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Chapter 1: Problem 68

Find a vector space \(X\) with two norms on it such that both of them are complete norms and they are not equivalent. Hint: Take a vector space \(V\) of linear dimension \(c\), and let \(T_{1}\) and \(T_{2}\) be linear bijections of \(V\) onto \(\ell_{2}\) and \(\ell_{4}\), respectively. Define norms on \(V\) by \(\|x\|_{1}=\left\|T_{1}(x)\right\|_{2}\) and \(\|x\|_{2}=\left\|T_{2}(x)\right\|_{4}\). Then \(\left(V_{1}\|\cdot\|_{1}\right)\) is isomorphic to \(\ell_{2}\) and \(\left(V_{i}\|\cdot\|_{2}\right)\) is isomorphic to \(\ell_{4} .\) Since \(\ell_{2}\) is not isomorphic to \(\ell_{4}\) (see exercises above), \(\|\cdot\|_{1}\) and \(\|\cdot\|_{2}\) are not equivalent.

Short Answer

Expert verified
The vector space \(V\) with norms \(\|x\|_1 = \|T_1(x)\|_2\) and \(\|x\|_2 = \|T_2(x)\|_4\) are not equivalent.

Step by step solution

01

- Understand the problem and identify the vector space

A vector space with two norms is needed such that both norms are complete, but not equivalent. The hint suggests using a vector space of linear dimension \( c \) and bijections from this vector space to function spaces \( \ell_{2} \) and \( \ell_{4} \).
02

- Define the vector space \( V \)

Consider a vector space \( V \) with linear dimension \( c \). \( T_{1} \) and \( T_{2} \) are linear bijections mapping \( V \) onto \( \ell_{2} \) and \( \ell_{4} \), respectively.
03

- Define the norms

Define two norms on \( V \): \( \|x\|_{1} = \|T_{1}(x)\|_{2} \) and \( \|x\|_{2} = \|T_{2}(x)\|_{4} \).
04

- Analyze completeness of norms

\( \ell_{2} \) and \( \ell_{4} \) are both complete metric spaces. Since \( T_{1} \) and \( T_{2} \) are bijections, \( (V, \| \cdot \|_{1}) \) is complete under \( \|\cdot\|_{1} \), and \( (V, \|\cdot\|_{2}) \) is complete under \( \|\cdot\|_{2} \).
05

- Check isomorphisms

\( (V, \| \cdot \|_{1}) \) is isomorphic to \( \ell_{2} \), and \( (V, \| \cdot \|_{2}) \) is isomorphic to \( \ell_{4} \).
06

- Analyze non-equivalence

Since \( \ell_{2} \) is not isomorphic to \( \ell_{4} \), the norms \( \| \cdot \|_{1} \) and \( \| \cdot \|_{2} \) on \( V \) are not equivalent.

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Key Concepts

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

Complete Norms
A normed vector space is called complete if every Cauchy sequence in the space converges to an element within the space. In simpler words, there are no 'holes' where a sequence could get infinitely close but never actually reach a point in the space. This is an essential property for many analytical tools to work correctly and is foundational in functional analysis.
In the provided problem, we have two norms defined on a vector space \( V \). The norms come from function spaces \( \ell_2 \) and \( \ell_4 \). These function spaces are examples of complete metric spaces. Therefore, the constructed norms \( \| \cdot \|_1 \) and \( \| \cdot \|_2 \) are also complete. This completeness implies that every Cauchy sequence in \( (V, \| \cdot \|_1) \) and \( (V, \| \cdot \|_2) \) converges within their respective spaces, making further analysis reliable.
Keep in mind, completeness ensures that the space can support limits of sequences, which is crucial in various mathematical and engineering contexts.
Isomorphisms of Function Spaces
Isomorphisms are mappings between spaces that preserve the structure of those spaces. In this context, an isomorphism between two normed vector spaces means there is a bijective linear map that also preserves the norm. For example, if a space \( V \) is mapped to \( \ell_2 \) through a bijection that maintains the norm, then \( V \) and \( \ell_2 \) are isomorphic as normed spaces.
The problem provides bijections \( T_1 \) and \( T_2 \), which map the vector space \( V \) to \( \ell_2 \) and \( \ell_4 \) respectively. This means \( (V, \| \cdot \|_1)\) is isomorphic to \( \ell_2 \) and \( (V, \| \cdot \|_2) \) is isomorphic to \( \ell_4 \).
This isomorphism ensures that the norm structures are inherently maintained under the mappings, making \( T_1 \) and \( T_2 \) crucial for our analysis. It's important to note that even though \( \ell_2 \) and \( \ell_4 \) are complete, and their norms are preserved, the two spaces' norms do not equate exactly, passing on these properties to \( V \).
Vector Space Dimensions
Vector space dimension refers to the number of basis vectors in the space, indicating its 'size' in terms of linear independence. A vector space \( V \) with linear dimension \( c \) implies \( V \) has \( c \) basis vectors, forming a coordinate system within which any vector in the space can be expressed.
The problem specifies using a vector space with linear dimension \( c \). The specific dimension doesn't change the result but suggests that any infinite-dimensional space will work if the mappings (bijections) to \( \ell_2 \) and \( \ell_4 \) can be established.
Understanding the concept of dimensions is crucial because it helps recognize why \( \ell_2 \) and \( \ell_4 \) are not isomorphic. While both spaces are infinite-dimensional, their structural properties differ. Hence, the norms from these mappings will not be equivalent on \( V \), ensuring that \( \| \cdot \|_1 \) and \( \| \cdot \|_2 \) are also non-equivalent, exactly as required by the exercise.

