Does an integral inequality imply a pointwise inequality?

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).

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 $x\in \mathbb R.$