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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.

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1 Answer 1

up vote 4 down vote accepted

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 $x$.

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WOW, so simple, so smart. Thank you very much! –  Liren Lin Jun 28 '13 at 16:10

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