# 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.

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Can you write the definition of dissipative, and what you assume on $\Omega$? –  Davide Giraudo Nov 24 '11 at 19:58

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$\Omega$. - Though I've solved the problem, still thanks for your answer! – newbie Nov 30 '11 at 19:55 You're welcome. I have a question: why did you assumed that$\Omega$is a Banach space? – Davide Giraudo Nov 30 '11 at 19:56 that's because in the definition of dissipative operator it has to lie in a Banach space. – newbie Nov 30 '11 at 20:09 I don't understand: the operator is suppose to be defined on a Banach space (here$L^p(\Omega)$), but$\Omega$can a priori be any set. – Davide Giraudo Nov 30 '11 at 20:13 The operator is defined as$M:D(\Omega)\rightarrow L^{p}(\Omega)$, where$D(\Omega)=\{f:mf\in L^{p}(\Omega)\}\$. –  newbie Dec 1 '11 at 13:43