Is it true that $f\in W^{-1,p}(\mathbb{R}^n)$, then $\Gamma\star f\in W^{1,p}(\mathbb{R}^n)$? I am trying to understand the following  paper. In page 1191, in the beggining of the proof of Theorem 2.9. the authors consider the convolution $$v=\Gamma\star f$$
They claim that $v\in W^{1.p}(\mathbb{R}^n)$. Does anyone knows why this is true with $f$ being only in $W^{-1,p}(\mathbb{R}^n)$?
Update: 
Maybe this can thrown some light upon the matter.
First we have that $f\in W^{-1,p}(\Omega)$. This implies the existence of $F\in L^{p'}(\Omega)$ such that $$\langle f,\varphi\rangle=\int_\Omega F\cdot\nabla\varphi,\ \forall\ \varphi\in W_0^{1,p}(\Omega) $$
Extend $F$ by zero outside $\Omega$, so $f\in W^{-1,p}(\mathbb{R}^3)$ and $$\langle f,\varphi\rangle=\int_\Omega F\cdot\nabla\varphi,\ \forall\ \varphi\in W_0^{1,p}(\mathbb{R}^3)$$
By definition $\Gamma\star f$ is a tempered distribution defined by $$\langle\Gamma\star f,\varphi\rangle=\langle \Gamma,\langle f,\varphi(x+\cdot)\rangle\rangle,\ \forall\ \varphi\in \mathcal{S}$$
Note that \begin{eqnarray}
 \langle\Gamma\star f,\varphi\rangle &=& \int_{\mathbb{R}^3}\Gamma(x)\int_{\mathbb{R}^3}F(y)\cdot\nabla\varphi(x+y)dydx      \nonumber \\
   &=& \int_{\mathbb{R}^3}\int_{\mathbb{R}^3}\Gamma (z-y)F(y)\cdot\nabla\varphi(z)dzdy \nonumber \\
   &=& \int_{\mathbb{R}^3}\left(\Gamma\star F\right)(z)\cdot\nabla\varphi(z)dz
 \\ &=& -\int \operatorname{div}(\Gamma\star F)(z)\varphi(z)dz \\
&=& \left\langle -\sum_{i=1}^3\frac{\partial\Gamma}{\partial x_i}\star F,\varphi\right\rangle,\ \forall\ \varphi\in\mathcal{S}
\end{eqnarray}
If my calculations are right, we can see $\Gamma\star f$ as the functions defined by that $$(\Gamma\star f)(x)=-\sum_{i=1}^3\left(\frac{\partial\Gamma}{\partial x_i}\star F\right)(x)$$
Are my calculations right? If so, can we derive the desired regularity from the above expression?
Thank you
 A: For $f \in W^{-1,p}$, I take it you mean the dual of $W^{1,q}$, which is to say that it is a bounded linear operator on functions lying in $W^{1,q}$ where $q$ is the conjugate Holder exponent of $p$.  
Now let's consider the action of $f$ as a distribution, which is to say, $f$ acts on test functions $\psi$ in a way that is linear and bounded by the $W^{1,q}$ norm of $\psi$. Using the $L^p$ duality to $L^q$, we conclude that there are functions $F_i$ and $F$ such that 
$$ \langle f, \psi \rangle = \sum_i \int F_i \frac{\partial \psi}{\partial x_i} $$
just as you have already done (since the $W^{1,q}$ norm of $\psi$ and the $L^q$ norm of $\nabla \psi$ are equivalent due to the Sobolev embedding). Furthermore,
$$\langle f,\psi \rangle \leq \|f\|_{-1,p} \|\psi\|_{1,q}$$
Using the usual rules, we see that 
$$ \langle f * \Gamma, \psi \rangle = \sum_i \int F_i(y) \Gamma(x - y) \frac{\partial \psi}{\partial x_i}(x) dx dy = \langle f, \Gamma*\psi\rangle$$
To find the $L^p$ norm of $f*\Gamma$, we take 
$$ \|f*\Gamma\|_p = \sup_{\|\phi\|_q = 1} \langle f*\Gamma, \phi \rangle \leq \sup_{\|\phi\|_q = 1} \|f\|_{-1,p} \|\Gamma * \phi\|_{1,q} \leq C_{n,q} \|f\|_{-1,p}$$
where at the last step we have used the Hardy-Littlewood-Sobolev inequality, that $\|\psi * \Gamma\|_{2,q} \leq C_{n,q} \|\psi\|_q$.
Similarly, to find the $L^p$ norm of $\frac{\partial \Gamma * f}{\partial x_i}$, we take
$$ \|\frac{\partial f*\Gamma}{\partial x_i}\|_p = \sup_{\|\phi\|_q = 1} \langle \frac{\partial f*\Gamma}{\partial x_i}, \phi \rangle = \sup_{\|\phi\|_q = 1} \langle f*\Gamma,\frac{\partial \phi}{\partial x_i} \rangle = \\ \sup_{\|\phi\|_q = 1}  \sum_j \int F_j(y) \Gamma(x-y) \frac{\partial^2 \phi}{\partial x_i \partial x_j}(x) dx \leq \\ \|f\|_{-1,p} \sup_{\|\phi\|_q = 1} \|\Gamma * \phi\|_{2,q} \leq C_{n,q} \|f\|_{-1,p}$$
