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You'll learn a lot more from Jochen's answer, but maybe I'll point out anyway that there is a very simple argument for this: The eigenfunction $u_0$ of the smallest eigenvalue is positive (see below), …

No. Consider a function $f\in L^1(\mathbb R)$, $f\ge 0$, with $f(x) = 2^{n^2}$ on $2^{-n}<x<2^{-n}+2^{-n^2-n}$ and essentially $f=0$ otherwise. Then $\int_{-r}^r f(x)\, dx \simeq \sum_{n\gtrsim (-\log …

This is really no different from $k=0$. Your kernel is the kernel of the resolvent $(-\Delta-k^2)^{-1}$ on $L^2(\mathbb R^3)$. This is a standard fact, though I'm having trouble now locating a useful …

Update: To elaborate some on my discussion with Willie in the comments, I think I can now do this in $d=1$, and this is perhaps more interesting than the original answer. $-u''-aVu=0$ has a bounde …

Alternatively, the self-adjoint operator $-\Delta$ on the domain $H^2\subseteq L^2$ has spectrum $[0,\infty)$ in any dimension. Since $-1$ is not in the spectrum, $(-\Delta+1)^{-1}$ is bounded on $L^2 …

That doesn't work because $H_0^1$ functions are small near the boundary, so testing against them won't detect bad behavior of $u$ near $\partial\Omega$. For a concrete example, take $\Omega$ as the u …

This is basically a comment on Dario's answer. I'm going to compare the Dirichlet problem on $B$ with the one on $B_0\equiv B\setminus\{ 0\}$ (though I'm not going to justify formally that this is wha …

Update: I intended this to be a complete answer originally, but my "counterexample" was based on a miscalculation. So this is now more a collection of remarks on what I think the question is about. W …

We can use the integral kernel of $R=(A-z)^{-1}$ and nothing much changes when $p\not= 2$. By the variation of constants formula, $(Rf)(x)=\int_0^1 G(x,t;z) f(t)\, dt$. Here $G$ could of course be wor …