# Minimizing a functional in a subset of the $L^2$ sphere

I am stuck on the problem I am stating below and I was wondering if anyone was able to help me out.

Let $$N\ge 1$$ be an integer. Let $$\omega$$ be a bounded, open and connected subset in $$\mathbb{R}^N$$ with Lipschitz continuous boundary. By $$\mathcal{D}(\omega)$$ we denote the space of $$\mathcal{C}^\infty$$ functions with compact support in $$\omega$$.

Define the set $$\mathcal{L}:=\{v \in L^2(\omega); v \ge 0 \mbox{ a.e. in } \omega \mbox{ and } \|v\|_{L^2(\omega)}=1\},$$ define the set $$\mathcal{A}:=\{\psi \in \mathcal{D}(\omega); \psi \ge 0 \mbox{ in } \overline{\omega} \mbox{ and } \|\psi\|_{L^2(\omega)}=1\},$$ and define the linear and bounded functional $$J:L^2(\omega)\to\mathbb{R}$$ by $$J(v):=\int_{\omega} v \,\mathrm{d}x.$$

It is clear that, if we restrict $$J$$ to the set $$\mathcal{A}$$, we have that $$J(\varphi) > 0$$ for all $$\varphi \in \mathcal{A}$$.

It is also clear that $$\mathcal{A} \subset \mathcal{L}$$, and that $$\mathcal{L}$$ is non-empty and strongly closed in $$L^2(\omega)$$. Moreover, we have that $$0 \not\in \mathcal{A} =\overline{\mathcal{A}}^{\|\cdot\|_{L^2(\omega)}}$$.

I would like to show that: $$\inf_{\varphi \in \mathcal{A}} J(\varphi) >0.$$

In my attempt, I reasoned by contradition assuming that there was a minimizing sequence $$\{\varphi_k\}_{k=1}^\infty \subset \mathcal{A}$$ for which $$J(\varphi_k) \to 0,\quad\mbox{ as } k \to\infty.$$

This is equivalent to saying that $$\|\varphi_k\|_{L^1(\omega)} \to 0$$ as $$k \to\infty$$ and so, up to passing to a subsequence, we have that $$\varphi_k \to 0$$ a.e. in $$\omega$$.

Here I am stuck and I cannot go on. Can anyone help me?

Thanks in advance.

## 1 Answer

This is not true. Take $$\omega = (0,1)$$. Define $$v_k := \chi_{(0,1/k)}\sqrt k$$. Then $$\|v_k\|_{L^2}=1$$ but $$\int_\omega v_k = \frac1{\sqrt k} \to 0$$.

• That is an excellent counterexample. Thank you very much. Dec 14, 2022 at 12:42