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We consider the following spaces $H^{k,p}(\mathbb R^d)$, $k \geq 1$ is integer, $p \geq 1$ (Bessel potential spaces): $$ H^{k,p}(\mathbb R^d) = \bigl\{ f \in L^p(\mathbb R^d) \colon \mathcal F^{-1}[(1+|\xi|^2)^{\frac k 2} \mathcal F f] \in L^p(\mathbb R^d) \bigr\}, $$ where $\mathcal F$ is the Fourier transform. We also consider the classical Sobolev spaces $$ W^{k,p}(\mathbb R^d) = \bigl\{ f \in L^p(\mathbb R^d) \colon D^\alpha f \in L^p(\mathbb R^d), \; |\alpha| \leq k \bigr\}, $$ where $D^\alpha$ is the derivative of multi-order $\alpha$.

It is written in Wikipedia that for $p>1$ we have $H^{k,p}(\mathbb R^d) = W^{k,p}(\mathbb R^d)$. My question is what is the relation between the spaces $H^{k,1}(\mathbb R^d)$ and $W^{k,1}(\mathbb R^d)$? Are they still equal?

For example, if $f \in W^{k,1}(\mathbb R^d)$, then all derivatives of $f$ up to order $k$ are in $L^1(\mathbb R^d)$ and hence $(1+|\xi|^2)^{\frac k 2}\mathcal F f \in L^\infty(\mathbb R^d)$. But it is not clear that the latter function is the Fourier transform of some function from $L^1(\mathbb R^d)$ (if it the case, we have $W^{k,1}(\mathbb R^d) \subseteq H^{k,1}(\mathbb R^d)$).

On the other hand, if $f \in H^{k,1}(\mathbb R^d)$ then $f \in L^1(\mathbb R^d)$ and $(1+|\xi|^2)^{\frac k 2}\mathcal F f \in L^\infty(\mathbb R^d)$. I am not sure that it implies that $f$ has distributional derivatives in $L^1(\mathbb R^d)$ up to order $k$ (it would imply $H^{k,1}(\mathbb R^d) \subseteq W^{k,1}(\mathbb R^d)$).

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  • $\begingroup$ It is probably also true for p = 1, since $L^p(\mathbb{R}^n) \hookrightarrow \mathcal{S}'(\mathbb{R}^n)$, for $1 \leq p \leq \infty$, and $\mathcal{F}: S'(\mathbb{R}^n) \longrightarrow \mathcal{S}'(\mathbb{R}^n)$ is continuous isomorphism. Intuitively I would say so, but should be formalized $\endgroup$ – user288972 Feb 25 '16 at 14:49
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The spaces $W^{k,1}(\mathbb{R}^{d})$ and $H^{k,1}(\mathbb{R}^{d})$ coincide for $k$ even and $d=1$, while $W^{k,1}(\mathbb{R}^{d})\subset H^{k,1} (\mathbb{R}^{d})$ for $k$ even and $d>1$. For $k$ odd there is no relation between $W^{k,1}(\mathbb{R}^{d})$ and $H^{k,1}(\mathbb{R}^{d})$. It's a guided exercise in Stein's book on singular integrals (see Exercise 6.6 on page 160). The notation is different (he uses $L^{k,p}(\mathbb{R}^{d})=W^{k,p} (\mathbb{R}^{d})$ and $\mathcal{L}^{k,p}(\mathbb{R}^{d})=H^{k,p} (\mathbb{R}^{d})$).

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