# Estimate of semigroup with dual norm?

Consider a semigroup $$(T(t))_{t\in\mathbb{R}^+}$$ generated by a densely defined strictly positive symmetric linear operator $$A: D(A) \subset X \to X$$, where $$X$$ is a Banach space with norm $$\|\cdot\|$$.

Besides, we introduce a Sobolev scale $$(X_n)_{n\in\mathbb {Z}}$$ induced by completion of $$D(A^\infty)$$ with respect to $$\|\cdot\|_n:=\|A^n\cdot\|$$, in particular $$\|\cdot\|_0 = \|\cdot\|$$.

My question is, under what conditions, there exists estimates such that

for $$t\in (0,T)$$, some $$a >0$$, $$\|T(t)x\|_0 \leq C \|x\|_{-a}$$, or $$c \|x\|_{-a}\leq\|T(t)x\|_0 \leq C \|x\|_{-a}$$.

In alternative, one may consider the weaker version, $$t\in (t_0,T)$$ with fixed $$t_0>0$$.

First, if $A$ is symmetric, then $X$ should be a Hilbert space, but I remain in a Banach space.

If $$\|T(t)x\| \leq C\|A^{-1} x\|$$ holds for all $x\in X$, then using the substitution $y=A^{-1}x$, you obtain $$\|AT(t)y\|\leq C\|y\|.$$

But by Theorem 5.3 in Section 2.5 of

A. Pazy, MR 710486 Semigroups of linear operators and applications to partial differential equations, ISBN: 0-387-90845-5.

this implies that $A$ is bounded. Hence, I do not see much of a chance for your inequalities to hold.

• Thanks for the answer. Which theorem do you mean? They are numbered within the chapter so there are multiple theorem 5.3. Dec 14, 2016 at 21:01
• @newbie: thanks. Edited the question accordingly. Dec 15, 2016 at 6:47
• That theorem actually requires $AT(t)$, inflated by $\frac{1}{t}$ to be bounded as $t\to 0$. I think it is stronger than the boundedness of $AT(t)$. Besides, according to Theorem 4.6, chapter II, from One-Parameter Semigroups for Linear Evolution Equations, in the proof (e) $\Rightarrow$ (c), $AT(t)$ is always bounded. Dec 15, 2016 at 14:58
• You misunderstand something quite. From what you cite, it follows that $\|AT(t)\|\leq C(t)$. You require $C$ to be independent of $t$. But for analytic semigroups, $C(t)\approx \frac{1}{t}$. Dec 15, 2016 at 17:40
• What I cite allows $\|AT(t)\|$ to go to $\infty$ like $\frac{1}{t}$, it is much weaker than boundeness. Dec 15, 2016 at 17:42