If $u_n \rightharpoonup u$ in $L^2(\Omega)$ and $u_n \geq 0$, is also $u \geq 0$?

Question

If $\Omega$ is the usual bounded domain and $u_n \rightharpoonup u$ in $L^2(\Omega)$ and $u_n \geq 0$ a.e, is also $u \geq 0$ a.e?

I know weak limits usually mess up things that one expects so I reckon this is probably not true but who knows?

Answer 1

Here is an alternative answer Denote $$ A:=\{ u \in L^2(\Omega): \ u(x)\ge 0 \text{ a.e.}\}. $$ Then it can be shown that

• $A$ is convex (trivial)
• $A$ is closed in $L^2(\Omega)$ (strong convergent sequences have pointwise-a.e. converging subsequences)

Then it follows that $A$ is weakly sequentially closed, which follows from Mazur's lemma.

Answer 2

Yes, this is true.

Let $A = \{u \le 0\}$, which is a measurable set of finite measure, so $1_A$ is a nonnegative function in $L^2(\Omega)$. Therefore $\int u_n 1_A \ge 0$ and hence $\int u 1_A = \lim \int u_n 1_A \ge 0$. As $u$ is nonpositive on $A$, it must be that $u = 0$ almost everywhere on $A$, which is to say $u \ge 0$ almost everywhere.

The same fact is true if you replace $\Omega$ by any measure space, though you have a little more work to do since $A$ might have infinite measure. (But the set $\{u < 0\}$ will be $\sigma$-finite...)