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

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28
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3answers
1k views

If $\sum a_n b_n <\infty$ for all $(b_n)\in \ell^2$ then $(a_n) \in \ell^2$

I'm trying to prove the following: If $(a_n)$ is a sequence of positive numbers such that $\sum_{n=1}^\infty a_n b_n<\infty$ for all sequences of positive numbers $(b_n)$ such that $\sum_{n=1}^\...
14
votes
2answers
3k views

The direct sum of two closed subspace is closed? (Hilbert space)

I know that if $X$ is a Banach space, then, the direct sum of two closed subspace $X_1$ and $X_2$ is not necessarily closed. But what if $X$ is Hilbert? I assume there is something to do with the ...
46
votes
3answers
5k views

Connections between metrics, norms and scalar products (for understanding e.g. Banach and Hilbert spaces)

I am trying to understand the differences between $$ \begin{array}{|l|l|l|} \textbf{vector space} & \textbf{general} & \textbf{+ completeness}\\\hline \text{metric}& \text{metric ...
15
votes
1answer
2k views

How to prove that an operator is compact?

Consider $T\colon\ell^2\to\ell^2$ an operator such that $Te_k=\lambda_k e_k$ with $\lambda_k\to 0$ as $k \to \infty$ how to prove that it is compact?
6
votes
1answer
2k views

Compactness of Multiplication Operator on $L^2$

Suppose we have an bounded linear operator A that operates from $L^2([a,b]) \mapsto L^2([a,b])$. Now suppose that $A(f)(t) = tf(t)$. Is A compact? Edit: I know $A = A^*$ but I'm not really sure ...
10
votes
5answers
1k views

How to show that this set is compact in $\ell^2$

Let $(a_n)_{n}\in\ell^2:=\ell^2(\mathbb{R})$ be a fixed sequence. Consider the subspace $$C=\{(x_n)_{n}\in\ell^2 : |x_n|\le a_n\text{ for all }n\in\mathbb{N}\}.$$ According to the book [Dunford and ...
19
votes
2answers
8k views

Finding the adjoint of an operator

This is from my homework, I'm totally lost as to how to proceed. Consider the operator $T: L^2([0,1]) \rightarrow L^2([0,1])$ defined by $(Tf)(x) = \int^x_0 f(s) \ ds$ What is the adjoint of $T$? ...
7
votes
3answers
611 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?
17
votes
1answer
882 views

Is there a constructive proof of this characterization of $\ell^2$?

I would like to revisit this question, which can be equivalently stated as: Proposition. Let $(a_n)$ be a sequence of real (or complex) numbers such that $\sum a_n b_n$ converges for every $(b_n) ...
15
votes
3answers
3k views

An orthonormal set cannot be a basis in an infinite dimension vector space?

I'm reading the Algebra book by Knapp and he mentions in passing that an orthonormal set in an infinite dimension vector space is "never large enough" to be a vector-space basis (i.e. that every ...
4
votes
2answers
686 views

A counterexample to theorem about orthogonal projection

Can someone give me an example of noncomplete inner product space $H$, its closed linear subspace of $H_0$ and element $x\in H$ such that there is no orthogonal projection of $x$ on $H_0$. In other ...
7
votes
1answer
652 views

Sum of Closed Operators Closable?

Let $A$ and $B$ be closed operators on a (separable complex) Hilbert space with dense domains $D(A)$ and $D(B)$ respecitvely. Then, we may define the operator $A+B$ on $D(A)\cap D(B)$. In general, ...
5
votes
1answer
3k views

$\ell_p$ is Hilbert space if and only if $p=2$

Can anybody please help me to prove this.. Let $p$ greater than or equal to $1$, show that the space of all $p$-summable sequences is an inner product space if and only if $p=2$.
1
vote
1answer
340 views

Proving the closed unit ball of a Hilbert space is weakly sequentially compact

I bumped into this statement in Hofer-Zehnder in the middle of proving a Hamiltonian field always has a periodic orbit over a level set of the hamiltonian if that set is a regular compact and strictly ...
0
votes
1answer
54 views

Spectral Measures: Pushforward

This thread is Q&A. Problem Given a Hilbert space $\mathcal{H}$. Consider a normal operator: $$N:\mathcal{D}(N)\subseteq\mathcal{H}\to\mathcal{H}:\quad N^*N=NN^*$$ And its spectral measure: $$...
5
votes
1answer
4k views

Volterra Operator is compact but has no eigenvalue

Volterra operator is defined as operator $V:L^2[0,1]\rightarrow L^2[0,1]$ by \begin{eqnarray} (V)(f(x))=\int_0^xf(y)dy \end{eqnarray} Would you help me to prove that this operator is compact but has ...
7
votes
3answers
368 views

Sequences are Cauchy depending on the norm?

