Is $-(I-\Delta)$ an dissipative operator in the context of $L^p$ spaces?

Question

\begin{align} \frac{d}{dt}u(t)&=-(I-\Delta)u,\quad t>0\\ u(0)&=f \end{align} with initial condition $f\in L^2$.

If $P:=-(I-\Delta):D(P):={H}^{2}\subset L^2\to L^2$ with $H^2:=\left\{u\in L^2: \int |(1+|\xi|^2)\widehat{u}(\xi)|^2\,d\xi<\infty\right\}$ (Bessel space or Sobolev space)

Using the Fourier transform, and noting that the symbol of the operator $-(I-\Delta)$ is $-(1+|\xi|^2)$, the operator takes the form $-(I-\Delta)u=-\mathcal{F}^{-1}((1+|\xi|^2)\widehat{u}(\xi))$.

¿Is $P:=-(I-\Delta)$ an dissipative operator?

My attempt: Let $\lambda>0$ and $u\in D(A)$, then \begin{align} \left\|(\lambda I-P)u\right\|_p&=\left\|-\mathcal{F}^{-1}((\lambda+(1+|\xi|^2))\widehat{u}(\xi))\right\|_p\\ &=\left|(\lambda+(1+|\xi|^2))\widehat{u}(\xi)\right\|_p\quad (\text{Plancherel theorem})\\ &\geq \lambda \left\|u\right\|_p\quad (\text{Triangle's inequality}) \end{align} Therefore, $P$ is dissipative.

Question 1. This proof is correct?

Question 2. Now, if $P:D(P)=H^{p}\subset L^p\to L^p$ with $H^p:=\left\{u\in L^p: \mathcal{F}^{-1}((1+|\xi|^2)\widehat{u}(\xi))\in L^p\right\}$ ($1<p<\infty$). In this case: Is $P$ an dissipative operator? I tried to do this but in $L^p$ Plancherel's theorem is not valid... Thanks.

Answer 1

For the case $p=2$, the proof you exhibited is actually correct.

The result still holds on $\mathrm{L}^p(\mathbb{R}^n)$, whenever $1<p<+\infty$, $p\neq2$, but as you mentionned, your proof is no longer valid.

Here is presented a simple argument, using elementary Fourier analysis, and basic $\mathrm{L}^p$ inequalities.

We argue by density provided $f\in\mathcal{S}(\mathbb{R}^n)$, we set for $\lambda\geqslant 0$ and all $x\in\mathbb{R}^n$, \begin{align*} u(x) := (\lambda\mathrm{I} + \mathrm{I}-\Delta)^{-1}f (x) := \frac{1}{(2\pi)^{n}}\int_{\mathbb{R}^n}\frac{1}{\lambda+1+\left\lvert\xi\right\rvert^2}\mathcal{F}f(\xi)e^{ix\cdot \xi}\mathrm{~d}\xi. \end{align*} One can even show that $u\in\mathcal{S}(\mathbb{R}^n)$.

• I] rewriting $u$ in a different form.

The main trick is the following identity: \begin{align*} \frac{1}{\lambda+1+\left\lvert\xi\right\rvert^2} = \int_0^{+\infty} e^{-\lambda t} e^{-t}e^{-t\left\lvert\xi\right\rvert^2} \mathrm{~d}t \text{.} \end{align*}

So we can replace then use the Fubini-Lebesgue Theorem, so that one obtains \begin{align*} u(x) &= \frac{1}{(2\pi)^{n}}\int_{\mathbb{R}^n}\left(\int_0^{+\infty} e^{-(\lambda+1) t} e^{-t\left\lvert\xi\right\rvert^2} \mathrm{~d}t\right)\mathcal{F}f(\xi)e^{ix\cdot \xi}\mathrm{~d}\xi\\ &=\frac{1}{(2\pi)^{n}}\int_0^{+\infty} e^{-(\lambda+1) t} \int_{\mathbb{R}^n} e^{-t\left\lvert\xi\right\rvert^2} \mathcal{F}f(\xi)e^{ix\cdot \xi}\mathrm{~d}\xi\mathrm{~d}t\text{ . } \end{align*}

Now we use the identity \begin{align*} \int_{\mathbb{R}^n} e^{-t\left\lvert\xi\right\rvert^2} \mathcal{F}f(\xi)e^{ix\cdot \xi}\mathrm{~d}\xi = G_t\ast f (x) \end{align*} where for all $t>0$, all $x\in\mathbb{R}^n$, \begin{align*} G_t(x) := \frac{1}{\sqrt{4\pi t}^n} e^{-\frac{x^2}{4t}} = \frac{1}{\sqrt{ t}^n}G_1\left(\frac{x}{\sqrt{t}}\right). \end{align*}

Thus we obtain the last identity valid for all $x\in\mathbb{R}^n$, \begin{align*} u(x) &=\int_0^{+\infty} e^{-(\lambda+1) t} \left[G_t\ast f\right](x)\mathrm{~d}t\text{.} \end{align*}

• II] The $\mathrm{L}^p$-bound on $u$.

