The weak limit of a sequence of truncated functions

Question

I got stuck on a problem about weakly convergent sequences of functions in $L^{1}([0,1])$. Let $\{ u_{n} \}_{n \in \mathbb{N}}$ be a sequence of functions in $L^{1}([0,1])$ such that $u_{n} \overset{L^1}{\rightharpoonup} u$ for some $u$ in $L^1([0,1])$. Let $K \in \mathbb{N}$ and $\{ u_{n}^{k} \}_{n \in \mathbb{N}}$ be defined by
$$ u_{n}^{k}(x):=\begin{cases}
K && u_{n}(x)>K \\
u_{n}(x) && \lvert u_{n}(x) \rvert \leq K \\
-K && u_{n}(x)<-K.
\end{cases}
$$
Conclude that $u_{n}^{k} \overset{L^1}{\rightharpoonup} v$ for some $v$ in $L^{1}([0,1])$ and moreover $\lvert v(x) \rvert \leq \lvert u^{k}(x) \rvert $ for almost every $x$ in $[0,1].$
This is my attempted solution:
To prove that the weak limit $v$ of the sequence $\{ u_{n}^{k} \}_{n\in \mathbb{N}}$ exists I should use the equintegrability of the sequence $\{ u_{n}\}_{n \in \mathbb{N}}$, but I miss a passage. The equintegrability of the sequence $\{ u_{n} \}_{n \in \mathbb{N}}$ easily implies the equintegrability of $\{ u_{n}^{k} \}_{n \in \mathbb{N}},$ so every subsequence of $\{ u_{n}^{k} \}_{n}$ admits another convergent subsequence. If I could prove that all the convergent subsequences of $\{ u_{n}^{k} \}_{n \in \mathbb{N}}$ have a unique limit then the whole sequence $\{ u_{n}^{k} \}$ would have to converge to that limit. Finally, about the bound on the weak limit $v$ I have been able to prove:
$$ \lvert v(x) \rvert \leq K \quad \text{for a.e} \, x \in [0,1].$$
I proceeded by absurd, if the set $A:=\{ \lvert v \rvert >K \}$ had positive measure then I could choose as a test function the characteristic function of $A,$ then I would obtain
\begin{gather*}
\int_{[0,1]} \mathbb{1}_{A}(x) \cdot u_{n}^{k} \to \int_{[0,1]} \mathbb{1}_{A}(x) \cdot v> \lvert \{v>K\} \rvert \cdot K.
\end{gather*}
So there exist $\nu \in \mathbb{N}$ such that
$$\int_{[0,1]} \mathbb{1}_{A}(x) \cdot u_{n}^{k}> K \cdot \lvert \{ \lvert v \rvert>K \} \rvert \quad \text{for all } n \geq \nu ,$$ but this is absurd because $\lvert u_{n}^{k}(x)\rvert \leq K,$ so I would have
$$ \lvert \{v>K\} \rvert \cdot K <\int_{[0,1]} \mathbb{1}_{A}(x) \cdot u_{n}^{k} \leq \lvert \{v>K\} \rvert \cdot K. $$
What is left to conclude is to prove the inequality $$\lvert v \lvert \leq \lvert u \lvert,$$ does anyone have any idea? I hope the explanation has been clear enough.

Answer 1

The last claim is not true. Define $$ w(x):= \begin{cases} 1 & \text{ if } x\in (0,\frac34)\\ -3& \text{ if } x\in [\frac34,1] \end{cases} $$ and extend it periodically to $\mathbb R$. Then define $u_n(x):= w(nx)$. This is an oscillating sequence that converges weakly to the integral mean of $w$, which is $u=0$.

Now set $K=1$. Then $u_n^K$ is also an oscillating sequence and converges weakly to the integral mean of the truncated $w$, which is $v=\frac12$.