4
$\begingroup$

I have an intriguing and probably simple question: reading the articles and books of Wolfgang Arendt on semigroups of linear operators, I found on many places properties of the Neumann Laplacian.

In W. Arendt, Semigroups and evolution equations: functional calculus, regularity and kernel estimates, Handbook of Differential Equations: Evolutionary Equations. Vol. 1. North-Holland, 2002, pages 1-85 (it can be seen here),

on page 69 we found an assertion that the Neumann Laplacian generates an ultracontractive semigroup. This means that the below property is satisfied only for any $1\geq t>0$ (see the definition from page 65).

Here is my question: How can we prove that the property is true for any $t\in (0,\infty)$? And why the definition of ultracontractivity is only for $t\in (0,1]$?

For an open, bounded, connected and with an uniform Lipschitz boundary $\Omega\subseteq\mathbb{R}^2$ consider the semigroup of linear operators $S(t)_{t\geq 0}$ generated by the Neumann Laplacian:

\begin{equation} \Delta_N:D(\Delta_N)\to L^2(\Omega),\ D(\Delta_N)=\left\{u\in H^1(\Omega)\ \big |\ \Delta u\in L^2(\Omega),\ \dfrac{\partial u}{\partial\nu}=0, \ \mathcal{H}^{1}\text{- a.e. on}\ \partial\Omega\right\}. \end{equation}

Then for any $1\leq p\leq q\leq +\infty$ there is a constant $c=c(\Omega,p,q)$ that possess the following property (called ultracontractivity):

\begin{equation} \Vert S(t)\phi\Vert_{L^q(\Omega)}\leq c t^{-\frac{N}{2}\left (\frac{1}{p}-\frac{1}{q}\right )}\Vert\phi\Vert_{L^p(\Omega)},\ \forall\ \phi\in L^p(\Omega),\ \forall\ t\geq 0 \end{equation}

P.S. It's a natural question, since in many other books like Cazenave & Haraux – An introduction to semilinear evolution equations (page 44) or Barbu Viorel - Analysis and control of nonlinear infinite dimensional systems (page 31) the above property is proved for any $t>0$ in the case of Dirichlet Laplacian.

$\endgroup$
5
  • 2
    $\begingroup$ Consider what happens for $\phi=1$. $\endgroup$ Aug 19, 2021 at 13:05
  • 2
    $\begingroup$ Since $S(t)1=1$ ultracontractivity holds only in bounded intervals in the form written above. $\endgroup$ Aug 19, 2021 at 13:12
  • $\begingroup$ Now I understand. So we can assume it on intervals of the type $(0,T]$ not just on $(0,1]$. There is nothing special with 1 then. Thanks a lot! $\endgroup$
    – Bogdan
    Aug 19, 2021 at 13:39
  • $\begingroup$ But when $p=q=+\infty$ then we can assume it on $(0,\infty)$, right? $\endgroup$
    – Bogdan
    Aug 19, 2021 at 13:42
  • $\begingroup$ Does it happen to know some place where I can find a proof of that fact? Thanks a lot! $\endgroup$
    – Bogdan
    Aug 19, 2021 at 14:11

1 Answer 1

7
$\begingroup$

It might be helpful to point out the following conceptual reasons why ultracontractivity estimates are mainly interesting for times close to $0$.

Let us consider the following general setting: We have a finite measure space $(\Omega,\mu)$, two integrability indices $1 \le p < q \le \infty$ and a $C_0$-semigroup $(S(t))_{t \ge 0}$ on $L^p(\Omega,\mu)$.

Generally speaking, we are interested in situations where one of the operators $S(t)$ (or all of them for $t > 0$) map $L^p$ into $L^q$ (which can be interpreted as somekind of "smooting" property). Ultracontractivity means that we have specific estimates of the operator norm from $L^p$ to $L^q$ for all $t > 0$.

Now here are some useful observations:

The long-time behaviour does not depend on the choice of spaces.

If, for some $t_0 > 0$, we have $S(t_0)L^p \subseteq L^q$, then there are constants $c_1,c_2,c_3,c_4 > 0$ such that \begin{align*} \tag{1} \|S(t)\|_{L^p \to L^p} & \le c_1 \|S(t)\|_{L^p \to L^q} \le c_2 \|S(t-t_0)\|_{L^q \to L^q} \\ & \le c_3 \|S(t-2t_0)\|_{L^q \to L^p} \le c_4 \|S(t-2t_0)\|_{L^p \to L^p} \end{align*} for all $t > 2t_0$. The first and the fourth inequality follow from the fact that $L^q$ embeds continuously into $L^p$ (since the measure space is finite), and the second and the third inequality follow from the semigroup law and the fact that $S(t_0)$ is a bounded operator from $L^p$ to $L^q$ due to the closed graph theorem.

Inequality (1) shows that estimates for the norm $\|S(t)\|_{L^p \to L^q}$ are not particularly interesting for large $t$, since they are not different from the estimates that one can prove for $\|S(t)\|_{L^p \to L^p}$. For instance, if the semigroup is generated by the Dirichlet Laplacian, then $\|S(t)\|_{L^p \to L^p}$ converges exponentially to $0$, and hence so does $\|S(t)\|_{L^p \to L^q}$ (with the same rate).

