All second partial derivatives of harmonic function are $0$ I am given this question as a homework assignment. 
Assume that $f$ is from $\mathbb R^2$ to $\mathbb R$ and has a strict local maximum at $(x_0, y_0)$. prove that all second partial derivatives of harmonic function $f$ at $(x_0,y_0)$ are $0$.
I tried to solve this question but it's really difficult and I don't get the idea... please help me.
 A: Suppose $(x_0,y_0) = (0,0).$ Since $u$ has a local max at $(0,0),$ $\nabla u (0,0) = (0,0).$ Thus the Taylor expansion of $u$ at $(0,0)$ has the form
$$u(x,y) = u(0,0) + ax^2 - ay^2 +bxy +O((x^2+y^2)^{3/2}).$$
The coefficients of $x^2,y^2$ have to add to $0$ because $u_{xx} + u_{yy} = 0.$ Now look along the lines $x=0,y=0,y= \pm x$ to see $a,b=0.$
(A much stronger result is true: $u$ is constant on $\mathbb R^2.$ But I assume you are just beginning with harmonic functions.)
A: I will leave here another way to approach this problem.
$f$ is harmonic means that
$$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0.$$
Since $f$ has a strict local maximum at $(x_0,y_0)$, its Hessian is negative semidefinite at that point, in particular, when evaluated on a vector $v = \begin{pmatrix}1\\ 1\end{pmatrix}$ it gives:
$$H(1,1)=\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial x\partial y} + \frac{\partial^2 f}{\partial y \partial x} + \frac{\partial f}{\partial y^2}\leq0.$$
Sylvester's criterion gives
$$\frac{\partial^2 f}{\partial x^2}\leq0$$
and
$$\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2} - \frac{\partial^2 f}{\partial x\partial y}\frac{\partial^2 f}{\partial y\partial x}\geq0.$$
Second order partial derivatives of harmonic function are continuous, thus mixed partial derivatives are equal. Now it's an algebraic exercise to conclude from all the equations above that all second order partial derivatives are zeros.
