Question

Strongly bounded set is weakly bounded

score 1.5k · Answer 1 · 2021-06-11 16:00:30

1.5k

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Strongly bounded set is weakly bounded

Published on 11 Jun 2021 - 16:00

score 31 · Answer 2 · 2021-06-13 17:15:16

31

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If $R$ is a closed subspace of a Banach space $X$ with $\text{codim } R = 1$, $\exists \psi \in X^*\backslash 0$ with $R = \ker{\psi}$

Published on 13 Jun 2021 - 17:15

score 428 · Answer 3 · 2026-02-23 08:36:03

428

Views

Regarding Lax-Milgram theorem (Operator theory)

Published on 23 Feb 2026 - 8:36

score 98 · Answer 4 · 2021-06-28 18:59:59

98

Views

Showing that a function $f$ has a fixpoint

Published on 28 Jun 2021 - 18:59

score 68 · Answer 5 · 2021-07-02 20:07:56

68

Views

A subset fundamental in a normed linear space $\Leftrightarrow $ its annihilator is $\{0\}$

Published on 02 Jul 2021 - 20:07

score 57 · Answer 6 · 2021-07-25 22:56:47

57

Views

Confusion about the application of Hahn-Banach extension Theorem in a duality argument

Published on 25 Jul 2021 - 22:56

score 118 · Answer 7 · 2021-07-29 18:43:11

118

Views

Given $x_1, x_2, \cdots $ and reals $\alpha_1,\alpha_2,\cdots $ prove that there is a linear functional s.t. $\varphi(x_i)=\alpha_i$

Published on 29 Jul 2021 - 18:43

score 51 · Answer 8 · 2021-09-08 10:46:30

51

Views

If $A$ and $B$ are convex and disjoint then there exist a continuous linear function such that $f(A)$ and $f(B)$ are dijoint.

Published on 08 Sep 2021 - 10:46

score 32 · Answer 9 · 2021-09-21 14:39:11

32

Views

If $E$ is separable, why can we find $(x'_n)\subseteq E'$ such that $\langle x,x'_n\rangle$ for all $n$ implies $x=0$?

Published on 21 Sep 2021 - 14:39

score 507 · Answer 10 · 2021-10-16 20:01:48

507

Views

Hahn-Banach theorem and closed subspaces

Published on 16 Oct 2021 - 20:01

score 96 · Answer 11 · 2021-10-22 11:51:59

96

Views

Brezis' Functional analysis 3.27 item 2.

Published on 22 Oct 2021 - 11:51

score 325 · Answer 12 · 2021-11-07 21:08:29

325

Views

Extend the linear functional on $\mathbb R^2$

Published on 07 Nov 2021 - 21:08

score 204 · Answer 13 · 2021-11-11 20:00:58

204

Views

Every infinite-dimensional Banach space not containing $l^1$ has a normalized, weakly null sequence.

Published on 11 Nov 2021 - 20:00

score 225 · Answer 14 · 2021-11-12 23:23:09

225

Views

Trouble applying Hahn-Banach theorem in exercise

Published on 12 Nov 2021 - 23:23

score 203 · Answer 15 · 2021-11-17 20:22:58

203

Views

Hahn-Banach Theorem and non bounded linear functionals.

Published on 17 Nov 2021 - 20:22

List Question

Strongly bounded set is weakly bounded

If $R$ is a closed subspace of a Banach space $X$ with $\text{codim } R = 1$, $\exists \psi \in X^*\backslash 0$ with $R = \ker{\psi}$

Regarding Lax-Milgram theorem (Operator theory)

Showing that a function $f$ has a fixpoint

A subset fundamental in a normed linear space $\Leftrightarrow $ its annihilator is $\{0\}$

Confusion about the application of Hahn-Banach extension Theorem in a duality argument

Given $x_1, x_2, \cdots $ and reals $\alpha_1,\alpha_2,\cdots $ prove that there is a linear functional s.t. $\varphi(x_i)=\alpha_i$

If $A$ and $B$ are convex and disjoint then there exist a continuous linear function such that $f(A)$ and $f(B)$ are dijoint.

If $E$ is separable, why can we find $(x'_n)\subseteq E'$ such that $\langle x,x'_n\rangle$ for all $n$ implies $x=0$?

Hahn-Banach theorem and closed subspaces

Brezis' Functional analysis 3.27 item 2.

Extend the linear functional on $\mathbb R^2$

Every infinite-dimensional Banach space not containing $l^1$ has a normalized, weakly null sequence.

Trouble applying Hahn-Banach theorem in exercise

Hahn-Banach Theorem and non bounded linear functionals.

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