Question

Hilbert spaces: Orthonormal set has a dense span iff the only vector orthogonal to it is zero.

score 1.4k · Answer 1 · 2026-03-26 22:53:22

1.4k

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Hilbert spaces: Orthonormal set has a dense span iff the only vector orthogonal to it is zero.

Published on 26 Mar 2026 - 22:53

score 90 · Answer 2 · 2026-03-26 08:00:43

90

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If $A$ is self-adjoint, how are $\inf\sigma(A)$ and $\inf_{\left\|x\right\|_H=1}\langle Ax,x\rangle_H$ related?

Published on 26 Mar 2026 - 8:00

score 140 · Answer 3 · 2026-04-02 07:03:18

140

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Is the weak derivative invertible from $H^1(\mathbb{R})\to L^2(\mathbb{R})$?

Published on 02 Apr 2026 - 7:03

score 46 · Answer 4 · 2019-12-31 09:32:22

46

Views

Is $A+A^{\perp}$, dense in $H$ for any $A$.

Published on 31 Dec 2019 - 9:32

score 26 · Answer 5 · 2026-04-06 08:03:07

26

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Does the closure make a difference with regards to an inner product?

Published on 06 Apr 2026 - 8:03

score 3.4k · Answer 6 · 2026-04-03 07:49:48

3.4k

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Unitary operators preserve the inner product between the vectors they act on.

Published on 03 Apr 2026 - 7:49

score 60 · Answer 7 · 2026-03-26 22:13:05

60

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Compact operators on $L^2([a,b])$: Are the dimensions of eigenspaces the same on dense subspaces?

Published on 26 Mar 2026 - 22:13

score 57 · Answer 8 · 2020-01-02 11:11:14

57

Views

Estimation of the $L^2$ norm of a function

Published on 02 Jan 2020 - 11:11

score 96 · Answer 9 · 2026-03-29 00:00:03

96

Views

$\langle u| \langle v| A |u\rangle |v\rangle \geq 0$ for any $|u\rangle \in \mathcal{H}_1$ and $|v\rangle \in \mathcal{H}_2$. Then $A$ is Hermitian?

Published on 29 Mar 2026 - 0:00

score 135 · Answer 10 · 2026-03-25 16:03:24

135

Views

Unique bounded operator constructed from sequence of given operators from orthogonal sum of Hilbert spaces to another Hilbert space.

Published on 25 Mar 2026 - 16:03

score 39 · Answer 11 · 2020-01-03 20:49:51

39

Views

Showing that $\mathcal{H}^{(n)}$ is complete

Published on 03 Jan 2020 - 20:49

score 52 · Answer 12 · 2026-04-02 14:19:57

52

Views

Proof method for Hilbert spaces

Published on 02 Apr 2026 - 14:19

score 1k · Answer 13 · 2026-04-02 14:19:11

1k

Views

A complex Banach space satisfying the parallelogram law is Hilbert

Published on 02 Apr 2026 - 14:19

score 181 · Answer 14 · 2020-01-04 13:08:28

181

Views

Complement of closed subspace in $L^2$

Published on 04 Jan 2020 - 13:08

score 380 · Answer 15 · 2020-01-06 17:28:58

380

Views

Prove that position operator is bounded

Published on 06 Jan 2020 - 17:28

List Question

Hilbert spaces: Orthonormal set has a dense span iff the only vector orthogonal to it is zero.

If $A$ is self-adjoint, how are $\inf\sigma(A)$ and $\inf_{\left\|x\right\|_H=1}\langle Ax,x\rangle_H$ related?

Is the weak derivative invertible from $H^1(\mathbb{R})\to L^2(\mathbb{R})$?

Is $A+A^{\perp}$, dense in $H$ for any $A$.

Does the closure make a difference with regards to an inner product?

Unitary operators preserve the inner product between the vectors they act on.

Compact operators on $L^2([a,b])$: Are the dimensions of eigenspaces the same on dense subspaces?

Estimation of the $L^2$ norm of a function

$\langle u| \langle v| A |u\rangle |v\rangle \geq 0$ for any $|u\rangle \in \mathcal{H}_1$ and $|v\rangle \in \mathcal{H}_2$. Then $A$ is Hermitian?

Unique bounded operator constructed from sequence of given operators from orthogonal sum of Hilbert spaces to another Hilbert space.

Showing that $\mathcal{H}^{(n)}$ is complete

Proof method for Hilbert spaces

A complex Banach space satisfying the parallelogram law is Hilbert

Complement of closed subspace in $L^2$

Prove that position operator is bounded

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