Question

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

score 131 · Answer 1 · 2019-03-28 16:19:39

131

Views

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

Published on 28 Mar 2019 - 16:19

score 45 · Answer 2 · 2019-03-30 10:20:50

45

Views

$X$ 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 30 Mar 2019 - 10:20

score 68 · Answer 3 · 2026-03-25 04:55:34

68

Views

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

Published on 25 Mar 2026 - 4:55

score 295 · Answer 4 · 2019-04-07 17:15:54

295

Views

Proof of the Alaoglu Theorem

Published on 07 Apr 2019 - 17:15

score 141 · Answer 5 · 2019-04-11 13:17:55

141

Views

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

Published on 11 Apr 2019 - 13:17

score 77 · Answer 6 · 2019-04-14 14:33:45

77

Views

convergence for the weak-* topology

Published on 14 Apr 2019 - 14:33

score 57 · Answer 7 · 2019-04-28 19:04:25

57

Views

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

Published on 28 Apr 2019 - 19:04

score 79 · Answer 8 · 2019-04-29 14:06:51

79

Views

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

Published on 29 Apr 2019 - 14:06

score 420 · Answer 9 · 2019-05-21 12:20:27

420

Views

Find a counterexample of weak convergence and weak* convergence

Published on 21 May 2019 - 12:20

score 98 · Answer 10 · 2019-05-21 20:26:55

98

Views

Characterization of weak compactness in a Banach space

Published on 21 May 2019 - 20:26

score 86 · Answer 11 · 2019-05-22 17:58:55

86

Views

Weak convergence - restriction

Published on 22 May 2019 - 17:58

score 69 · Answer 12 · 2019-06-04 12:04:28

69

Views

The weak dual coincides with the strong dual

Published on 04 Jun 2019 - 12:04

score 74 · Answer 13 · 2019-06-11 13:10:01

74

Views

Alaoglu theorem over the $p$-adics

Published on 11 Jun 2019 - 13:10

score 181 · Answer 14 · 2019-06-15 14:36:51

181

Views

A sequence of functionals without a weak-* convergent subsequence

Published on 15 Jun 2019 - 14:36

score 262 · Answer 15 · 2019-06-23 15:30:05

262

Views

Multiplicative Linear Functionals on a Banach Algebra

Published on 23 Jun 2019 - 15:30

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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