Reflexivity and Separability of $L^p(0,T; X)$ spaces

Question

I am familiar with the separability and reflexivity conditions for $L^p$ spaces. How do they generalize for Bouchner spaces

$$L^p(0,T; X)$$

Do they follow from separability and reflexivity of $X$? That is, if $X$ is reflexive, so is $L^p(0,T; X)$ for $1< p < \infty$; and if $X$ is separable, so is $L^p(0,T; X)$ for $1\leq p < \infty$.

Answer 1

Yes that is true. Suppose that $(\Omega,\Sigma, \mu)$ is a measure space.

$L^p(\Omega;X)$ is reflexive for $1<p<+\infty$ whenever $X$ is reflexive.

Separability, however, is a bit tricker. Consider on $\Sigma $ the semimetric $d_\mu(A,B) = \mu (A \Delta B)$ where $A \Delta B$ is the symmetric difference $A\Delta B = (A \setminus B) \cup (B\setminus A)$. Consider the equivalence relation $\sim$ on $\Sigma$ where $A \sim B $ if $ \mu(A \Delta B)=0$. Then $d_\mu$ is a metric on $\Sigma(\mu):=\Sigma /\sim$.

If the metric space $(\Sigma(\mu),d_\mu)$ is separable and $X$ is separable then $L^p(\Omega;X)$ is separable for $1\le p <+\infty.$

If $\Omega=(0,T)$ then $(\Sigma(\lambda),d_\lambda)$ is separable so $L^p(0,T;X)$ is separable for $1\le p<+\infty$ whenever $X$ is separable. In general, if $\Sigma = \sigma(\mathcal L)$ is generated by a countable family $\mathcal L$ then $(\Sigma(\mu),d_\mu)$ is separable. Moreover, $\Sigma(\mu)$ is separable iff $L^1(\Omega,\Sigma,\mu)$ is separable.