For question involving Hilbert spaces, that is, complete normed spaces whose norm comes from an inner product.

learn more… | top users | synonyms

14
votes
0answers
151 views
+500

Over ZF, does “every Hilbert space have a basis” imply AC?

I know there is a similar result due to Blass [1] that over ZF, "every vector space has a (Hamel) basis" implies AC. Looking around, however, I can't find any results on the question for Hilbert ...
0
votes
1answer
19 views

Spectral Measures: Riemann-Lebesgue

Given a Hilbert space $\mathcal{H}$. Consider a Borel spectral measure $E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$. Denote its associated measures by: $$\mu_{x,z}(A):=\langle ...
-2
votes
1answer
23 views

a question about contractions on Hilbert spaces

Let $\cal{H}$ be a Hilbert space, $T_1,T_2\in\cal{B(H)}$, $\|T_1(h_1)+T_2(h_2)\|^2\leq\|h_1\|^2+\|h_2\|^2$ for all $h_1,h_2\in\cal{H}$. $T_1T^\ast_1+T_2T^\ast_2\leq I$. Then can we verify that 1 ...
1
vote
0answers
17 views

Closure in a Hilbertspace

Define for a pure contraction $S$ (remember: $\|S\|\leq1$ and $\pm1\notin\sigma_p(S))$ the following set: $C_c^*(S):=\{g(S):g\in C_c(\hat{\sigma}(S))\}$ with $\hat{\sigma}(S)=\sigma(S)\cap(-1,1)$. Now ...
0
votes
0answers
31 views

Hilbert Subspaces: ONB

This might be a duplicate. If so, then please let me know - I will close this thread then. Thanks! Given a Hilbert space $\mathcal{H}$. Consider a dense subspace $\overline{Z}=\mathcal{H}$. Then it ...
0
votes
0answers
7 views

A Fredholm alternative for nonlinear operators?

There is a Fredholm alternative of the form: Let $K$ be a compact linear operator. Then $(I + K)u = f$ has a solution $u$ for every $f$ if and only if $$\text{$(I+K)u=0 \implies u=0$.}$$ Is ...
1
vote
0answers
12 views

Reducing Subspaces: Nonexample?

Given a Hilbert space $\mathcal{H}$. Consider an operator $T:\mathcal{D}(T)\to\mathcal{H}$. Suppose there exists a closed subspace $Z\leq\mathcal{H}$: $$TZ\subseteq Z,TZ^\perp\subseteq Z^\perp$$ ...
0
votes
1answer
7 views

Ordered Projections: Range

Given a Hilbert space $\mathcal{H}$. Consider two orthogonal projections $P,Q$. Then: $$P\leq Q\implies\mathcal{R}(P)\subseteq\mathcal{R}(Q)$$ The ordering being induced by: ...
3
votes
1answer
41 views

Normal Operator: Everywhere defined implies bounded?

Given a Hilbert space $\mathcal{H}$. Consider a normal operator $N:\mathcal{H}\to\mathcal{H}$. If its domain is the whole Hilbert space then is it necessarily bounded? The point is that I'm trying ...
2
votes
0answers
65 views

Making a complex inner product symmetric

Let $(V, (\cdot, \cdot))$ be a complex inner product space, say a space of complex-valued functions, with $(\cdot, \cdot)$ linear in the second position and sesquilinear in the first. Assume that $V$ ...
2
votes
2answers
58 views

Is $\|x\| = \| \overline{x} \|$ in an inner product space?

