Find a sequence $f_n,$ bounded in $L^1([0,1]),$ and $f\in L^1,$ such that $$\lim_{n\to \infty}\int_0^x f_n=\int_0^x f$$ for all $x\in [0,1],$ and such that$\{f_n\}$ does not converge weakly to $f$ in $L^1$
Below is my attempt:
let $$f_n(x)=2n\chi_{[0,1/n]}(x)$$ then $f_n \in L^1([0,1])$, for all $x$ with $\int_{[0,1]} f_n =2$
define $f$ to be identically $0$, ie. $f\equiv 0$.
then $f_n \to f$ pointwise a.e
So since $f_n \geq 0, \forall n $ and $f_n$ is increasing, by monotone convergence theorem $$\lim_{n\to \infty}\int_0^x f_n=\int_0^x f=\int_0^x 0 = 0$$
Below is Royden's definition of weak convergence
Let $E$ be measurable set, and $1 \leq p< \infty$ and $q$ a conjugate of $p$, then $\{f_n\}$ is said to converge weakly to $f$ in $L^p(E)$ if and only if $$\lim_{n\to \infty}\int_E g.f_n=\int_E g.f , \forall g \in L^q(E)$$
As per this, I define $g=\chi_{[0,1]}(x)$, then clearly $g\in L^{\infty}([0,1])$, but $$\lim_{n\to \infty}\int_0^x g\cdot f_n=\int_0^xf_n =2$$ and $$\lim_{n\to \infty}\int_0^x g\cdot f=\int_0^x 0=0$$
hence $f_n$ does not converge weakly to $f$ in $L^1([0,1])$
Is this meaningful? if not can someone help me come up with one that works. thank you
We can choose $a_ 2 > a_3 >\cdots \to 0$ such that $(1/n-a_n,1/n+a_n), n=2,3, \dots$ are pairwise disjoint subintervals of $[0,1].$ For each such $n,$ define
$$f_n = \frac{1}{a_n}\chi_{(1/n, 1/n + a_n)} - \frac{1}{a_n}\chi_{(1/n-a_n, 1/n)}.$$
Then $\|f_n\|_1 = 2$ for all $n,$ and $\int_0^x f_n \to 0$ for each $x\in [0,1].$
However $f_n$ does not converge weakly to $0.$ To see this, define
$$g(x) = \sum_{n=2}^{\infty}\chi_{(1/n, 1/n + a_n)} - \chi_{(1/n-a_n, 1/n)}.$$
Then $g\in L^\infty[0,1],$ and $\int_0^1 f_n g =2$ for all $n.$