0
$\begingroup$

On a bounded domain $\Omega \subset \mathbb{R}^n$ with smooth boundary, consider the Laplacian $-\Delta$ with either the Dirichlet or Neumann boundary conditions. More generally, one can also consider a strongly elliptic self-adjoint operator $L$, where $L$ is negative definite. Now consider the heat semigroup $e^{tL}$, which is $L^2$ contractive. I am curious whether $e^{tL}$ is bounded on $L^1(\Omega)$, that is, could we say something like $$ \Vert e^{tL}f\Vert_{L^1(\Omega)} \leq C\Vert f\Vert_{L^1(\Omega)}, f \in L^1(\Omega) ?$$ If yes, will the constant $C$ be independent of $t$? Most books list these types of results when $p \in (1, \infty)$. I would be really grateful if someone can provide a reference. If results like the above hold only under special conditions on the operator $L$, I would like to know that too. Thanks.

$\endgroup$
  • 1
    $\begingroup$ For positive $f$ and neumann BC for the laplacian, you can prove the conservation of the $L^1$ norm by using the divergence theorem. $\endgroup$ – guacho Apr 29 '15 at 5:54
  • $\begingroup$ This can be done on $C(\bar\Omega)$ by using the maximum principle and then on $L^1$ by duality. See for instance Theorem 3.10 in Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations. $\endgroup$ – Michael Renardy Apr 29 '15 at 13:43
  • $\begingroup$ @MichaelRenardy Thanks, I just checked the theorem. Is anything known for Neumann boundary conditions? $\endgroup$ – mathgirl Apr 29 '15 at 15:54
  • $\begingroup$ That is left as an exercise to the reader. $\endgroup$ – Michael Renardy Apr 29 '15 at 16:03

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.