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Let \(\left(e_{n}: n=1,2, \ldots\right)\) be an orthonormal sequence in the Hilbert space \(V\). Then the following statements are equivalent. (i) The sequence \(\left(e_{n}: n=1,2, \ldots\right)\) is complete. (ii) For any \(f \in V\) we have \(\|f\|^{2}=\sum\left|\left(f, e_{n}\right)\right|^{2}\). (iii) For any \(f \in V\) the series \(\sum\left(f, e_{n}\right) e_{n}\) converges to \(f\), i.e., \(\| f-\) \(\sum_{1}^{n}\left(f, e_{k}\right) e_{k} \| \rightarrow 0\) as \(n \rightarrow \infty\).

Short Answer

Expert verified
The statements (i), (ii), and (iii) are equivalent because each implies the others.

Step by step solution

01

Understand the Definitions

An orthonormal sequence in a Hilbert space means that for any two distinct elements \(e_i\) and \(e_j\), we have \( (e_i, e_j) = 0 \) and \( \|e_i\| = 1 \). Completeness of a sequence in this context means that any element in the Hilbert space can be expressed as a series involving these sequence elements.
02

Prove (i) implies (ii)

Assume the sequence \((e_{n})\) is complete. For any \(f \in V\), since the sequence is complete, it forms an orthonormal basis for \(V\). By Parseval's identity, we have \(\|f\|^2 = \sum_{n=1}^{\infty} |(f, e_n)|^2\). This proves statement (ii).
03

Prove (ii) implies (iii)

Assume statement (ii) is true. For any \(f \in V\), \(\|f\|^2 = \sum_{n=1}^{\infty} |(f, e_n)|^2\). The Fourier series \(\sum_{n=1}^{\infty} (f, e_n)e_n\) converges to \(f\) in the norm of the Hilbert space, which implies \(\|f - \sum_{1}^{n} (f, e_k)e_k\| \to 0\) as \(n \to \infty\). Thus, statement (iii) is true.
04

Prove (iii) implies (i)

Assume statement (iii) is true. For any \(f \in V\), the series \(\sum (f, e_n) e_n\) converges to \(f\). This indicates that the sequence is an orthonormal basis for \(V\) because every element in \(V\) can be represented as a linear combination of the elements \(e_n\). Therefore, \((e_{n})\) is complete.

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Key Concepts

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

Orthonormal sequence
In the world of Hilbert spaces, an orthonormal sequence is a sequence of vectors that have two key properties: orthogonality and norm equal to one. Orthogonality means that each vector in the sequence is perpendicular to all other vectors, which is mathematically expressed as - \[ (e_i, e_j) = 0 \] for any two distinct vectors \( e_i \) and \( e_j \). The second property is that each vector is normalized, meaning - \[ \| e_i \| = 1 \].This means they form a very special and balanced set of vectors in the space. An orthonormal sequence serves as a foundation for analyzing and constructing other vectors in the Hilbert space. Like building blocks, these vectors provide a structured way to approach problems involving Fourier series and other complex applications.
Completeness
Completeness is a critical property when discussing sequences in Hilbert spaces. When we say that an orthonormal sequence is complete in its Hilbert space, it means that any vector within the space can be constructed as a limit of some series formed using these orthonormal vectors. This idea is crucial because it ensures that the sequence can represent any function or vector in the space fully. It's analogous to having all the pieces of a puzzle; you can complete the picture without missing any parts. Mathematically, this means if a sequence can generate every possible vector in the space, it is complete and acts as a basis for that space. Being complete allows the sequence to fully describe the Hilbert space, which is the core idea behind many mathematical and engineering concepts, including signal processing.
Parseval's identity
Parseval's identity is an essential theorem that links a sequence's coefficients to its norm in a Hilbert space. The concept is simple but powerful. According to Parseval's identity, if - \[ \text{\((e_{n})\)} \] is an orthonormal sequence and \( f \) is in the Hilbert space, then - \[ \|f\|^2 = \sum_{n=1}^{\infty} |(f, e_n)|^2 \]. This equation essentially demonstrates how one can express the total energy or the square of the norm of \( f \) as the sum of the squares of its coefficients in the sequence.Parseval's identity plays a fundamental role in understanding Fourier series and other forms of series expansions by showing how different components contribute to the overall structure of a vector or function.
Fourier series
The Fourier series is a way to represent a function as a sum of sines and cosines, which can be very helpful in solving various problems in physics and engineering. In the context of a Hilbert space with an orthonormal sequence, the Fourier series helps break down any function into components that make up the whole.A Fourier series of a function \( f \) is expressed as - \[ \sum_{n=1}^{\infty} (f, e_n)e_n \], where \((f, e_n)\) are the coefficients that show how much each sine and cosine function contributes to the shape of \( f \).Fourier series enable us to work with complex functions in an easier way by simplifying them into sums of simple periodic functions, allowing for applications in signal processing, acoustics, and heat distribution, among others. By breaking the function into orthonormal components, we see clearer insights into the underlying behavior of \( f \).

