2
$\begingroup$

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 $t_0>0$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks for the answer. Which theorem do you mean? They are numbered within the chapter so there are multiple theorem 5.3. $\endgroup$ – newbie Dec 14 '16 at 21:01
  • $\begingroup$ @newbie: thanks. Edited the question accordingly. $\endgroup$ – András Bátkai Dec 15 '16 at 6:47
  • $\begingroup$ 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. $\endgroup$ – newbie Dec 15 '16 at 14:58
  • $\begingroup$ 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, $C(t)\approx \frac{1}{t}$. $\endgroup$ – András Bátkai Dec 15 '16 at 17:40
  • $\begingroup$ What I cite allows $\|AT(t)\|$ to go to $\infty$ like $\frac{1}{t}$, it is much weaker than boundeness. $\endgroup$ – András Bátkai Dec 15 '16 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.