If we consider $u-\Delta u=f$ in $\mathcal{S}'$, we obtain $(1-\Delta)^{-1} f=\mathcal{F}^{-1} ((1+|\xi|^2)^{-1} \hat f)$. By means of a functional calculus on the multiplier we may define $$(1-\Delta)^{-s/2} f=\mathcal{F}^{-1} ((1+|\xi|^2)^{-s/2} \hat f), \quad f\in \mathcal{S}'.$$
For $s>0$ one can show that $G_s:=\mathcal{F}^{-1} (1+|\xi|^2)^{-s/2}$ is a regular distribution and in $\mathrm{L}^1$. Hence $$(1-\Delta)^{-s/2} f=G_s\ast f, \quad f\in \mathcal{S}$$ is bounded from $\mathrm{L}^p$ into itself for $1\leq p\leq \infty$. Therefore it extends to an operator on $\mathrm{L}^p$ for $1\leq p < \infty$.
We may define potential spaces via $$\tilde{\mathrm{H}}^{s,p}=\{(1-\Delta)^{-s/2}f: f\in \mathrm{L}^p\}, \quad s\geq0, 1\leq p < \infty.$$
Now it is also common to define $$\mathrm{H}^{s,p}=\{f\in \mathrm{L}^p: \mathcal{F}^{-1}((1+|\xi|^2)^{s/2})\in \mathrm{L}^p \}, \quad s\in \mathbb{R}, 1\leq p\leq \infty.$$
Since $s\mapsto (1-\Delta)^{-s/2}$ is an operator group on $\mathcal{S}$, we see that $\tilde{\mathrm{H}}^{s,p}\cap \mathcal{S}=\mathrm{H}^{s,p}\cap \mathcal{S}$ for $s\geq 0, 1\leq p<\infty$.
Now my questions are:
1) How can we conclude that both definitions coincide for $p$ and $s$ such that the variant via potentials is defined? A priori it is not clear that $\mathcal{S}$ is dense, right?
2) Can we extend the potential definition to $p=\infty$ in some way despite that $\mathcal{S}$ is not dense in there?
3) Is it possible to give a potential characterization for $s<0$, and if yes, how would one show that it coincides with the intrinsic definition?