Let $T(t)$ be a $C_0$semigroup on Banach space $X$, and $A$ its generator. By LumerPhilipps theorem we know that if $A$ is densely defined and mdissipative operator then it generates a $C_0$semigroup of contractions, i.e., $$\T(t)\ \leq 1, \quad \forall t \geq 0.$$ My question here: is there any theorem which gives conditions to generate a strictly contraction $C_0$semigroup? That is, $$\T(t)\ < 1, \quad \forall t > 0.$$ For example, for $A=\Delta$ the Dirichlet Laplacian, is the associated semigroup strictly contractive? Can we calculate the norm $\T(t)\$ in this case? Or at least, Is $IT(t)$ invertible? If this is true for the Dirichlet laplacian, can we obtain the same result for second order elliptic operator?

$\begingroup$ It can't be true for $t = 0$ or you don't have a semigroup... $\endgroup$– Nate EldredgeOct 20, 2018 at 14:46

1$\begingroup$ For the Laplacian, assuming you only want it for $t > 0$, this is basically equivalent to showing that the lowest eigenvalue $\lambda_1$ of $\Delta$ is strictly positive, i.e. there is a spectral gap. Then you immediately get $\T(t)\ \le e^{\lambda t}$ by the spectral theorem. $\endgroup$– Nate EldredgeOct 20, 2018 at 14:51

2$\begingroup$ See also the HilleYosida theorem and try to apply it with $\omega < 0$. $\endgroup$– Nate EldredgeOct 20, 2018 at 14:52

$\begingroup$ Yeah, sure for t>0. I understand that it's true for the $\Delta$, is it true for $A$ where $A$ is a second order elliptic operator ? $\endgroup$– S. MathsOct 20, 2018 at 15:07

$\begingroup$ For HilleYosida theorem, why there exists $\omega<0$ in this case ? $\endgroup$– S. MathsOct 20, 2018 at 15:08
1 Answer
Your conditions is for contraction semigroups equivalent to have uniform exponential stability, i.e., to have growth bound less than zero, see Proposition V.1.7. in
Engel, KlausJochen; Nagel, Rainer, Oneparameter semigroups for linear evolution equations, Graduate Texts in Mathematics. 194. Berlin: Springer. xxi, 586 p. (2000). ZBL0952.47036.
Assuming you work in a Hilbert space, you can reformulate the Lumer.Phillips theorem for your case as $$\Re \langle Ax,x\rangle \leq \omega<0$$ to get a sufficient condition. For many examples you can test this by applying some version of Green's theorem.
For second order elliptic operators on bounded domains it boils down to look at the eigenvalues. If they are negative, you have exponential stability.
ADDED after comments: For uniformly elliptic operator it all depends on the boundary conditions. Neumann boundary conditions imply that you do not have all negative eigenvalues.
If you are interested in the spectral theory of differential operators, there are many good books. I liked the one
Davies, E. B., Spectral theory and differential operators, Cambridge Studies in Advanced Mathematics. 42. Cambridge: Cambridge Univ. Press. ix, 182 p. (1995). ZBL0893.47004.
very much, it is a good start.

1$\begingroup$ @S.Cho: yes, we do. I added a reference for this. $\endgroup$ Oct 20, 2018 at 18:01

1$\begingroup$ @AndrásBátkai: I'm not sure I understand you correctly. Exponential stability is of course equivalent to having $\T(t_0)\ < 1$ for at least one time $t_0 > 0$. However, I think the OP asks for a characterization of the property $\T(t)\ < 1$ for all $t > 0$  which, I suppose, is a much more subtle question. $\endgroup$ Oct 21, 2018 at 9:03

1$\begingroup$ Hi, @JochenGlueck :) You are right of course. But I perceived from the comments a different problem.I tried to give sufficient conditions, not characterization. Characterization is impossible at the moment... $\endgroup$ Oct 21, 2018 at 9:08

2$\begingroup$ @S.Cho: As András Batkai pointed out in his answer, this depends on the boundary conditions and on the domain $\Omega \subseteq \mathbb{R}^n$ which you consider. Examples: Let your operator be the Laplace operator. (i) If $\Omega$ is bounded and you consider Neumann boundary conditions, then $1$ is an eigenvalues of $T(t)$ for each $t > 0$. (ii) If $\Omega$ is bounded and you consider Dirichlet boundary conditions, then $\T(t)\ < 1$ for all $t > 0$. [to be continued] $\endgroup$ Oct 21, 2018 at 17:00

2$\begingroup$ @S.Cho: [continuation] (iii) If $\Omega$ is unbounded and you consider Dirichlet boundary conditions, then it can happen (depending on the precise shape of the $\Omega$) that $1$ is a spectral value of each operator $T(t)$. [By the way, here I considered the semigroup on $L^2(\Omega)$, but similar results are true on $L^p(\Omega)$.] $\endgroup$ Oct 21, 2018 at 17:01