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Chapter 12: Problem 11

For any two functions \(f\) and \(g\) in \(C([a, b], \mathbb{R}),\) define $$ d^{*}(f, g)=\int_{a}^{b}|f(x)-g(x)| d x $$ a. Prove that this defines a metric on \(C([a, b], \mathbb{R})\). b. Prove the following inequality relating this metric and the uniform metric: $$ d^{*}(f, g) \leq(b-a) d(f, g) $$ c. Compare the concepts of convergence of a sequence of functions in this metric and in the uniform metric.

Short Answer

Expert verified
The function \(d^{*}(f, g)\) defines a metric because it satisfies non-negativity, definiteness, symmetry, and the triangle inequality. The inequality \(d^{*}(f, g) \leq (b-a) d(f, g)\) holds by the properties of the integral and the supremum. Convergence in the metric \(d^{*}\) does not imply uniform convergence, but uniform convergence does imply convergence in the metric \(d^{*}\).

Step by step solution

01

- Prove Non-negativity

To show that \(d^{*}(f, g)\) is a metric, we first need to prove that the distance between any two functions \(f\) and \(g\) is non-negative. Since the absolute value function \(|f(x) - g(x)|\) is always non-negative for all \(x\) in \([a, b]\), the integral of a non-negative function is also non-negative. Thus we have \(d^{*}(f, g) = \int_{a}^{b} |f(x)-g(x)| dx \geq 0\).
02

- Prove Definiteness

Next, we need to prove that \(d^{*}(f, g) = 0\) if and only if \(f = g\) on \([a, b]\). First, if \(f = g\), then \(|f(x) - g(x)| = 0\) for all \(x\) in \([a, b]\), and thus the integral of zero over any interval is zero. Conversely, if \(d^{*}(f, g) = 0\), it means that the integral of \(|f(x) - g(x)|\) is zero. Given that the absolute value function is non-negative, it must be the case that \(f(x) = g(x)\) almost everywhere on \([a, b]\). Since \(f\) and \(g\) are continuous, they must be equal everywhere on \([a, b]\).
03

- Prove Symmetry

The metric must be symmetric, which means that \(d^{*}(f, g) = d^{*}(g, f)\). This property follows from the fact that the absolute value function is symmetric: \(|f(x) - g(x)| = |g(x) - f(x)|\) for all \(x\) in \([a, b]\). Hence, the integral will be the same regardless of the order of \(f\) and \(g\).
04

- Prove Triangle Inequality

To complete the proof that \(d^{*}\) is a metric, we need to establish the triangle inequality: for any functions \(f, g, h\) in \(C([a, b], \mathbb{R})\), \(d^{*}(f, h) \leq d^{*}(f, g) + d^{*}(g, h)\). This follows from the triangle inequality for the absolute value: \(|f(x) - h(x)| \leq |f(x) - g(x)| + |g(x) - h(x)|\). Taking the integral of both sides over \([a, b]\) gives the required inequality.
05

- Prove the Metric Inequality for \(d^{*}(f, g)\)

To prove \(d^{*}(f, g) \leq (b-a) d(f, g)\), we make use of the definition of the uniform metric \(d(f, g) = \sup_{x \in [a, b]}|f(x) - g(x)|\) and the fact that \(|f(x) - g(x)| \leq d(f, g)\) for all \(x\) in \([a, b]\). Integrating both sides of this inequality from \(a\) to \(b\), we get \(d^{*}(f, g) \leq \int_{a}^{b} d(f, g) dx = (b-a) d(f, g)\).
06

- Compare Convergence in \(d^{*}\) and Uniform Metric

A sequence of functions \(\{f_n\}\) converges to a function \(f\) in the metric \(d^{*}\) if \(\lim_{n \to \infty} d^{*}(f_n, f) = 0\). In the uniform metric, convergence of \(\{f_n\}\) to \(f\) requires \(\lim_{n \to \infty} d(f_n, f) = 0\), meaning \(f_n\) converges uniformly to \(f\). Since \(d^{*}(f_n, f)\) is the integral of the pointwise absolute difference over an interval, convergence in \(d^{*}\) does not necessarily imply uniform convergence. However, if \(f_n\) converges uniformly to \(f\), it does converge in \(d^{*}\) because the integral of a function that converges uniformly to zero over a bounded interval also converges to zero.

