If $A$ is bounded and dissipative, is $\lambda \mathbb 1-A$ invertible for $\lambda>0$?

Question

Let $X$ be a Banach space and $j$ a map (not necessarily linear or anti-linear!) from $X$ to $X'$ so that

$$j_x(x)=\|x\|^2 \qquad \forall x \in X \text{ and }\qquad \|j_x\|_{X'}=\|x\|$$

We say that an operator $A$ is dissipative (wrt to $j$) if $j_x(Ax)$ has negative real part for all $x$.

It is easy to show that $\lambda\mathbb 1 - A$ is injective and bounded from below for all $\lambda>0$, since if $x_n$ is a sequence in $X$ with $\|x_n\|=1$ and $\lambda x_n-A x_n \to 0$ then

$$j_{x_n}(\lambda x_n-Ax_n)=\lambda - j_{x_n}(A x_n)$$

has real part $≥\lambda>0$. But since $\|j_{x_n}\|=1$ and $(\lambda x_n - Ax_n)\to 0$ we get that the expression must go to zero, a contradiction.

I would like to see whether the map is actually invertible, meaning (in the case of bounded $A$) that the resolvent of $\lambda \mathbb 1 -A$ is all of $X$.

This book seems to suggest that it is true, since Theorem 2.1 states (paraphrased)

A bounded operator generates a semi-group of contractible operators iff it is dissipative

And Theorem 3.1:

A densely defined operator $A$ generates a strongly continuous semi-group of contractible operators iff it is dissipative and the resolvent of $\mathbb 1 - A$ is $X$.

However in the proofs I can see no way to see that the resolvent of $\mathbb 1 - A$ is all of $X$. Indeed from the way Theorem 3.1 is proven it seems like as if this property should be clear/is trivial.

How can I prove that the resolvent of $\mathbb 1-A$ is all of $X$?

Answer 1

I suggest the book by Pazy on Semigroups of Linear Operators. It's one of the most elegant and readable books in Functional Analysis I've encountered.

On page 14, there is a theorem:

Theorem: A Linear operator is dissipative if and only if $$ \|(\lambda I-A)x\| \ge \lambda \|x\|,\;\; x\in\mathcal{D}(A),\; \lambda >0. $$

His setting for the chapter is a Banach space, and the operator of the theorem is not assumed to be densely defined. Those assumptions are stated in the introduction to the section.

Then he goes on to prove the Lumer-Philips theorem.

Theorem [Lumer-Philips]: Let $A$ be a linear operator with dense domain $\mathcal{D}(A)$ in $X$.

(a) If $A$ is dissipative and there is a $\lambda_0 > 0$ such that the range, $\mathcal{R}(\lambda_0 I-A)$, of $\lambda_0 I-A$ is $X$, then $A$ is the infinitesimal generator of a $C_0$ semigroup of contractions in $X$.

(b) If $A$ is the infinitesimal generator of a $C_0$ semigroup of contractions on $X$, then $\mathcal{R}(\lambda I -A)=X$ for all $\lambda > 0$ and $A$ is dissipative. Moreover, for every $x\in\mathcal{D}(A)$ and every $x^*\in F(x)$, $\Re\langle Ax,x^*\rangle \le 0$.

I'm sure you can guess what $F(x)$ is in the context of dissipative.

The next example shows that you cannot omit the assumption of surjectivity, even when working in Hilbert space.

EXAMPLE [Differentiation]: Consider the operator $A = \frac{d}{dx}$ on the domain $\mathcal{D}(A) \subset L^2[0,\infty)$ consisting of all absolutely continuous functions $f$ in $L^2[0,\infty)$ for which $f'\in L^2$ and $f(0)=0$. Then \begin{align} \Re (Af,f) & = \Re \int_{0}^{\infty}f'(t)\overline{f(t)}dt \\ & = \int_{0}^{\infty}\{f'(t)\overline{f(t)}+f(t)\overline{f'(t)}\}dt \\ & = \int_{0}^{\infty}\frac{d}{dt}|f(t)|^2dt \\ & = |f(t)|^2|_{0}^{\infty} = 0. \end{align} However, $A-\lambda I$ is not surjective for $\lambda > 0$ because \begin{align} ((A-\lambda I)f,e^{-\lambda t}) & = \int_{0}^{\infty}(f'(t)-\lambda f(t))e^{-\lambda t}dt \\ & = \int_{0}^{\infty}\frac{d}{dx}(e^{-\lambda t}f(t))dt = 0. \end{align} (Note: $A$ and $-A$ are dissipative in this case.)