8 июн. 2014 г. · Theorem: Every weakly convergent sequence in X is bounded. Let {xn} be a weakly convergent sequence in X. Let Tn∈X∗∗ be defined by Tn(ℓ)=ℓ( ...

14 авг. 2020 г. · In this MSE post, the author stated that as a consequence of uniform boundedness principle, a weak-star convergent sequence is norm bounded.

20 июл. 2019 г. · Let X be normed vector space. If (xn)n is a weakly convergent sequence in X it follows that (xn)n is bounded.

6 мар. 2015 г. · To show the weak convergence of the bounded sequence (xn) assume first that H is separable and let {x′1,x′2,…} be a dense set in the dual space.

1 февр. 2022 г. · A simpler proof that bounded sequence has a convergent subsequence in weak topology ... I'm proving Theorem 3.18 in Brezis's book of Functional ...

30 апр. 2021 г. · We can prove that in Hilbert space, bounded space has a weakly convergent subsequence, as we can see here or there. Does the converse hold?

21 июл. 2015 г. · Generally, without completeness, you can't deduce that a weak∗ convergent sequence is bounded. Let X=c00 be the space of sequences with only ...

12 июн. 2018 г. · No, a weakly convergent sequence on a normed (even non-Banach such as c00) space is always bounded. This follows from the uniform ...

17 сент. 2012 г. · To show that a weakly convergent sequence is norm-bounded, usually the uniform boundedness principle is applied.

3 июн. 2022 г. · First let's recall some definitions. The sequence (xn)⊂ℓp(N) weakly converges to x∈ℓp(N) iff ϕ(xn)→ϕ(x) for all functional ϕ∈ℓp(N)∗.