Counterexample of non-negative sequence weakly converging in $\mathscr{M}^1$ but not $L^1$

Question

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

Answer 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 $\int_I f = 0 = \int_I f_k$.

Answer 2

For $n>100$ let $F_n = (k/n)_{k=1}^{n -1}$ and for $x= k/n $ in $F_n$ let $I_{k,n}$ be a symmetric interval around $x$ having length $n^{-n}/(n-1)$. Set $f_n = n^n \sum_{k=1}^{n-1} 1_{I_{k,n}}$. It is clear that $f_n$ converges weak$^*$ to $1_{[0,1]}$. But the $(f_n)$ are essentially disjointly supported and hence are equivalent to the unit vector basis for $\ell_1$.

Answer 3

I think the idea of Bill Johnson's counterexample would have been more easily understandable if expressed e.g. as follows: For $i\ge 0$ a natural number, let $U_i=\bigcup_{\,k=0}^{\,i}J_{\,ik}$ where $J_{\,ik}={\,]\,}{-1}+2\,k\,(i+1)^{-1},-1+2\,k\,(i+1)^{-1}+(i+1)^{-3}{[}\ $. Putting $U=\bigcup_{\,i=0}^{\,\infty}U_i$ and $K=[{-1},1]\setminus U$, then $U$ is open, and $K$ is closed with positive measure since $U$ has measure at most $\sum_{\,i=0}^{\,\infty}\,(i+1)^{-2}< 1+\int_{\,1}^{+\infty}x^{\,-2}\,{\rm d\,}x=2$. Defining $f_i$ on $[-1,1]$ by $t\mapsto 2\,(i+1)^{\,2}$ for $t\in U_i$, and $f_i(t)=0$ otherwise, and $f:[-1,1]\owns t\mapsto 1$, it is clear that $\int_{-1}^1(f_i\cdot g)\to\int_{-1}^1(f\cdot g)$ for every continuous $g$. However, this fails if we take as $g$ the indicator function of $K$ since then $\int_{-1}^1(f_i\cdot g)=0$ but $\int_{-1}^1(f\cdot g)$ equals the measure of $K$.