I would like to know the proof of the following theorem:



> 
> Consider two Banach spaces $X\hookrightarrow Y$ and $1<p,q\leq\infty$. Let $(f\_n)\_{n\geq 0}$ be a bounded sequence in $L^q(I,Y)$ and let $f:I\mapsto Y$ be such that $f\_{n}(t)\rightharpoonup f(t)$ in $Y$ for a.a. $t\in I$. If $f\_n$ is bounded in $L^p(I,X)$ and if $X$ is reflexive, then $f\in L^p(I,X)$ and $\|f\|\_{L^p(I,X)}\leq\liminf\_{n\rightarrow}\|f\_n\|\_{L^p(I,X)}$.
> 
> 
> 


It is stated as Theorem 1.2.5 in Cazenave's *Semilinear Schroedinger Equations* but is given without proof. 



A useful criterion in vector integration
Here is an answer when $p<\infty$ and $I$ bounded. The argument might be adapted to $p=\infty$ by replacing some weak $L^p$ convergences below by the weak-\* $L^{\infty}$ one, and I really believe $I$ bounded is not an issue (otherwise argue locally on any $J\subset I$)


Just a few obvious remarks first: if $X$ is reflexive and $f\_n$ bounded in $L^p(I,X)$ with $p<\infty$ (resp. $p=\infty$) then by the Banach-Alaoglu theorem we can extract a subsequence $f\_{n\_k}\rightharpoonup \tilde{f}$ in this topology (resp. 
$f\_{n\_k}\overset{\*}{\rightharpoonup} \tilde{f}$). Also, $X\hookrightarrow Y$ implies $ Y'\hookrightarrow X'$.


1. **Claim: $f\_n\rightharpoonup f$ in $L^p(I,Y)$**. To see this, fix a test function $\varphi\in L^{p'}(I,Y')$ and let
$$
h\_n(t):=\left<f\_n(t),\varphi(t)\right>\_{Y,Y'}.
$$
By the OP's assumptions we have
$$
h\_n(t)\to h(t)=<f(t),\varphi(t)>\_{Y,Y'}\quad \text{a.e. }t\in I,
$$
but also since $f\_n\in L^{p}(I,X)\Rightarrow f\_n(t)\in X$ a.e. and $Y'\hookrightarrow X'$
$$
h\_n(t)=\left<f\_n(t),\varphi(t)\right>\_{X,X'}\in L^1(I).
$$
My claim would immediately follow from $h\_n\to h$ in $L^1(I)$, which we can get by Vitali's convergence theorem. In order to apply the latter we only need to check: (i) that $|I|<\infty$ (OK since I'm considering bounded $I$), (ii) $h\_n(t)\to h(t)$ for a.e. $t\in I$ (OK by the OP's assumption), (iii) that $|h(t)|<\infty$ for a.e. $t$ (still OK because $f\_n(t)\rightharpoonup f(t)$ in $Y$ and the test function is fixed so $\varphi(t)\in Y'$ for a.e. $t$), and finally (iv) that $\{h\_n\}\_{n}$ is uniformly integrable. The only delicate point is to check (iv), which by the Dunford-Pettis is equivalent to showing that $\{f\_n\}\_{n}$ is relatively compact for the weak $\sigma(L^1,L^{\infty})$ topology. In order to check this relative compactness, recall from my preliminary remark that $f\_{n\_k}\rightharpoonup \tilde{f}$ in $L^{p}(I,X)$. For any fixed $\lambda\in L^{\infty}(I)$ we have $\lambda\varphi\in L^{p'}(I,Y')\hookrightarrow L^{p'}(I,X')$ so along this subsequence
\begin{align\*}
\int\_I h\_{n\_k}(t)\lambda(t) & =\int\_I\left<f\_{n\_k}(t),\varphi(t)\right>\_{X,X'}\lambda(t)\\
& =\int\_I\left<f\_{n\_k}(t),\lambda\varphi(t)\right>\_{X,X'}\\
&\underset{k\to\infty}{\to} \int\_I\left<\tilde{f}(t),\lambda\varphi(t)\right>\_{X,X'}\\
& =\int\_I \tilde{h}(t)\lambda(t)
\end{align\*}
with $\tilde{h}(t)=\left<\tilde{f}(t),\varphi(t)\right>\_{X,X'}$. This means that $\tilde{h}\in L^{1}(I)$ is a cluster point for the $\sigma(L^1,L^{\infty})$ topology. Thus $\{h\_n\}\_{n}$ is weakly $L^1$ relatively compact hence also uniformly integrable, and my claim follows.
2. **Improvement to $L^p(I,X)$.** From step 1 and my preliminary remark we have now
$$
f\_n\rightharpoonup f\text{ in }L^{p}(I,Y),\qquad f\_{n\_k}\rightharpoonup \tilde{f}\text{ in }L^{p}(I,X).
$$
Note that the whole sequence converges for the weak $Y$ norm, but only up to a subsequence for the stronger $X$ norm. Restricting to test-functions $\varphi\in L^{p'}(I,Y')\subset L^{p'}(I,X')$ and by uniqueness of the limit it follows that $f=\tilde{f}$ in $L^{p}(I,Y)$. In particular $f\in L^{p}(I,X)$, and because $\tilde{f}$ was obtained extracting a weakly $L^p(I,X)$ converging subsequence we get
$$
||f||\_{L^p(I,X)}=||\tilde{f}||\_{L^p(I,X)}\leq \liminf\limits\_{n\to\infty} ||f\_n||\_{L^p(I,X)}
$$
as desired and the proof is complete.


Note that I never used the $L^q(I,Y)$ bound, so this assumption can probably be removed. The pointwise convergence in $Y$-weak a.e. $t$ is a quite strong assumption to start with, so this is not surprising.


