Find a sequence $\{f_n\} \in L^1([0,1])$ such that $\int_0^x f_n \to \int_0^x f$ for all $0<x<1$ and $||f_n|| < M$ for some $M$ but $f_n $ doesn't converge weakly to $ f$ in $L^1([0,1])$

My Try:

Define the functions $f_n$ on $[0,1]$ by setting:

$$f_n(x)= \begin{cases} 1 \ \text{ for } \frac{k}{2^n} +\frac{1}{2^{2n+1}} \leq x < \frac{(k+1)}{2^n} \text{ and } 0 \leq k \leq 2^n -1\\ 1- 2^{n+1} \text{ elsewhere on }[0,1] \end{cases}$$ for $n$ natural number and define $f \equiv 0$ on $[0,1]$

Note that $||f_n||_1 \leq 2$ therefore $\{f_n\}$ is bounded in $L^1([0,1])$ Also note that $\int_0^x f_n \to \int_0^x f $ for all $x \in (0,1)$

Now I am trying to find a function $g$ for which the following corollary will fail therefore my sequence will not be weakly converget.

Here is the Corollary: $f_n \rightharpoonup f$ in $L^p(E) 1 \leq p < \infty \Leftrightarrow \int g \cdot f_n \to \int g \cdot f$ for all $g \in L^q$

Again: any help finding a function $g$ for which above corollary will fail or is there a simple sequence of functions which respect my problem condition?

  • $\begingroup$ At the top, it is only required that $\int_0^x f_n \to \int_0^x f$ for $x \in (0,1)$, not $x\in [0,1]$. That makes things much easier. $\endgroup$ – Daniel Fischer Jan 18 '16 at 14:47
  • $\begingroup$ I think you are right, i will make the change $\endgroup$ – Lucas Jan 18 '16 at 14:52

or is there a simple sequence of functions which respect my problem condition?

There is a simple such sequence. Since we only look at $x < 1$ in the condition $\int_0^x f_n \to \int_0^x f$, if we let the support of $f_n$ shrink towards $1$, we will have $\int_0^x f_n = 0$ for all large enough $n$. So let's look for example at

$$g_n(x) = \begin{cases} 0 &, x \leqslant \frac{n-1}{n} \\ n &, \frac{n-1}{n} < x.\end{cases}$$

Clearly $\lVert g_n\rVert_{L^1} = 1$ for all $n$, so the sequence is bounded. And it's not hard to find $h \in L^{\infty}([0,1])$ with $\int_0^1 g_n\cdot h = 1$ for all $n$.

  • $\begingroup$ much simple function, then $ h=\frac{1}{n}$ would work here correct? $\endgroup$ – Lucas Jan 18 '16 at 15:16
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    $\begingroup$ $h$ can't depend on $n$. $h\equiv 1$ does it. $\endgroup$ – Daniel Fischer Jan 18 '16 at 15:18

What about $$ f_n(x) = n\chi_{[1-\frac1n,1]}(x) $$ ($\chi_E$ being the indicator function of the set $E$)? $f_n\to 0$ almost everywhere, therefore if the weak limit exists it should equal $0$. However, you have $\int_0^1f_n = 1$ for all $n$ (and therefore $f_n\not\rightharpoonup 0$, but for all $x\in(0,1)$, for $n$ large enough $\int_0^xf_n = 0$.

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    $\begingroup$ I guess this is exactly the same answer by @DanielFisher, who won on time though.. :( $\endgroup$ – AndreasT Jan 18 '16 at 15:10

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