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Let $\mathcal{H}$ be a separable Hilbert space (e.g. $L^{2}(\mathbb{R}^{d}))$, and let $\mathcal{D}_{1}\subset\mathcal{H}$ be a dense subspace (e.g. $\mathcal{S}(\mathbb{R}^{d})$). Suppose that an operator $A:\mathcal{D}_{1}\rightarrow \mathcal{D}_{1}$ is essentially self-adjoint. By Stone's theorem, we know that there exists a strongly continuous one-parameter group $e^{itA}$ with infinitesimal generator $A$, i.e.

$$\lim_{t\rightarrow 0} \frac{-i}{t}(e^{itA}-Id)f = Af, \qquad f\in Dom(A).$$

My question is the following.

Question (attempt 1). When does the group $e^{itA}$ preserve the space $\mathcal{D}_{1}$, which is invariant under $A$?

As Nate Elredge pointed out in the comments, the preceding question is ill-posed as stated. For example, the Laplacian $\Delta: C_{c}^{\infty}(\mathbb{R}^{d}) \rightarrow C_{c}^{\infty}(\mathbb{R}^{d})$ is an essentially self-adjoint operator on $L^{2}(\mathbb{R}^{d})$, but $e^{it\Delta}$ does not preserve compact support.

Stone's theorem tells us that for any $t\in\mathbb{R}$, $e^{itA}$ preserves $Dom(A)$, but in general $Dom(A)$ will be a larger space than $\mathcal{D}$. For example, consider the Laplacian $\Delta$ on $L^{2}(\mathbb{R})$ with $\mathcal{D}=\mathcal{S}(\mathbb{R})$, the Schwartz space.

If $A$ is any Fourier multiplier with smooth, polynomially bounded symbol, it's easy to see from Fourier analysis that $e^{itA}$ preserves $\mathcal{D}=\mathcal{S}(\mathbb{R}^{d})$. Similarly, if $A$ is a multiplication operator with polynomially bounded symbol, then $A$ also preserves $\mathcal{S}(\mathbb{R}^{d})$. I suspect that my question is at least true for pseudodifferential operators $\sigma(X,D)$ with symbols satisfying the estimates

$$\|\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\sigma(x,\xi)\| _{\infty} \lesssim_{\alpha,\beta} \langle{x}\rangle^{\mu}\langle{\xi}\rangle^{m-\beta}, \qquad \alpha,\beta\in \mathbb{N}_{0}^{d}$$

for some $\mu,m\in\mathbb{R}$. I would be content with this special case of my question.

Question (attempt 2). If $a(X,D): \mathcal{S}(\mathbb{R}^{d})\rightarrow \mathcal{S}(\mathbb{R}^{d})$ is a pseudodifferential operator with symbol satisfying the above estimates for some $m,\mu$, which is essentially self-adjoint on $L^{2}(\mathbb{R}^{d})$, does the group $\left\{e^{ita(X,D)}\right\}_{t\in\mathbb{R}}$ preserve the Schwartz space?

As commented above, the answer is yes if $a(x,\xi) = a(x)$ or $a(x,\xi)=a(\xi)$. I can also answer my question affirmatively if $a(X,D)$ is an order $1$ operator or if $a(X,D)$ is elliptic along with some other cases, but I am still unable to answer the general case.

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    $\begingroup$ Seems like you would need some more hypotheses to be sure you are choosing the "right" domain. For instance, the Laplacian is also essentially self-adjoint on the domain $\mathcal{D} = C^\infty_c(\mathbb{R})$. The operator leaves this domain invariant but the semigroup does not? $\endgroup$ – Nate Eldredge Mar 21 '18 at 15:09
  • $\begingroup$ @NateEldredge: Thank you for your comment. I am not sure from where the hypotheses on the domain would come. I picked the Schwartz space because it is quite natural from the point of view of Fourier analysis/microlocal analysis. Perhaps, my question is too hopelessly general. $\endgroup$ – Matt Rosenzweig Mar 21 '18 at 15:20

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