For real-valued $f$ the failure of commutativity is expressed as the presence of an imaginary term in the formal partials.