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Chapter 1: Problem 50

Show that the set \(A=\left\\{x \in \ell_{2} ; \sum\left(1+\frac{1}{i}\right) x_{i}^{2} \leq 1\right\\}\) does not contain an element with norm equal to \(\sup \\{\|x\| ; x \in A\\}\). Hint: If \(x \in A\), then \(\|x\|^{2}=\sum x_{i}^{2}<\sum\left(1+\frac{1}{i}\right) x_{i}^{2} \leq 1=\sup _{A}\|x\|\).

Short Answer

Expert verified
No element in the set \( A \) has a norm equal to 1.

Step by step solution

01

- Understand the Set

The set is defined as \[A = \{ x \in \ell_2 ; \sum(1 + \frac{1}{i}) x_i^2 \leq 1 \} \].This means any vector \( x = (x_1, x_2, x_3, ...) \) in the set must satisfy the inequality involving the weighted sum of the squares of its elements.
02

- Supremum of Norms

We need to show that no element in the set \( A \) has a norm equal to \[ \sup _{A} \| x \| \].First, compute the norm of an element \( x \in \ell_2 \):\[ \| x \|^2 = \sum x_i^2 \].
03

- Relation Between Norm and Given Inequality

Given \[ \| x \|^2 = \sum x_i^2 < \sum (1 + \frac{1}{i}) x_i^2 \leq 1, \]we infer the inequality involving the norm:\[ \| x \|^2 < 1 \].
04

- Supremum Value

The supremum of the norms of elements in \( A \) is \[ \sup _{A} \| x \| = 1 \].However, by the previous inequality, \( \| x \| \) is always less than 1 for any \( x \in A \).
05

- Conclusion

Since \( \| x \| \) is strictly less than 1 for any \( x \in A \), no element in \( A \) has a norm exactly equal to 1. Hence, \[ \sup _{A} \| x \| \] is not achieved by any element in \( A \).

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Key Concepts

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

normed spaces
In functional analysis, a normed space is a vector space equipped with a function called the norm which assigns a length to each vector. For a given vector space V over a field 饾斀 (usually the real numbers 鈩� or complex numbers 鈩�), the norm is denoted as \(\big\forall x \in V, \|x\| \in [0, \infty)\).\br\br This norm satisfies three important properties:\br\br
  • **Non-negativity**: \(\big\forall x \in V, \|x\| \geq 0\)\br
    **Definiteness**: \(\big\forall x \in V, \|x\| = 0 \Rightarrow x = 0\)
    • \br
  • **Triangle Inequality**: \(\big\forall x, y \in V, \|x + y\| \leq \|x\| + \|y\|\)
supremum
The supremum (or least upper bound) of a set S of real numbers is the smallest real number that is greater than or equal to every number in S.\br\brThe concept of supremum is vital in functional analysis and plays an important role in understanding bounded sets.\br\brFor example, if S = \(\big\{ x \in \bbr ; x \leq 5 \}\), then the supremum of S is 5. It is the maximum value in the set.\br\brIn our exercise, we say \(\textrm{need to show that the set } A \{\textrm{ no element in } A \textrm{ has a norm equal to its supremum, } 1 \ \textrm{. This is crucial for proving that the set does not contain an element with the maximum norm.}\)
inequalities in norms
Inequalities involving norms are essential in functional analysis. They help in understanding how different elements relate to each other in terms of their length or size.\br\brConsider the inequality given in the exercise: \(\big\forall x \in A, \|x \|^2 = \sum x_i^2 < \sum (1 + \frac{1}{i}) x_i^2 \leq 1\).\br\brThis shows that for every element in A, its norm squared is always strictly less than 1. Thus, no element in A can achieve the maximum allowable value of its weighted sum of squares, which is exactly 1.\br\brBy understanding such inequalities, one can better grasp the constraints and properties of different sets in normed spaces.
Hilbert spaces
A Hilbert Space is a special kind of normed space that is complete with respect to the norm induced by an inner product. In simpler terms, a Hilbert space has an inner product that allows for the generalization of Euclidean space to infinite-dimensional spaces.\br\brKey properties of Hilbert spaces include:\br
  • **Inner Product**: This function takes two vectors and returns a scalar (real or complex), satisfying certain properties such as linearity, symmetry, and positivity.
    • \br\br
  • **Norm Induction**: The norm in a Hilbert space is induced by the inner product, and it is defined as \(\big\forall x \in \textrm{Hilbert space }, \|x\| = \sqrt{}.\)
    • \br\br
  • **Completeness**: Every Cauchy sequence (a sequence where elements get arbitrarily close to each other) in a Hilbert space converges to a limit within the space.
    • \br\br A common example of a Hilbert space is the space \(\bll^2\), which consists of all infinite sequences of real or complex numbers whose series of squares is convergent.\br\brUnderstanding Hilbert spaces is fundamental in many areas of math and physics, particularly in the study of quantum mechanics and Fourier analysis.

