# Dissipativity for Hilbert spaces

I want to prove that an operator $A:D(A)\to X$ is dissipative $\iff$ $\text{Re}\langle Ax,x\rangle\le 0$ $\forall x\in D(A)$.

The proof for this is actually sketched on the Wikipedia page for Dissipative operators.

So if we suppose that $\text{Re}\langle Ax,x\rangle\le 0$ $\forall x\in D(A)$ then \begin{aligned} \|x-Ax\|^{2}& =\|x\|^{2}+\|Ax\|^{2}-2\text{Re}\langle Ax,x\rangle \\ &\ge\|x\|^{2}+\|Ax\|^{2}+2\text{Re}\langle Ax,x\rangle=\|x+Ax\|^{2}\end{aligned}

Thus $\|x-Ax\|\ge\|x+Ax\|$.

Since $I-A$ has an inverse (why?), we have that $(I+A)(I-A)^{-1}$ is a contraction (why?), or more generally, $(\lambda I+A)(\lambda I-A)^{-1}$ is a contraction for every $\lambda>0$.

If this operator is a contraction for some $\lambda>0$, then $A$ is dissipative (why?).

Now we want to prove for the other direction.

Define $J(x):=\{x^{\ast}\in X^{\ast}:\|x^{\ast}\|^{2}_{X^{\ast}}=\|x\|^{2}_{X}=\langle x^{\ast},x\rangle\}\subset X^{\ast}$.

The Hahn-Banach theorem implies that for a given $x$ there exists a continuous linear functional $\Lambda$ such that $\Lambda(x)=\|x\|$ with $\|\Lambda\|=1$. We identify $\|x\|_{\Lambda}$ with $x^{\ast}$. Hence the set $J(x)$ is non-empty. The canonical duality between a Hilbert space and its dual implies that $J(x)$ consists of a single element $x$. Using this notation, $A$ is dissipative if and only if $\forall x\in D(A)$ there exists $x\in J(x)$ such that $\text{Re}\langle Ax,x\rangle\le 0$.

I didn't fully understand this final paragraph of the proof either so I'd really appreciate it if someone would provide some clarity and/or provide a more constructive argument.

Your first equation, in particular, implies that $\| x- Ax\| \geq \|x\|$. From that we conclude $(I-A)x = 0$ implies $x=0$ (injectivity).

I think we can deduce surjectivity from this as well.

For the Caley-transform $T=(I+A)(I - A)^{-1}$ of $A$ being a contraction, the equation $$\|x-Ax\| \leq \|x+ Ax\|$$ can help us. Let $y \in X$ and let $x \in X$ such that $y = (I-A)x$ (this can be done because of surjectivity). Then $Ty = (I+A)(I-A)^{-1}(I - A)x = x + Ax$, i.e. the equation above yields

$$\|Ty\|\leq \|y\|,$$

i.e. $T$ is indeed a contraction.

• How exactly is $\|x-Ax\|\ge\|x\|$ deduced? – Jason Born Feb 13 '16 at 16:08
• We have $\|x-Ax\|^2 = \|x\|^2 + (\|Ax\|^2 - \Re \langle Ax,x \rangle)$ Since $\Re \langle Ax,x \rangle \leq 0$, everything in the bracket is $>0$ and we get $\|x-Ax\|^2 \geq \|x\|^2$. – Roland Feb 13 '16 at 16:11
• So $I-A$ surjective means that $I-A$ indeed has an inverse. Then you prove that the Cayley transform is indeed a contraction (then it would follow that $A$ is dissipative)? – Jason Born Feb 13 '16 at 16:49
• @user3482534 Yes, but my answer is partial in this regards that: a) I dont show the surjectivity and b) I dont show how to get from $T$ being a contraction that $A$ is dissipative. – Roland Feb 13 '16 at 18:16
• Also the second part of the proof, with the $J(x)$? – Jason Born Feb 14 '16 at 13:21