If $V=\{ u\in H^{1}_{0}(\Omega) : u^2\in H^{1}_{0}(\Omega)\}$ where $H^{1}_{0}(\Omega)$ is sobolev space (whose functions are vanishes near boundary) and $\Omega$ is bounded domain in ${\rm I\!R}^{n}$ and n $H^{1}_{0}(\Omega)$ , $\|u\|=(\int_{\Omega}|\nabla{u}|^{2} dx)^{\frac{1}{2}}$, then for $u\in V$ can we say, $\|u^2\|\leq \|u\|^2$ ?
where $|\nabla{u}| $ is usual ${\rm I\!R}^{n}$ norm.
No. To see why, suppose the inequality did hold. Since $\nabla u^2 = 2u\nabla u$ we would have
$$\left(\int_\Omega 4u^2|\nabla u|^2 dx\right)^\frac{1}{2} \leq \int_\Omega |\nabla u|^2 dx$$
for all $u \in V$. Now consider replacing $u$ by $u+C$ for a constant $C$. Since $\nabla (u+C)=\nabla u$ and $u+C \in V$ we would have
$$\left(\int_\Omega 4(u+C)^2|\nabla u|^2 dx\right)^\frac{1}{2} \leq \int_\Omega |\nabla u|^2 dx$$
for every $u$ and all constants $C$. Since only the left hand side depends on $C$, we can make $C$ large to violate the inequality (unless $u$ is constant so $\nabla u \equiv 0$.). So the inequality cannot hold.