Let $\Omega$ be a bounded regular domain of $\mathbb{R}^d$. Define the following space :
$X:=\{\nabla h, \ h \in W^{1,s}(\Omega)\} \subset (L^s(\Omega))^3$.
Can I say that the dual space of $X$ is the space $B=\{\nabla h, \ h \in W^{1,p}(\Omega) \}$ where $p$ is the conjugate of $s$ ($1= \frac{1}{p}+ \frac{1}{s}$). Obviously I have one inclusion ($B \subset X'$) since for any $\nabla h \in B$, I can see it as a continuous linear map $f \in X$ by setting :
$$f(\nabla a) = \int_{\Omega} \nabla a \cdot \nabla h \ \mathrm{d}x, \quad \forall \ u=\nabla a \in X.$$ It is continuous using Holder inequality : $|f(\nabla a)| \leq ||\nabla h||_{(L^s(\Omega))^3}||\nabla a||_{(L^p(\Omega))^3}$.
But I don't know if the other inclusion is true. Furtermore it feels weird since I also have $(L^p(\Omega))^3 \subset X'$ using the exact same construction, replacing $\nabla h$ by any vector $v \in (L^p(\Omega))^3$. Does anyone have any clue on this problem ?