There is a popular result, which proves, that strong convergence in $L^2(0,T;V)$ implies strong convergence almost everywhere in $[0,T]$.
Is there an analoguous result for weak convergence? Thanks, FFoDWindow
Weak convergence in $L^2(0,T;V)$ implies weak convergence a.e.
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integration
functional-analysis
convergence
bochner-spaces
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0The typical example of sequence weakly converging to zero but not strongly is $f_n(x)=\sin(nx)$, which does not converge a.e., even up to subsequences – 2017-01-16
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0Maybe you understood my question wrong. I was asking, if weak convergence in $L^2(0,T;V)$ implies weak convergence a.e. in $V$ ? – 2017-01-16
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0Oh sorry. Now I see that the functions are $V$-valued. In any case, if $V=\mathbb R$ then strong and weak convergence coincide with the usual convergence of real numbers, so does the same example work? – 2017-01-16
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0Yeah, this counterexample holds. Unfortunately. Thank you! – 2017-01-16