I was trying to solve prove the following inequality:
For all compactly supported smooth functions on the $d$-dimensional unit cube $u\in C^\infty([0,1]^d)$ (where if $\Omega\subset\mathbb{R}^d$ is open, we define $C^\infty(\overline{\Omega})=\{u\in C^\infty(\Omega)\mid u \text{ has > a continuous extension to } \overline{\Omega}\}$), we have
$$\int_{\Omega}|u|^{2} d x \leq \alpha \int_{\Omega} \operatorname{dist}(x, \partial \Omega)^{2} \left(|u|^{2}+\|\nabla u\|^{2} \right) d x, $$
where $\Omega=(0,1)^d$ and if $A, B\subset\mathbb{R}^d$, then $\operatorname{dist}(A,B)=\inf_{a\in A,b\in B}d(a,b)$.
To prove it, I thought I better study how the function $\operatorname{dist}(x,\partial\Omega)$ behaves on $\Omega$. For that, I determined the following open subsets of $\Omega$: for each $1\leq i\leq d$, $$ U_i^+=\{x\in\Omega:x_i>\max\{x_j,1-x_j\mid 1\leq j\leq n,\;j\neq i\}\},\\ U_i^-=\{x\in\Omega:x_i<\min\{x_j,1-x_j\mid 1\leq j\leq n,\;j\neq i\}\}. $$
These sets (which on $d=1$ are just $(0,1/2)$ and $(1/2,0)$, on $d=2$ can be seen to be some triangles, and on $d=3$ some piramids) satisfy the property $$ \begin{align} x\in U_i^+&\Rightarrow d(x,\partial\Omega)=1-x_i,\\ x\in U_i^-&\Rightarrow d(x,\partial\Omega)=x_i. \end{align} $$ Furthermore, if $\Omega'=\bigcup_{i=1}^d U_i^+\cup U_i^-$, then $\Omega\setminus\Omega'$ is a null set (i.e., the $d$-dimensional Lebesgue measure equals zero).
From here, I was thinking of using some partition of unity which involved the $U_i^\pm$'s to "break down" the integral in different manageable parts. But I could not find out exactly which and how to apply it.
I was even trying to prove it on $d=1$ but I could not get the proof to work. Trying to get the bound taking a compactly supported function $\theta$ on $(0,1)$ with values $0\leq\theta\leq 1$ and which evaluates $1$ on $[1/n,1/2-1/n]\cup[1/2+1/n,1-1/n]$ and trying to bound $\int_0^1\theta udx$. But I could not control the value of the derivative of $\theta$.
Any thoughts on the problem will be appreciated :)