Does $\|f_n *g-g\|_1\rightarrow 0$ imply $\int f_n \rightarrow 1$?

Question

Let $f_n \in L^1$ such that $\forall g \in L^1$: $\|f_n *g-g\|_1 \rightarrow 0$.

Does this imply that $\int f_n \rightarrow 1$?

I tried few examples and could not find a counterexample, but I don't know how to show this.

If anything is unclear, please let me know.

Answer 1

The Fourier transforms satisfy $\widehat{f_n} \widehat{\ g\ } = \widehat{f_n \ast g} \to \widehat{\ g\ }$ at least pointwise. Taking any $g\in L^1$ with $\widehat{\ g\ }(0)\neq 0$ you get $$\int f_n =\widehat{f_n}(0)\to 1.$$