# Constructive definitions of the Sobolev trace

I have a function $u\in H^1_0(\Omega)$, $\partial\Omega$ being bounded and as nice as you need, and I need to prove that $u^k\in H^1_0(\Omega)$ for a fixed $k>1$ (integer if this can help).

Certainly, this requires additional assumptions on $u$. For instance, if I know that $u$ is continuous up to the boundary, then this is also true of $u^k$ and gives the necessary conclusion. But expressing this condition in the Sobolev scale requires to assume that $u$ belongs to $H^s$ for $s$ larger than $n/2$. Are there weaker additional conditions on $u$ which ensure that $u^k\in H^1_0(\Omega)$?

• What would be wrong with the obvious condition $u\in L^\infty\cap H^1_0$ ? I can't think of anything weaker. Apr 8 '18 at 13:18

For $u\in W^{1,\frac{2nk}{n+2k-2}}$ we have using Young's and Gagliardo-Nirenberg-Sobolev inequality:
$$|u^k|^2_{W^{1,2}}\leq \int_\Omega |u(x)|^{2(k-1)} |\nabla u(x)|^2dx\leq C \int_\Omega |u(x)|^\frac{2nk}{n-2}+|\nabla u(x)|^\frac{2nk}{n+2k-2}\leq = C\left(\left(\|u\|_{L^\frac{2nk}{n-2}}\right)^\frac{2nk}{n-2} +\left(|u|_{W^{1,\frac{2nk}{n+2k-2}}}\right)^{\frac{2nk}{n+2k-2}} \right) \leq C \left(\left(|u|_{W^{1,\frac{2nk}{n+2k-2}}}\right)^\frac{2nk}{n-2} +\left(|u|_{W^{1,\frac{2nk}{n+2k-2}}}\right)^{\frac{2nk}{n+2k-2}} \right).$$ On the other hand the example $u(x)=|x|^\frac{2-n}{2k},\ \Omega=B_1(0)$ shows that $u\in W^{1,\frac{2nk}{n+2k-2}}$ is necessary.