If $A \colon H^s \to H^{s+t}$, is $A^\alpha \colon H^s\to H^{s+\alpha t}$?

Question

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?

Answer 1

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 $$ I\left( \frac{a+b}{2}\right)^2 =\left\langle B^{\frac{a+b}{2}} f\Big| B^{\frac{a+b}{2}}f\right\rangle=\langle B^a f| B^b f\rangle\le \|B^a f\|\|B^b f\|=I(a) I(b).\ \square$$

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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. $\square$