Let $-A$ be the generator of a $C^0$-semigroup $S$. How can we show that $S(t)x\in\mathcal D(A^α)$ for all $t\ge 0$ and $x\in\mathcal D(A^α)$?

Question

Let $(H,\langle\;\cdot\;,\;\cdot\;\rangle)$ be a $\mathbb R$-Hilbert space and $(\mathcal D(A),A)$ be a linear operator.

Assume $(e_n)_{n\in\mathbb N}\subseteq\mathcal D(A)$ is 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 $$\lambda_{n+1}\ge\lambda_n\;\;\;\text{for all }n\in\mathbb N\;.\tag 2$$

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

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

Let $t\ge 0$ and $x\in\mathcal D(A^\alpha)$. How can we show that $S(t)x\in\mathcal D(A^\alpha)$?

I know how this can be proved in the case $\alpha=1$. However, the proof uses $$-Ax=\lim_{h\to0+}\frac{S(h)x-x}h\;\;\;\text{for all }x\in\mathcal D(A)\tag 3$$ and $$\mathcal D(A)=\left\{x\in E:\lim_{h\to0+}\frac{S(h)x-x}h\text{ exists in }H\right\}\tag 4$$ and hence I don't see an immediate generalization.

Answer 1

Note that $$\begin{align} S(t)x\in D(A^\alpha)\quad&\Longleftrightarrow \quad\sum_{n\in\mathbb N}\lambda_n^{2\alpha}\left|\langle S(t)x,e_n\rangle_H\right|^2<\infty\\ &\Longleftrightarrow \quad\sum_{n\in\mathbb N}\lambda_n^{2\alpha}\left|\left\langle \sum_{j\in\mathbb N}e^{-t\lambda_j}\langle x,e_j\rangle e_j,e_n\right\rangle_H\right|^2<\infty\\ &\Longleftrightarrow \quad\sum_{n\in\mathbb N}\lambda_n^{2\alpha}\left|\sum_{j\in\mathbb N}e^{-t\lambda_j}\langle x,e_j\rangle\left\langle e_j,e_n\right\rangle_H\right|^2<\infty\\ &\Longleftrightarrow \quad\sum_{n\in\mathbb N}\lambda_n^{2\alpha}\left|e^{-t\lambda_n}\langle x,e_n\rangle\right|^2<\infty\\ \end{align}\tag{1}$$ And, as $x\in D(A^\alpha)$, $$\sum_{n\in\mathbb N}\lambda_n^{2\alpha}\left|e^{-t\lambda_n}\langle x,e_n\rangle\right|^2\leq \sum_{n\in\mathbb N}\lambda_n^{2\alpha}|e^{-t\lambda_n}|\left|\langle x,e_n\rangle\right|^2<\infty\tag{2}$$

From $(1)$ and $(2)$ we get the desired result.

Answer 2

Because you have formulated the problem with an ONB in which $A$ is diagonalised its not necessary (also in the case $\alpha=1$) to do a consideration with $-A=\partial_t S(t)\lvert_{t=0}$.

Rather use that $e^{-\lambda_n t}≤1$ for all $t≥0$ and from that follow if: $$\sum_n \lambda_n^{2\alpha}|\langle x, e_n\rangle_H|^2<\infty\tag{1}$$ Then also $$\sum_n e^{-\lambda_n t}\lambda_n^{2\alpha}|\langle x,e_n\rangle_H|^2<\infty\tag{2}.$$ Because $S(t)x=\sum_k e^{-\lambda_k t}\langle x,e_k\rangle e_k$ you have $$\langle S(t)x,e_n\rangle=\sum_k e^{-\lambda_k t}\langle x,e_k\rangle \langle e_k,e_n\rangle = e^{-\lambda_n t}\langle x,e_n\rangle\tag{3}.$$

For $S(t)x$ to lie in $D(A^\alpha)$ you must substitute $S(t)x$ for $x$ in condition $(1)$. But $(3)$ shows that this results in equation $(2)$ which is true if $x$ satisfies condition $(1)$. So if $x\in D(A^\alpha)$ you have $S(t)x\in D(A^\alpha)$.

Answer 3

Let me provide the proof of a stronger result for the case $\alpha>0$. Using $$\theta^\alpha e^{-\theta}\le\left(\frac\alpha e\right)^\alpha=:C_\alpha\;\;\;\text{for all }\theta\ge 0\;,\tag 1$$ we obtain $$\sum_{n=1}^N\left(\lambda_n^\alpha e^{-t\lambda_n}\right)^2\left|\langle x,e_n\rangle\right|^2\le\left(\frac{C_\alpha}{t^\alpha}\right)^2\sum_{n=1}^N\left|\langle x,e_n\rangle\right|^2\xrightarrow{N\to\infty}\left(\frac{C_\alpha}{t^\alpha}\right)^2\left\|x\right\|^2\tag 2$$ for all $t>0$ and $x\in H$. Using Pedros answer, we can conclude $$S(t)x\in\mathcal D(A^\alpha)\;\;\;\text{for all }t>0\text{ and }x\in H\;.\tag 3$$