I was looking for some help about a question in my Advanced Calculus course, about a function $u(\vec{x}):\mathbb{R}^{2} \rightarrow \mathbb{R}$ such that the trace of the Hessian of the function is $\geq 0$ for all $\vec{x} \in \mathbb{R}^2$.
The question is about showing that the maximum of such a function restricted to the closed unit disc must take its maximum at the boundary.
I looked at this answer: https://math.stackexchange.com/a/2643384/489963
It's a good answer and it makes the whole idea clear, except for one thing. The crux of the argument is to assume that a slightly modified version of $u$, $$u^{\epsilon}=u(\vec{x})+\epsilon||u(\vec{x})||^{2}$$ assumes its maximum on the closed unit disc in the INTERIOR of the disc, and then to find a contradiction.
I don't know what I'm missing here, but I don't see where the argument uses the fact that the assumed maximum $(x_{0}, y_{0})$ is inside the disc. Couldn't you assume that the maximum was anywhere at all in the closed unit disc, and then get the same contradiction?
The point is I feel like you could just as easily use the logic of the proof to assume a max of $u^{\epsilon}$ on the boundary and show that’s impossible too. Using the logic of the proof, let’s assume $u^{\epsilon}$ takes its max on the boundary—then $\Delta u^{\epsilon}= 4 \epsilon >0$ by construction of $u^{\epsilon}$. But if this is a maximum point $\Delta u^{\epsilon} \leq 0$. Therefore $u^{\epsilon}$ must assume its maximum on unit closed disc somewhere in the interior. But it doesn’t...
Please help me see the error of my ways...
The proof shows that $u^\epsilon$ cannot attain maximum in the interior, it doesn't show that $u$ cannot attain maximum in the interior.
If $u^\epsilon$ attains a maximum in the interior, it satisfies $\Delta u^\epsilon \le 0. $
It is possible when $u$ is a constant and you won't find a contradiction if you assume it is in the interior.