Let $B:=B\_1\cap B\_2\cap...\cap B\_n$, where each $B\_j$ is a *reflexive* Lebesgue space or Sobolev space (such as $L^4$, $H^1$, etc.) on a domain in $\mathbb{R}^d$. Then $B$ is a Banach space endowed with the norm 
$$\|\cdot\| = \|\cdot\|\_{B\_1}+...+\|\cdot\|\_{B\_n}.$$ 
Let $\{f\_n\}$ be a sequence in $B$ and $f\in B$. 



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> Is the assertion that $f\_n$ weakly converges to $f$ in $B$ equivalent to the assertion that $f\_n$ weakly converges to $f$ in each $B\_j$? 
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Note that it's easy to show that the weak convergence in 
$B$ implies that in each $B\_j$. It's the converse that is not obvious. Does anyone know the answer or some reference for it?


**Remark.** My question is closely related to the following: 



> 
> Is the reflexivity of every $B\_j$ implies the reflexivity of $B$?
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To see this, assume $f\_n$ converges weakly to $f$ in each $B\_j$. Then $f\_n$ is a bounded sequence in each $B\_j$, and hence is bounded in $B$. If we can prove that $B$ is reflexive, then $f\_n$ has weakly convergent subsequence in $B$. It's easy to show that every weak convergent subsequence of $f\_n$ in $B$ must have weak limit $f$, and hence the sequence $f\_n$ itself converges weakly to $f$. Conversely, if for every sequence $f\_n$ in $B$, $f\_n$ converges weakly in each $B\_j$ implies that $f\_n$ converges weakly in $B$, then $B$ must be reflexive. I omit the proof of this assertion.



Weak convergence in the intersection of Lebesgue spaces or Sobolev spaces
The answer to both questions is "yes". One way of seeing this is to observe that $B$ embeds isometrically into $B\_1 \oplus\_1 B\_2 \dots \oplus\_1 B\_n$ via $b \mapsto (b,b,\dots, b)$. Another way is to use the (obvious) fact that a sequence in a Banach space converges weakly to $x$ if and only if for every subsequence $y\_k$ there are convex combinations of $y\_k$ that converge in norm to $x$.


