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

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 $$u\in C^1_0(\Omega)$$

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

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

• Thank you @Pedro – Siki May 21 at 8:14