Riesz potential and homogeneous Sobolev spaces

Question

Consider the Bessel potential and Riesz potential spaces $$L^{s,p}=\{f:f=(I-\Delta)^{-s/2}g,\, g\in L^p\},$$ $$\dot{L}^{\alpha,p}=\{f:f=(-\Delta)^{-s/2}g,\, g\in L^p\},$$ where $(I-\Delta)^{s/2}$ and $(-\Delta)^{s/2}$ are the Fourier multipliers with symbol $(1+|\xi|^2)^{s/2}$ and $|\xi|^s$, respectively.

There is a result by A.P. Calderón which states that $L^{s,p}\cong W^{s,p}$ when $s\in\mathbb{N}$ and $1<p<\infty$. My question is, does a similar result hold for $\dot{L}^{s,p}$ and $\dot{W}^{s,p}$?

Actually, the only thing I need to know right now is whether $\dot{L}^{1,2}\cong \dot{H}^{1}$. Formally, it makes all the sense in the world to me, considering Plancherel theorem and how the Fourier transform interacts with derivatives, but I often get surprised by how much the devil is in the details when studying functional analysis.

Answer 1

Sure. You can even deduce the result for the dotted norms by the non-dotted one, via scaling. Anyway, the Fourier transform argument is sufficient in this case. Namely, if $u$ is Schwartz class then of course you have $$ \|u\|_{\dot H^1}^2\simeq \|\xi\widehat{u}\|_{L^2}\simeq \||\xi|\widehat{u}\|_{L^2}\simeq \|u\|_{\dot L^{1,2}}. $$ Then by density you obtain the equivalence of spaces. The only devilish detail I can think of is when you try to do the same for higher order Sobolev spaces; if you consider $\dot H^s$ with $s>n/2$ you run into the difficulty that the space $\dot H^s$ is not naturally embedded into temperate distributions and you must quotient away polynomials (but the previous equivalence of norms for good functions still holds).