Maximum condition for second order differential operators Let $A$ be a second order differential operator such that 
$$Af(x) = \sum_{ij} a_{ij}(x) \big(\partial_i \partial_j f(x)\big) + \sum_j b_j(x) \partial_j f(x) $$
Assume that  $x \in B(0,r)\Rightarrow Af(x)>0$. 
How do I see that $f$ cannot attain a maximum inside of $B(0,r)$?
Here $f : \Bbb{R}^n \to \Bbb{R}$ is $C^2$ and for every $x\in \Bbb{R}^n$$$\sum_{ij}x_ia_{ij}x_j \geq 0$$
 A: You cannot, unless you put additional assumptions on the second-order coefficients $a_{ij}$.
Suppose $f$ attained a maximum at $x_0\in B(0,r)$. Then $x_0$ is a critical point for $f$, so at $x_0$ we have $\partial_j f(x_0) = 0$ for all $j$. Then
$$
Af(x_0) = \sum_{i,j}a_{ij}\partial_i\partial_j f(x_0)>0.
$$
If $(a_{ij})$ had some additional conditions, e.g. they determine a positive-definite symmetric bilinear form (as is usually the case when $A$ is elliptic in some sense), then one could argue as follows: diagonalize the matrix $(a_{ij})$, and rewrite $f$ in terms of the new coordinates. (This is where the assumption on $(a_{ij})$ is required.) Then use the fact that at an interior local maximum $y_0$ for any $C^2$ function $F$, $\partial_{ii}F(y_0)\leq 0$.
Without additional conditions on $(a_{ij})$, the conclusion fails. For instance, in one variable consider
$$
Af = -f''\hspace{2em}\text{in}~(-\frac{\pi}{2},\frac{\pi}{2}).
$$
Then $f(x) = \cos x$ satisfies
$$
Af(x) = \cos(x) > 0\hspace{2em}\text{in}~(\frac{\pi}{2},\frac{\pi}{2}).
$$
However, $f(x)$ attains a global maximum at $0$. Notice that the matrix of coefficients $(a_{ij}) = [-1]$ does not determine a positive-definite symmetric form. On the other hand, if you repeat this exercise with $(a_{ij}) = [+1]$ then the claim reduces to the single-variable second derivative test, and in higher dimensions this is also the direction you can work toward.
