Tag: measure

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 $Bin mathcal{L}_{X}^{1}$ 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)_{ngeq 1} subset mathcal {L}_{X}^1$ is a seqsequence with : $$sup_n int_{E}{|f_n| dmu} < infty .$$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.