Question

$x_n \to 0$ weakly $\iff$ $Tx_n \to 0$ for all $T$ in $X^*$

score 66 · Answer 1 · 2026-04-29 12:43:06

66

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$x_n \to 0$ weakly $\iff$ $Tx_n \to 0$ for all $T$ in $X^*$

Published on 29 Apr 2026 - 12:43

score 84 · Answer 2 · 2026-04-29 12:43:00

84

Views

Weak limit in $L^p$

Published on 29 Apr 2026 - 12:43

score 144 · Answer 3 · 2026-05-02 20:09:46

144

Views

How does this sequence of distributions converge?

Published on 02 May 2026 - 20:09

score 68 · Answer 4 · 2026-04-29 12:43:09

68

Views

If $F_n$ converges weakly then $\exists X_n$ that converges almost shurely theorem proof clarification

Published on 29 Apr 2026 - 12:43

score 297 · Answer 5 · 2026-05-02 23:46:39

297

Views

Unit ball of the adjoint space of a separable Banach space is second-countable in the weak* topology.

Published on 02 May 2026 - 23:46

score 1.4k · Answer 6 · 2026-04-29 12:42:48

1.4k

Views

Bounded sequence in probability admits converging subsequence in probability?

Published on 29 Apr 2026 - 12:42

score 1.3k · Answer 7 · 2026-04-29 12:42:44

1.3k

Views

Understanding weak convergence of probability measures

Published on 29 Apr 2026 - 12:42

score 258 · Answer 8 · 2026-04-29 12:43:07

258

Views

Equivalence of asymptotic uniform integrability and convergence of mean

Published on 29 Apr 2026 - 12:43

score 2.2k · Answer 9 · 2026-04-29 12:43:03

2.2k

Views

Weak convergence and convergence of norms imply strong convergence in Hilbert space

Published on 29 Apr 2026 - 12:43

score 251 · Answer 10 · 2026-04-29 12:42:58

251

Views

Is the Space of Borel Probability Measures over R connected?

Published on 29 Apr 2026 - 12:42

score 207 · Answer 11 · 2026-04-29 12:42:45

207

Views

Let $X$ be a normed space. If $X$ is weakly sequentially complete, then $X$ is reflexive.

Published on 29 Apr 2026 - 12:42

score 77 · Answer 12 · 2026-04-29 12:42:45

77

Views

What does $\sigma(X^,(X^,wk^)^)$ stand for?

Published on 29 Apr 2026 - 12:42

score 42 · Answer 13 · 2026-05-07 19:14:27

42

Views

If each subseries of $\sum_{n=1}^\infty x_n$ is weakly convergent, then $\varphi\mapsto(\varphi(x_n))_{n\in\mathbb N}$ is a closed operator

Published on 07 May 2026 - 19:14

score 87 · Answer 14 · 2026-04-29 12:43:12

87

Views

$f_n(x)$ does not converge to $1$ for any $x\in [0,1]$

Published on 29 Apr 2026 - 12:43

score 669 · Answer 15 · 2026-04-29 12:42:55

669

Views

Letting $x_n\to x$ weakly in a normed vector space $X$, prove the following statement.

Published on 29 Apr 2026 - 12:42

List Question

$x_n \to 0$ weakly $\iff$ $Tx_n \to 0$ for all $T$ in $X^*$

Weak limit in $L^p$

How does this sequence of distributions converge?

If $F_n$ converges weakly then $\exists X_n$ that converges almost shurely theorem proof clarification

Unit ball of the adjoint space of a separable Banach space is second-countable in the weak* topology.

Bounded sequence in probability admits converging subsequence in probability?

Understanding weak convergence of probability measures

Equivalence of asymptotic uniform integrability and convergence of mean

Weak convergence and convergence of norms imply strong convergence in Hilbert space

Is the Space of Borel Probability Measures over R connected?

Let $X$ be a normed space. If $X$ is weakly sequentially complete, then $X$ is reflexive.

What does $\sigma(X^,(X^,wk^)^)$ stand for?

If each subseries of $\sum_{n=1}^\infty x_n$ is weakly convergent, then $\varphi\mapsto(\varphi(x_n))_{n\in\mathbb N}$ is a closed operator

$f_n(x)$ does not converge to $1$ for any $x\in [0,1]$

Letting $x_n\to x$ weakly in a normed vector space $X$, prove the following statement.

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