Is Schwartz space $\mathbb R^n$ contained in every fractional Sobolev space on $\mathbb R^n$? [closed]

Question

Can someone kindly confirm that the Schwartz space $\mathcal S(\mathbb R^n)$ made of all infinitely-differentiable functions $f:\mathbb R^n \to \mathbb R$ with rapidly decreasing derivatives of all orders is contained in any fractional Sobolev space $H^k(\mathbb R^n)$ ? Thanks in advance.

After all, at least for the case $k=1$, it holds for any $f \in \mathcal S(\mathbb R^n)$ and $m>0$ that

$$ \|f\|_{H^1(\mathbb R^n)}^2 = \int_{\mathbb R^n}Lap(f)(x)\,dx \le C\int_{\mathbb R^n}(1+\|x\|)^{-m}\,dx \le C'\int_0^\infty (1+r)^{-m}r^{n-1}\,dr. $$ Thus, taking $m$ sufficiently larger than $n$, we see that $f \in H^1(\mathbb R^n)$.

Answer 1

One way to approach such questions is to start with an abstract situation (Hilbert scales). If $T$ is an unbounded s.a. operator with $T>1$ on a separable Hilbert space (in your case, the Schrödinger operator on the usual $L^2$-space), then it embeds in a natural way into an increasing family $H^\alpha$ (with $\alpha \in ]-\infty,\infty[$) of Hilbert spaces, with limiting spaces $H^{\infty}$, their intersection, and $H^{-\infty}$, their union. These are a Fréchet space and a $DF$ space, resp. They are both nuclear under natural conditions on the spectrum of $T$.

As indicated above the Sobolev spaces arise from the Schrödinger operator. I think that this answers your question. For references, just google "Hilbert scales". Most of the interesting spaces of distributions, test functions and Sobolev-type spaces arise in this manner by choosing the classical operators (Laplace-Beltrami, Legendre, Laguerre ...).

Added as an edit to clear up the confusion indicated in the comment below. Given the two quotes from my answer "in your case, the Schrödinger operator" and "As indicated above the Sobolev spaces arise from the Schrödinger operator", I am taken aback by the fact that my answer was misconstrued as the ridiculous claim that the Schwartz space is generated by the Laplace operator on the line. The latter space is, of course, $\cal D'_{L^2}$ which (surprise, surprise) was also introduced by Schwartz.

As regards the raised question of nuclearity, this is very transparent in the abstract situation--it is the case precisely when the given operator has discrete spectrum with eigenvalues which satisfy the growth condition $\Sigma \frac 1 {\lambda^\alpha}<\infty$ for some positive $\alpha$. This immediately clears up the question of nuclearity for the two spaces mentioned here.