Assume that we have two Banach spaces $B_1, B_2$ with their underlying sets being identical. Is it possible that a Cauchy sequence in one of the spaces would fail to be Cauchy in the other? I am ...
4
votes
3answers
2k views

Compact operator

If $H$ and $K$ are Hilbert spaces,show that if $T:H\longrightarrow K$ is a compact operator and $\{e_{n}\}$ is any orthonormal sequence in $H$ then $\|Te_{n}\|\to0$.Is the converse true? thanks.
11
votes
1answer
444 views

Every Hilbert space operator is a combination of projections

I am reading a paper on Hilbert space operators, in which the authors used a surprising result Every $X\in\mathcal{B}(\mathcal{H})$ is a finite linear combination of orthogonal projections. The ...
4
votes
1answer
786 views

For positive invertible operators $C\leq T$ on a Hilbert space, does it follow that $T^{-1}\leq C^{-1}$?

I need the following result. I think it's quite obvious but I don't know how to prove that: Let $C, T : \mathcal{H} \rightarrow \mathcal{H}$ be two positive, bounded, self-adjoint, invertible ...
1
vote
1answer
97 views

Summary: Spectrum vs. Numerical Range

This thread is only Q&A! Given a Hilbert space $\mathcal{H}$. Consider operators: $$T:\mathcal{D}(T)\to\mathcal{H}$$ Denote for shorthand: $$\Omega\subseteq\mathbb{C}:\quad\langle\Omega\rangle:=...
35
votes
1answer
549 views

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 ...
10
votes
4answers
1k views

Measure on Hilbert Space

On $\mathbb{R}^n$, we of course have the usual Lebesgue meausre. In many ways, separable, infinite-dimesional Hilbert space is the most natural generalization of $\mathbb{R}^n$ to infinite-dimensions,...
9
votes
3answers
4k views

Orthogonal complement of a Hilbert Space

I have this problem: Let $S$ be a subset of a Hilbert $H$ and let $M$ be the closed subspace generated by $S$. Show that $M^{\perp} = S^{\perp}$ $M = (S^{\perp})^{\perp}$ if $V$ is a subspace of $H$...
8
votes
2answers
631 views

Is compactness a stronger form of continuity?

Let $H$ be a Hilbert space. We say that a linear operator $T \colon H \to H$ is compact if it maps bounded sets to precompact ones, that is, if for every bounded sequence $(a_n)$ in $H$, $(Ta_n)$ has ...
5
votes
1answer
1k views

Proof Complex positive definite => self-adjoint

I am looking for a proof of the theorem that says: A is a complex positive definite endomorphism and therefore is A self-adjoint. Does anybody of you know how to do this?
5
votes
1answer
954 views

Direct sum of orthogonal subspaces

I'm working on the following problem set. Let $\mathcal{H}$ be a Hilbert space and $A$ and $B$ orthogonal subspaces of $\mathcal{H}$. Prove or disprove: 1) $A \oplus B$ is closed, then $A$ and $B$ ...
9
votes
0answers
498 views

On the weak and strong convergence of an iterative sequence

I have some difficulties in the following problem. I would like to thank for all kind help and construction. Let $H$ be an infinite dimensional real Hilbert space and $F: H\rightarrow H$ be a ...
5
votes
1answer
556 views

The sup norm on $C[0,1]$ is not equivalent to another one, induced by some inner product

Let $\mathrm{C}[0,1]$ be the space of continuous functions $[0,1]\rightarrow \mathbb{R}$ endowed with the norm $||x||_{\infty}=\mathrm{max}_{t\in [0,1]}|x(t)|$. It is easy to verify that this norm is ...
4
votes
1answer
78 views

Hilbert vs. De Morgan

Problem Given a Hilbert space $\mathcal{H}$. Then it holds: $$\overline{\left\langle\bigcap_{\lambda\in\Lambda}A_\lambda^\perp\right\rangle}=\left(\bigcup_{\lambda\in\Lambda}A_\lambda\right)^\...
14
votes
1answer
1k views

Different versions of Riesz Theorems

In Wikipedia, there are three versions of Riesz theorems: 1 The Hilbert space representation theorem for the (continuous) dual space of a Hilbert space; 2 The representation theorem for ...
5
votes
2answers
2k views

Every Hilbert space has an orthonomal basis - using Zorn's Lemma

The problem is to prove that every Hilbert space has a orthonormal basis. We are given Zorn's Lemma, which is taken as an axiom of set theory: Lemma If X is a nonempty partially ordered set with the ...
4
votes
1answer
102 views

Normal Operators: Meet

Given a Hilbert space $\mathcal{H}$. Normal Operators: $$\mathcal{N}(\mathcal{H}):=\{N:N^*N=NN^*\}$$ Borel Calculus: $$\mathcal{B}(N):=\{\eta({N}):\eta\in\mathcal{B}(\mathbb{C},\mathbb{C})\}$$ ...
2
votes
3answers
2k views

Question on weak convergence ( Example).

