3
$\begingroup$

In several sources (Choquet-Bruhat & Christodoulou 1981, Nirenberg-Walker 1973) estimates for elliptic partial differential equations on a noncompact manifold are derived in weighted Sobolev spaces. These usually go along the lines: if $L$ is an elliptic operator of order $m$, then $$ \|u\|_{H_{s,\delta}} \leq C(\|Lu\|_{H_{s - m, \delta + m}} + \|u\|_{H_{0,\delta}}). $$ My questions is: is such an estimate type valid in the case of "standard" (non-weighted) Sobolev spaces $H^s$?

Edit: The manifold in question, though noncompact, may be assumed to have various convenient properties at infinity (for instance, asymptotically flat, bounds on the volume growth, etc.).

$\endgroup$
7
  • 2
    $\begingroup$ Generally not: do a basic computation where you solve $\Delta u = f$ on $\mathbb{R}^n$ for some smooth compactly supported nonnegative $f$. You can check explicitly that it behaves like the fundamental solution at infinity in arbitrarily strong topology, i.e. like $|x|^{2-n}$. But this function is not square integrable at infinity in some dimensions ($n \leq 4$), and its derivative is not square integrable in some dimensions $n \leq 2$. If you have a higher-order operator on a manifold, the same is possible in higher dimensions too (and depends on volume growth at infinity, etc.) $\endgroup$
    – user378654
    Commented Jul 6, 2023 at 19:03
  • $\begingroup$ That makes sense - but I don't see how it's a counterexample to this inequality. If $u \sim |x|^{2-n}$ near infinity, then the right-hand side will be infinite due to the presence of the $H^0$ norm, so there is no contradiction. Also I've edited the question regarding the geometry at infinity of the manifold. $\endgroup$
    – Chris
    Commented Jul 6, 2023 at 19:36
  • 1
    $\begingroup$ This holds provided that $L$ has suitably regular coefficients. If you look at Gilbarg and Trudinger Theorem 9.11, it contains most of what you need for the proof; all you need is some additional control on the regularity of $L$ as $|x|\to\infty$ to ensure that you get a uniform bound as $\Omega'$ increases toward the whole space. $\endgroup$ Commented Jul 10, 2023 at 16:51
  • 2
    $\begingroup$ Actually, in the case where you have the whole space, if $L$ has sufficiently regular coefficients you have: given any unit ball $B$, let $B'$ denote the ball of the same center but twice the radius, then $\|u\|_{H^k(B)} \leq C ( \| Lu \|_{H^{k-m}(B')} + \|u\|_{L^2(B')})$ by standard bounded elliptic estimate, with $C$ a uniform constant depending only on the properties of $L$. Next you just sum over a collection of balls with finite overlap and you are done. $\endgroup$ Commented Jul 10, 2023 at 17:02
  • $\begingroup$ @WillieWong What if one has a bounded smooth domain and not the whole space? Any references that state this result (if it is true)? I assume that boundary conditions have to be considered (Dirichlet, Neumann) and will affect the right-hand side of the inequality. CONTEXT: I intend to use this with homogenous boundary conditions to define “equivalent norms” of the $H^k$ spaces given that the homogenous boundary conditions are satisfied. $\endgroup$ Commented Aug 25 at 20:25

0

You must log in to answer this question.

Browse other questions tagged .