The following Theorem was described in a text I was reading as a compactness result. The proof is probably too advanced for me but I was just wondering with respect to what topology we have compactness or relatively compactness here. The first one that came to my mind is the product topology $X^{(0,T)}$ where $X$ is endowed with the weak convergence topology. However in this topology, convergence of the $f_{j_k}$ would mean $$\langle \psi,f_{j_k}(t)-f(t)\rangle\to 0\text{ when }k\to\infty\text{ for all }\psi\in X'\text{(dual of $X$) and all }t\in[0,T]$$ so it seems to be another topology that we consider.
Following is the theorem I mention...
Theorem : Let $X$ be a reflexive and separable Banach space and $T\in\mathbb{R}_{\ge 0}$. Let $(f_j)_{ j\in\mathbb{N}}$ be a sequence of functions in $L^q((0,T),X)$ with $1 \le q < +\infty$. Let us also assume that there exists a map $m ∈ L_q (0,T)$ such that $\|f_j(t)\|_X\le m(t)$ for almost all $t ∈ [0,T]$. Then there exist a subsequence $(f_{j_k} )_{k\in\mathbb{N}}$ and a function $f\in L^q((0, T ), X)$ such that $$\lim\limits_{k\to\infty}\int_0^T\langle\phi(t),f_{j_k}(t,\cdot)-f(t,\cdot)\rangle dt=0$$ for all $\phi ∈ L^{q'} ((0, T ), X' )$, with $q'$ the conjugate exponent of $q$, and $$\int_{t_1}^{t_2}f_{j_k}(t)dt\stackrel{}{\rightharpoonup}\int_{t_1}^{t_2}f(t)dt,\quad\text{ when }k\to\infty\text{ for all } t_1,t_2\in[0,T]$$ where "$\stackrel{}{\rightharpoonup}$" denotes the weak convergence in $X$ and the integrals are in the sense of Bochner.
