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

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?

• Hint: Let $A \subset \Omega$ with $\mu(A) < \infty$. Can $\int_A u$ ever be negative? What does this imply about whether or not u is a.e. non-negative? – Justthisguy Jun 26 '14 at 17:55
• In general, I've found that as long as you have a few simple examples of weakly converging sequences, you can make educated guesses about what should be true and why. You asked about a "local" property of the weak limit, so the local example that can go wrong is wild oscillation, and this doesn't do anything to positivity. Another example to keep in mind are solutions going off to infinity in physical space (instead of oscillating wildly which is going off to infinity in Fourier space). There's a sense in which these are the only bad examples. – felipeh Jun 26 '14 at 18:00
• Thanks both. @user835747 that's good advice, I should memorise these functions. – LapLace Jun 27 '14 at 13:04

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.
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...)