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Does $L^p$-contractivity of an operator semigrop imply the $L^p$-dissipativity of the operator?

Please note the definition of $L^p$-dissipativity:

$(Au, |u|^{p-2}u)\leq 0$ for all $u\in C^1_0(\Omega)$

Thank you in advance !

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  • $\begingroup$ Lumer Phillips theorem is the answer. en.wikipedia.org/wiki/Lumer–Phillips_theorem $\endgroup$ – Nathanael Skrepek May 18 at 19:08
  • $\begingroup$ Thank for the feed back. The lumer philips theorem concerns dissipativity, does it still work for the L^p dissipativity ? $\endgroup$ – Siki May 18 at 19:31
  • $\begingroup$ on wikipedia the theorem is stated for a general Banach space. Since $L^p$ is a Banach space this theorem also covers the situation in $L^p$. $\endgroup$ – Nathanael Skrepek May 18 at 19:55
  • $\begingroup$ I agree with you but the definition of dissipativity and L^p dissipativity are rather différents $\endgroup$ – Siki May 18 at 20:07
  • $\begingroup$ Okay maybe you can provide the definition in your question because i have never heard about this concept. $\endgroup$ – Nathanael Skrepek May 19 at 5:02
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$L^p$-contractivity implies $L^p$-dissipativity?

I suppose that "$L^p$-contractivity" means $$\|T(t)\|_{\mathcal L}\leq 1,\quad\forall\ t\geq 0$$ where $T(t):L^p\to L^p$ is the semigroup generated by $A:D(A)\subset L^p\to L^p$.

In this case, according to your definition of $L^p$-dissipativity, the answer is yes provided that $ C^1_0(\Omega)\subset D(A)$.

Proof:

The $L^p$-contractivity implies that

$$(T(t)u, |u|^{p-2}u)\leq \underbrace{\|T(t)\|}_{\leq 1}\|u\|_{L^p}\underbrace{\||u|^{p-2}u\|_{L^{p/(p-1)}}}_{= \|u\|_{L^p}^{p-1}}\leq\|u\|^p_{L^p},\quad\forall\ u\in L^p.$$ Therefore, $$(T(t)u-u, |u|^{p-2}u)=(T(t)u, |u|^{p-2}u)-(u, |u|^{p-2}u)\leq \|u\|^p_{L^p}-\|u\|_{L^p}^p=0,\quad\forall\ u\in L^p$$ and thus, multiplying by $\frac{1}{h}$, we obtain $$\left(\frac{T(t)u-u}{h}, |u|^{p-2}u\right)\leq 0,\quad\forall\ u\in L^p,\; h>0.$$ Taking the limit as $h\to 0^+$, we conclude that $$(Au, |u|^{p-2}u)\leq 0,\quad\forall\ u\in D(A).\;\square$$

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  • $\begingroup$ Thank you @Pedro $\endgroup$ – Siki May 21 at 8:14

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