Let $(E,\mathcal{A},\mu)$ be a finite measure space and $(X,\|.\|)$ be a reflexive Banach space. The en set of all Bochner-integrable function from $E$ to $X$ is denoted by $\mathcal{L}_{X}^{1}$.
$(i)$ $F \subset \mathcal {L}^0_X $ is a closed in measure if $ F $ is empty or any sequence $ \{f_n \} $ of elements of $ F $ which converges in measure to an element $ f \in \mathcal{L}^0_X $ has its limit $ f $ in $ F $.
$(ii)$ $ \phi: \mathcal{L}^0_X \to \overline {\mathbb{R}} $ is said to be lower semi-continuous in measure if: $$ \forall a \in \mathbb {R} ~: ~ \{f \in \mathcal {L}^0_X ~: ~ \phi (f) \leq a \} \text {is a closed in measure}. $$
Theorem:
Suppose that $I:\mathcal{L}_{X}^{1}\to (-\infty,+\infty]$ is a convex functional which is lower semi-continuous in measure and that $B\in \mathcal{L}_{X}^{1}$ is convex, closed in measure and uniformly bounded in $\mathcal{L}^{1}$-norm.
Show that $I$ attains its minimum on $B$.
Hint. Apply the following theorem :
Theorem:
Suppose that $ (f_n)_{n\geq 1} \subset \mathcal {L}_{X}^1$ is a seqsequence with : $$\sup_n \int_{E}{\|f_n\| d\mu} < \infty .$$Then there exist $ h _{\infty} \in \mathcal {L}_{X}^1 $ and a subsequence $ (g_k)_k $ of $(f_n)_n $ such that for every subsequence $ (h_m)_m $ of $(g_k)_k$ : $$ \frac{1}{n}\sum_{j=1}^{n}{h_j}\underset{n}{\to} h _{\infty} ~~~~~~\text{weakly in}~X~~\text{ a.e.}$$
My effort:
We set $ {\displaystyle m: = \inf_ {f \in B} I (f)} $. Suppose that $ m <+ \infty $, according to the definition of the lower bound, we have: $$ \forall n \in \mathbb{N} ~,~ \exists f_n \in B \text{ such as: } I (f_n) <m + \frac{1}{n} $$ as $ B $ is uniformly bounded in $ \mathcal{L}_{X}^1$, in particular, we have: $$ \sup_n \int_{E} {\| f_n \| d \mu} <\infty. $$ According to the previous theorem, there exists $ g_\infty \in \mathcal{L}_{X}^1 $ and a subsequence $ \{g_n\} $ of $ \{f_n \} $ such that: $$ \frac{1}{k} \sum_{i = 1}^{k} {g_{i}} \overset {\sigma (X, X^{*})} {\longrightarrow} g_{\infty} \qquad a.e $$ therefore $ {\displaystyle \frac{1}{k} \sum_{i = 1}^{k} {g_ {i}}} $ converges in measure to $ g_{\infty} $. Since $ B $ is convex therefore: $$ \frac{1}{k} \sum_{i = 1}^{k} {g_ {i}} \in B $$ $ B $ is closed in measure therefore: $$ g_{\infty} \in B $$ as $ I $ is semi-continuous inferior in measure and convex, therefore: $$ I (g_{\infty}) \leq \liminf_{n} I (\frac{1}{k} \sum_{i=1}^{k}{g_{i}}) \leq \liminf_{n} \frac{1}{k} \sum_{i=1}^{k}{I (g_{i})} \qquad \text{ en mesure } $$ so : $$ I (g_{\infty}) \leq \liminf_{k} \frac{1}{k} \sum_ {i = 1}^{k} {\big (m + \frac{1}{i} \big) } = m + \liminf_{k} \frac {\sum_{i = 1}^{k}{\frac{1}{i}}}{k} \leq m + \liminf_{k} \frac {1+ \log(k)}{k} $$ like $ {\displaystyle \liminf_{k} \frac {1+ \log (k)}{k} = 0} $, so $ I (g _ {\infty}) \leq m $. from where : $$ I (g_{\infty}) = \inf_{f \in B} I (f). $$
My problem: Do I have the right to say that:$ \frac{1}{k} \sum_{i = 1}^{k} {g_{i}} \overset {\sigma (X, X^{*})} {\longrightarrow} g_{\infty}$ a.e, imply that $\frac{1}{k} \sum_{i = 1}^{k} {g_{i}} $ converges in measure to $ g_{\infty} $? If the answer is no, please have an idea.