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 $H^{k,1}(\mathbb R^d) \subseteq W^{k,1}(\mathbb R^d)$).