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Chapter 3: Problem 11

Show that every infinite, orthonormal sequence in a Hilbert space converges weakly to 0 .

Short Answer

Expert verified
Every infinite orthonormal sequence in a Hilbert space converges weakly to 0 by Bessel's inequality.

Step by step solution

01

Define the Terms

First, we need to understand the terms used. An orthonormal sequence \( \{ e_n \} \) in a Hilbert space \( H \) means that \( \langle e_n, e_m \rangle = 0 \) for \( n eq m \) and \( \| e_n \| = 1 \) for every \( n \). Weak convergence in a Hilbert space implies that for a sequence \( x_n \) to converge weakly to \( x \), we must have \( \langle x_n, y \rangle \to \langle x, y \rangle \) for all \( y \) in the Hilbert space.
02

Establish the Goal

We are tasked with showing that any infinite orthonormal sequence \( \{ e_n \} \) converges weakly to 0, which means showing that \( \langle e_n, y \rangle \to 0 \) for all \( y \in H \).
03

Consider the Inner Product

Take an arbitrary \( y \in H \) and note that it can be expressed using the orthonormal system \( \{ e_n \} \) by the Parseval's identity, which involves expressing \( y = \sum_{n=1}^{\infty} \langle y, e_n \rangle e_n \). The coefficients \( \langle y, e_n \rangle \) are the key to weak convergence.
04

Apply Bessel's Inequality

Use Bessel's inequality which states that \( \sum_{n=1}^{\infty} | \langle y, e_n \rangle |^2 \leq \| y \|^2 \). This implies that each \( |\langle y, e_n \rangle| \) must tend to 0, otherwise the sum could not converge as \( n \to \infty \).
05

Show Weak Convergence

Conclude that as \( n \to \infty \), \( \langle e_n, y \rangle = \langle y, e_n \rangle \to 0 \) due to Bessel's inequality ensuring the sum converges. Therefore, \( e_n \) converges weakly to 0 for any fixed \( y \).
06

Summarize the Proof

To summarize, we showed that for any \( y \in H \), the inner product \( \langle e_n, y \rangle \to 0 \) as \( n \to \infty \) due to Bessel's inequality and the properties of the inner product and orthonormal sequence. Hence, the sequence \( \{ e_n \} \) converges weakly to 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Orthonormal Sequence
An orthonormal sequence in a Hilbert space is a very special sequence of vectors. These vectors have two key properties that distinguish them from just any sequence:
  • Orthogonal: Any two distinct vectors in the sequence are perpendicular to each other. Mathematically, this means their inner product is zero (\( \langle e_n, e_m \rangle = 0 \text{ for } n eq m \)).
  • Normalized: Every vector in the sequence has a length (or norm) of one (\( \| e_n \| = 1 \)).
These properties make orthonormal sequences particularly useful in mathematical analysis and applications such as signal processing and quantum mechanics. They essentially provide a standardized 'coordinate system' within the infinite-dimensional world of Hilbert spaces.
Hilbert Space
A Hilbert space is a generalization of Euclidean space to possibly infinite dimensions. It is a complete inner product space, meaning two things:
  • Inner Product: It has an operation that combines two vectors to produce a scalar, somewhat akin to dot product in 3D space.
  • Complete: Every Cauchy sequence of vectors within the space converges to a limit that is also within the space.
These properties provide a rich framework to handle problems in functional analysis and quantum mechanics. Hilbert spaces are crucial in studying 'continuous' phenomena in infinite dimensions where familiar geometric intuition often falls short. They support analyses of waves, heat distribution, and more, through decomposing functions and finding solutions.
Bessel's Inequality
Bessel's Inequality is an important result related to orthonormal sequences in a Hilbert space. It provides a bound for the sum of squares of the coefficients of an expansion of a vector in terms of an orthonormal sequence.When you express a vector \( y \) in a Hilbert space using an orthonormal sequence \( \{ e_n \} \), the coefficients \( \langle y, e_n \rangle \) relate how much "influence" each vector in the sequence has on \( y \). Bessel's inequality states:\[\sum_{n=1}^{ ext{infinity}} |\langle y, e_n \rangle|^2 \leq \| y \|^2\]This inequality tells us that the collective 'weight' of these coefficients, squared and summed up, can never exceed the squared length of the vector \( y \). This insight is crucial in various mathematical proofs and applications, especially when dealing with convergent series and expansions.
Parseval's Identity
Parseval's Identity is a fundamental concept that ties together orthonormal sequences and Hilbert spaces. It can be seen as an extended version of Bessel's inequality that holds when you have a complete sequence (also known as an orthonormal basis).For a vector \( y \) in a Hilbert space, if \( \{ e_n \} \) is a complete orthonormal sequence, Parseval's Identity states:\[\sum_{n=1}^{ ext{infinity}} |\langle y, e_n \rangle|^2 = \| y \|^2\]This means that when using a complete orthonormal sequence, you can perfectly reconstruct the vector \( y \) by summing the contributions of each sequence element. Parseval's identity ensures that the total energy or 'power' of the vector \( y \) (indicated by its squared norm) is exactly partitioned among its components. This result finds applications in signal processing, quantum mechanics, and many field equations, providing a way to precisely decompose and analyze complex phenomena.

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Most popular questions from this chapter

(Weighted shifts.) Given a bounded sequence \(\left(\lambda_{n}\right)\) in \(\mathbb{C}\) define an operator \(S\) in \(\mathbf{B}\left(\ell^{2}\right)\) by \((S x)_{1}=0\) and $$ (S x)_{n}=\lambda_{n} x_{n-1}, \quad n>1, \quad \text { for } x=\left(x_{n}\right) \text { in } \ell^{2} . $$ Find the polar decomposition of \(S\), and characterize those sequences \(\left(\lambda_{n}\right)\) in \(\ell^{\infty}\) for which \(S\) is compact.

