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

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 $L^1-w$.

Ayman

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Notice order intervals are weakly sequentially compact. So take you convergent function $f$. Fix $m$, the sequence $f_n\wedge m(f+1)\in [0,m(f+1)]$ has a subsequence converges weakly in $L_1$ to some $f_m\leq f$ (note that lattice operations are not weakly continuous). Keep on doing this for $m$ and see what you get. – Rabee Tourky May 22 '13 at 12:29
Thanks for the comment Rabee. Could you detail a bit more (what is the lattice operation ?), or give a reference about order intervals ? Thanks. – Ayman Moussa May 22 '13 at 12:43
$f\wedge g$ is is just $\inf{f,g}$, state by stat a.e. I'm using notation from Banach lattice theory. My conjecture is that $f_n$ converges to $f$ weakly because order intervals $\{g: 0\leq g\leq m(f+1)$ are weakly compact in $L-1$ and thus sequentially compact. – Rabee Tourky May 22 '13 at 12:55
Aliprantis and Border "infinite dimensional analysis" is perhaps the best for this for you as it caters for applied mathematicians from a pure math perspective and treats order structures of $L_1$ seriously because they are important in economics. In particular, the weak compactness of $L_1$ order intervals. You can get it in pdf form somehow for the internet. – Rabee Tourky May 22 '13 at 13:03
If $f\neq 0$, then we can work with probability measures. We have a sequence of probability measures $\mu_n$ which converges in law to $\mu$, and all these measures are absolutely continuous with respect to Lebesgue measure. By portmanteau theorem, we know that $\mu_n(A)\to \mu(A)$ provided that $\mu(\partial A)=0$. And the question is whether this holds when $A$ is an arbitrary measurable set. – Davide Giraudo May 23 '13 at 13:09

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

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Thank you Gerald for this detailed and very well written answer. I still have one question though. Could you detail a bit the equality of the integrals on $I$, for $k>n$ ? $\alpha$ is defined as a limit and appears in the definition of $f$, so I can't understand how this equality may be true for $n$ "large enough" (meaning, wuthiut any passage to the limit). Also, your counterxample, such as the one of Bill, suggests that my proof above is wrong (it is supposed to work in the case where the limit function is bounded), any hint for the mistake is also welcome ! – Ayman Moussa May 24 '13 at 16:32
I may won't have access often to internet in the coming days, but I thank you all for you answers and good ideas, and will carefully read them asap. – Ayman Moussa May 24 '13 at 16:34

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 $\ell_1$.

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Okay, I do understand the idea but I don't see how you manage to insure that the intervalls $I_{k,n}$ do not overlap. Also, this seems to be in contradiction with the proof that I have written above (which is supposed to work in the case where the limit is continuous - here the limit is $\mathbf{1}_{[0,1]}$), but there is probably a mistake in my reasoning ! – Ayman Moussa May 24 '13 at 11:48
The intervals do intersect, but the functions are a small perturbation of a disjoint sequence of functions (multiply $f_n$ by the characteristic function of the complement of the union of the supports of $f_m$ for $m>n$). – Bill Johnson May 24 '13 at 12:11
Well, one mistake in your proof is that the final inequality in Step 3 only holds for $n$ sufficiently large, where "sufficiently large" depends on $x$. – Bill Johnson May 24 '13 at 12:24
But if I do this, I feel that I am losing the approximation of the Dirac mass in each $x\in F_n$, at least, it is not clear for me ! – Ayman Moussa May 24 '13 at 12:43
For $x$ in $D$ let $N(x)$ be the set of $n$ s.t. $f_n(x) < f(x) + 1$. Since $D$ is infinite, $\cap_{x \in D} N(x)$ can be empty. – Bill Johnson May 24 '13 at 16:20

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 $K$.

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Thanks TaQ for the precision. – Ayman Moussa May 27 '13 at 6:31