Does an integral inequality imply a pointwise inequality?

Question

$\newcommand{\R}{\mathbb{R}}$ Let $(f_n)_n \subset L^1(\R^N)$. Suppose that for any nonnegative function $\phi \in C_c^{\infty}(\R^N)$, we have:

$$ 0 \leq \liminf_{n \rightarrow \infty} \int f_n \, \phi $$

Can we conclude that $0 \leq \liminf_{n \rightarrow \infty} f_n(x)$ for almost every $x \in \R^N$?

My first idea is to use the definition of $\liminf$ to get an estimate of $\inf_{m \geq N} \int f_m \, \phi \geq -\epsilon$, where $N$ depends on $\epsilon$. However, I'm not sure where to go from there.

Another fact that might be useful is that if $\int f \, \phi \geq 0$ for every nonnegative $\phi \in C_c^{\infty}(\R^N)$, then $f \geq 0$ a.e. (this is proven similarly to corollary 4.24 in Brezis' Functional Analysis). We might be able to use this for $f = \liminf f_n$, but I haven't been able to get it to work.

Answer 1

For $N=1$, let us consider the functions $$ f_n(x) := \begin{cases} \sin(nx), & x\in [0,2\pi],\\ 0, & \text{otherwise}. \end{cases} $$ Then, for every $\phi\in C_c(\mathbb{R})$, by the Riemann-Lebesgue lemma one has $\lim_{n\to+\infty} \int f_n \phi = 0$. On the other hand, $\liminf_n f_n(x) = -1$ for a.e. $x\in (0,2\pi)$ (and $0$ otherwise).

Answer 2

On $\mathbb R,$ define the sequence $f_n$ as

$$-\mathbb {1}_{[0,1]},-\mathbb {1}_{[0,1/2]}, -\mathbb {1}_{[1/2,1]}, -\mathbb {1}_{[0,1/3]},-\mathbb {1}_{[1/3,2/3]},-\mathbb {1}i_{[2/3,1]}, \dots$$

Then $\int f_n\phi \to 0$ for every $\phi \in L^1,$ yet $\liminf f_n(x) = -1$ for every $x\in [0,1].$

With a little extra care we could get $\liminf f_n(x) = -1$ for every $x\in \mathbb R.$