$\pm A$ is sectorial implies it is bounded

banach-spacesfunctional-analysisoperator-theorysemigroup-of-operatorsspectral-theory

The question is as follows: suppose $(A,D(A))$ is a sectorial operator on a Banach space $X$ such that $-A$ is sectorial as well, prove that $A$ must be bounded.

Since I’ve found there are many more or less equivalent conventions on sectoriality I’ll state the one I’m more familiar with:
$A$ is sectorial if the sector

$\Sigma(\omega,\theta):=\left\{ z\in\mathbb{C}| \;\;|\arg(\omega-z)|<\theta\right\}$

is contained in the resolvent set $\rho(A)$, where $\omega\in\mathbb{R}$ and $\theta\in\left(\frac {\pi}2,\pi\right)$ and

$\forall z\in\Sigma(\omega,\theta)\;\;\; ||R(z,A)||\leq M|z|^{-1},$

for some positive constant $M$.

It is straightforward to show that if $\pm A$ is sectorial then the spectrum $\sigma(A)$ is bounded. Now, I have two strategies in my mind that I cannot conclude:

  1. Showing that the generated Analytic Semigroup is actually Uniformly Continuous, since I know that then such semigroup is of the form $e^{tA}$, for bounded $A$ and since the generator is unique we can conclude. It may help to use the fact that, if we call $T(t)$ the generated semigroup, we now that

$$\forall x\in X\;\;\; A\int_0^s T(s)xds=T(t)x-x,$$

since showing uniform continuity amounts to show that $||T(t)-I|| \rightarrow 0^+$ in the operator norm, however this puts me in a vicious cycle since what we need is precisely a bound on the norm of $A$.

  1. Another property I’ve seen is that the image of $T(t)$ is contained in $D(A)$, thus if we show that $T(t)$ is surjective (in particular invertible) we can conclude immediately. Now, in the context of $C_0$-semigroups I would simply note that $A$ actually generates a group, i.e. $-A$ generates the semigroup $T(-t)=T(t)^{-1}$ but, after doing some research, I was surprised to not found a single piece of literature on “Analytic Groups of Operators”. This makes me think that this approach won’t work, at least not that straight-forwardly, to show that $T(t)$ is invertible (which it must be, since a uniform semigroup is actually a group of operators).

Do you have any idea on how to conclude one of this approaches? A new approach would of course be welcome too!

Best Answer

Let $(S_t)$ and $(T_t)$ be the semigroups generated by $A$ and $-A$, respectively. As both of them are analytic, they map $X$ into $D(A)$ for $t>0$ and commute with $A$. If $\xi\in X$, then $$ \frac{d}{dt}S_tT_t\xi=AS_t T_t\xi-S_tAT_t\xi=0 $$ for $t>0$. Thus $S_tT_t=\mathrm{id}_X$. The proof of $T_t S_t=\mathrm{id}_X$ is analogous. Thus $S_t$ is invertible for every $t>0$. Since $S_t(X)\subset D(A)$, this implies $D(A)=X$. By the closed graph theorem, $A$ must hence be bounded.

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