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.
Best Answer
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.)