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    Most popular questions from this chapter

    Let \(K, C\) be subsets of a normed space \(X\). (i) Show that if \(K, C\) are closed, \(K+C\) need not be closed. (ii) Show that if \(K\) is compact and \(C\) is closed, then \(K+C\) is closed. Is \(\operatorname{con}(K \cup C)\) closed? (iii) Show that if \(K\) is compact and \(C\) is bounded and closed, then the set \(\operatorname{conv}(K \cup C)\) is closed. Hint: (i): Consider \(K=\\{(x, 0) ; x \in \mathbf{R}\\}\) and \(C=\left\\{\left(x, \frac{1}{x}\right) ; x>0\right\\}\). (ii): If \(x_{n}=k_{n}+c_{n} \rightarrow y\) for \(k_{n} \in K, c_{n} \in C\), then by compactness assume \(k_{n} \rightarrow k\); then also \(c_{n}=x_{n}-k_{n} \rightarrow(y-k)\) and use that \(C\) is closed. For a negative answer to \(\operatorname{conv}(K \cup C)\), see the previous exercise. (iii): If \(x_{n}=\lambda_{n} k_{n}+\left(1-\lambda_{n}\right) c_{n} \rightarrow x\) for \(k_{n} \in K, c_{n} \in C, \lambda \in[0,1]\) find a subsequence \(n_{i}\) such that \(k_{n,} \rightarrow k\) and \(\lambda_{n_{i}} \rightarrow \lambda\). If \(\lambda=1\), then by boundedness \(\left(1-\lambda_{n_{i}}\right) c_{n_{1}} \rightarrow 0\), and \(x=k \in K .\) If \(\lambda \neq 1, c_{n_{i}} \rightarrow \frac{x-\lambda k}{1-\lambda} \in C\) by closedness of \(C\).

    Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). If there is \(\delta>0\) such that \(\|T(x)\| \geq \delta\|x\|\) for all \(x \in X\), then \(T(X)\) is closed in \(Y .\) Moreover, \(T\) is an isomorphism from \(X\) into \(Y\). Hint: Let \(\left\\{T\left(x_{n}\right)\right\\} \subset T(X)\) be such that \(T\left(x_{n}\right) \rightarrow y\). Then \(\left\\{T\left(x_{n}\right)\right\\}\) is Cauchy and hence also \(\left\\{x_{n}\right\\}\) is Cauchy. By completeness, \(x_{n} \rightarrow x\), so \(T\left(x_{n}\right) \rightarrow T(x)\), by the uniqueness of the limit we get \(y=T(x)\). The inequality we assume implies that \(T\) is one-to-one, so the inverse \(T^{-1}: T(X) \rightarrow X\) is defined and \(\left\|T^{-1}\right\| \leq \frac{1}{\delta}\)

    Let \(X\) be the normed space obtained by taking \(c_{0}\) with the norm \(\|x\|_{0}=\sum 2^{-i}\left|x_{i}\right| .\) Show that \(X\) is not a Banach space. Note that this shows that \(\|\cdot\|_{0}\) is not an equivalent norm on \(c_{0}\). Hint: The sequence \(\\{(1,1, \ldots, 1,0, \ldots)\\}_{n=1}^{\infty}\) is Cauchy and not convergent since the only candidate for the limit would be \((1,1, \ldots) \notin c_{0}\).

    Show that a Banach space \(X\) is separable if and only if \(S_{X}\) is separable. Hint: If \(\left\\{x_{n}\right\\}\) is dense in \(S_{X}\), consider \(\left\\{r_{k} x_{n}\right\\}_{k, n}\) for some dense sequence \(\left\\{r_{k}\right\\}\) in \(\mathbf{K}\). If \(\mathcal{S}\) is countable and dense in \(X\), consider \(\left\\{\frac{x}{\|x\|} ; x \in \mathcal{S}\right\\}\).

    Show that \(\overline{\operatorname{span}}(L)=\overline{\operatorname{span}(L)}\) and \(\overline{\operatorname{conv}}(M)=\overline{\operatorname{conv}(M)}\). Hint: By the definition, \(\overline{\operatorname{span}}(L)\) is the intersection of all closed subspaces containing \(L\), and \(\overline{\operatorname{span}(L)}\) is one such subspace. From this, one inclusion follows. The other inclusion follows similarly because \(\operatorname{span}(L)\) is the intersection of all subspaces containing \(L .\)
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