Suppose $\{f_n\}$ is a sequence of functions on $\mathbb R$ that is an approximate identity. In particular, it satisfies for every real $c>0$,
$$\int_{-\infty}^{-c} |f_n(x)|dx\to0\qquad\int_c^\infty |f_n(x)|dx\to0$$
Does it follow that for every $c>0$, $f_n\to0$ uniformly on $(-\infty,-c]\cup[c,+\infty)$?
My intuition tells me that the answer must be positive but I am unable to prove why it is so.
My attempt at proof:
Seeking contradiction, assume that there is $c>0$ and $\epsilon>0$ so that for each $N\in\mathbb N$, there is $n>N$ and $x\in(-\infty,-c]\cup[c,+\infty)$ such that $|f_n(x)|\ge\epsilon$...
I am not sure what to do next. My idea is to show something like $\int_c^\infty |f_n|dx<\epsilon$ and $\int_c^\infty |f_n|dx\ge\epsilon$ to obtain a contradiction.
The claim is false. Here is a counter-example.
Define the function $$h = \begin{cases} 1 - |x| & |x| \leq 1 \\ 0 & \text{otherwise} \end{cases}$$ and let $f_n(x) = nh\left ( n^2(x-1) \right )$.
Geometrically, $f_n$ is a hat function centered around $x=1$ with height $n$ and base $\frac{2}{n^2}$. From this it follows that $$\int_{-\infty}^\infty|f_n| dx = \frac{1}{n}$$ which implies that the sequence $f_n$ satisfies the conditions.
However $f_n$ fails to uniformly converge to 0 on $(-\infty,-c]\cup[c,\infty)$ for any $c < 1$. In fact it doesn't even converge pointwise to 0.
At a high level, uniform convergence and pointwise convergence are much stronger types of convergence than integral-based convergences (like the one you are asking about). It is rarely the case that an integral condition can imply uniform or pointwise convergence without imposing additional conditions.