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Hi.

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

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

EDIT

Okay, I think I manage to get a proof in the case where $f \in\mathscr{C}^0([-1,1])$ (or almost everywhere continuous). Maybe this wil help to get an answer in the general case !

1. First Step : we may assume (even in the general case) that each $f_n$ is continuous. Indeed, consider a sequence of non-negative and pair mollifiers $(\rho_n)_n$ and replace $f_n$ by $f_n\star \rho_n$. Also, using a classical argument of uniqueness of accumulation points, it is enough to show that $(f_n)_n$ is relatively weakly sequentially compact in $L^1$.

2. Second Step : The set $\{x\,:\, \liminf\,f_n \geq f+1\}$ has an empty interior. Indeed, if it contained a non-empty open interval $I$, one could find a non-negative and non-zero continuous fonction $\varphi$ with support in $I$ and the convergence $(1)$ would lead to the following, using Fatou's lemma \begin{align*} \int_{-1}^1 (f+1) \varphi \leq \int_{-1}^1 \liminf \,f_n\varphi \leq \liminf \int_{-1}^1 f_n \varphi = \int_{-1}^1 f \varphi, \end{align*} whence a contradiction.

3. Third Step : $\{x\,:\, \liminf\,f_n \geq f+1\}$ is hence of empty interior and therefore so is \begin{align*} A := \bigcup_{n\in\mathbb{N}} \bigcap_{p \geq n} \{x\,:\,f_p \geq f+1\}. \end{align*} The complement of $A$ in $[-1,1]$ is therefore dense and is precisely given by the formula \begin{align*} [-1,1]\backslash A := \bigcap_{n\in\mathbb{N}} \bigcup_{p\geq n} \{x\,:\,f_p < f+1\}. \end{align*} Because of the previous formula, it is easy to see that for each $x\in [-1,1]\backslash A$, there exists an subsequence $(f_{\sigma_x(n)})_n$ for which $f_{\sigma_x(n)} < f(x)+1$ for all $n$. Now pick a coutable dense subset $D\subset [-1,1]\backslash A$, and extract diagonally along $D$ an extraction (still labeled) $(f_n)_n$ such as, for all $x\in D$, $f_n(x) < f(x) +1$.

4. Fourth Step : The inequality $f_n(x) < f(x)+1$ is true on a dense subset and may be hence extended (replacing $<$ by $\leq$) to the whole intervall for the chosen subsequence. One may hence conclude using Dunford-Pettis theorem.

In fact the same proof applies when $f$ is only assumed to be in $L^\infty$ (replace $f+1$ by $\|f\|_\infty+1$). I tried to use Luzin's theorem but it failed : the set $D$ is dense but not necessarily dense in the subsets in which $f$ is continuous.

Any advice is welcome, thanks again !

Ayman