If $A \colon H^s \to H^{s+t}$, is $A^\alpha \colon H^s\to H^{s+\alpha t}$? Let $D \subset \mathbb R^d$ be a bounded domain with smooth boundary and consider a bounded invertible real linear operator
$$
  A \colon H^s(D) \to H^{s+t}(D), \quad t > 0. \tag{1}
$$
Suppose that $A$ is symmetric, when considered as a compact operator in $H^s(D)$. Then $A$ admits the eigendecomposition:
$$
  A = \sum_{k=0}^\infty \lambda_k \varphi_k \otimes \varphi_k.
$$
Suppose also that $\lambda_k > 0$ for all $k$. Then we can define
$$
   A^\alpha = \sum_{k=0}^\infty \lambda^\alpha \varphi_k \otimes \varphi_k, \quad \alpha > 0.
$$
It is possible to say, in general, that $A^\alpha$ is a bounded invertible operator
$$
   A^\alpha \colon H^{s}(D) \to H^{s+\alpha t}(D)? \tag{2}
$$
Probably, I should mention this answer, which states that the result is in fact true if $A$ is an elliptic pseudodifferential operator.
Remark. If (1) holds for all $s$, then (2) holds for integer $\alpha$, does it also hold for other alpha?
 A: 
NOTATION. We let $\langle f|g\rangle:=\int_D f(x) \overline{g(x)}\, dx$ and $\|f\|^2:=\langle f|f\rangle$. 

With the notation introduced above, we have that 
$$\|f\|_{H^s}= \| (1-\Delta)^{s/2} f\|, $$
so we can reformulate the problem as follows: 

PROBLEM. Given that the operator $A$ satisfies 
  $$\| (1-\Delta)^{t/2} A f\|\le C\|f\|$$
  for a certain $t>0$, is it true that 
  $$\|(1-\Delta)^{\alpha t/2}A^\alpha f\|\le C_\alpha\|f\|,\quad \forall \alpha \in [0, 1]\ ?$$

The answer is affirmative provided that this extra assumption is satisfied: 
$$\tag{Extra}
(1-\Delta)^{\alpha t/2} A^\alpha = [ (1-\Delta)^{t/2} A]^\alpha.$$ 
Indeed, this assumption enables the use of the following lemma. 

LEMMA. Let $B$ be a self-adjoint nonnegative operator on $L^2(D)$ and let $f\in L^2(D)$. Define 
  $$\tag{1}I(\alpha):=\| B^\alpha f\|.$$ 
  Then $I\colon [0, \infty)\to [0, \infty)$ is log-convex, that is, $$\tag{2}I((1-\theta)a+\theta b)\le I(a)^{1-\theta}I(b)^\theta.$$

Proof of Lemma. Since $I$ is continuous, it suffices to prove (2) with $\theta=\frac12$. Using self-adjointness of $B$ we write 
$$
I\left( \frac{a+b}{2}\right)^2 =\left\langle B^{\frac{a+b}{2}} f\Big| B^{\frac{a+b}{2}}f\right\rangle=\langle B^a f| B^b f\rangle\le \|B^a f\|\|B^b f\|=I(a) I(b).\ \square$$

Solution to the Problem. Let $B:=(1-\Delta)^{t/2} A$, where $t>0$ is given in the statement of the Problem. By assumption, the function $I$ given by (1) satisfies 
$$
I(0)=\|f\|, \qquad I(1)\le C\|f\|.$$
By the Lemma we have that 
$$I(\alpha)\le I(0)^{1-\alpha}I(1)^\alpha= C^\alpha \|f\|.$$ 
If (Extra) is satisfied, the left hand side of the last inequality equals $\|(1-\Delta)^{\alpha t/2} A^\alpha f\|$, and so the proof is complete. $\square$
