Sobolev inequalities for $W^{1,s}(Q)$ and $W^{2,s}(Q)$ Let $Q$ a bounded regular domain of $\mathbb{R}^3$, and $s>3$.
Let $u \in W^{1,s}(Q)$ (i.e $u, \nabla u$ are in $L^s(Q)$).
I know I have the following sobolev continuous embedding : $W^{1,s}  \hookrightarrow C^{0,1-3/s}(Q)$ and therefore I can write :
$$
\sup_{x \in Q} |u| \leq C \|u\|_{W^{1,s}(Q) }
$$
However, it is unclear for me, denoting $m=\frac{1}{|Q|}\int_{Q} u \ \mathrm{d}x$, that i have the following inequality :
$$
\sup_{x \in Q} |u-m| \leq C_1 \|\nabla u\|_{L^s(Q) } \label{1}\tag{1}
$$
It seems like some Poincaré inequality was used but I don't really see how. Do you have any ideas on how to proof this inequality ?
Additional questions
If I take $u \in W^{2,s}(Q)$, can i proof the following inequality ?
$$\sup_{x \in Q} |u-m| \leq C_1 \|\nabla^2 u\|_{L^s(Q) } \label{2}\tag{2}$$
Do I need to add any hypothesis on $u$ to prove \eqref{1} and \eqref{2}? I was thinking something like $\gamma_0 (u)=0$ on $\partial Q$...
Thanks for your help.
 A: Let me prove \eqref{1} for functions $u$ satisfying $\int_\Omega u=0$. Assume the claim does not hold. Then for each $n$ there is $(u_n)$ in $W^{1,s}$ such that
$$
\|u_n\|_{L^\infty} > n \|\nabla u_n\|_{L^s}.
$$
Hence, $u_n\ne0$, and we can assume
$\|u_n\|_{L^\infty}=1$. This implies $\nabla u_n\to 0$. In addition, $(u_n)$ is bounded in $W^{1,s}$. Thus $(u_n)$ is bounded in that Hoelder norm. By Arzela-Ascoli, we have after - possibly extracting a subsequence - $u_n \to u$ in $C(\bar Q)$. By the convergence of $\nabla u_n$, we have $\nabla u=0$, so $u$ is constant. However, along the sequence $\int_\Omega u_n=0$, so $u=0$. This implies $\nabla u_n \to 0$ in $L^s$ and $u_n\to 0$ in $W^{1,s}$, which is a contradiction to $\|u_n\|_{L^\infty}=1$.
The second claim is not true: take $Q=B_1(0)$, $u(x) = c^Tx$ with $c\ne 0$. Then $m=0$, $\nabla ^2u=u$, but $\|u\|_{L^\infty}= |c|$. If one would try the above proof, then one finds that the limit is not a constant function but piecewise linear. The condition $m=0$ only fixes one degree of freedom, but there are $3$ more. One needs additional conditions to conclude $u=0$.
