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



Counterexample of non-negative sequence weakly converging in $\mathscr{M}^1$ but not $L^1$
Let's build a "fat Cantor set". Start with $A\_0 = [0,1]$ with measure $\alpha\_0=1$.
Then remove a short open interval centered at $1/2$, leaving a set $A\_1 \subset A\_0$ of measure $\alpha\_1 < \alpha\_0$. So $A\_1$ is made up of $2$ closed intervals of length $\alpha\_1/2$. Let $B\_1$ be the removed interval with length $1-\alpha\_1 = \alpha\_0-\alpha\_1$. 


Next remove a short open interval from the center of each of the intervals of $A\_1$, to leave $A\_2 \subset A\_1$ of measure $\alpha\_2<\alpha\_1$. And $A\_2$ is made up of $4$ closed intervals of length $\alpha\_2/4$. Let $B\_2$ be made up of the $2$ removed intervals, each of length $(\alpha\_1-\alpha\_2)/2$.


Continue in this way. $A\_n \subset A\_{n-1}$ has measure $\alpha\_n < \alpha\_{n-1}$, and $A\_n$ is made up of $2^n$ closed intervals each of length $\alpha\_n/2^n$. $B\_n$ consists of the $2^{n-1}$ newly removed open intervals, each of length $(\alpha\_n-\alpha\_{n-1})/2^{n-1}$


Let $A = \bigcap\_{n=1}^\infty A\_n$. Choose the lengths of the intervals removed so that $\alpha>0$, where $\alpha = \lim\_{n \to \infty} \alpha\_n$. (This is what makes it a "fat" Cantor set.) Of course $m(A) = \lim\_{n \to \infty} m(A\_n) = \alpha > 0$, where $m$ is Lebesgue measure. 


Our limit function is
$$
f = \frac{1}{\alpha} \mathbf1\_A
$$
where $\mathbf1\_A$ denotes the indicator function of set $A$. For $n\ge 1$ define
$$
f\_n = \frac{1}{(\alpha\_n-\alpha\_{n-1})}\mathbf1\_{B\_n}
$$
(I used Bill Johnson's idea of making an $l^1$ basis. But now these really are disjoint, so you don't have to do estimates to show they are "close enough" to being disjoint.)
Now we claim: 



> 
> (1) $\int f\_n g$ converges to $\int f g$ for all continuous $g$; 
> 
> 
> (2) there is $h \in L^\infty[0,1]$ such that $\int f\_n h$ does not converge to $\int fh$. 
> 
> 
> 


**(1)** 


Let $g$ be a continuous function.
Let $\pi\_n$ be the partition of $[0,1]$ made up of the $2^{n+1}$ endpoints of the set $A\_n$. Note that for each interval $I$ of partition $\pi\_n$ we have
$$
\int\_I f\_{n+1} = \int\_I f
\tag{\*}
$$
and more generally
$$
\int\_I f\_k = \int\_I f
\tag{\*\*}
$$
for all $k > n$. (In technical language, we have a "martingale".) As $n \to \infty$, the lengths of these intervals goes to $0$. And $g$ is uniformly continuous. So we will conclude that $\int f\_n g \to \int f g$.


**(2)** 


Let $h = \mathbf1\_A$. Then $f\_nh=0$ so $\int f\_nh = 0$ for all $n$. But $fh=f$ a.e. and $\int fh = \int f = 1$.


**added:** more on $({}^\ast)$ and $({}^{\ast\ast})$ 


For $\pi\_0$. Note $\int\_{[0,1]} f = 1 = \int\_{[0,1]} f\_k$ for all $k$. 


For $\pi\_1$. If $I$ is one of the two intervals that make up $A\_1$, then $\int\_I f = 1/2 = \int\_I f\_k$ for all $k \ge 2$. If $I = B\_1$, the removed middle interval, then $\int\_I f = 0 = \int\_I f\_k$ for all $k \ge 2$.


For $\pi\_2$. If $I$ is one of the $4$ intervals that make up $A\_2$, then $\int\_I f = 1/4 = \int\_I f\_k$ for all $k \ge 3$. If $I$ is one of the intervals that make up $B\_2$ or the interval $B\_1$, then $\int\_I f = 0 = \int\_I f\_k$ for all $k \ge 3$.


For general $\pi\_n$. If $I$ is one of the $2^n$ intervals that make up $A\_n$, then $\int\_I f = 1/2^n = \int\_I f\_k$ for all $k \ge n+1$. If $I$ is one of the intervals that make up any of $B\_1,\dots,B\_n$, then $\int\_I f = 0 = \int\_I f\_k$.


