Let $d\in\mathbb N$, $p\in[1,d)$, $$p^\ast:=\frac{dp}{d-p}$$ and $\Lambda\subseteq\mathbb R^d$ be bounded and open. It's well known that $$W_0^{1,\:p}(\Lambda)\subseteq L^q(\Lambda)\;\;\;\text{for all }q\in[1,p^\ast]\;.\tag 1$$
So, if $u\in W_0^{1,\:p}(\Lambda)$, we know that $u$ is not only an element of $L^p(\Lambda)$, but even of $L^{p^\ast}(\Lambda)$. But what can we say about the weak derivative $\nabla u$ of $u$? By definition, of $W_0^{1,\:p}(\Lambda)$, $\nabla u\in L^p(\Lambda,\mathbb R^d)$. Can we show more?
Maybe we can show that the weak partial derivatives $\frac{\partial u}{\partial x_i}$ of $u\in W_0^{1,\:p}(\Lambda)$ are elements of $W_0^{1,\:p-1}(\Lambda)$ and hence reapply $(1)$. Is that the case?