Suppose $X$ is a complex inner product space of complex valued functions that is closed under conjugation. Is it true that $\|x\| = \| \overline{x} \|$ for all $x$? If not, is there a simple ...
0
votes
1answer
6 views

Evaluating the $L_2[-1, 1]$ inner product on rescaled Legendre polynomials

Let $z_n(t) = \sqrt{\frac{2n+1}{2}} \frac{1}{2^n n!} \frac {d^n}{dt^n} (t^2-1)^n$, a rescaled Legendre polynomial. As an intermediate step of a larger problem, I need to show that in terms of the ...
3
votes
2answers
53 views

Selfadjointness of the Dirac operator on the infinite-dimensional Hilbert space

I am a physicist, so my background in functional analysis is limited only to basics. However, I would like to prove that the free Dirac operator is selfadjoint (or Hermitian, or neither). The free ...
1
vote
1answer
59 views

Hilbert Space: Weak Convergence implies Strong Convergence

This probably might be a duplicate - let me know if so. I read the following in Graf's notes on quantum mechanics - can you give me a hint for the proof. In Hilbert spaces weak convergence in a way ...
2
votes
1answer
77 views

$\langle Tx,x\rangle =0$ , then T is zero

I just wanted to be sure about something. The implication $\langle Tx,x\rangle =0$ , then T is zero , holds only if $T$ is self-adjoint right? If $T$ is an arbitrary operator, we need to have $\langle ...
1
vote
1answer
15 views

Isometric Operators: Common Core

Given a Hilbert or Banach space $\mathcal{H}$. Consider two closed operators $S:\mathcal{D}(S)\to\mathcal{H}$ and $T:\mathcal{D}(T)\to\mathcal{H}$. Suppose they're isometric on a common core ...
1
vote
1answer
22 views

Bounding a linear functional in $L_2[0, 1]$

For each f in $L_2[0, 1]$ let $\phi(t)$ be the solution of $y' + ay = f$ that satisfies $\phi(0) = 0$, where a is a constant. Define $l: L_2[0,1] \to \mathbb{C}$ by $l(f) = \int_0^1 \phi(t) dt.$ ...
0
votes
0answers
11 views

Continuous extension of $(\cdot,\cdot)_H:H \times V \to \mathbb{R}$ to $\langle \cdot, \cdot \rangle_{V^*, V}$

Let $V \subset H$ be dense and continuous where $V$ is a reflexive Banach space and $H$ is a Hilbert space. Can someone explain the continuous extension to me? I thought it had something to do with ...
0
votes
1answer
13 views

Gelfand triple: what happens if we don't identify the pivot Hilbert space with its dual?

People usually say $V \subset H = H^* \subset V^*$ is a Gelfand triple if $V$ is continuously and densely embedded in $H$ and $H$ is identified with its dual. Sometimes they do not mentioned that ...
1
vote
1answer
40 views

Why is the infinite dimensional vector space with only finitely many nonvanishing components incomplete?

Define a complex vector space $V$ such that any element $\{a_i\}=(a_1,a_2,\dots)\in V$ has only finitely many components $a_i\ne 0$. The inner product is defined as $$(\{a_i\},\{b_j\})=\sum_i^\infty ...
3
votes
1answer
30 views

Family of sequences in a Hilbert space with certain property

Suppose $\mathcal{F}$ is a family of sequences on the unit sphere of $l_2$ with the following property: For any sequence $\varepsilon_n\downarrow 0$ but which is not eventually identically $0$, there ...
2
votes
3answers
43 views

Hermitian and self-adjoint operators on infinite-dimensional Hilbert spaces

I am a physicist and I am trying to get a grasp on the following terms from functional analysis: As I understand, an operator is Hermitian if it is symmetric and bounded (domains of A and A* don't ...
3
votes
2answers
25 views

Finding the maximum of an integral of a function with given constraints.

This comes from Rudin's Real Analysis text. The first part of the problem asks us to compute $\displaystyle\min_{a,b,c}\int_{-1}^1|x^3-a-bx-cx^2|dx$ (which I have done). Now it asks us to find ...
0
votes
1answer
10 views

on Limits and sequences proofs of a closed subspace of the hilbert space.

