Weak convergence in intersection of Bochner spaces First off, my question has some similarities to Weak convergence in the intersection of Lebesgue spaces or Sobolev spaces (on MathOverflow).
To be specific, we have a Gelfand triple $V \subset H \subset V^*$ and a complete probability space $([0,T]\times\Omega,\mathcal{B}([0,T])\otimes\mathcal{F},\text{Lebesgue}\otimes\mathbb{P})$. Then, we have the Bochner spaces $L^{\alpha}([0,T]\times\Omega;V)$ (where $\alpha>1$) and $L^2([0,T]\times\Omega;H)$. Let $p:=\min\{\alpha,2\}>1$. Then, both of these spaces can be continuously and linearly embedded into the reflexive space $L^p([0,T]\times\Omega;H)$, and so, $K:=L^{\alpha}([0,T]\times\Omega;V)\cap L^2([0,T]\times\Omega;H)$ is a Banach space, when endow with the norm 
$$
\|f\|_{K}:=\max\left\{\|f\|_{L^{\alpha}([0,T]\times\Omega;V)},\|f\|_{L^{2}([0,T]\times\Omega;Hil)}\right\}
$$
for all $f\in K$. I would like to verify that the following statement holds.

We consider a bounded sequence $(f_n)_{n\in\mathbb{N}} \subset K$ and we want to finish by concluding that, by possibly taking some subsequence, we have 
  $f_n \overset{\text{wk}}{\to}f$ in $L^{\alpha}([0,T]\times\Omega;V)$ and $f_n \overset{\text{wk}}{\to}f$ in $L^2([0,T]\times\Omega;H)$ for some $f\in K$. To emphasize: the $f$ is the same in both cases.

At first, I showed that weak convergence in $K$ implies weak convergence to both respective spaces and thought I was done, since $(f_n)$ is a bounded sequence in $K$. Later, I realized that it might be possible that $K$ is not reflexive, so the boundedness of the sequence in $K$ does not directly imply weak convergence.
Then, I recalled that closed subspaces of reflexive spaces are reflexive as well, but $K$ is endowed with a different norm, so that statement does not apply.
I searched online for some results, and I found a paper by Liu and Wang, Sums and Intersection of Lebesgue Spaces, in which it is showed that the intersection of Lebesgue-spaces $L^{p}$ and $L^{q}$ (where $p,q>1$) is indeed reflexive (Corollary 2 on page 245), so I am inclined to believe such a statement can hold in this setting as well.
However, the reflexivity is not necessary, I just want to verify that such a result holds. I believe it might be possible with interpolation theory, but I am not really familiar in that branch of mathematics.

[ADDED] I have yet another unfruitful attempt to show this result. I tried combining Reflexivity of intersection of two $L^p$ space with Uniform convexity of equivalent intersection norm, which would have given a positive answer, if the Banach space $V$ would have been uniformly convex, which is not assumed. Namely, if that were the case, Theorem 2 in M. M. Day's Some more uniformly convex spaces would yield that $L^{\alpha}([0,T]\times\Omega;V)$ is indeed uniformly convex and we could apply Uniform convexity of equivalent intersection norm and rest our heads.
 A: The solution is based on this comment in the linked question Reflexivity of intersection of two $L^p$ space.

Define $p:=\min\{\alpha,2\}$. We can continuously and linearly embed the Banach spaces $L^{\alpha}([0,T]\times\Omega;V)$ and $L^{2}([0,T]\times\Omega;H)$ in the Banach space $L^{p}([0,T]\times\Omega;H)$. Hence, the intersection $K := L^{\alpha}([0,T]\times\Omega;V)\cap L^{2}([0,T]\times\Omega;H)$ is a Banach space as well, when endowed with the norm
$$
\left\|\Xi\right\|_{K} := \max\left\{\left\|\Xi\right\|_{L^{\alpha}([0,T]\times\Omega;V)}, \left\|\Xi\right\|_{L^{2}([0,T]\times\Omega;H)}\right\}
$$
for any $\Xi \in K$. 
Moreover, the Banach spaces $L^{\alpha}([0,T]\times\Omega;V)$ and $L^{2}([0,T]\times\Omega;H)$ are reflexive. This means that the product space $\mathcal{P} := L^{\alpha}([0,T]\times\Omega;V)\times L^{2}([0,T]\times\Omega;H)$, endowed with the norm 
$$
\|(\phi,\psi)\|_{\mathcal{P}} := \max\left\{\|\phi\|_{L^{\alpha}([0,T]\times\Omega;V)},\|\psi\|_{L^{2}([0,T]\times\Omega;H)}\right\}
$$
is a reflexive Banach space as well. If we look at the "diagonal" of this product space $\mathcal{D} := \{(\phi,\psi)\in \mathcal{P} : \phi = \psi\}$, it immediately follows that this linear subspace of the product space is closed. As a consequence, we see that $\left(\mathcal{D}, \|\cdot\|_{\mathcal{P}}\right)$ is a reflexive Banach space.
Consider the bounded linear operator $\kappa: K \to\mathcal{P} : \Xi \mapsto (\Xi,\Xi)$. Then, we see that the image of $\kappa$ is equal to the diagonal $\mathcal{D}$ of the product space $\mathcal{P}$ and moreover, we find that $K$ and $\mathcal{D}$ are isometrically isomorphic. This means that $K$ is reflexive.
In particular, this means that the bounded sequence $\{f_n:n\geq1\}\subset K$ has a weakly convergent subsequence that we will again denote by $\{f_n:n\geq1\}$. Then, as "weak convergence in K implies weak convergence to both respective spaces" (by restricting the functionals, which are again functionals , but now on $K$), we are done.
