Consider a a sequence of non-negative functions $(f_n)_n$, bounded in $L^1([-1,1])$ and weakly$-\star$ converging in $\mathscr{M}^1([-1,1])$ to some $f\in L^1([-1,1])$. What I mean by this convergence is that for any continuous function $\varphi\in\mathscr{C}^0([-1,1])$,

\begin{align*} \int_{-1}^1 f_n \varphi \;\operatorname*{\longrightarrow}_{n\rightarrow +\infty}\; \int^{1}_{-1} f \varphi. \qquad (1) \end{align*}

It is easy to see that $f$ is necessarily also non-negative.

**Question** : do we also have $(f_n)_n \rightharpoonup f$ in $L^1([-1,1])-w$, *i.e.* can we replace in the previous convergence the continuous function $\varphi$ by any element of $L^\infty([-1,1])$ ?

I am quite sure that the answer is no (because there is no density of regular functions in $L^\infty$), but I did not manage to find a counterexample.

Two remarks :

1) Without the assumption of non-negativeness one may consider $f_n:= n\mathbf{1}_{[0,1/n]} - n\mathbf{1}_{[-1/n,0]}$, which, in some sense , approximates $ " \delta_{0^+}-\delta_{0-} "$, and hence $(f_n)_n \operatorname*{\rightharpoonup}^{\mathscr{M}-\star} 0$, but it may be checked easily that $(f_n)_n$ do not converge to $0$ weakly in $L^1$ : the sequence "concentrates" the eventual discontinuity in $0$.

2) Keeping the non-negativeness but working on the open set $(-1,1)$ instead of $[-1,1]$ simplifies also the problem, since the mass may then concentrate to the boundary : $f_n:= n \mathbf{1}_{[1-1/n,1[}$ tends to $0$ in $\mathscr{M}^1(-1,1)-w\star$ (test functions in $\mathscr{C}^0_c(-1,1)$) but clearly not in $L^1-w$.

Thanks in advance for any advice !

Ayman