It is also worthwhile to note that the long-term behaviour of $\|S(t)\|_{L^p \to L^p}$ (and hence of $\|S(t)\|_{L^p \to L^q}$) can be changed by rescaling the semigroup: if one substracts $a\operatorname{id}$ (for a real number $a$) from the semigroup generator, then the semigroup gets multiplied by $e^{-ta}$, so we are always able to enforce exponential decay of the $\|\cdot\|_{L^p \to L^q}$-norm as $t \to \infty$, simply by a scalar shift of the generator.

The short-time behaviour depends heavily on the choice of spaces.

This is probably easiest to explain by considering an explicit example, say the Neumann-Laplace operator $\Delta_N$ on a bounded domain, which generates a semigroup $(S(t))_{t \ge 0}$.

The semigroup satisfies $\|S(t)\|_{L^\infty \to L^\infty} \le 1$ for all $t \ge 0$ (this follows, for instance, from the positivity of the semigroup and the fact that the constant one-function is a fixed point of the semigroup) as well as $\|S(t)\|_{L^1 \to L^1} \le 1$ for all $t \ge 0$ (this follows, for instance, by duality from the aforementioned $L^\infty$-estimate). Hence, by interpolation, $\|S(t)\|_{L^p \to L^p} \le 1$ for all $p \in [1,\infty]$ and all $t \ge 0$. In other words, we have boundedness of the semigroup for all times when we consider the operator norm on a fixed $L^p$-space.

However, the situation changes significantly if we consider the norm $\|\cdot\|_{L^p \to L^q}$ for $p < q$. To see this, fix a function $f$ which is in $L^p$ but not in $L^q$. By the strong continuity of the semigroup on $L^p$ we have $S(t)f \to f$ in $L^p$ as $t \downarrow 0$, but we must have $\|S(t)f\|_{L^q} \to \infty$ since $f \not \in L^q$. This proves that $\|S(t)\|_{L^p \to L^q} \to \infty$ as $t \downarrow 0$.

In other words, while the "smoothing effect" caused by the operator $S(t)$ is still present for small $t$ (i.e., $S(t)L^p \subseteq L^q$ for all $t > 0$), the effect becomes quantitively weaker in the sense that the operator norm from $L^p$ to $L^q$ explodes as $t \downarrow 0$.

So, the interesting thing about an ultracontractivity estimate is that it gives us an explicit quantititve bound for the blow-up of $\|S(t)\|_{L^p \to L^q}$ as $t \downarrow 0$.

Relation to Sobolev embeding theorems.

I think the following relation to Sobolev embedding theorems is quite illuminating. As pointed out in the question, for the semigroup $(S(t))_{t \ge 0}$ generated by the Neumann Laplace operator, one has the ultracontractivity estimate $$ \tag{2} \|S(t)\|_{L^p \to L^q} \le c t^{-\frac{N}{2}(\frac{1}{p} - \frac{1}{q})} \quad \text{for } t \in (0,1], $$ where $N$ is the dimension of the the underlying domain.

Now, if $\lambda > 0$ is a real number, then the resolvent of $\Delta_N$ on $L^p$ is given by the integral $$ R(\lambda,\Delta_N)f = \int_0^\infty e^{-\lambda t} S(t) f \, dt, $$ for each $f \in L^p$; the integral converges as a Bochner integral in $L^p$. Now let us analyse whether the integral converges also converges in $L^q$, even if $f$ is only in $L^p$. To this end, we split the integral into the integrals $\int_0^1$ and $\int_1^\infty$:

  • It follows from $(1)$ that $\int_1^\infty \|e^{-\lambda t} S(t)f\|_{L^q} \, dt < \infty$ for each $f \in L^p$.

  • It follows from the ultracontractivity estimate $(2)$ that $\int_0^1 \|e^{-\lambda t} S(t)f\|_{L^q} \, dt < \infty$ for each $f \in L^p$ if the additional condition $\frac{N}{2}(\frac{1}{p}-\frac{1}{q}) < 1$ (equivalently, $\frac{2}{N} > \frac{1}{p} - \frac{1}{q}$) is satisfied.

Hence, if $\frac{2}{N} > \frac{1}{p} - \frac{1}{q}$, then the resolvent $R(\lambda,\Delta_N)$ maps $L^p$ into $L^q$; since the range of the resolvent equals the domain of $\Delta_N$, this means that the domain of the Neumann Laplace operator on $L^p$ is contained in $L^q$ if $\frac{2}{N} > \frac{1}{p} - \frac{1}{q}$.

If the underlying domain $\Omega$ in $\mathbb{R}^N$ has sufficiently smooth boundary, the domain of $\Delta_N$ in $L^p$ is the Sobolev space $W^{2,p}$; so in this case, what we have shown before amounts to saying that the Sobolev embedding theorem $W^{2,p} \subseteq L^q$ holds if $\frac{2}{N} > \frac{1}{p} - \frac{1}{q}$. Since all we needed for the proof was the ultracontractivity property $(2)$, we conclude that ultracontractivity gives rise to an abstract version of the Sobolev embedding theorem.

More information in this direction can, for instance, be found in Chapter 2 of Davies' book "Heat kernels and spectral theory" (1989).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.