The following definition and Theorem are given in the book "A short course on operator semigroup" by the author "K-J Engel and R Nagel".

Sectoral operator: A closed linear operator $(A,D(A))$ in Banach space $X$ is called sectoral (of angle $\delta$) if there exists $0<\delta\leq\frac{\pi}{2}$ such that sector $\sum_{\frac{\pi}{2}+\delta}:=\{\lambda\in \mathbb{C}:| arg\lambda |<\frac{\pi}{2}+\delta\} \backslash\{0\}$ is contained in the resolvent set $\rho(A),$ and if for each $\epsilon\in (0,\delta)$, there exists $M_{\epsilon}\geq1$ such that$$\|R(\lambda,A)\|\leq\frac{M_{\epsilon}}{|\lambda|},$$ for $0\neq\lambda\in \overline{\sum_{\frac{\pi}{2}+\delta-\epsilon}}. $ Again, we know the following theorem

Theorem An operator $(A,D(A))$ on a Banach space $X$ generates a bounded analytic semigroup $\Big\{T(z):z\in\sum_{\delta}\cup\{0\}\Big\}$ on $X$ if and only if A is densly defined and sectoral operator.

Where $T(0)=I$ and for $z\in\sum_{\delta}$, $$T(z)=\frac{1}{2\pi i}\int_{\gamma}e^{\mu z}R(\mu,A)d\mu$$ where $\gamma$ is any piecewise smooth curve in $\sum_{\frac{\pi}{2}+\delta}$ going from $\infty e^{-i(\frac{\pi}{2}+\delta')}$ to $\infty e^{i(\frac{\pi}{2}+\delta')}$, for some $\delta'\in(|argz|,\delta)$.

In the chapter 5 of the book "R. deLaubenfels, Existence Families, Functional calculi and Evollution Equations, springer-verlag, Berlin,1994" at page no 69, it is given that $T(z)$ are one to one and it's range is dense in $X$.

My question: For $t>0$, What is the inverse operator of $T(t)$ from it's range set in term of contour integration? Please help me.

I was expecting $$T(-t)=\frac{1}{2\pi i}\int_{\gamma}e^{-\mu t}R(\mu,A)d\mu$$ would be the inverse of T(t) but what i found , it can not be. please help me.

  • $\begingroup$ The heat equation is not reversible in time. That corresponds to the classical analytic semigroup where, at any positive time, $T(t)f$ is in the domain of every positive power of the generator. The Poisson semigroup has the same property. $\endgroup$ – DisintegratingByParts Aug 12 '15 at 6:41
  • $\begingroup$ @TrialAndError : You want to say that inverse can not define in term of contour integration? The property you mention that is the property of analytic semigroup for positive time. $\endgroup$ – Ajoy Jana Aug 12 '15 at 6:53
  • $\begingroup$ @TrialAndError: I want Inverse of the linear operator $T(t)$, may not bounded. I am looking for form of inverse of $T(t)$. $\endgroup$ – Ajoy Jana Aug 12 '15 at 6:57
  • $\begingroup$ How do you know that the form you have is not correct? It's tough to recognize the domain of the inverse, isn't it? How many functions are you going to find that are in the domain of the inverse for all $t$? Can you find even one? $\endgroup$ – DisintegratingByParts Aug 12 '15 at 15:53
  • $\begingroup$ @TrialAndError: The domain of the inverse is the range of the operator T(t). I was started doing by taking one element from the range but I got some problem. $\endgroup$ – Ajoy Jana Aug 12 '15 at 18:06

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