Necessary and sufficient condition of being dissipative I want to know a necessary and sufficient condition on $m:\Omega \mapsto\mathbb{C}$ such that the multiplication operator $M_{m}$ is dissipative in $L^{p}(\Omega)$, where $\Omega$ is a Banach space.
It shouldn't be so difficult and I suspect it has to play with the real part of $M_{m}$. However sadly I don't know how to tackle the problem...
And in case the term is not so popular (that I assume.. :S), one can find the definition of the term dissipative in the book A short course on Operator Semigroups by Engel and Nagel: Link to Google Books. Other relevant terms also can be found in that book.
Any suggestions?
Cheers.
 A: I think I have an answer when $(\Omega,\mathcal F,\mu)$ is a $\sigma$-finite measure space. 

For $1\leq p<\infty$, $M_m$ is dispersive if and only if we have $\Re m(x)\leq 0$ for almost every $x$.

We show the result when $(\Omega,\mathcal F,\mu)$ is finite. We assume that $M_m$ is dispersive. We put for $\lambda\in\mathbb Q_+^*$ and $n\in\mathbb N^*$, $A_n^{\lambda}:=\left\{x\in \Omega\mid|\lambda -m(x)|^p-|\lambda|^p\leq -\frac 1n\right\}$. We have for each $f\in L^p(\Omega)$ and $\lambda\in\mathbb Q_+^*$
$$\int_{\Omega}|\lambda-m(x)|^p|f(x)|^pd\mu x\geq \lambda^p\int_{\Omega}|f(x)|^pd\mu,$$
and in particular, for $f=\mathbf 1_{A_n^{\lambda}}$, which is integrable 
$$0=\int_{A_n^{\lambda}}(|\lambda-m(x)|^p-|\lambda|^p)d\mu\leq -\frac 1n\mu\left(A_n^{\lambda}\right),$$
hence $\mu\left(A_n^{\lambda}\right)=0$ and the set $N:=\bigcup_{n>0,\lambda\in\mathbb Q^*_+}A_n^{\lambda}$ has measure $0$. 
Hence we have for each $x\in \omega\setminus N$ and $\lambda\in\mathbb Q_+^*$, $|\lambda-m(x)|\geq \lambda$ and by continuity this inequality is true for $\lambda \in\mathbb R$. Writing $m(x)=f(x)+ig(x)$ where $f$ and $g$ are real functions, we get $(\lambda-f(x))^2+g(x)^2\geq \lambda^2$ hence $-2\lambda f(x)+f(x)^2+g(x)^2\geq 0$ and $f(x)\leq \frac{f(x)^2+g(x)^2}{2\lambda}$ so $f(x)\leq 0$ for each $x\in \Omega\setminus N$. 
Conversely, we assume that $f(x)\leq 0$ for almost every $x$. We have 
for each $\lambda>0$ and for almost every $x$$$|\lambda-f(x)-ig(x)|^2-\lambda^2=-2\lambda f(x)+f(x)^2+g(x)^2\geq 0$$
which yields the inequality $\lVert (\lambda I-M_m)f\rVert_p\geq \lVert f\rVert_p$ for all $f\in L^p(\Omega)$.
To jump from the finite case to the $\sigma$-finite one, write $\Omega$ as a countable disjoint union of set of finite measure $\Omega_n$, and apply what was done to this case to get that we should have almost everywhere on $\Omega_n$, $\Re m(x)\leq 0$, hence almost everywhere on $\Omega$.
