If $\int_\Omega (u_n/\delta)^2=1$ then $(u_n)_{n=1}^{\infty}$ is bounded in $H_{0}^{1}(\Omega)$

Question

I am reading a paper ('Hardy's inequalities revisited' by Brezis and Marcus). In one of the proofs, they write the following:

Here, $\Omega$ is a bounded smooth domain, $u_n$ is a sequence in $H_0^1(\Omega)$ and $\delta(x)$ returns the distance of the point $x$ from $\partial \Omega$.

My question is - why is ${u_n}$ bounded in $H_0^1$? I definitely agree that $\int_\Omega u_n^2$ is bounded (due to the normalization of $u_n/\delta$), but why is $\int_{\Omega}|\nabla u_n|^2$ bounded? The sequence $u_n$ is chosen to approach the following infimum:

Where $\lambda \in \mathbb R$ is fixed.

I thought that maybe the result is some consequence of the Poincare inequality, but it seems to only give me the opposite inequality of what I want.

Thanks in advance.

Answer 1

Indeed, it directly follows from the definition of the functional.

Denote $$J_\lambda (u)=\frac{\int_\Omega |\nabla u|^2-\lambda \int_\Omega u^2}{\int_\Omega(u/\delta)^2},\qquad u\in H_0^1(\Omega).$$ Then $J_\lambda^\Omega=\inf_{u\in H_0^1(\Omega)} J_\lambda(u)<\infty.$ If $\{u_n\}$ is a minimizing sequence, then $J_\lambda(u_n)\to J_\lambda^\Omega$, and in particluar, $\{J_\lambda(u_n)\}$ is bounded. By nomarlization, we have known that $\int_\Omega u_n^2$ and $J_\lambda(u_n)=\int_\Omega |\nabla u_n|^2-\lambda \int_\Omega u_n^2$ are both bounded. Hence $$\int_\Omega |\nabla u_n|^2 =\lambda\int_\Omega u_n^2+J_\lambda(u_n)$$ is also bounded. Therefore, $\{u_n\}$ is bounded in $H_0^1(\Omega)$.