Editing Maximum principle (section)

===Summary of proof===
Let {{mvar|M}} be an open subset of Euclidean space. Let <math>u:M\to\mathbb{R}</math> be a twice-differentiable function which attains its maximum value {{mvar|C}}. Suppose that
:<math>a_{ij}\frac{\partial^2u}{\partial x^i\,\partial x^j}+b_i\frac{\partial u}{\partial x^i}\geq 0.</math>

Suppose that one can find (or prove the existence of):
* a compact subset {{mvar|Ω}} of {{mvar|M}}, with nonempty interior, such that {{math|''u''(''x'') < ''C''}} for all {{mvar|x}} in the interior of {{mvar|Ω}}, and such that there exists {{math|''x''<sub>0</sub>}} on the boundary of {{mvar|Ω}} with {{math|''u''(''x''<sub>0</sub>) {{=}} ''C''}}.
* a continuous function <math>h:\Omega\to\mathbb{R}</math> which is twice-differentiable on the interior of {{mvar|Ω}} and with
::<math>a_{ij}\frac{\partial^2h}{\partial x^i\,\partial x^j}+b_i\frac{\partial h}{\partial x^i}\geq 0,</math>
: and such that one has {{math|''u'' + ''h'' ≤ ''C''}} on the boundary of {{mvar|Ω}} with {{math|''h''(''x''<sub>0</sub>) {{=}} 0}}
Then {{math|''L''(''u'' + ''h'' − ''C'') ≥ 0}} on {{mvar|Ω}} with {{math|''u'' + ''h'' − ''C'' ≤ 0}} on the boundary of {{mvar|Ω}}; according to the weak maximum principle, one has {{math|''u'' + ''h'' − ''C'' ≤ 0}} on {{mvar|Ω}}. This can be reorganized to say
:<math>-\frac{u(x)-u(x_0)}{|x-x_0|}\geq \frac{h(x)-h(x_0)}{|x-x_0|}</math>
for all {{mvar|x}} in {{mvar|Ω}}. If one can make the choice of {{mvar|h}} so that the right-hand side has a manifestly positive nature, then this will provide a contradiction to the fact that {{math|''x''<sub>0</sub>}} is a maximum point of {{mvar|u}} on {{mvar|M}}, so that its gradient must vanish.