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Let $(M,g)$ be a complete Riemannian manifold with Laplacian $\Delta:C^{\infty}_{c}(M)\to C^{\infty}_{c}(M)$ (think of $\mathbb{R}^{d}$ if you wish). This operator is essentially self-adjoint in $L^{2}(M)$ and hence, its closure $\overline{\Delta}:\mathcal{D}(\overline{\Delta})\to L^{2}(M)$ is self-adjoint. This allows us to define for any positive $s\in\mathbb{R}$ the operator $\overline{\Delta}^{s}$ by the spectral theorem of self-adjoint unbounded operators.

How to show that $C^{\infty}_{c}(M)$ is contained in the domain $\mathcal{D}(\overline{\Delta}^{s})$ for any $s$?

By the spectral theorem, the domain $\mathcal{D}(\overline{\Delta}^{s})$ is the set of all functions $f\in L^{2}(M)$ such that

$$\int_{\sigma(\overline{\Delta})}\lambda^{2s}\,\mathrm{d}\langle P(\lambda)f,f\rangle_{L^{2}}<\infty$$

where $P$ denotes the spectral measure of $\overline{\Delta}$. Now, for $f\in C^{\infty}_{c}(M)$, I don't see how to use its support of $f$ (information which is contained in the complex measure $\langle P(\lambda)f,f\rangle_{L^{2}}$) to show that the above integral is finite.

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  • $\begingroup$ In questions of this type, it is often expedient to use the spectral theorem in the form that every (unbounded) self-adjoint operator can be regarded as multiplication by a messurable function on an $L^2$. They then sometimes become quite transparent, as here. In the specisl case you mention we have the bonus that this diagonalisation is effected by the Fourier transform.. $\endgroup$
    – terceira
    Oct 18, 2023 at 17:34

2 Answers 2

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If $0<s\le 1$, then $D(T^s)\supseteq D(T)$ for any self-adjoint $T$, as is obvious from the description of the domain that you quote at the end of your post.

If $s=n+t$ with $n\in\mathbb N$ and $0<t\le 1$, then $T^s=T^tT^n$. In your case, $T^n=\overline{\Delta}^n$ maps $C_0^{\infty}$ back to itself, and, as just observed, $C_0^{\infty}\subseteq D(T^t)$. Hence $C_0^{\infty}\subseteq D(T^s)$.

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I have always found this kind of things very confusing myself, but now that I have some ideas I'll try to address it. You can find a proof of a stronger statement in Proposition 7.7 in this work of mine with Luca Gennaioli, but part of the proof is hidden in Proposition 7.5. Note that we only assume that $(M,g)$ is stochastically complete, and this is not even really needed (see the Note below). The main point is the following. Let me call $$ (-\Delta)^{s}_{\rm B} \, u := \frac{1}{\Gamma(-s)} \int_{0}^{\infty} (e^{t\Delta}u-u)\frac{dt}{t^{1+s}}$$ the Bochner's fractional Laplacian, this may seem unrelated but for me it is the key to understand how this works. Moreover, note that for $\lambda \ge 0$ and $s \in (0,1)$ there holds $$\varphi(\lambda) := \lambda^{s} = \frac{1}{\Gamma(-s)} \int_0^\infty (e^{-\lambda t}-1)\frac{dt}{t^{1+s}} . $$ This is an equality between real numbers and has nothing to do with the Laplacian. Then, from here spectral theory does most of the work.

I claim the following fact, which answers your question in great generality.