(Unitary dilation.) For each \(T\) in \(\mathbf{B}(\mathfrak{S})\) with \(\|T\| \leq 1\) define the operator \(U\) in \(\mathbf{B}(\mathfrak{\mathcal} \oplus \mathfrak{5})\) by $$ U=\left(\begin{array}{cc} T & \left(I-T T^{*}\right)^{1 / 2} \\ \left(I-T^{*} T\right)^{1 / 2} & -T^{*} \end{array}\right) $$ Show that \(T\left(I-T^{*} T\right)^{1 / 2}=\left(I-T T^{*}\right)^{1 / 2} T\), and use this to prove that \(U\) is unitary.

(Fourier transform.) In this and the next two exercises we use the normalized Lebesgue integral on \(\mathbb{R}\) [i.e. the usual one divided by \(\left.(2 \pi)^{1 / 2}\right]\). One consequence of this is that \(\int g_{0}(x) d x=1\), where \(g_{0}(x)=\) \(\exp \left(-\frac{1}{2} x^{2}\right) .\) We shall use the abbreviated notation \(L^{p}\) for \(L^{p}(\mathbb{R})\). For each \(f\) in \(L^{1}\) define \(f(y)=\int f(x) \exp (-\mathrm{i} x y) d x\). (a) Show that if \(f \in L^{1}\) and id \(\cdot f \in L^{1}[\) where \(i d(x)=x]\), then \(f\) is a differentiable function and \(f^{\prime}(y)=-\mathrm{i}(i d \cdot f)^{\wedge}(y)\). (b) Show that if \(f\) is differentiable and both \(f\) and \(f^{\prime}\) belong to \(L^{1}\), then \(\left(f^{\prime}\right)^{\wedge}(y)=\mathrm{i} y f(y)\). (c) Let \(\mathscr{S}\) (the Schwartz space) denote the space of \(C^{\infty}\)-functions on \(\mathbb{R}\) such that \((i d)^{n} f^{(m)} \in L^{1}\) for all \(n\) and \(m\). Use (a) and (b) to show that \(f \in \mathscr{S}\) for every \(f\) in \(\mathscr{S}\). (d) Show that the map \(F: f \rightarrow f\) belongs to \(\mathbf{B}\left(L^{1}, C_{0}(\mathbb{R})\right)\). Hints: (a) Use Lebesgue's theorem on majorized convergence (6.1.15) on the difference quotient of \(f .\) (b) Use integration by parts and the fact that \(f \in C_{0}(\mathbb{R})\). (d) For any polynomial \(p\) we have \(p g_{0} \in \mathscr{S}\). Use this to show that \(\mathscr{S}\) is dense in \(L^{1}\).

Let \(\left\\{e_{n} \mid n \in \mathbb{N}\right\\}\) be an orthonormal basis for the Hilbert space \(\mathfrak{y}\), and define for each \(T\) in \(\mathbf{B}(\mathfrak{)})\) the doubly infinite matrix \(A=\left(\alpha_{n m}\right)\), by letting \(\alpha_{n m}=\left(T e_{m} \mid e_{n}\right) .\) Show that every row and every column in \(A\) is square summable (i.e. belongs to \(\ell^{2}\) ). Use this to prove that the matrix product \(A B=C, C=\left(\gamma_{n m}\right)\), where \(\gamma_{n m}=\sum_{k} \alpha_{n k} \beta_{k m}\), is welldefined when the matrices \(A\) and \(B\) correspond to operators \(T\) and \(S\) in \(\mathbf{B}(\mathfrak{)})\). Show that \(C\) is the matrix corresponding to the operator \(T S\). Also find the matrices corresponding to the operators \(S+T\) and \(T^{*}\).

(Fourier transform.) In this and the next two exercises we use the normalized Lebesgue integral on \(\mathbb{R}\) [i.e. the usual one divided by \(\left.(2 \pi)^{1 / 2}\right]\). One consequence of this is that \(\int g_{0}(x) d x=1\), where \(g_{0}(x)=\) \(\exp \left(-\frac{1}{2} x^{2}\right) .\) We shall use the abbreviated notation \(L^{p}\) for \(L^{p}(\mathbb{R})\). For each \(f\) in \(L^{1}\) define \(f(y)=\int f(x) \exp (-\mathrm{i} x y) d x\). (a) Show that if \(f \in L^{1}\) and id \(\cdot f \in L^{1}[\) where \(i d(x)=x]\), then \(f\) is a differentiable function and \(f^{\prime}(y)=-\mathrm{i}(i d \cdot f)^{\wedge}(y)\). (b) Show that if \(f\) is differentiable and both \(f\) and \(f^{\prime}\) belong to \(L^{1}\), then \(\left(f^{\prime}\right)^{\wedge}(y)=\mathrm{i} y f(y)\). (c) Let \(\mathscr{S}\) (the Schwartz space) denote the space of \(C^{\infty}\)-functions on \(\mathbb{R}\) such that \((i d)^{n} f^{(m)} \in L^{1}\) for all \(n\) and \(m\). Use (a) and (b) to show that \(f \in \mathscr{S}\) for every \(f\) in \(\mathscr{S}\). (d) Show that the map \(F: f \rightarrow f\) belongs to \(\mathbf{B}\left(L^{1}, C_{0}(\mathbb{R})\right)\). Hints: (a) Use Lebesgue's theorem on majorized convergence (6.1.15) on the difference quotient of \(f .\) (b) Use integration by parts and the fact that \(f \in C_{0}(\mathbb{R})\). (d) For any polynomial \(p\) we have \(p g_{0} \in \mathscr{S}\). Use this to show that \(\mathscr{S}\) is dense in \(L^{1}\).
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