We first recall that, for all $t>0$, \begin{align*} \lVert G_t \rVert_{\mathrm{L}^1(\mathbb{R}^n)} = \lVert G_1 \rVert_{\mathrm{L}^1(\mathbb{R}^n)} =1. \end{align*}

By Minkwoski's integral inequality (see in the appendix below) \begin{align*} \left\lVert u \right\rVert_{\mathrm{L}^p(\mathbb{R}^n)} &= \left\lVert x\mapsto \int_0^{+\infty} e^{-(\lambda+1) t}\, G_t\ast f(x) \mathrm{~d}t\right\rVert_{\mathrm{L}^p(\mathbb{R}^n)}\\ &\leqslant \int_0^{+\infty} e^{-(\lambda+1) t}\, \left\lVert G_t\ast f \right\rVert_{\mathrm{L}^p(\mathbb{R}^n)}\mathrm{~d}t\\ &\leqslant \left(\int_0^{+\infty} e^{-(\lambda+1) t} \left\lVert G_t \right\rVert_{\mathrm{L}^1(\mathbb{R}^n)}\mathrm{~d}t\right) \left\lVert f \right\rVert_{\mathrm{L}^p(\mathbb{R}^n)}\\ &\leqslant \left(\int_0^{+\infty} e^{-(\lambda+1) t} \mathrm{~d}t\right) \left\lVert f \right\rVert_{\mathrm{L}^p(\mathbb{R}^n)} \end{align*} The third line above follows from Young's inequality for the convolution. Finally, we can compute the last integral to obtain the desired result, \begin{align*} \lambda \left\lVert u \right\rVert_{\mathrm{L}^p(\mathbb{R}^n)} \leqslant (\lambda +1) \left\lVert u \right\rVert_{\mathrm{L}^p(\mathbb{R}^n)} \leqslant \left\lVert f \right\rVert_{\mathrm{L}^p(\mathbb{R}^n)}. \end{align*}

We conclude with few remarks.

1.Notice the the same proof applies removing the $+1$ here and there to obtain, $\mathrm{L}^p$-dissipativity of the (negative) Laplacian it-self for all $p\in[1,+\infty]$. However, the domain is much more difficult to compute when $p=1,+\infty$, and is not the "expected natural one".

2.In fact, even for $p\in(1,+\infty)$, having the exact description of the domain $\mathrm{W}^{2,p}(\mathbb{R}^n)$ for the Laplacian is a non-trivial (but well documented) question. (This relies on proving the fact that the space $H^p$ you defined and $\mathrm{W}^{2,p}(\mathbb{R}^n)$ are equals with equivalent norms).

3.Notice that your notation $H^p$ for Bessel potential Sobolev space is not conventionnal, one usually prefer $\mathrm{H}^{2,p}(\mathbb{R}^n)$.

• APPENDIX

Theorem (Minkowski's Integral inequality) ([BCD11,Proposition 1.3])

Let $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$ be two $\sigma$-finite measure space and $F$ a non-negative real valued function $(\mathcal{A}\otimes\mathcal{B})$-measurable. Then for all $p\in[1;+\infty)$, one has : \begin{align*} \left(\int_X\left(\int_Y F(x,y)\,\mathrm d\nu(y)\right)^p\,\mathrm d\mu(x)\right)^{\frac1p}\le\int_Y\left(\int_X F(x,y)^p\,\mathrm d\mu(x)\right)^{\frac1p}\,\mathrm d\nu(y). \end{align*}

Proposition (Young's inequality for the convolution) ([BCD11,Lemma 1.4])

Let $p,q,r\in[1,+\infty]$ such that $$1+\frac{1}{r}=\frac{1}{p}+\frac{1}{q}.$$

Then for all $f\in\mathrm{L}^p(\mathbb{R}^n)$, $g\in\mathrm{L}^q(\mathbb{R}^n)$, the convolution of $f$ and $g$ given formally by $$f\ast g (x) :=\int_{\mathbb{R}^n} f(x-y)g(y) ~\mathrm{d}y,$$ yields actually a well defined measurable function, such that $f\ast g\in\mathrm{L}^r(\mathbb{R}^n)$. Moreover, one has the estimate \begin{align*} \lVert f\ast g \rVert_{\mathrm{L}^r(\mathbb{R}^n)}\leqslant \lVert f \rVert_{\mathrm{L}^p(\mathbb{R}^n)}\lVert g \rVert_{\mathrm{L}^q(\mathbb{R}^n)}. \end{align*}

[BCD11] Bahouri, Hajer; Chemin, Jean-Yves; Danchin, Raphaël, Fourier analysis and nonlinear partial differential equations, Grundlehren der Mathematischen Wissenschaften 343. Berlin: Heidelberg (ISBN 978-3-642-16829-1/hbk; 978-3-642-16830-7/ebook). xvi, 523 p. (2011). ZBL1227.35004.