Claim. Let $X \subset L^2(M)$ be a linear subspace with the following property: whenever $u \in X$ then $(-\Delta)_B^s \, u $ exists in the sense of Bochner (that is, the right-hand side in its definition converges absolutely in $L^2(M)$). Then $X\subseteq {\rm Dom}((-\Delta)^{s}_{\rm Spec})$ and $$ (-\Delta)_{\rm Spec}^s u := \int_{\sigma(-\Delta)} \lambda^{2s} d\langle E_\lambda u, \cdot \rangle = (-\Delta)_B^s \, u \,.$$

Note that the last equality alone implies $X\subseteq {\rm Dom}((-\Delta)^{s}_{\rm Spec})$ as $$\int_{\sigma(-\Delta)} \lambda^{2s} d\langle E_\lambda u, u \rangle = \| (-\Delta)_{\rm Spec}^s u \|^2_{L^2(M)} = \| (-\Delta)_B^s \, u \|^2_{L^2(M)} \\= \left\| \int_0^\infty (e^{t\Delta}u-u) \frac{dt}{t^{1+s/2}} \right\|_{L^2(M)}^2 \le \left( \int_0^\infty \| e^{t\Delta}u-u \|_{L^2(M)} \frac{dt}{t^{1+s/2}} \right)^2 <+\infty$$ since $u\in X$ and we have used Minkowski's Integral Inequality.

Basically you need the property in the hypothesis in the claim with $X=C_c^\infty(M)$ and in my work with Luca we prove it for $X=H^{2s+\epsilon}(M)$ (for every $\epsilon \ll 1$) on every stochastically complete $(M,g)$. I believe this to be true with $H^{2s}(M)$ as it is know in bounded domains $\Omega \subset \mathbb{R}^n$, but in a setting so general as every complete Riemannian manifold one needs to be careful. Every proof I have seen of this for $\Omega \subset \mathbb{R}^n$ heavily uses the discreteness of the spectrum and interpolation theory, and Fourier transform the proofs for the entire $\mathbb{R}^n$.

The fact that the hypothesis is true for $X=C_c^\infty(M)$ is easy, you just need to prove that the integral in $(-\Delta)_B^s \, u$ is absolutely convergent near $0$ and $\infty$. At the origin just using $\|e^{t\Delta} u-u \|_{L^2(M)} \le t \|u\|_{C^2(M)}$ gives convergence, and at $\infty$ use that $ \|e^{t\Delta} u-u \|_{L^2(M)} \le \|e^{t\Delta} u \|_{L^2(M)} + \|u \|_{L^2(M)} \le 2\|u \|_{L^2(M)} $.

It only remains to prove the claim, and the proof goes as follows. For every $v\in L^2(M)$ by standard spectral theory (see e.g. equation (A.49) in A. Grigor'yan's book "Heat Kernel and Analysis on Manifolds") $$ \langle (-\Delta)_{\rm Spec}^s u, v \rangle_{L^2} = \int_{0}^\infty \lambda^s d\langle E_\lambda u,v \rangle = \int_{0}^\infty \varphi(\lambda) d\langle E_\lambda u,v \rangle \\ = \frac{1}{\Gamma(-s)} \int_0^\infty \left( \int_0^\infty (e^{-\lambda t}-1)\frac{dt}{t^{1+s}} \right) d\langle E_\lambda u,v \rangle \,.$$ Since the integrand is always negative by Tonelli's theorem you can exchange the order of integration and you get $$ \langle (-\Delta)_{\rm Spec}^s u, v \rangle_{L^2} = \int_0^\infty \langle e^{t\Delta}u-u,v \rangle_{L^2} \frac{dt}{t^{1+s}} = \langle (-\Delta)_B ^s\, u,v \rangle_{L^2} <+\infty \,, $$ and this is finite and all the integrals converge since $u\in X$ satisfies the hypothesis in the claim. This proves $(-\Delta)_{\rm Spec}^s u = (-\Delta)_{B}^s \, u$ (equality in $L^2(M)$) and we are done.

Note: indeed we assume $(M,g)$ being stochastically complete because we want, at the same time, to prove that the Bochner's fractional laplacian and the Singular Integral fractional Laplacian coincide, but you are uninterested in this. The fact that Bochner's fractional Laplacian is in $L^2(M)$ is still true even without stochastical completeness, and this what you really need.

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