$L^p$-contractivity implies $L^p$-dissipativity?

Question

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 !

Answer 1

$L^p$-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 $$(Au, |u|^{p-2}u)\leq 0,\quad\forall\ u\in D(A).\;\square$$