# Show that $\frac{\int_\Omega|\nabla u|^2+\int_\Omega\alpha|u|^2}{\int_\Omega|u|^2}$ attains a minimum in $W_0^{1,2}(\Omega)$

Let

• $\Omega\subseteq\mathbb{R}^n$ be a bounded domain
• $H:=W_0^{1,2}(\Omega)$ be the Sobolev space
• $|\;\cdot\;|_p$ be the seminorm $$|u|_p^p:=\int_\Omega|\nabla u|^p\;d\lambda^n\;\;\;\text{for }u:\Omega\to\mathbb{R}\;\text{weakly differentiable}$$ on $L^p:=L^p(\Omega)$
• $\left\|\;\cdot\;\right\|_p$ be the $L^p$-norm

From the basic theory of the eigenvalue problem of the Laplacian, one knows that $$R(u):=\frac{|u|_2^2}{\left\|u\right\|_2^2}\;\;\;\text{for }u\in H\setminus\left\{0\right\}\tag{1}$$ attains its infimum in $H\setminus\left\{0\right\}$. Now, I would like to show, that $$\tilde{R}(u):=R(u)+\frac{\left\|\sqrt{\alpha}u\right\|_2^2}{\left\|u\right\|_2^2}\;\;\;\text{for }u\in H\setminus\left\{0\right\}\tag{2}\;,$$ for some $\alpha\in L^\infty$, has a minimum, too.

We may note, that we can assume, that $R$ attains its minimum $\lambda_1$ in $u_1\in H$ with $\left\|u_1\right\|_2^2=1$. Then, $$\tilde{R}(u_1)=\lambda_1+\left\|\sqrt{\alpha} u_1\right\|_2^2\;.$$ However, I don't see how I need to proceed for me. Maybe this is not the right track and we need to use the Poincaré inequality $$\left\|u\right\|_2^2\le C|u|_2^2\;\;\;\text{for all }u\in H\;,$$ for some $C>0$, instead.

• Can't you just use the usual proof, i.e. that an ($L^2$ normalized) minimising sequence admits a weakly convergent subsequence, which converges strongly in $L^2$, so that we get a minimiser by weak lower semicontinuity of the functional? – PhoemueX Jun 28 '15 at 20:44
• @PhoemueX I know that we can show, that each minimizing sequence contains a subsequence, which converges to $u_1$ in $L^2$. How does this help to show the desired statement. Maybe you can provide an answer. – 0xbadf00d Jun 28 '15 at 21:10
• Note that your $\tilde R$ is the question are not the same as that in the title. Can you clarify? – user99914 Jun 29 '15 at 10:49
• @JohnMa There was a little typo. Now, it should be the same. – 0xbadf00d Jun 29 '15 at 20:13

As remarked in the comment, the construction is identical to the case where one minimizes $R(u)$. First of all, observe that

$$\tilde R(u) = R(u) + \frac{\int_M \alpha |u|^2 }{\|u\|^2_{2}} \ge R(u) - \|\alpha\|_\infty \ge \lambda_1 - \|\alpha\|_\infty.$$

Thus $\tilde R$ is bounded below. So we can take $u_1, u_2, \cdots \in H$, so that $\|u_i\|_2 = 1$ and

$$\tilde R(u_i) \to \inf_{u\in H\setminus\{0\}} \tilde R(u):= C$$

In particular, we can assume that $\tilde R(u_i)$ is bounded independent of $i$. As

$$\int_M|\nabla u_i|^2\,d\lambda \le |\tilde R(u_i)|+\| \alpha\|_\infty ,$$

$u_i$ has uniformly bounded $W^{1, 2}_0(\Omega)$ norm. Thus by Rellich–Kondrachov theorem, there is $u \in H$ so that $u_i \to u$ strongly in $L^2(\Omega)$ and weakly in $W^{1, 2}_0(\Omega)$. By the strong $L^2$-convergence $u_i \to u$, we have $$\|u\|_{L^2} = \lim_{i\to \infty} \|u_i\|_{L^2} = 1,\ \ \ \ \int_M \alpha |u|^2 \,d\lambda = \lim_{i\to \infty}\int_M \alpha |u_i|^2 \,d\lambda.$$

By the weak convergence of $u_i \to u$ in $W^{1, 2}_0(\Omega)$, we have

$$\int_M |\nabla u|^2 \,d\lambda \le \liminf_{i\to \infty}\int_M |\nabla u_i|^2 \,d\lambda.$$

Summing up, we have

$$\tilde R(u) \le \liminf_{i\to \infty} \tilde R(u_i) = C$$

and so $u$ minimizes $\tilde R$.

• Nice. Maybe you should note (to emphasize why strong convergence is needed) that $\Vert u \Vert_2 = 1$, since $u_i \to u$ strongly. – PhoemueX Jun 30 '15 at 19:12
• @PhoemueX : Thanks for your comment. I have expanded the answer to make it clearer. – user99914 Jul 1 '15 at 3:06
• Why do we've got $$\tilde{R}(u)\ge R(u)-\left\|\alpha\right\|_\infty\;?$$ – 0xbadf00d Jul 1 '15 at 19:31
• @PhoemueX : Since $\int_\Omega \alpha |u|^2 \ge -\|\alpha\|_\infty \int_\Omega |u|^2$. – user99914 Jul 2 '15 at 3:11
• @0xbadf00d : That's Rellich–Kondrachov theorem. – user99914 Jul 24 '15 at 3:06