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Chapter 7: Problem 12

Let \(C^{1}([a, b], \mathbb{R})\) be the set of once continuously differentiable functions on \([a, b] .\) Define $$d(f, g):=\|f-g\|_{u}+\left\|f^{\prime}-g^{\prime}\right\|_{u}$$ where \(\|\cdot\|_{u}\) is the uniform norm. Prove that \(d\) is a metric.

Short Answer

Expert verified
Metric is verified as it satisfies non-negativity, symmetry, identity of indiscernibles, and triangle inequality.

Step by step solution

01

Definition of a Metric Space

A metric space requires a set \(X\) and a metric \(d: X \times X \to \mathbb{R}\) that satisfies:1. Non-negativity: \(d(x, y) \geq 0\) with equality if and only if \(x = y\).2. Symmetry: \(d(x, y) = d(y, x)\).3. Triangle Inequality: \(d(x, z) \leq d(x, y) + d(y, z)\) for all \(x, y, z \in X\).4. Identity of indiscernibles: \(d(x, x) = 0\).
02

Non-negativity and Identity of Indiscernibles

We first show that \(d(f, g) = \|f-g\|_u + \|f'-g'\|_u \geq 0\) because the uniform norm is always non-negative. \(d(f, g) = 0\) only if \(\|f-g\|_u = 0\) and \(\|f'-g'\|_u = 0\), which implies \(f = g\) and \(f' = g'\) by the properties of the uniform norm.
03

Symmetry Property

Since the uniform norm satisfies symmetry, we have \(\|f-g\|_u = \|g-f\|_u\) and \(\|f'-g'\|_u = \|g'-f'\|_u\). Therefore, \(d(f, g) = \|f-g\|_u + \|f'-g'\|_u = \|g-f\|_u + \|g'-f'\|_u = d(g, f)\), satisfying symmetry.
04

Triangle Inequality

For the triangle inequality, we need \(d(f, h) = \|f-h\|_u + \|f'-h'\|_u \leq (\|f-g\|_u + \|f'-g'\|_u) + (\|g-h\|_u + \|g'-h'\|_u)\). Applying the triangle inequality of the uniform norm, \(\|f-h\|_u \leq \|f-g\|_u + \|g-h\|_u\) and \(\|f'-h'\|_u \leq \|f'-g'\|_u + \|g'-h'\|_u\), confirm this property as well.

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Key Concepts

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

Continuously Differentiable Functions
In mathematics, a continuously differentiable function is one that has a derivative, and that derivative is continuous. This means these functions not only change smoothly but also without any abrupt changes in their behavior or slope. They are particularly important in calculus and analysis because they allow the use of powerful tools like Taylor series expansion.

Formally, a function \( f \) is continuously differentiable on an interval \( [a, b] \), and can be written as \( C^1([a, b], \mathbb{R}) \). Here's what this means:
  • The function \( f \) is defined and differentiable for all points \( x \) in the interval \( [a, b] \).
  • The derivative of \( f \), denoted as \( f'(x) \), is continuous on \( [a, b] \).
Understanding continuously differentiable functions allows us to ensure smooth transitions in models and simulations, making them a cornerstone of mathematical analysis.
Uniform Norm
The uniform norm is a way to measure the size of a function with respect to how far it deviates from zero over its entire domain. For a function \( f \) defined on an interval \([a, b]\), the uniform norm is denoted by \( \|f\|_u \) and is defined as:
  • \( \|f\|_u = \sup_{x \in [a, b]} |f(x)| \)
This means taking the supremum (or the least upper bound) of the absolute values of the function over the interval.

Consider the uniform norm like measuring the highest peak of a mountain range鈥攊t tells you everything you need to understand the potential extremities or worst-case scenarios that a function might reach. In a functional context, it ensures that we're evaluating the function under its most demanding conditions, making it critical for assessing convergence and stability.
Triangle Inequality
The triangle inequality is a crucial property of a metric that ensures the sum of the lengths of two sides of a triangle is always greater than or equal to the length of the remaining side. This keeps the notion of 'distance' in a metric space consistent with our intuitive understanding of distance.

