Question

$u_n$ converges weakly to $u$ $\Leftrightarrow$ $\{u_n\}$ is bounded and $\langle x^, u_n \rangle \to \langle x^, u \rangle$

score 134 · Answer 1 · 2026-04-14 16:34:32

134

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$u_n$ converges weakly to $u$ $\Leftrightarrow$ $\{u_n\}$ is bounded and $\langle x^, u_n \rangle \to \langle x^, u \rangle$

Published on 14 Apr 2026 - 16:34

score 48 · Answer 2 · 2026-04-15 18:28:34

48

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$X$ Banach, $u_n \to u$ and $x^_n \xrightarrow{w^} x^$ implie that $\langle x^_n, u_n\rangle \to \langle x^*,u\rangle$.

Published on 15 Apr 2026 - 18:28

score 71 · Answer 3 · 2026-05-07 08:25:29

71

Views

Is the set $\overline{\text{conv}}^{w^} C$ weakly compact?

Published on 07 May 2026 - 8:25

score 298 · Answer 4 · 2026-04-12 13:11:56

298

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Proof of the Alaoglu Theorem

Published on 12 Apr 2026 - 13:11

score 144 · Answer 5 · 2026-04-14 00:46:23

144

Views

reference request: $K$ is relatively weakly compact in $C(X)$ iff $K$ is relatively pointwise compact in $C(X)$

Published on 14 Apr 2026 - 0:46

score 81 · Answer 6 · 2026-04-13 17:42:57

81

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convergence for the weak-* topology

Published on 13 Apr 2026 - 17:42

score 60 · Answer 7 · 2026-04-15 23:28:08

60

Views

$B \subseteq X$ is bounded $\Leftrightarrow$ $\forall U \in \mathcal{U}(0) ~ \exists \lambda_U: ~B \subseteq \lambda_u U $

Published on 15 Apr 2026 - 23:28

score 83 · Answer 8 · 2026-04-15 22:21:55

83

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$A \in \mathcal{L}(X,Y) \implies A^* \in \mathcal{L}(Y^_{w^}, X^_{w^})$

Published on 15 Apr 2026 - 22:21

score 423 · Answer 9 · 2026-04-15 10:56:39

423

Views

Find a counterexample of weak convergence and weak* convergence

Published on 15 Apr 2026 - 10:56

score 101 · Answer 10 · 2026-04-13 21:39:44

101

Views

Characterization of weak compactness in a Banach space

Published on 13 Apr 2026 - 21:39

score 90 · Answer 11 · 2026-04-15 07:10:51

90

Views

Weak convergence - restriction

Published on 15 Apr 2026 - 7:10

score 72 · Answer 12 · 2026-04-11 08:58:55

72

Views

The weak dual coincides with the strong dual

Published on 11 Apr 2026 - 8:58

score 76 · Answer 13 · 2026-04-15 07:17:40

76

Views

Alaoglu theorem over the $p$-adics

Published on 15 Apr 2026 - 7:17

score 184 · Answer 14 · 2026-04-14 14:22:13

184

Views

A sequence of functionals without a weak-* convergent subsequence

Published on 14 Apr 2026 - 14:22

score 267 · Answer 15 · 2026-04-17 05:15:19

267

Views

Multiplicative Linear Functionals on a Banach Algebra

Published on 17 Apr 2026 - 5:15

List Question

$u_n$ converges weakly to $u$ $\Leftrightarrow$ $\{u_n\}$ is bounded and $\langle x^, u_n \rangle \to \langle x^, u \rangle$

$X$ Banach, $u_n \to u$ and $x^_n \xrightarrow{w^} x^$ implie that $\langle x^_n, u_n\rangle \to \langle x^*,u\rangle$.

Is the set $\overline{\text{conv}}^{w^} C$ weakly compact?

Proof of the Alaoglu Theorem

reference request: $K$ is relatively weakly compact in $C(X)$ iff $K$ is relatively pointwise compact in $C(X)$

convergence for the weak-* topology

$B \subseteq X$ is bounded $\Leftrightarrow$ $\forall U \in \mathcal{U}(0) ~ \exists \lambda_U: ~B \subseteq \lambda_u U $

$A \in \mathcal{L}(X,Y) \implies A^* \in \mathcal{L}(Y^_{w^}, X^_{w^})$

Find a counterexample of weak convergence and weak* convergence

Characterization of weak compactness in a Banach space

Weak convergence - restriction

The weak dual coincides with the strong dual

Alaoglu theorem over the $p$-adics

A sequence of functionals without a weak-* convergent subsequence

Multiplicative Linear Functionals on a Banach Algebra

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