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Chapter 6: Problem 14

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).

Short Answer

Expert verified
Since \(T\), the linear transformation defined by \((e_k - v_k)\), is invertible, \((v_n)\) forms a basis for \(V\).

Step by step solution

01

Understand the Problem Statement

We are given a Banach space \(V\) with a known basis \((e_n)\). There is another sequence \((v_n)\) in the same space that satisfies a condition involving a constant \(\lambda\), where \(0 \leq \lambda < 1\). We need to prove that the sequence \((v_n)\) also forms a basis for \(V\).
02

Definition of a Basis

Recall that a basis in a Banach space \(V\) is a sequence \((e_n)\) such that any element in \(V\) can be uniquely written as a convergent series of scalar multiples of elements of \((e_n)\). Our goal is to show that \((v_n)\) meets this criterion.
03

Consider the Given Inequality

For any finite set of coefficients \(c_k\), we have:\[\left\|\sum_{m}^{n} c_{k}\left(e_{k}-v_{k}\right)\right\| \leq \lambda\left\|\sum_{m}^{n} c_{k} e_{k}\right\|.\]This implies that \(\sum_{m}^{n} c_{k} v_{k}\) is close to \(\sum_{m}^{n} c_{k} e_{k}\) for sufficiently small \(\lambda\).
04

Express Elements in Terms of (v_n)

Since \((e_n)\) is a basis for \(V\), any vector \(x\in V\) can be expressed as \(x = \sum a_k e_k\). We aim to write \(x\) similarly using \((v_n)\), demonstrating that \(x = \sum b_k v_k\) for some scalars \(b_k\).
05

Observing Bounded Linear Transformation

Let \(T: V \rightarrow V\) be the transformation such that \(T(e_k) = e_k - v_k\). Given that the norm of \(T\) is strictly less than 1, i.e., \(\|T\| < 1\), \(T\) is an invertible operator, meaning \(T^{-1}\) exists. We seek \(b_k\) such that:\[x = T^{-1}\left(\sum a_k e_k\right) = \sum b_k v_k.\]
06

Conclusion of the Basis Perturbation Result

The condition \(\|T\| < 1\) ensures \(T\) is invertible, allowing every vector \(x\) in \(V\) to be uniquely represented with respect to \((v_n)\). Hence, \((v_n)\) itself is a basis for \(V\) due to the same invertibility condition.

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Key Concepts

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

Basis
In the context of a Banach space, a basis is a fundamental concept. It allows us to express every element of the space uniquely as a series of scalar multiples of the basis elements. Think of a basis as a set of building blocks. Just like in geometry, where any point in space can be reached using x, y, and z coordinates, a basis in a Banach space allows every possible vector in that space to be constructed. The sequence \(e_n\) in the problem is such a set of building blocks, where each vector in space can be decomposed into a series of these elements multiplied by some scalar coefficients. This ensures that every vector in the space is represented entirely and uniquely using this set.
Sequence
A sequence in a Banach space, like \(v_n\) or \(e_n\), is an ordered set of elements from within the space. Think of a sequence like a list that goes on either finitely or infinitely, where the order matters significantly. For the basis under consideration, both sequences \(e_n\) and \(v_n\) are crucial.
  • In the problem, \(v_n\) needs to satisfy a specific inequality which relates it to \(e_n\).
  • This inequality ensures that the sequence \(v_n\) approximates \(e_n\) closely enough that it can serve as another basis for the space.
These sequences are pivotal because they allow for the precise construction of any vector within the Banach space via finite or infinite sums. In essence, they are like continuous maps that extend throughout the space, touching every possible vector.
Invertibility
Invertibility is an interesting property when dealing with linear transformations. It refers to the ability of a function to be "reversed," so that each input is uniquely paired with an output, and vice versa. In the context of this problem, the significance of invertibility arises with the bounded linear transformation \(T(e_k) = e_k - v_k\).
  • The inequality provided in the exercise, \[\|T\| < 1\], tells us that the transformation \(T\) is continuous and bounded below a specific threshold.
  • This means \(T\) is invertible, allowing us to recover any original vector from its image through \(T\).
Knowing that \(T\) is invertible is crucial because it signifies that the sequence \(v_n\) doesn't "lose" any information compared to the original basis \(e_n\), and can itself serve as a complete basis for the Banach space.
Linear Transformation
A linear transformation is a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. Essentially, if you apply the transformation to a combination of vectors, it's the same as applying it to each one individually and then combining the results. In the exercise, the transformation \(T\) is created from \(e_k\) and \(v_k\) as \(T(e_k) = e_k - v_k\). Here's why
  • The operation ensures that each vector from the original sequence gets translated to a corresponding one in the new sequence while preserving its essential properties.
  • This kind of transformation allows us to work seamlessly across different bases without losing the fundamental structure of vectors in the Banach space.
Thus, understanding linear transformations helps in grasping how transformations can seamlessly alter bases without compromising the space's underlining structure, a key necessity when proving \(v_n\) as a basis here.

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Most popular questions from this chapter

In the inner product space \(V\) we have $$ \|f+g\| \leq\|f\|+\|g\| $$ for all \(f, g \in V\), with equality if and only if one at least of \(f\) and \(g\) is a non-negative multiple of the other one.

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 \(f \in V\).

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, \(T A f=f\).

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 the Hilbert space \(V\) and let \(\left(c_{n}: n=1,2, \ldots\right)\) be a sequence of complex numbers. Then \(\sum_{1}^{\infty}\left|c_{n}\right|^{2}\) converges if and only if there exists an element \(f \in V\) such that \(c_{n}=\left(f, e_{n}\right)\) for every \(n\).
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