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Most popular questions from this chapter

Show that \(\ell_{4}\) is not isomorphic to a subspace of \(\ell_{2}\). Hint: Show that \(\ell_{4}\) is not of cotype 2 by considering the standard unit vectors.

Let \(C\) be a convex set in a normed space \(X\), assume \(\operatorname{Int}(C) \neq \emptyset\) (recall that \(\operatorname{Int}(C)\) denotes the interior of \(C\) ). Show that \(\overline{\operatorname{Int}(C)}=\bar{C}\) and \(\operatorname{Int}(\bar{C})=\operatorname{Int}(C)\) Hint: Use the "cone" argument. There is a point \(x_{0}\) and an open ball \(B_{b}^{\circ}\) such that \(x_{0}+B_{\delta}^{O} \subset C .\) Note that if we take any \(c \in C\), then by the convexity of \(C\), the cone with the vertex \(c\) and the base \(x_{0}+B_{\delta}^{O}\) is a subset of \(C\), and all points in this cone but the vertex \(c\) are its interior points (draw a picture!). Therefore, given \(c \in C\), we can find points from \(\operatorname{Int}(C)\) arbitrarily close to \(c\), showing that \(\bar{C} \subset \overline{\operatorname{Int}(C)}\) Given \(c \in \operatorname{Int}(\bar{C})\), there is \(\varepsilon>0\) such that \(c+\varepsilon B_{X} \subset \bar{C} .\) Let \(d=\) \(c+\varepsilon\left(c-x_{0}\right) .\) Then \(d \in \bar{C}\), so for every \(\nu>0\) there is \(y \in C\) with \(\|d-y\|<\nu\) If \(\nu\) is small (namely \(\left.\nu \leq \frac{\varepsilon \delta}{\sqrt{\delta^{2}+\left\|x_{0}-c\right\|^{2}}}\right), c\) is in the cone with the base \(x_{0}+B_{\delta}^{O}\) and vertex \(y\), so \(c \in \operatorname{Int}(C)\).

Let \(T\) be a one-to-one bounded linear operator from a normed space \(X\) into a normed space \(Y\). Show that \(T\) is an isometry onto \(Y\) if and only if \(T\left(B_{X}\right)=B_{Y}\) if and only if \(T\left(S_{X}\right)=S_{Y}\) if and only if \(T\left(B_{X}^{O}\right)=B_{Y}^{O}\), where \(B_{X}^{O}\) is the open unit ball in \(X\). Hint: By homogeneity, \(T\) is an isometry onto \(Y\) if and only if \(T\left(S_{X}\right)=S_{Y}\). Assume that \(T\left(B_{X}\right)=B_{Y}\). If there is \(x \in S_{X}\) such that \(\|T(x)\|=C<1\) then \(\|x / C\|>1\) and \(\|T(x / C)\|=1 .\) But there must be \(y \in B_{X}\) such that \(T(y)=T(x / C)\), a contradiction with \(T\) being one-to-one.

Let \(X\) be a Banach space and \(G\) be a subspace of \(X\) that is a \(G_{\delta}\) set in \(X\). Show that \(G\) is closed in \(X\). Hint: Let \(S=\bar{G} \backslash G\), where \(\bar{G}\) denotes the closure of \(G .\) Since \(G\) is a \(G_{\delta}\) set in \(X, G\) is \(G_{\delta}\) in \(\bar{G}\). Hence \(G=\bigcap G_{n}\), where \(G_{n}\) are open subsets in \(\bar{G}\). Therefore, \(S=\bigcup\left(\bar{G} \backslash G_{n}\right)\) and each \(G_{n}\) is dense in \(\bar{G}\) because it contains \(G .\) Thus, each \(\bar{G} \backslash G_{n}\) is nowhere dense in \(\bar{G}\), so \(S\) is of first category in \(\bar{G}\). We will show that \(S\) is an empty set. If \(x_{0} \in S\), consider the set \(G^{*}=\left\\{x_{0}+x ; x \in G\right\\}\). Note that \(G^{*} \subset S\). Indeed, any point \(x_{0}+x, x \in G\), is in \(\bar{G}\) since \(\bar{G}\) is a linear set. If for some \(x \in G\) we have \(x_{0}+x \in G\), then by linearity of \(G\) we have \(x_{0} \in G_{1}\) a contradiction. Therefore, \(G^{*} \subset S\) and \(G^{*}\) is of first category in \(\bar{G} .\) Thus, the shift \(G\) of \(G^{*}\) is also of first category in \(\bar{G}\). Hence, the whole \(\bar{G}\) as a union of \(G\) and \(S\), which are both of first category in \(\bar{G}\), is of first category in itself. This is a contradiction, since \(\bar{G}\) is a complete metric space.

Let \(A\) and \(B\) be two convex sets in a normed space \(X\). Show that \(\operatorname{conv}(A \cup B)=\\{\lambda x+(1-\lambda) y ; x \in A, y \in B, \lambda \in[0,1]\\}\) Hint: Show first that the set on the right-hand side is convex.
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