# Curious Laplacian inequality ($\Delta f>0 \Rightarrow f$ has no maxima). Additional premises just to confuse students?

In a homework question, we've been given the following things:

• Let $B=\overline{B(0,1)} \subseteq \mathbb R^n$ be the closed unit ball, $f : B\to \mathbb R^n$ a $\mathcal C^2$ function and let some $a_i : B \to \mathbb R$ be continuous for $i=1,\ldots,n$.

• For all $x\in B$, we have $$\Delta f(x)+\sum_{i=1}^n a_i(x)\frac{\partial f}{\partial x_i}(x) > 0$$ where $\Delta f$ is the Laplacian of $f$.

Then show that $f$ has no local maximum in $Int(B)$.

I'm kind of confused by the question. We're used to having questions where all the premises matter, but this doesn't seem to be the case here, or is it?

I mean, let $x_0$ be the maximum whose existence we want to contradict. Then $x_0$ has to be a critical point and the sum in the inequality vanishes anyway. But $\Delta f$ needs to be negative in maxima, right? Hence contradiction.

So, what's the point of the $a_i$, why make them continuous? Do we really need the contraint on the unit ball, or if not, what property of $B$ matters? Convexity, compactness?

Are all these premises just to confuse the students and we actually don't need them? Or am I severely overlooking something? If so, could you give a hint please on what to keep in mind for the proof?

Thank you.

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The Laplacian should be non positive. And yes, continuity of $a_i$ is irrelevant and domain can be arbitrary. Perhaps the course is about classical solutions, so... – Andrew Jun 16 '12 at 11:25

The $a_i$ are not additional assumptions, but additional constraints. The statement is, of course, true, if the $a_i$ are all identically zero. The point is that you may even have this additional term in the equation and the statement still holds true.

And the Laplacian is not necessarily negative, only nonpositive, in $x_0$ such that $f(x_0)$ is maximal.

And, finally, if the $a_i$ were not continuous you'd run into problems assuming $f\in C^2$ (usually one considers solutions of the type $\Delta f + \sum a_i \frac{\partial f}{\partial x_i} = g(x)$) But you are right, to just prove the above statement the assumption can be weakened.

(Edit: note though, that it is not clear that the statement remains true if the $a_i$ have, say, a pole at $x_0$. If then, e.g., $\lim_{x\rightarrow x_0}a_i \frac{\partial f}{\partial x_i} < 0$, how would you reason in that case? For this reason I'd say it is, for an exercise, safe and reasonable to assume the $a_i$ are continuous.)

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Thanks. Regarding your comment on the continuity of the $a_i$: As I see it, for infering $x_0$ is not a maximum, all we need is the local information that $\Delta f(x_0) > 0$, not anything from an environment of $x_0$, right? Could you elaborate please why I ran into problems if $a_i$ weren't continuous? – H.Kroeger Jun 16 '12 at 14:16
@H.Kroeger It is, in general, not sufficient to look at this pointwise. If, as I suggested, $a_i$ is unbounded and $a_i\frac{\partial f}{\partial x_i} > M >0$, say, when $x$ tends to a critical point of $f$, then you will not be able to conclude that $\Delta f >0$, but only $\Delta F > - M$, if at all. So, if $a_i$ is not continuous you need at least to know that $a_i \frac{\partial f}{\partial x_i}\rightarrow 0$ if $x$ tends to a critical point of $f$ (e.g. cause $a_i$ is assumed to be bounded). (Or you need additional reasoning, which I, right now, could not provide). – user20266 Jun 16 '12 at 14:31
Ah okay, I'll have a try with this approach, thank you. – H.Kroeger Jun 16 '12 at 14:40
@H.Kroeger: maybe Andrew can comment on that (he who commented on your question). As pointed out in my answer, though, I don't think that your exercise is aiming at a discussion like this. It's intention, for sure, is rather that you figure out the reasoning you actually found, so, for simplicity's sake, they just assumed $a_$ to be continuous. – user20266 Jun 16 '12 at 14:53