Can anybody tell me why $\sin(nx)$ converges weakly in $L^2(-\pi,\pi)$. I can't see how $\sin(nx)$ can converge? Explanation with any other example will be nice as well.
2
votes
3answers
831 views

Compact operators and uniform convergence

Suppose $T: H \rightarrow H$ is a compact operator, $H$ is a Hilbert space, and let $(A_n)$ be a sequence of bounded linear operators on $H$ converging strongly to $A$. Show that $A_nT$ converges in ...
2
votes
1answer
139 views

Operator: not closable!

Is there an operator between Banach spaces with the following properties: $$T:\mathcal{D}(T)\subseteq X\to Y:\text{ injective, dense range, continuously invertible, not closable!}$$ (Note that the ...
2
votes
3answers
168 views

Normal Operators: Polar Decomposition (Rudin)

On page 332 theorem 12.35b) of Rudin functional analysis is show that if T is normal then it has a polar decomposition $T=UP$. Does he mean that $P=|T|$? He's a bit ambiguous as to how he defines ...
0
votes
3answers
291 views

show that $l^2$ is a Hilbert space

Let $l^2$ be the space of square summable sequences with the inner product $\langle x,y\rangle=\sum_\limits{i=1}^\infty x_iy_i$. (a) show that $l^2$ is H Hilbert space. To show that it's a ...
0
votes
1answer
318 views

Nested sequence of sets in Hilbert space [duplicate]

How can I prove that nested sequence of non-empty bounded closed convex sets in Hilbert space have nonempty intersection? I just don't know where to start. Thanks
0
votes
1answer
87 views

Spectral Measures: Reducibility

Given a Hilbert space $\mathcal{H}$. Consider a normal operator: $$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$ And its spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\...
2
votes
1answer
98 views

Question about SOT and compact operators

I need some help with functional analysis / Hilbert space theory. If you have a favorite text to recommend, please let me know~ Here is my question: Given $v_t$ be the "squeeze operator" on $H=L^2[0,...
1
vote
1answer
1k views

double Orthogonal complement is equal to topological closure

So I'm in an advanced Linear Algebra class and we just moved into Hilbert spaces and such, and I'm struggling with this question. Let $A$ be a nonempty subset of a Hilbert space $H$. Denote by $\...
0
votes
1answer
122 views

Normal Operators: Transform

Given a Hilbert space $\mathcal{H}$. Consider a normal operator: $$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$ Construct the operator: $$Q:=(1+N^*N)^{-1}:\quad Z:=N\sqrt{Q}$$ Then it is ...
15
votes
4answers
1k views

What really is ''orthogonality''?

I know that we can define two vectors to be orthogonal only if they are elements of a vector space with an inner product. So, if $\vec x$ and $\vec y$ are elements of $\mathbb{R}^n$ (as a real ...
7
votes
1answer
959 views

Does there exist a real Hilbert space with countably infinite dimension as a vector space over $\mathbb{R}$?

Essentially what the title says - where to me a Hilbert space is a complete (Hermitian) inner product space, am I safe to assume every such real Hilbert space is of uncountable dimension over $\mathbb{...
13
votes
2answers
225 views

Is there a concept of a “free Hilbert space on a set”?

I am looking for a "good" definition of a Hilbert space with a distinct orthonormal basis (in the Hilbert space sense) such that each basis element corresponds to an element of a given set $X$. Before ...
11
votes
1answer
578 views

orthonormal system in a Hilbert space

Let $\{e_n\}$ be an orthonormal basis for a Hilbert space $H$. Let $\{f_n\}$ be an orthonormal set in $H$ such that $\sum_{n=1}^{\infty}{\|f_n-e_n\|}<1$. How do I show that $\{f_n\}$ is also an ...
10
votes
1answer
721 views

An approximate eigenvalue for $ T \in B(X) $.

This is a problem from Conway’s Functional Analysis: Definition An approximate eigenvalue for $ T \in B(X) $ is a scalar $ \lambda $ such that there is a sequence of unit vectors $ x_{n} \in X $ ...
8
votes
2answers
176 views

Prove of inequality under a Hilbert space.

Let $x\neq y$ when $x,y\in H$ and H is a Hilbert space which satisfy $\|x\|=\|y\|=r$. Show that $\|\frac{x+y}{2}\|<r$. Actually in my question r=1 but as far as i could understand there is a way ...
5
votes
1answer
241 views

If $\sum (a_n)^2$ converges and $\sum (b_n)^2$ converges, does $\sum (a_n)(b_n)$ converge?

Could someone help me to solve this or at least give me a hint?, I have tried using Cauchy's criterion, the Dirichlet test for convergence, etc, but I can´t prove it.Honestly I don´t know where to ...