Existence of minimum in $H^{1,2}(\Omega)$ I am considering a functional 
$$\mu(\Omega) = \min \{ u \in H^{1,2}(\Omega),  \frac{\alpha \int_{\partial \Omega} u^2 ds + \int_{\Omega} |\nabla u|^2}{\int_{\Omega} u^2 dx} \}$$
I want to show the existence of minimum , can someone help me ? 
 A: The problem in proving existence is that standard techniques do not apply in an obvious way: The function to be minimized is not convex and its domain of definition is not convex either. 
I assume $\alpha>0$ here.
Denote
$$ J(u) := (\alpha \|u\|_{L^2(\partial \Omega)}^2 + \|\nabla u \|_{L^2(\Omega)}^2 ) \|u\|_{L^2(\Omega)}^{-2}.
$$
Then $J(u)$ is defined for $u\ne 0$. Moreover, it is bounded from below:
Using the inequality
$$
\|u\|_{L^2(\Omega)} \le c\left(  \|u\|_{L^2(\partial \Omega)} + \|\nabla u \|_{L^2(\Omega)}\right) \quad \forall u\in H^1(\Omega),
$$
it follows that $J(u)\ge c'>0$. Let now $(u_n)$ be a minimizing sequence, i.e.
$$
\lim_{n\to \infty}J(u_n) = \inf_{u\ne 0}J(u).
$$
Since $J(\lambda u)=J(u)$ for all $\lambda>0$, we can rescale the sequence, such that $\|u_n \|_{L^2(\Omega)}^2 =1$.
Since $J(u_n)$ is converging, it is bounded from above, which implies that
$(u_n)$ is a bounded sequence in $H^1(\Omega)$.
Hence,  we can extract a weakly converging subsequence (denoted again by $(u_n)$), $u_n \rightharpoonup u$ in $H^1(\Omega)$. After extracting another subsequence (again denoted by $(u_n)$) we find $u_n\to u$ in $L^2(\Omega)$.
Since $\|u_n\|_{L^2(\Omega)}=1$ and $u_n\to u$ in $L^2(\Omega)$ it follows $\|u\|_{L^2(\Omega)}=1$, in particular $u\ne 0$.
It remains to prove $J(u) \le \lim\inf J(u_n)$. Due to weak lower semicontinuity of norms, we have
$$
\alpha \|u\|_{L^2(\partial \Omega)}^2 + \|\nabla u \|_{L^2(\Omega)}^2
\le \lim\inf_{n\to\infty}(\alpha \|u_n\|_{L^2(\partial \Omega)}^2 + \|\nabla u_n \|_{L^2(\Omega)}^2) .
$$
Moreover, $0<\|u\|_{L^2}^2= \lim_{n\to \infty}\|u_n\|_{L^2}^2$. Hence it follows,
$$
\lim\inf_{n\to \infty} J(u_n)=
\lim\inf_{n\to \infty}\frac{\alpha \|u_n\|_{L^2}^2 + \|\nabla u_n \|_{L^2}^2}{\|u_n\|_{L^2}^2} 
= \frac{\lim\inf_{n\to \infty}(\alpha \|u_n\|_{L^2}^2 + \|\nabla u_n \|_{L^2}^2)}{\lim_{n\to \infty}\|u_n\|_{L^2}^2} \ge J(u).
$$
Hence, $u$ is a solution of the problem.