$M$ is a closed subspace of the Hilbert space $H$ and $ x\in H$ My book states these two claims. (1) If $d = \inf_{y \in M} \|x - y \|^2 $then there is a sequence of elements $\{y_n\}$ of $M$ such ...
0
votes
1answer
27 views

Spectral Measures: Support vs. Norm

Given a complex Hilbert space $\mathcal{H}$. Consider a spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$ and its associated normal operator: $$T:=\int_\mathbb{C}zdE(z)$$ ...
0
votes
2answers
39 views

Spectral Measures: Property

Given a Hilbert space $\mathcal{H}$ and spectral a measure $E:\Sigma(\Omega)\to\mathcal{B}(\mathcal{H})$. Can you give me a hint for: $$E(A)E(B)=E(A\cap B)$$ So far for disjoints I checked: ...
0
votes
1answer
50 views

Spectral Measures: Integration

Given a Hilbert space $\mathcal{H}$ and spectral a measure $E:\Sigma(\Omega)\to\mathcal{B}(\mathcal{H})$. How to define the integral for unbounded measurable functions: ...
0
votes
2answers
29 views

Spectral Measures: Support vs. Spectrum

Given a complex Hilbert space $\mathcal{H}$. Consider a spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$ and its associated normal operator: $$T:=\int_\mathbb{C}zdE(z)$$ ...
2
votes
2answers
54 views

Spectral Measures: Support vs. Concentration

The support of a Borel spectral measure is defined by: $$\lambda\in\mathrm{supp}E:\iff E(U)>0\quad\lambda\in U\in\mathcal{T}$$ (See the german wikipedia article: Spektralmaß) Now, consider a Borel ...
1
vote
1answer
61 views

Spectral Measures: Integration of Product

Given a Hilbert space $\mathcal{H}$ and spectral a measure $E:\Sigma(\Omega)\to\mathcal{B}(\mathcal{H})$. Define the integral of simple functions by: $$\int_\Omega ...
1
vote
3answers
878 views

Hilbert Space is reflexive

A normed space $X$ is reflexive iff $X^{**}=\{g_x:x\in X\}$ where $g_x$ is bounded linear functional on $X^*$ defined by $g_x(f)=f(x)$ for any $f\in X^*$. Let $X$ be a Hilbert space, would you help ...
0
votes
0answers
21 views

Proving that Riesz map is bijection [closed]

1) Prove that Riesz map is bijection 2) Prove that Riesz map is monomorphism
1
vote
1answer
18 views

An empty subdifferential

Can you give me an example of function $f$ defined on an Hilbert space, real valued (extended with $+ \infty$), lower semi continuous, convex and proper for which $\operatorname{dom}(\partial f)= ...
0
votes
0answers
14 views

Every Hilbert Space has an orthonormal basis [duplicate]

I'd be really grateful if someone could tell me what steps I should take (ie. what books to read) before I can prove the statement in the title. I currently have taken rigorous courses in Linear ...
1
vote
1answer
30 views

$\{ x \in H: x=\sum_{k=1}^{\infty}c_{k}u_{k}$, $|c_{k}| \leq \frac{1}{k}\}$ is compact

Let $H$ be a complex inner product space that is also a complete metric space with respect to the distance induced by the inner product. Assume $\{u_{k}\}_{k=1}^{\infty}$ be an orthonormal set in ...
1
vote
1answer
35 views

Showing $||L|| = \sup_{x,\, y\, \in H,\, x,\, y\, \neq 0} \frac{|\langle Ax,y\rangle|}{||x||\cdot ||y||}.$

Prove that, for a Hilbert space $H$ and a linear bounded operator $L:H \to H$ such that the domain of $L$ is $H$, $$||L|| = \sup_{x,\, y\, \in H}_{ x,\, y\, \neq 0} \frac{|\langle ...
1
vote
1answer
35 views

Identifying a Hilbert space with its dual

Let $H$ be a Hilbert space. Often people say "we identify $H$ with its dual $H^*$ with the Riesz representation theorem". I know there is a map $R:H^* \to H$ which is ismometric and isomorphic. So by ...
1
vote
3answers
37 views

ONB: Fourier Series

Given the Hilbert space $L^2([-\pi,\pi])$. Consider the orthonormal system: $$\mathcal{S}:=\{\frac{1}{\sqrt{2\pi}}e^{ikx}:k\in\mathbb{Z}\}$$ This is an ONB. How do I prove this? I guess, I could try ...
1
vote
1answer
33 views

Nonseperable Hilbert space: Explicit ONB?