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Most popular questions from this chapter

Let \(\left(e_{n}\right)\) be a basis in the Banach space \(V\) and let the sequence \(\left(v_{n}\right)\) in \(V\) have the property that there exists a number \(\lambda(0 \leq \lambda<1)\) such that, for all \(m \leq n\), $$ \left\|\sum_{m}^{n} c_{k}\left(e_{k}-v_{k}\right)\right\| \leq \lambda\left\|\sum_{m}^{n} c_{k} e_{k}\right\| $$ for all possible choices of the complex numbers \(c_{k}(k=m, \ldots, n)\). Show that \(\left(v_{n}\right)\) is a basis in \(V\). This criterion to perturb a basis so that it remains a basis is due to R.E.A.C. Paley and N. Wiener (1934).

Let \(\left(e_{n}\right)\) be a basis in the Banach space \(V\) and let \(T \in \mathcal{B}(V)\) satisfy \(\|I-T\|<1\). Write \(v_{n}=T e_{n}\) for all \(n\). Show that \(\left(v_{n}\right)\) is a basis in \(V\). For any \(f \in V\) denote the coefficients in the expansion of \(f\) with respect to \(\left(e_{n}\right)\) by \(f_{n}^{\wedge}\), i.e., \(f=\sum f_{n}^{\wedge} e_{n} .\) Denote the inverse operator of \(T\) by \(T^{-1}\). Given \(f \in V\), let \(g=T^{-1} f\). Then \(g=\sum g_{n}^{\wedge} e_{n}\), so $$ f=T g=\sum g_{n}^{\wedge} T e_{n}=\sum g_{n}^{\wedge} v_{n}=\sum\left(T^{-1} f\right)_{n}^{\wedge} v_{n} $$

Let \(\left(e_{n}: n=1,2, \ldots\right)\) be an orthonormal sequence in \(V\). Then $$ \sum_{1}^{\infty}\left|\left(f, e_{n}\right)\right|^{2} \leq\|f\|^{2} $$ for any \(f \in V\).

The trigonometric sequence \(\left(e^{i n x}: n=0, \pm 1, \pm 2, \ldots\right)\) is complete in \(L_{2}(\Pi, \mu), \mu\) Lebesgue measure. Now, let \(\Delta\) be an interval in \(\mathbb{R}\) of length greater than \(2 \pi\). Show that the trigonometric sequence is not complete in \(L_{2}(\Delta, \mu) .\) Note that it may be assumed that \(\Delta=(-\pi-\epsilon, \pi+\epsilon)\) for some \(\epsilon>0\).

Let \(V\) be a Banach space. As defined in section 3 , the linear operator \(T: V \rightarrow V\) is said to be bounded if there exists a number \(M \geq 0\) such that \(\|T f\| \leq M\|f\|\) for all \(f \in V\). The infimum of all \(M\) satisfying this condition is called the norm of \(T\) and denoted by \(\|T\|\). The space \(\mathcal{B}(V)\) of all bounded \(T: V \rightarrow V\) is a normed vector space with respect to \(\|T\|\) as the norm of \(T\). In fact, \(\mathcal{B}(V)\) is a normed algebra. It was shown in section 3 that $$ \|T\|=\sup (\|T f\| /\|f\|: f \neq 0)=\sup (\|T f\|:\|f\| \leq 1) $$ (i) Show that \(\mathcal{B}(V)\) is a Banach space. (ii) Let \(I\) be the identity operator in \(V\) and let \(T \in \mathcal{B}(V)\) satisfy \(\lambda=\) \(\|I-T\|<1\). Write \(S=I-T\), so \(\|S\|<1\). Show that for any \(f \in V\) the series \(f+S f+S^{2} f+\ldots\) converges in \(V\) and the sum \(A f\) satisfies \(\|A f\| \leq(1-\lambda)^{-1}\|f\| .\) Hence \(A \in \mathcal{B}(V) .\) Show that \(A\) is the inverse of \(T\) in \(\mathcal{B}(V)\), i.e., \(A T=T A=I .\) Hint: For (i), let \(\left(T_{n}\right)\) satisfy \(\left\|T_{m}-T_{n}\right\| \rightarrow 0\) as \(m, n \rightarrow \infty\). Then \(\left\|T_{m} f-T_{n} f\right\| \rightarrow 0\) as \(m, n \rightarrow \infty\) for every \(f \in V\), so \(\lim _{n} T_{n} f\) exists for every \(f \in V\). Write \(T f=\lim _{n} T_{n} f\). Show that \(T\) is a linear operator on \(V\) and \(T\) is bounded (since \(\left\|T_{n}\right\|\) is bounded, say \(\left\|T_{n}\right\| \leq M\) for all \(n\), we have \(\|T f\|=\lim \left\|T_{n} f\right\| \leq M\|f\|\) for every \(f\) ). Finally, for any \(n\) and any \(f\) with \(\|f\| \leq 1\), we have \(\left\|\left(T-T_{n}\right) f\right\|=\lim _{m \rightarrow \infty}\left\|\left(T_{m}-T_{n}\right) f\right\| \leq \sup _{m \geq n}\left\|T_{m}-T_{n}\right\|\), which is arbitrarily small for sufficiently large \(n\). For (ii), observe that $$ \left\|\sum_{m+1}^{n} S^{k} f\right\| \leq \sum\left\|S^{k} f\right\| \leq \sum \lambda^{k}\|f\| \rightarrow 0 $$ as \(m \rightarrow \infty\), so \(f+S f+S^{2} f+\ldots\) converges. Finally, for any \(f\), $$ A T f=A(I-S) f=A f-A S f=\left(f+S f+S^{2} f+\ldots\right)-\left(S f+S^{2} f+\ldots\right)=f . $$ Similarly, \(T A f=f\).

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