$ \|u_k-v_k\|_2\leq \min \bigg(\inf_{n\geq k}{\|f_n1_{\{|f_n|\leq k\}}\|_2},\frac{\epsilon_{k-1}}{4k}\bigg) $

Question

Let $(E,\mathcal{A},\mu)$ be a finite measure space and $\{f_n\}_n$ be a sequence of simple functions such that: \begin{align*} f_n1_{\{\lvert f_n\rvert\leq k\}}\overset{\sigma(L^2,L^2)}{\underset{n}{\longrightarrow}} u_k, &\qquad\forall k\geq 1 \\ \|u_k\|_2\leq 2\|f_n1_{\{\lvert f_n\rvert\leq k\}}\|_2,&\qquad \forall n\geq k. \end{align*} Put $\epsilon_k=\frac{1}{2^k}$ $(k\geq 1)$. Why does there exist, for each $k\geq 1$, a simple function $v_k$ such that: $$ \|u_k-v_k\|_2\leq \min \bigg(\inf_{n\geq k}{\|f_n1_{\{\lvert f_n\rvert\leq k\}}\|_2},\frac{\epsilon_{k-1}}{4k}\bigg) $$

Answer 1

$\newcommand\de{\delta}$ $\newcommand\ep{\epsilon}$ $\newcommand\al{\alpha}$
Fix any natural $k$. Let $$\de_k:=\inf_{n\ge k}\|f_n 1_{\{|f_n|\le k\}}\|_2$$ and $$\eta_k:=\ep_{k-1}/(4k).$$ We have $$\|u_k\|_2\le2\de_k \tag{1}$$ and $$\eta_k>0.$$ We want to show that there exists a simple function $v_k$ such that $$ \|u_k-v_k\|_2\le\al_k:=\min(\de_k,\eta_k). \tag{2} $$ If $\de_k=0$, then, by (1), $\|u_k\|_2=0$, and hence we may let $v_k:=0$, to have (2).

Otherwise, $\al_k>0$. By the condition $f_n 1_{\{|f_n|\le k\}} \overset{\sigma(L^2,L^2)}{\underset n\longrightarrow} u_k$, we have $|u_k|\le k$ $\mu$-almost everywhere ($\mu$-a.e.). So, we can find a simple function $v_k$ such that $|u_k-v_k|\le\al_k/\sqrt{1+\mu(E)}$ $\mu$-a.e. Then we will have the inequality in (2), as desired.