Every Hilbert space admits an ONB by axiom of choice. For separable Hilbert spaces this can in fact be constructed by Gram-Schmidt. For nonseparable Hilbert spaces there can be no general construction ...
0
votes
1answer
20 views

Norm of $A$ is zer0 when $\Bbb H$ is Complex Hiblert Space

If $\Bbb H$ is a $\Bbb C$-Hilbert space and $A\in \Bbb B(\Bbb H),$i.e. bounded linear operator on $\Bbb H$ such that $\langle Ah,h\rangle=0$ for all h in $\Bbb H$, then $A=0$ For the proof of this ...
1
vote
1answer
26 views

Radon-Nikodym: Integrability?

Let $\lambda:\Sigma\to\mathbb{R}_+$ and $\kappa:\Sigma\to\mathbb{R}_+$ be finite measures on $\Omega$. Then by Radon-Nikodym: $$\kappa(E)\leq L\cdot\lambda(E)\quad(\forall ...
0
votes
1answer
27 views

Norms for which every subset of closed unit ball containing the open unit ball is convex

It can be shown without much difficulty that any Euclidean norms satisfies the following condition :$$(P) \quad B \subset X \subset B' \Rightarrow X \, \text{is convex}$$ where $B=\{x \in E / \|x\| ...
0
votes
3answers
21 views

When a symmetric densely defined operator is an adjoint operator?

I am wondering if it is possible to say that if a symmetric differential operator is densely defined then the operator is self-adjoint indeed? More Precisely, Let $A:D(A)(\subset H)\to H$ a densely ...
0
votes
1answer
11 views

Calculating the form domain of an operator

I am reading the book "Mathematical Methods in Quantum Mechanics" by Gerald Teschl and just came across the concept of a form domain. It is defined for non-negative operators i.e $<\phi,A \phi> ...
0
votes
0answers
13 views

Prove that every reproducing kernel is a positive matrix (and vice versa)

Let $\mathcal{H}$ be a functional hilbert space (defined over a set $S$) with a reproducing kernel K. Prove that: a) $K$ is a positive matrix means the queadtric form is positive, i.e ...
-1
votes
0answers
28 views

prove of inner product space [on hold]

Show that$$ L^1 (R^n ) $$under the operator $$〈.,.〉:L^1×L^1→R$$ such that $$〈f,g〉=∫_(R^n)▒〖f(x) (g(x)) ̅ 〗$$ for all$$ f,g∈L^1 (R^n )$$forms an inner product space
0
votes
1answer
20 views

Proving for each seperatble hilbert space exist complete sequence

Let $H$ be a separable Hilbert Space. Prove that exists orthonormal complete sequence and give example for one non-orthonormal sequence. I thought taking orthonormal basis for $H$ denoted by ...
1
vote
1answer
25 views

Proving that if $\sum\|f_n-e_n\|^2< 1$, $\{f_n\}$ is a complete sequence

Let $\{e_n\}$ be a complete orthonormal sequence in an Hilbert space $H$ and let $\{f_n\}$ be an arbitrary sequence of elements in $H$ s.t $$\sum_{n=1}^\infty\|f_n-e_n\|^2<1$$Show that ...
0
votes
0answers
25 views

dense subspace of $C(0,T)$

I want to prove that the space H of functions which are continuous in [0,T] with weak derivative in $L^2[0,T]$ and their value in 0 is 0, is dense in the space of continuous functions in [0,T] with ...
2
votes
1answer
49 views

What is the image of operator exponential?

Given a Banach space $V$ and a bounded linear operator $A:V\to V$, the operator $e^A$ is bounded and invertible. When $V$ is finite dimensional, every invertible operator is of the form $e^B$ (one can ...