Let $1\leq p < \infty$ and consider a sequence $(x_{n})_{n}\subseteq L^{p}[0,1]$. Show the equivalence of:
$1.$ $x_{n} \xrightarrow{ w} 0$
$2.$ $\sup\limits_{n \in \mathbb N} \vert \vert x_{n}\vert\vert_{p}<\infty $ and $\int_{A}x_{n}(t)dt\xrightarrow{n \to \infty} 0$ for any borel sets $A$ on $[0,1]$.
for $1. \Rightarrow 2.$
note that for any $\ell \in (L^{p})^{*}$ we have $\ell(x_{n})\xrightarrow{n \to \infty} 0$ and hence:
$\sup\limits_{n \in \mathbb N}\vert \ell(x_{n})\vert<\infty$ for any $\ell \in (L^{p})^{*}$ but now for $1<p < \infty$ we know that $L^{p}$ is reflexive, so we can write: $\sup\limits_{n \in \mathbb N} \vert \vert x_{n}\vert\vert_{p}=\sup\limits_{n \in \mathbb N} \vert \vert Jx_{n}\vert\vert_{*}<\infty$ by the uniform boundedness principle and the fact that $\sup\limits_{n \in \mathbb N} \vert \ell (x_{n})\vert= \sup\limits_{n \in \mathbb N}\vert Jx_{n}(\ell)\vert$ by reflexiveness. But we have now only shown this for $1< p < \infty$, since $L^{1}$ is not reflexive. How do we show this when $p=1$?
For "$\int_{A}x_{n}(t)dt\xrightarrow{n \to \infty} 0$ for any borel sets $A$ on $[0,1]$", consider any $1 \in L^{q}[0,1]$, note that because of Riesz there has to exist a $\ell \in (L^{p})^{*}$ so that $\ell(x)=\int 1 x(t)dt $ for any $x\in L^{q}$. Then it is clear that $\ell(x_{n})=\int_{0}^{1}x_{n}(t)dt\xrightarrow{n \to \infty}0$ by assumption. But how can I show that $\int_{A}x_{n}(t)dt\xrightarrow{n \to \infty}0$ for any Borel set $A$? Particularly since I have no assumption on whether the $(x_{n})_{n}$ are positive functions?
for $2. \Rightarrow 1.$
note that for any $\ell \in (L^{p})^{*}$ we can find a unique $y \in L^{q}$ where $\frac{1}{p}+\frac{1}{q}=1$ so that:
Note that $\vert \ell(x_{n})\vert =\vert\int_{0}^{1}y(t)x_{n}(t)dt\vert\leq \sup\limits_{n \in \mathbb N} \vert \vert x_{n}\vert\vert_{p}\vert\vert y\vert\vert_{q}<\infty$ and can I state from $\int_{A}x_{n}(t)dt\xrightarrow{n \to \infty} 0$ for any borel sets $A$ that $x_{n} \to 0$ almost everywhere, but I am not sure as the assumption of positivity of $x_{n}$ is once again missing. If I could use this and dominated convergence, then
$\lim\limits_{n\to \infty}\ell(x_{n})=\lim\limits_{n\to \infty}\int_{0}^{1}y(t)x_{n}(t)dt=0$ and we are done.