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

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Identifying a Hilbert space with its dual, given that it's related to another Hilbert space

Let $X$ be a Hilbert space with inner product $(,)_X$, and let $Y$ be another Hilbert space with inner product $(,)_Y$. Suppose there is a bijective continuous linear operator $F:X \to Y$ between the ...
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Normed Space and Hibert Space Problem

Anyone could describe me, why this is True? Suppose $(H, \|.\|) $ is a normed space. the norm $\|.\|$ induced by an inner product if and only if Parallelogram law is valid. Regards.
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Equivalent definitions of the trace of a Hilbert-Schmidt operator

I am currently reading the book Spectral Methods in Automorphic Forms, and Iwaniec defines the trace operator in a different way than I am accustomed to. Throughout, assume that everything converges ...
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1answer
14 views

One Note about One to one and Surjective of linear functional [on hold]

I read a note that: if $ f \neq 0$ is a linear functional on H, then f is onto (surjective) and it is not one to one (injective) in general. Why this is true? i think it need advance ...
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57 views

Closure in a Hilbertspace

Define for a self-adjoint pure contraction $S$ (remember: $\|S\|\leq1$ and $\pm1\notin\sigma_p(S))$ on a Hilbert space $\mathcal{H}$ the following set: $C_c^*(S):=\{g(S):g\in C_c(\hat{\sigma}(S))\}$ ...
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2answers
34 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: $$T:=\int_\mathbb{C}zdE(z)$$ ...
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27 views

Spectral Measures: Spectral Subspaces

Given a Hilbert space $\mathcal{H}$ and let the Lebesgue measure be $\lambda$. Consider a normal operator $N:\mathcal{D}\to\mathcal{H}$. Denote its associated Borel spectral measure by: ...
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Composition of projections has a fixed point in a Hilbert space

Let Let H be a Hilbert space with an inner product ⟨⋅,⋅⟩ : H×H→R, and induced norm $∥⋅∥ : H→R_+$ Let $C_1$ and $C_2$ be closed, convex, nonempty, disjoint subsets of $H$ with at least one of ...
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1answer
42 views

Spectral Measures: Riemann-Lebesgue

Given a Hilbert space $\mathcal{H}$ and let the Lebesgue measure be $\lambda$. Consider a Borel spectral measure $E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$. Denote its associated ...
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305 views

A complete orthonormal system contained in a dense sub-space.

Let H be a separable complex Hilbert space. Let A be a dense sub-space of H. Is it possible to find a complete orthonormal system for H that is contained in A?
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39 views

Spectral Measures: Support vs. Spectrum

Given a complex Hilbert space $\mathcal{H}$. Consider a Borel spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$ and its associated normal operator: ...
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24 views

If $z$ is in an inner product space $X$, show that $f(x)=\langle x,z \rangle$ defines a bounded linear functional $f$ on $X$.

If $z$ is any fixed element of an inner product space $X$, show that $f(x)=\langle x,z \rangle$ defines a bounded linear functional $f$ on $X$, of norm $||z||$. If the mapping $X\to X'$ given $z\to f$ ...
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1answer
41 views

Hilbert Subspaces: ONB

This might be a duplicate. If so, then please let me know. Thanks! Given a Hilbert space $\mathcal{H}$. Consider a dense subspace $\overline{Z}=\mathcal{H}$. Then it provides an ONB: ...
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1answer
15 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 $Z\leq\mathcal{H}$: $$TZ\subseteq Z,TZ^\perp\subseteq Z^\perp$$ ...
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A bounded set in a hilbert space with a compact domain is equicontinuous?

I came across this line in a book "As bounded sets in $H^1(\Omega)$ are equi-continuous and $\Omega$ is compact..." It goes on to prove a result from this but what has me stuck is I don't see is ...
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+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 ...
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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 ...
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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 ...
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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: ...
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43 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 ...
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66 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 $V$ ...
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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 ...
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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 ...
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60 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 ...
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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 ...
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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 ...
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1answer
18 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 ...
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1answer
23 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 $l(f) = \int_0^1 \phi(t) dt.$ ...
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1answer
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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 ...
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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 ...
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1answer
31 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 ...
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3answers
51 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 ...
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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 ...
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1answer
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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 ...
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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: ...
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1answer
51 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: ...
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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 ...
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62 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 ...
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3answers
880 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 ...
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22 views

Proving that Riesz map is bijection [closed]

1) Prove that Riesz map is bijection 2) Prove that Riesz map is monomorphism
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19 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)= ...
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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 ...
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31 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 ...
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37 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 ...
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1answer
38 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 ...
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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 ...
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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 ...
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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 ...
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1answer
27 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 ...
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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\| ...