Let $A$ be the infinitesimal generator of a contraction semigroup $(T(t))_{t\ge 0}$ on the Hilbert space $X$, and $D\in\mathcal{L}(X)$.

I want to show that the operator $A+D-2\|D\|I$ with domain $D(A)$ generates a contraction semigroup on $X$.

By the Lummer-Phillips Theorem, we have that if $A$ is a linear operator with domain $D(A)$ on a Hilbert space $X$, then $A$ is the infinitesimal generator of a contraction semigroup $(T(t))_{t\ge 0}$ if and only if $A$ is dissipative and $ran(I-A)=X$.

We have that a linear operator $A:D(A)\subset X\to X$ is dissipative if and only if

$$\|(\alpha I-A)x\|\ge\alpha\|x\|$$

for $x\in D(A)$, $\alpha>0$.

So for $\alpha>0$, we have

$\|(\alpha I-(A+D-2\|D\|I)x\|=\|(\alpha I-A-D+2\|D\|I)x\|$

Where do I go from here?

  • $\begingroup$ What is $D$? What properties does it have? $\endgroup$ – DisintegratingByParts May 21 '15 at 10:53
  • $\begingroup$ @T.A.E. $D\in\mathcal{L}(X)$. $\endgroup$ – Jason Born May 21 '15 at 12:04

Assuming you're working on a complex Hilbert space, the range condition and the following condition are equivalent to dissipative: $$ \Re(Ax,x) \le 0,\;\;\; x \in \mathcal{D}(A). $$ This last condition persists if you add another bounded operator $C$ satisfying the same condition. In your case, because $D$ is bounded, $$ |(Dx,x)| \le \|D\|\|x\|^{2},\\ \Re(Dx,x) \le \|D\|\|x\|^{2}, \\ \Re((D-\|D\|I)x,x) \le 0. $$ Let $$ B = A+(D-\|D\|I) $$ Because $\Re(Bx,x) \le 0$, it follows that $B$ is dissipative iff $B-\alpha I$ is surjective for some real $\alpha > 0$ because $\mathcal{R}(B-\alpha I)$ is the same for all $\alpha > 0$. Notice that \begin{align} B-\alpha I & = (A-\alpha I)+(D-\|D\|I) \\ & = \{I+(D-\|D\|I)(A-\alpha I)^{-1}\}(A-\alpha I) \end{align} The operator in braces is invertible for large enough $\alpha$ because $$ \|(D-\|D\|I)(A-\alpha I)^{-1}\| \le 2\|D\|/\alpha < 1 $$ for large enough $\alpha > 0$. Hence, $B-\alpha I$ is surjective because the operator on the right is surjective for large enough $\alpha$. And that gives you what you want: $B$ is the generator of a contractive $C_{0}$ semigroup. The same is true of $B-\|D\|I$, which is the operator you wanted to know about.

  • $\begingroup$ $$ \Re(Dx,x) \le \|D\|\|x\|^{2}, \\ \Re((D-\|D\|I)x,x) \le 0. $$ Could you explain this step please? $\endgroup$ – Jason Born May 26 '15 at 10:35
  • $\begingroup$ @user3482534 : The real part of the inner product is bounded by the absolute value of the inner product, which, by Cauchy-Schwarz, is bounded by the products of norms of vectors. $\endgroup$ – DisintegratingByParts May 26 '15 at 10:42
  • $\begingroup$ I was talking about how you derived $\Re((D-\|D\|I)x,x)\le 0$ from the inequality above? $\endgroup$ – Jason Born May 26 '15 at 15:44
  • $\begingroup$ @user3482534 : $\Re(Dx,x) \le \|D\|\|x\|^{2}=\|D\|\Re(x,x) = \Re ((\|D\| I)x,x)$ $\endgroup$ – DisintegratingByParts May 26 '15 at 16:55
  • $\begingroup$ For 'Because $\Re(Bx,x) \le 0$, it follows that $B$ is dissipative iff $B-\alpha I$ is surjective for some real $\alpha > 0$' Do you mean that it is a contraction semigroup (rather than 'dissipative' (since this would be the Lummer-Phillips theorem and $\Re(Bx,x) \le 0$ suggests that $B$ is dissipative by definition anyway)? $\endgroup$ – Jason Born May 29 '15 at 14:58

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