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.

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.

Consider the Bessel potential and Riesz potential spaces $$L^{\alpha,p}=\{f:f=(I-\Delta)^{s/2}g,\, g\in L^p\},$$ $$\dot{L}^{\alpha,p}=\{f:f=(-\Delta)^{s/2}g,\, g\in L^p\},$$ 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 $L^{\alpha,p}\cong W^{\alpha,p}$ when $\alpha\in\mathbb{N}$ and $1<p<\infty$. My question is, does a similar result hold for $\dot{L}^{\alpha,p}$ and $\dot{W}^{\alpha,p}$?

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.

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.