Suppose we have a compact Riemmanian manifold $(M^n,g)$ without boundary, and some class $\{g_u\}$ of conformal metrics $g_u:=e^{2u}g$, where $\{u\}\subset C^\infty(M^n)$. For some $k,p$ suppose there exists a constant $C_1=C_1(g)$ such that each $u$ in this class has corresponding $W^{k,p}(M^n,g_u)$-norm bounded by $C_1$: \begin{equation} ||u||_{W^{k,p}(M^n,g_u)} \leq C_1. \end{equation}

Can we conclude there exists $C_2=C_2(g)$ such that each $w$ in this class has $W^{k,p}(M^n,g)$-norm bounded by $C_2$, i.e. \begin{equation} ||u||_{W^{k,p}(M^n,g)} \leq C_2? \end{equation}

UPDATE: The only thing I can think of is if $p$ is high enough then from Sobolev embeddings we get a bound on the Holder norms, but is this uniform in $g$? I should think uniform bounds on the Sobolev norms does imply uniform bounds on the Holder norms. If this is the case, then in particular we get uniform bounds on all the functions $u$, in which case just doing conformal transformations in the integrals (see the comments) answers in the affirmative.

  • $\begingroup$ Your two inequalities seem to be identical - can you clarify the difference? $\endgroup$ – Anthony Carapetis Mar 27 '18 at 22:51
  • $\begingroup$ Preliminary comment: I think this would follow by rewriting the volume element of $g$ in terms of that of $g_u$, then writing the $L^p$ norms of $u$ and its gradient using the change of variable formula. I would then use the compactness and the fact that these $L^p$-norms with respect to the volume element of $g_u$ are bounded. I haven't done the computations but maybe this argument would be enough. EDIT: In my mind I read $K=1$ :( $\endgroup$ – Sak Mar 27 '18 at 22:51
  • $\begingroup$ @AnthonyCarapetis yes sorry, the second one shouldn't have a subscript $u$. I have edited. $\endgroup$ – kt77 Mar 27 '18 at 22:53
  • 1
    $\begingroup$ @Sak Yes each $u$ would be bounded, but that doesn't imply a uniform bound on all $u$ in our class (which may be infinite)? Unless I'm missing something very obvious. $\endgroup$ – kt77 Mar 27 '18 at 22:56
  • 1
    $\begingroup$ @AnthonyCarapetis Hopefully I've edited it to make it a little clearer :) Perhaps this is something more suited to MathOverflow? It's late in the day and I've lost the ability to tell whether I'm missing something obvious $\endgroup$ – kt77 Mar 27 '18 at 23:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.