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Let

  • $H$ be a $\mathbb R$-Hilbert space
  • $(\mathcal D(A),A)$ be a linear operator
  • $(e_n)_{n\in\mathbb N}\subseteq\mathcal D(A)$ be an orthonormal basis of $H$ with $$Ae_n=\lambda_ne_n\;\;\;\text{for all }n\in\mathbb N\tag 1$$ for some $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $$\lambda_{n+1}\ge\lambda_n\;\;\;\text{for all }n\in\mathbb N\tag 2$$

Moreover, let $$e^{-tA}x:=\sum_{n\in\mathbb N}e^{-t\lambda_n}\langle x,e_n\rangle_He_n\;\;\;\text{for }t\ge 0\text{ and }x\in H\;.$$ We can show that $$S(t):=e^{-tA}\;\;\;\text{for }t\ge 0$$ is a uniformly continuous semigroup on $H$ and $-A$ is the infinitesimal generator of $S$.

Now, let $$\mathcal D(A^\alpha):=\left\{x\in H:\sum_{n\in\mathbb N}\lambda_n^{2\alpha}\left|\langle x,e_n\rangle_H\right|^2<\infty\right\}$$ and $$A^\alpha x:=\sum_{n\in\mathbb N}\lambda_n^\alpha\langle x,e_n\rangle_He_n\;\;\;\text{for }x\in\mathcal D(A^\alpha)$$ for $\alpha\in\mathbb R$.

I've read (in An Introduction to Computational Stochastic PDEs, Lemma 3.22-iii) that we can show that for all $\alpha\in[0,1]$ $$\left\|A^{-\alpha}\left(\operatorname{id}_H-S(t)\right)\right\|_{\mathfrak L(H)}\le C_\alpha t^\alpha\;\;\;\text{for all }t\ge 0\tag 3$$ for some $C_\alpha\ge 0$. However, $(3)$ isn't well-defined, unless $$x-S(t)x\in\mathcal D(A^{-\alpha})\;\;\;\text{for all }t\ge 0\text{ and }x\in H\tag 4\;.$$

Let $t>0$ and $x\in H$. Then, $$x-S(t)x\in\mathcal D(A^{-\alpha})\Leftrightarrow\sum_{n\in\mathbb N}\lambda_n^{-2\alpha}\left|\langle x-S(t)x,e_n\rangle_H\right|^2<\infty\;.\tag 5$$ Using $$e^\theta<1\;\;\;\text{for all }\theta<0\tag 6$$ and $$e^\theta\ge1+\theta\;\;\;\text{for all }\theta\in\mathbb R\;,\tag 7$$ we obtain $$\left|1-e^{-\lambda_nt}\right|\le\lambda_nt\;\;\;\text{for all }n\in\mathbb N\tag 8\;.$$ Thus, $$\sum_{n\in\mathbb N}\lambda_n^{-2\alpha}\left|\langle x-S(t)x,e_n\rangle_H\right|^2=\sum_{n\in\mathbb N}\lambda_n^{-2\alpha}\left|\left(1-e^{-\lambda_nt}\right)\langle x,e_n\rangle_H\right|^2\le t^2\sum_{n\in\mathbb N}\lambda_n^{2(1-\alpha)}\left|\langle x,e_n\rangle_H\right|^2\tag 9\;.$$ However, this only yields $$y-S(t)y\in\mathcal D(A^{-\alpha})\;\;\;\text{for all }y\in\mathcal D(A^{1-\alpha})\;.\tag{10}$$

So, either my estimate was to weak or something is wrong with $(3)$.

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    There is an assumption (3.19) that $0 < \lambda_1 \le \lambda_2 \le \lambda_3 \le \cdots$. Therefore $A^{-\alpha}$ is bounded for $\alpha > 0$. – 2017-01-01
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    @TrialAndError Doesn't that mean that we've got $\mathcal D(A^\beta)\subseteq H$, i.e. $\mathcal D(A^\beta)=H$, for all $\beta\le 0$? – 2017-01-02
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    It would seem so. And you probably suspected that the result you wanted couldn't hold unless something like that were tacitly assumed. And it is because of the assumption 3.19 on the ordering of the eigenvalues with a minimum one. – 2017-01-02
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    @TrialAndError I've described my *real* problem in [a new question](http://mathoverflow.net/questions/258687/regularity-of-the-mild-solution-of-a-semilinear-evolution-equation). – 2017-01-03
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    By a uniformly continuous semigroup, you can't mean that $t\mapsto S(t)$ is continuous in the operator norm because that would mean a bounded generator. So, what do you mean? – 2017-01-03
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    @TrialAndError You're right. I've changed the other question. The $S$ their should be the same $S$ as here. – 2017-01-04

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