In my class we learnt about the decay estimate of the Poisson equation. The statement is:
However, I do not understand the proof given in the notes. The key part of the theorem since to be that $u(x) \rightarrow 0$ as $|x| \rightarrow \infty$. But the proof uses this fact in a way that I don't understand. The proof is as follows:
What I don't understand is:
- Why can we assume that $f$ vanishes outside the ball $B_R(0)$. From the statement of the theorem, it seems to me that $u(x)$ vanishes outside some ball of radius $R$, rather than $f$.
2)If $f$ or $u$ vanish outside of the ball, then surely we are interested in what happens inside the ball, rather than outside the ball? So how come we only care about what happens outside the ball in the proof?
For 1: Since $f\in C^\infty_0(\mathbb R^n)$, there exists a closed, bounded set $K \subset \mathbb R^n$ such that $f=0$ outside $K$. In the first line of the proof, the author is saying that they can choose some $R>0$ sufficiently large such that $K \subset B_R$, so $f=0$ outside $B_R$.
It is not true that $u$ vanishes outside some $B_R$. Indeed, if $f\in C^\infty_0(\mathbb R^n)$, $f\geqslant 0$, and $f\neq 0$ and we believe (1.0.4) then we can immediately see that $u<0$ in $\mathbb R^n$.
For 2: It's not so much that we are "only interested in what happens inside/outside the ball" since we actually care about every value of $x\in \mathbb R^n$ (since (1.0.5) is supposed to hold in all of $\mathbb R^n$). Here's the reason why I believe the author says that it suffices to take $\vert x \vert >2R$. For simplicity, I'll take $n\geqslant 3$. If $\vert x \vert \leqslant 2R$ then $y \mapsto f(x-y)$ is zero outside (say) $B_{4R}$. Hence, $$ \vert u(x) \vert \leqslant \frac 1 {\omega_n} \int_{B_{4R}} \frac{\vert f(x-y)\vert }{\vert y \vert^{n-2}}\, dy \leqslant C \| f\|_{L^\infty(\mathbb R^n)}\int_{B_{4R}} \frac{1 }{\vert y \vert^{n-2}} \, dy \leqslant C\| f\|_{L^\infty(\mathbb R^n)}. $$ Then in $B_{2R}$ it follows that $$ \vert u(x) \vert \leqslant C\| f\|_{L^\infty(\mathbb R^n)} \cdot \frac{\vert x \vert^{n-2}}{\vert x \vert^{n-2}} \leqslant C\| f\|_{L^\infty(\mathbb R^n)} \cdot \frac{1}{\vert x \vert^{n-2}}. $$ But note there was nothing special about $\vert x \vert^{n-2}$ in the inequality above: in fact, if $g$ is any positive, bounded function then by the above argument $\vert u(x) \vert \leqslant C/g(x)$ in $B_{2R}$. So the reason we get (1.0.5) and not some other inequality has to be because of how $u$ behaves when $\vert x \vert$ is large. It is possible that this is the reason the author chose not to do the computation for $x\in B_{2R}$ (though it's hard to know for certain). It is also possible that they consider the above argument 'easy enough to skip', though this would be a belief that is not shared by other prominent textbooks where the full computation is usually done.