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Chapter 2: Problem 14

If \(d\) is a metric on a vector space \(X \neq\\{0\\}\) which is obtained from a norm, and \(\bar{d}\) is defined by $$ \tilde{d}(x, x)=0, \quad \vec{d}(x, y)=d(x, y)+1 \quad(x \neq y) $$ show that \(\bar{d}\) cannot be obtained from a norm.

Short Answer

Expert verified
\( \bar{d} \) cannot be obtained from a norm because it fails the triangle inequality.

Step by step solution

01

Understanding the Problem

We need to determine if \( \bar{d} \) can be derived from a norm. A key property of a norm-derived metric \( d \) is that it is positive definite and satisfies the triangle inequality \( d(x, y) \leq d(x, z) + d(z, y) \) for all \( x, y, z \). We will demonstrate whether \( \bar{d} \) satisfies these properties.
02

Considering the Properties of \( \bar{d} \)

Given \( \bar{d}(x, x) = 0 \), this satisfies one property of a norm-derived metric. However, for \( x eq y \), \( \bar{d}(x, y) = d(x, y) + 1 > 0 \), ensuring positive definiteness. We must now check the triangle inequality.
03

Checking the Triangle Inequality

Consider three distinct points \( x, y, z \) in the vector space. The triangle inequality should hold: \( \bar{d}(x, y) \leq \bar{d}(x, z) + \bar{d}(z, y) \). Replacing \( \bar{d} \), we need: \[ d(x, y) + 1 \leq (d(x, z) + 1) + (d(z, y) + 1) = d(x, z) + d(z, y) + 2. \]
04

Analyzing the Violated Condition

Rewriting, the inequality becomes \( d(x, y) + 1 \leq d(x, z) + d(z, y) + 2 \), reduced to \( d(x, y) \leq d(x, z) + d(z, y) + 1 \). However, if \( d(x, z) + d(z, y) = d(x, y) \), the original metrics should satisfy \( d(x, y) = d(x, z) + d(z, y) \), without the addition of 1. Therefore, \( \bar{d} \) fails the triangle inequality.
05

Conclusion

Since \( \bar{d} \) does not satisfy the triangle inequality, it cannot be obtained from a norm. Thus, \( \bar{d} \) is not a valid metric derived from any norm on the vector space.

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Key Concepts

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

Norm
A norm is a fundamental concept in mathematics, especially in vector spaces. It is essentially a function that assigns a non-negative length or size to each vector in the space.
Naturally, the properties of a norm are crucial to understanding the structure of a vector space:
  • **Non-negativity**: For any vector \( v \), the norm \( \|v\| \geq 0 \), and \( \|v\| = 0 \) if and only if \( v \) is the zero vector.
  • **Scalar multiplication**: For any scalar \( c \) and any vector \( v \), \( \|cv\| = |c| \cdot \|v\| \).
  • **Triangle inequality**: For any vectors \( u \) and \( v \), \( \|u + v\| \leq \|u\| + \|v\| \).
These properties are what make norms useful in creating a well-defined metric on a vector space, helping us measure "distances" or "lengths." When a metric is derived from a norm, such a metric is often called a 'norm-induced' metric.
Triangle Inequality
The triangle inequality is an essential component of metric spaces and norms. It effectively states how the sum of the lengths of two sides of a triangle must be at least the length of the third side.
In formal terms, given a norm \( \| \cdot \| \), the triangle inequality asserts that for any vectors \( x \), \( y \) in a vector space, the inequality \( \|x + y\| \leq \|x\| + \|y\| \) always holds. For metrics, this translates into:
  • \(d(x, y) \leq d(x, z) + d(z, y)\)
The breakdown often comes when this inequality is violated, as seen with \( \bar{d} \).
Here, a constant addition to the distance function complicates adherence to the triangle inequality, revealing why \( \bar{d} \) fails to be norm-derived.
Vector Space
A vector space is a mathematical structure formed by a collection of vectors. Vectors in this space can be added together and multiplied by scalars to produce another vector within the same space.
To comprehend a vector space, we must acknowledge its defining features:
  • **Closed under addition and scalar multiplication**: If you add two vectors from the space, the result is still within that space, and similarly for scalar multiplication.
  • **Contains a zero vector**: This is the identity element for vector addition, the vector that doesn't change others when added.
  • **Associative and commutative properties**: Addition in vector spaces behaves in a user-friendly way.
By combining these properties, vector spaces support deep explorations of linear algebra and geometry, where norms and metrics come into play to measure and assess various properties.
Positive Definiteness
Positive definiteness is a key property for norms and related metrics, which ensures that distances and lengths are measured correctly and without ambiguity.
A metric or norm is considered positively definite if:
  • For any vector \( x \), \( \|x\| \geq 0 \), with equality if and only if \( x \) is the zero vector.
  • This ensures that non-zero vectors cannot have a zero length, maintaining meaningful measurements.
In the context of the exercise, although \( \bar{d} \) seems to showcase positive definiteness at first due to \( \bar{d}(x, x) = 0 \) and \( \bar{d}(x, y) = d(x, y) + 1 > 0 \) for \( x eq y \), the triangle inequality violation disrupts the overall acceptance as a norm-derived metric.
Thus, while individual properties like positive definiteness might hold, failing any core aspect such as the triangle inequality disqualifies it from being derived from a norm.

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Most popular questions from this chapter

If \(f\) is a bounded linear functional on a complex normed space, is \(\bar{f}\) bounded? Linear? (The bar denotes the complex conjugate.)

If \(X\) is a normed space and \(\operatorname{dim} X=\infty\), show that the dual space \(X^{\prime}\) is not identical with the algebraic dual space \(X^{*}\).

If \(x\) and \(y\) are different vectors in a finite dimensional vector space \(X\), show that there is a linear functional \(f\) on \(X\) such that \(f(x) \neq f(y)\).

Show that the closure \(\bar{Y}\) of a subspace \(Y\) of a normed space \(X\) is again a vector subspace.

(Space \(\left.C^{\prime}[a, b]\right)\) The space \(C^{1}[a, b]\) or \(C^{\prime}[a, b]\) is the normed space of all continuously differentiable functions on \(J=[a, b]\) with norm defined by $$ \|x\|=\max _{t \in J}|x(t)|+\max _{t \in J}\left|x^{\prime}(t)\right| $$ Show that the axioms of a norm are satisfied. Show that \(f(x)=x^{\prime}(c)\), \(c=(a+b) / 2\), defines a bounded linear functional on \(C^{\prime}[a, b] .\) Show that \(f\) is not bounded, considered as a functional on the subspace of \(C[a, b]\) which consists of all continuously differentiable functions.
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