Let $B$ be a Banach-space. And let $L^p\left(0,T;B\right)$ be the space of all Lebesgue integrable functions $f:[0,T] \rightarrow B$. I want to show: If $F \subset L^p\left(0,\frac{T}{2};B\right)$ is relatively compact in $L^p\left(0,\frac{T}{2};B\right)$ and if $F$ is relatively compact in $L^p\left(\frac{T}{2},T;B\right)$, then $F$ is relatively compact in $L^p(0,T;B)$.
I tried to prove it by using that in Banach-spaces total boundedness is equivalent to relative compactness. So because $F$ is relatively compact in $L^p\left(0,\frac{T}{2};B\right)$ it holds that:
For every $\varepsilon>0$ one finds a finite subset $A := \{f_i : 1 \leq i \leq J\} \subset F$, such that for every $f \in F$ there exists a $f_i \in A$, such that $\Vert f - f_i \Vert_{L^p\left(0,\frac{T}{2};B\right)} \leq \frac{\varepsilon}{2^{\frac{1}{p}}}$.
Because $F$ is relatively compact in $L^p\left(\frac{T}{2},T;B\right)$ it holds that:
For every $\varepsilon>0$ one finds a finite subset $B := \{g_j : 1 \leq j \leq K\} \subset F$, such that for every $f \in F$ there exists a $g_j \in B$, such that $\Vert f - g_j \Vert_{L^p\left(\frac{T}{2},T;B\right)} \leq \frac{\varepsilon}{2^{\frac{1}{p}}}$.
Now let $\varepsilon >0$ be arbitrary and take the finite subset $E := A \cup B \subset F $ and define $l_k \in E$ as follows: \begin{equation*} l_k(x) :=\begin{cases} f_i(x) & \text{for } x \in [0,\frac{T}{2}) \\ g_j(x) & \text{for } x \in [\frac{T}{2},T] \end{cases} \end{equation*} Then we have that for every $ f\in F$ there exists a $l_k \in E$, such that \begin{align*} \left(\int_0^T \Vert f-l_k \Vert_B^p \ dt \right)^{\frac{1}{p}} &= \left(\int_0^{\frac{T}{2}} \Vert f-f_i \Vert_B^p \ dt+\int_{\frac{T}{2}}^T \Vert f-g_j \Vert_B^p \ dt \right)^{\frac{1}{p}} \\&\leq \left(\frac{\varepsilon^p}{2}+\frac{\varepsilon^p}{2}\right)^{\frac{1}{p}} = \varepsilon. \end{align*}
The problem is that I do not know wether $l_k \in F$. So the proof is not correct. Does anyone know how to save the proof or has a different approach?
An equivalent definition of relative compactness of a subset $F$ of a metric space $(X,d)$ is that for all positive $\varepsilon$, there exists a finite subset $F'$ of $X$ such that $F\subset \bigcup_{x\in F'}B(x,\varepsilon)$.
Indeed, if $F$ satisfies the later definition, for all $x\in F'$, pick $y_x\in B(x,\varepsilon)\cap F$ (we can assume that this set is not empty, otherwise we simply remove this $x\in F'$ without changing veracity of the inclusion $\bigcup_{x\in F'}B(x,\varepsilon)$). Then $F\subset \bigcup_{x\in F'}B\left(y_x,2\varepsilon\right)$: indeed, let $z\in F$. There exists $x\in F$ such that $d(x,z)\lt \varepsilon$. And $d\left(x,y_x\right)\lt \varepsilon$.
Therefore, in the context of your problem, you do not have to worry whether $l_k$ is in $F$ or not.