Vtyjkyuk

$newcommandRmathbbR$
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.

For $N=1$, let us consider the functions
$$
f_n(x) :=
begincases
sin(nx), & xin [0,2pi],\
0, & textotherwise.
endcases
$$
Then, for every $phiin C_c(mathbbR)$, by the Riemann-Lebesgue lemma one has $lim_nto+infty int f_n phi = 0$.
On the other hand, $liminf_n f_n(x) = -1$ for a.e. $xin (0,2pi)$ (and $0$ otherwise).

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

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

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

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