Let $(E,\mathcal{A},\mu)$ be a finite measure space. Let $ X $ be a Banach space and $ H $ a Hilbert space.
For $ t \in E $, we set $ F_a (f)(t) = f (t) 1_{\| f \| \leq a} (t)$
Lemma: Let $ \{x_n \} $ be a sequence of $ X $ weakly converging to $ x_\infty \in X $. Then there is an integer $ N \geq 1 $ such that: $$\|x_\infty\|\leq 2\inf_{n\geq N}\|x_n\|$$
Let $\{f_n\}$ be a bounded sequenece in $\mathcal{L}^1_H(:=\{f:E\to H: f \text{ Bochner-integrable function}\})$
My problem is to show that: there exists a subsequence of $\{f_n^{'}\}$ of $\{f_n\}$ for all $ k\geq 1 $ it exists $ u_k \in \mathcal {L}_H^2 $ such that: $$ F_k(f_n^{'}) \overset {\sigma (\mathcal{L}_H^2, \mathcal{L}_H ^ 2)} {\underset {n} {\longrightarrow}} u_k \qquad \forall k \geq 1 \qquad \text{and} \qquad \| u_k \|_2 \leq 2 \| F_k (f_n^{'}) \|_2, \qquad \forall n \geq k. $$
My effort
Let $ k \geq 1 $, the sequence $ \{F_k (f_n) \} $ is bounded in $ \mathcal {L}_H ^2 $, and therefore it is relatively weakly compact. There then exists (for each $ k $) a subsequence $ \{f_n^k\} $ of $ \{f_n \} $ and $ u_k \in \mathcal{L}_H^2 $ so that $ f_n^{k + 1} $ is a subsequence of $ \{f_n ^ k \} $ and $$ F_k (f_n ^ k) \overset {\sigma (\mathcal{L}_H^2, \mathcal{L}_H^2)} {\underset{n}{\longrightarrow}} u_k, $$ and according to the previous lemma we have: $$ \| u_k \|_2 \leq 2 \| F_k (f_n ^ k) \|_2, \qquad \forall n \geq 1, $$ because $ \{F_k (f_n ^ k) \} $ is a weakly convergent sequence in the Banach space $ (L^2_H, \|. \| _2) $.
We set $ f_n^{'} = f_n^n $ $ (n \geq 1) $. So $$ F_k (f_n^{'}) \overset {\sigma (\mathcal {L}_H^2, \mathcal {L}_H ^ 2)} {\underset {n} {\longrightarrow}} u_k \qquad \forall k \geq 1 \qquad \text {and} \qquad \| u_k \|_2 \leq 2 \| F_k (f_n ^ {'}) \|_2, \qquad \forall n \geq k. $$
Questions:
I can say that : because $ \{F_k (f_n ^ k) \} $ is a weakly convergent sequence in the Banach space $ (L^2_H, \|. \| _2) $.
I can say that : according to the previous lemma we have: $$ \| u_k \|_2\leq 2 \| F_k (f_n ^ k) \|_2, \qquad \forall n \geq 1.$$
An idea please.