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 1


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

  • $\begingroup$ That is an excellent counterexample. Thank you very much. $\endgroup$
    – user333943
    Dec 14, 2022 at 12:42

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