Show that $I$ attains its minimum on $B$.

Question

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 $\mathcal{L}_{X}^{1}$

Suppose that $I:\mathcal{L}_{X}^{1}\to (-\infty,+\infty]$ is a quasi-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 :

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 $(g_k)_k$ : $$ \frac{1}{n}\sum_{j=1}^{n}{h_j}\underset{n}{\to} h _{\infty} ~~~~~~\text{weakly in}~X~~\text{ a.e.}$$

An idea please

Answer 1

Let $m:= \inf_{f\in B}I(f)$. We know that by definition of the infimum, for all positive integer $n$, there exists a function $f_n\in B$ such that $I(f_n)\geqslant m+n^{-1}$. The assumptions on $B$ show that $\sup_n \int_{E}{\|f_n\| d\mu} < \infty .$ Then it remains to check that the function $h_\infty$ given by the hint does the job, that is, that $h_\infty$ is in $B$ and $I(h_\infty)=m$.