Question

Surjectivity of operator $(Vf)(t)= \begin{cases} f(t-1) & \text{if $x\geq1 $ } \\ 0 & \text{if $x<1 $} \end{cases}$in $L_2(0, \infty)$

score 28 · Answer 1 · 2019-10-19 05:40:31

Surjectivity of operator $(Vf)(t)= \begin{cases} f(t-1) & \text{if $x\geq1 $ } \\ 0 & \text{if $x<1 $} \end{cases}$in $L_2(0, \infty)$

score 490 · Answer 2 · 2026-03-25 19:17:34

490

Views

Example for $\mathbb{X}/\mathbb{M}$ separable but $\mathbb{X}$ is not.

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score 92 · Answer 3 · 2026-03-26 16:06:48

92

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Natural map from $\mathbb{X}$ to $\mathbb{X}/\mathbb{M}$, that is not closed

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score 179 · Answer 4 · 2019-10-20 01:55:42

179

Views

Identify $c/c_0$

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score 105 · Answer 5 · 2026-03-26 15:16:39

105

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a affine hyperplane in normed space

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score 97 · Answer 6 · 2026-03-30 12:01:54

97

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Highly recommended books for bounded linear operators on banach spaces

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score 262 · Answer 7 · 2019-10-20 18:22:03

262

Views

On the Completeness of Projective Tensor Products

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score 147 · Answer 8 · 2019-10-20 20:01:56

147

Views

A problem on Uniform Boundedness Principle

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score 182 · Answer 9 · 2019-10-22 09:22:47

182

Views

Let $U$ a Banach subspace of a normed vector space $V$. Show that if $V/U$ is Banach then it is also $V$

Published on 22 Oct 2019 - 9:22

score 27 · Answer 10 · 2019-10-22 23:49:38

27

Views

Let $(V,\left \langle .,. \right \rangle)$ be a space of Hilbert and $F\in V'$. Then $U=\{v\in V: F(v)=0\}$ is a closed subspace of $V$.

Published on 22 Oct 2019 - 23:49

score 164 · Answer 11 · 2026-03-26 17:32:14

164

Views

(simplified version of ) Uniform Convexity of Banach Space

Published on 26 Mar 2026 - 17:32

score 78 · Answer 12 · 2019-10-24 14:08:50

78

Views

Proving that $f( x ):=\sum_{i=1}^{\infty}x( i ) / i $ is continuous functional in sequence space l2

Published on 24 Oct 2019 - 14:08

score 260 · Answer 13 · 2019-10-25 22:11:02

Prove that the linear Functional $\phi:C[0,1]\to\mathbb{F}$ such that $\phi(f)=\int\limits_0^{1/2}f-\int\limits_{1/2}^1f$ Doesn't achieve its norm.

score 53 · Answer 14 · 2026-03-02 20:48:59

53

Views

The bidual of a Banach space

Published on 02 Mar 2026 - 20:48

score 380 · Answer 15 · 2019-10-26 17:07:46

380

Views

prove that the mapping $\Phi:\ell^1\to c^*$ is an isometric isomorphism

Published on 26 Oct 2019 - 17:07

List Question

Surjectivity of operator $(Vf)(t)= \begin{cases} f(t-1) & \text{if $x\geq1 $ } \\ 0 & \text{if $x<1 $} \end{cases}$in $L_2(0, \infty)$

Example for $\mathbb{X}/\mathbb{M}$ separable but $\mathbb{X}$ is not.

Natural map from $\mathbb{X}$ to $\mathbb{X}/\mathbb{M}$, that is not closed

Identify $c/c_0$

a affine hyperplane in normed space

Highly recommended books for bounded linear operators on banach spaces

On the Completeness of Projective Tensor Products

A problem on Uniform Boundedness Principle

Let $U$ a Banach subspace of a normed vector space $V$. Show that if $V/U$ is Banach then it is also $V$

Let $(V,\left \langle .,. \right \rangle)$ be a space of Hilbert and $F\in V'$. Then $U=\{v\in V: F(v)=0\}$ is a closed subspace of $V$.

(simplified version of ) Uniform Convexity of Banach Space

Proving that $f( x ):=\sum_{i=1}^{\infty}x( i ) / i $ is continuous functional in sequence space l2

Prove that the linear Functional $\phi:C[0,1]\to\mathbb{F}$ such that $\phi(f)=\int\limits_0^{1/2}f-\int\limits_{1/2}^1f$ Doesn't achieve its norm.

The bidual of a Banach space

prove that the mapping $\Phi:\ell^1\to c^*$ is an isometric isomorphism

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