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If for some strongly continuous contraction semigroup $\{T(t)\}$ it holds that $\forall f \in \mathcal{D}(A): T(t)f-f = \int_0^t T(s) Af ds$ where:

  • $\mathcal{D}(A)$ is dense in $L$
  • $A$ is dissipative
  • $\mathcal{R}(\lambda - A) = L$ for all $\lambda > 0$

then $A$ is the generator of $\{T(t)\}$. I know it holds that if $A$ is the generator of $\{T(t)\}$, this property holds but I don't see how to show the other direction?

Here $\mathcal{D}(A)$ is the domain of $A$.

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Suppose $T, A$ satisfy the properties you stated. Start with the identity $$ T(t)f = f+\int_{0}^{t}T(s)Afds. $$ Then $T(t)f$ has a strong right derivative at $t=0$ for all $f\in\mathcal{D}(A)$. So the generator $B$ of $T$ has a domain that includes $\mathcal{D}(A)$, and $$ Bf = \lim_{h\downarrow 0}\frac{1}{h}(T(t)f-f) = T(0)Af = Af,\;\; f\in\mathcal{D}(A). $$ So the generator $B$ of $T$ either equals $A$ or is a proper extension of $A$. It cannot be a proper extension because $(B-\lambda I)$ is surjective for the generator $B$, which means that if $x \in \mathcal{D}(B)\setminus\mathcal{D}(A)$, then $(A-\lambda I)y = (B-\lambda I)x$ for a unique $y\ne x$, which would be impossible because $(B-\lambda I)y=(B-\lambda I)x$ would then hold, and the invertibility of $B-\lambda I$ would then give a contradiction. So $A=B$ must hold.

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  • $\begingroup$ Awesome, totally clear, thanks! $\endgroup$ – HolyMonk Oct 25 '17 at 16:42
  • $\begingroup$ Ps: love your name $\endgroup$ – HolyMonk Oct 25 '17 at 16:42
  • $\begingroup$ @HolyMonk : Thanks. $\endgroup$ – DisintegratingByParts Oct 25 '17 at 17:19

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