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 $f\in B$.

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 $B_j$?

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 $B$?

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.


2 Answers 2


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 $x$.

  • $\begingroup$ WOW, so simple, so smart. Thank you very much! $\endgroup$
    – Liren Lin
    Jun 28, 2013 at 16:10

The answer is not so simple in the case of $H^1$ over a bounded smooth domain.In other words,considering its continuous imbedding in any of the Lebesgue spaces $L^q$ for $q\in[1,p*]$, $p*=np/(n-p)$ for $p\in[1,n)$, it is possible to define an incomplete metric on $H^1$ and what is more, Cauchyness if ever seen for $q<p*$ is transferred to $p*$ as well by Cauchy continuity. For an understanding, you may refer to the well known finite measure theoretic result of A.Villani, together with an obvious 'uniform gauge condition' on any fixed $H^1$ bounded sequence, noting $1-n/p + n/p* = 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.