Let $\Delta_R:D(\Delta_R)\to L^2(\Omega)$ the Robin Laplacian defined on:

$$D(\Delta_R)=\left\{u\in H^1(\Omega)\ \big |\ \Delta u\in L^2(\Omega),\ \dfrac{\partial u}{\partial\nu}+bu=0 \ \text{on}\ \partial\Omega\right\}$$,

where $b\in L^{\infty}(\partial\Omega)$ (can be taken positive if needed). Denote by $T(t)_{t\geq 0}$ the semigroup generated by $\Delta_R$. Here $\Omega\subseteq\mathbb{R}^N$ is an open and bounded set with uniform Lipschitz boundary.

I read in some article that it can be shown that for any $1\leq q\leq p\leq +\infty$ there is a constant $C=C(\Omega,p,q)>0$ (depending only on $\Omega,p,q$) such that following estimate hold:

$$\Vert T(t)\phi\Vert_{L^p(\Omega)}\leq C t^{-\frac{N}{2}\left (\frac{1}{q}-\frac{1}{p}\right )}\Vert\phi\Vert_{L^q(\Omega)},\ \forall\ \phi\in L^q(\Omega).$$

How can we prove that inequality?

I read the proof for Dirichlet boundary conditions in T. Cazenave & A. Haraux - An introduction to Semilinear Evolution Equations,1998, page 44. But how can it be done for Neumann or Robin boundary conditions?

  • $\begingroup$ The exponent $-N/2\times (1/p-1/q)$ should be replaced by $-N/2 \times (1/q-1/p)$? $\endgroup$
    – sharpe
    Commented Feb 21, 2021 at 12:37
  • $\begingroup$ Of course. Sorry. $\endgroup$
    – Bogdan
    Commented Feb 21, 2021 at 13:46
  • $\begingroup$ At least, if $p=\infty, q=1$, and $\Omega$ is a bounded Lipschitz domain, your inequality should follow from a Sobolev type inequality. $\endgroup$
    – sharpe
    Commented Feb 21, 2021 at 14:13

1 Answer 1


The estimate the OP is looking for is called an ultracontractivity estimate. A characterisation of semigroups that satisfy such an estimate can be found in the Theorem on page 65, Subsection 7.3.2, of the following survey article:

Wolfgang Arendt: Semigroups and Evolution Equations: Functional Calculus, Regularity and Kernel Estimates in the Handbook of differential equations: Evolutionary equations. Vol. I. (Link to publisher, Link to zbMATH)

Since the Robin Laplacian is associated to a bilinear form with form domain $H^1(\Omega)$, part (v) of the theorem is useful to answer the question: it tells us - in the case $N > 2$ - that we have ultra contractivity of the semigroup if and only if $H^1(\Omega)$ embeds into $L^{2N/(N-2)}(\Omega)$, i.e., the question reduces to a Sobolev embedding theorem. Such an embedding theorem is satisfied if the domain $\Omega$ isn't too rough; more precisely:

Assume that the domain $\Omega$ has the extension property (which is, for instance, satisfied if $\Omega$ is bounded and has Lipschitz boundary). If $N > 2$, then the embedding $H^1(\Omega) \hookrightarrow L^{2N/(N-2)}(\Omega)$ is true, so we get the desired ultracontractivity [op. cit., Subsection 7.3.6].

  • $\begingroup$ Thanks for your answer! I read the theorem that you have mentioned above. I work mainly in spaces of dimension $N=2$. But from what I see on page 69 (Neumann Laplacean) in the survey it seems to work even if $N=2$ (if we have the extension property satisfied). So similarly we can deduce that it holds for Robin bc. Am I right? $\endgroup$
    – Bogdan
    Commented Feb 21, 2021 at 15:03
  • $\begingroup$ @Bogdan: Good question; right now I'm somewhat confused by the case $N=2$, but this is most likely just due to my ignorance. I'll let you know in case that my confusion fades away... $\endgroup$ Commented Feb 21, 2021 at 17:04
  • $\begingroup$ @Bogdan: A few days ago, I briefly discussed the case $N=2$ with a colleague. Could you contact me via email about this (since the information I got is somewhat vague, I'd prefer not to discuss this publicly)? (Please see my profile for a link to my webpage, where my email address can be found.) $\endgroup$ Commented Mar 21, 2021 at 16:52

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