For a metric \( d(f, g) \) defined using the uniform norms for functions and their derivatives, the triangle inequality is expressed as:
  • \( d(f, h) \leq d(f, g) + d(g, h) \)
Breaking it down, if we have functions \( f, g, \) and \( h \), this inequality ensures that the direct 'distance' from \( f \) to \( h \) is not greater than taking a detour via \( g \). It ensures consistency and reliability, which makes it an indispensable tool for both theoretical exploration and practical application in metric spaces.
Symmetry Property
The symmetry property in the context of metric spaces dictates that the distance between two points should be the same irrespective of the direction you measure it. This reflects the intuitive understanding that distance is independent of direction.

For the metric \( d(f, g) \) defined with uniform norms, symmetry can be shown as follows:
  • \( d(f, g) = d(g, f) \)
This means switching the functions being compared doesn鈥檛 change the 'distance' between them.

The symmetry property ensures fairness and objectivity when measuring. Imagine trying to plot a journey鈥攊t's like saying the number of steps it takes to walk from your house to the park should be the same as walking from the park back home. This non-biased nature makes it vital to any definition of a metric in mathematical analysis.

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Most popular questions from this chapter

Suppose \((X, d)\) is a metric space and \(\varphi:[0, \infty) \rightarrow \mathbb{R}\) is an increasing function such that \(\varphi(t) \geq 0\) for all \(t\) and \(\varphi(t)=0\) if and only if \(t=0 .\) Also suppose \(\varphi\) is subadditive, that is, \(\varphi(s+t) \leq\) \(\varphi(s)+\varphi(t) .\) Show that with \(d^{\prime}(x, y):=\varphi(d(x, y)),\) we obtain a new metric space \(\left(X, d^{\prime}\right)\).

Take \(\mathbb{Q}\) with the standard metric, \(d(x, y)=|x-y|,\) as our metric space. Prove that \(\mathbb{Q}\) is totally disconnected, that is, show that for every \(x, y \in \mathbb{Q}\) with \(x \neq y\), there exists an two open sets \(U\) and \(V\), such that \(x \in U, y \in V, U \cap V=\emptyset,\) and \(U \cap V=\mathbb{Q}\).

Let \(A\) be a connected set in a metric space. a) Is \(\bar{A}\) connected? Prove or find a counterexample. b) Is A \(^{\circ}\) connected? Prove or find a counterexample. Hint: Think of sets in \(\mathbb{R}^{2}\).

Take \(\mathbb{R}^{*}=\\{-\infty\\} \cup \mathbb{R} \cup\\{\infty\\}\) be the extended reals. Define \(d(x, y):=\left|\frac{x}{1+|x|}-\frac{y}{1+|y|}\right|\) if \(x, y \in \mathbb{R},\) define \(d(\infty, x):=\left|1-\frac{x}{1+|x|}\right|, d(-\infty, x):=\left|1+\frac{x}{1+|x|}\right|\) for all \(x \in \mathbb{R},\) and let \(d(\infty,-\infty):=2\) a) Show that \(\left(\mathbb{R}^{*}, d\right)\) is a metric space. b) Suppose \(\left\\{x_{n}\right\\}\) is a sequence of real numbers such that for every \(M \in \mathbb{R},\) there exists an \(N\) such that \(x_{n} \geq M\) for all \(n \geq N .\) Show that \(\lim x_{n}=\infty\) in \(\left(\mathbb{R}^{*}, d\right)\) c) Show that a sequence of real numbers converges to a real number in \(\left(\mathbb{R}^{*}, d\right)\) if and only if it converges in \(\mathbb{R}\) with the standard metric.

Finish the proof of Proposition 7.2.11. Suppose \((X, d)\) is a metric space and \(Y \subset X .\) Show that with the subspace metric on \(Y\), if a set \(U \subset Y\) is open \((\) in \(Y)\), then there exists an open set \(V \subset X\) such that \(U=V \cap Y\)
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