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Key Concepts

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

Non-negativity of Metric
Understanding the concept of non-negativity in metric spaces is fundamental in functional analysis. Non-negativity assures that the distance measured by the metric between any two elements, in this case functions, is always greater than or equal to zero. When dealing with functions, we consider the integral of the absolute difference between them, denoted by \(d^{*}(f, g)\). Since absolute values are inherently non-negative and the integral accumulates these values, the result cannot be negative. This property is essential as it aligns with our physical intuition about distance 鈥� it cannot be less than zero.
Definiteness of Metric
The definiteness of a metric refers to a simple yet significant property: the distance between two identical points (or functions) must be zero, and if the distance is zero, then the functions must be identical almost everywhere on the interval. This property of the integral metric \(d^{*}(f, g)\) ensures that two functions are considered the same (in the context of the metric space) if they are equal at every point within the interval \([a, b]\). This fosters a rigorous approach to considering when functions are effectively equivalent in a metric space defined on a set of continuous functions.
Symmetry of Metric
Symmetry in a metric space is indicative of the metric being indifferent to the order of the objects it compares. In simpler terms, the distance from point A to B is the same as from B to A. For functions, the metric \(d^{*}(f, g)\) demonstrates symmetry because the absolute difference between functions \(f(x)\) and \(g(x)\) is the same regardless of their order. This ensures that the metric is consistent and fair, treating all functions in the space in a standardized manner.
Triangle Inequality in Metric Spaces
The triangle inequality is a pivotal concept in metric spaces guaranteeing that the direct route is always the shortest. It states that the sum of the distances from point A to B and from B to C is at least as long as the distance directly from A to C. Applied to the space of continuous functions, for any three functions \(f, g, h\), we have \(d^{*}(f, h)\) is less than or equal to \(d^{*}(f, g) + d^{*}(g, h)\). This inequality is a cornerstone in establishing the viable structure and coherence of the metric space.
Uniform Metric vs Integral Metric
Exploring the relationship between the uniform metric and the integral metric is crucial for understanding different modes of function convergence. The uniform metric gauges the maximum difference between functions at any point in an interval. In contrast, the integral metric sums up the absolute difference across the entire interval, resulting in a more cumulative measure of discrepancy. The inequality \(d^{*}(f, g) \leq (b-a) d(f, g)\) relates these two by showing the bounded nature of the integral metric, which indicates that while different in approach, these metrics are not entirely disconnected and have a consistent way to compare them.
Convergence of Functions
The final concept often stirs curiosity: convergence of functions within a metric space. Specifically, how a sequence of functions gets closer to a single function according to the metrics in use. Convergence in the integral metric, \(d^{*}\), means that the integral of the pointwise absolute differences of the functions in the sequence approaches zero. On the flip side, convergence in the uniform metric implies uniform convergence which is stronger; it requires the pointwise differences to become uniformly small. Understanding this distinction is vital in appreciating various nuances of function convergence in different contexts.

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Most popular questions from this chapter

Suppose that \(X\) is a metric space and that the mapping \(f: \mathbb{R} \rightarrow X\) is continuous. Let \(C\) be a closed subset of \(X\) and let \(f(x)\) belong to \(C\) if \(x\) is rational. Prove that \(f(\mathbb{R}) \subseteq C\).

Let \(X\) be a metric space and let \(K \subseteq X\) be a sequentially compact subspace. Let \(p\) be a point in \(X \backslash K\). Prove that there is a point \(q_{0}\) in \(K\) that is closest to \(p\) of all points in \(K\) in the sense that $$d\left(p, q_{0}\right) \leq d(p, q)$$ for all points \(q\) in \(K\).

We have defined what it means for a subset \(X\) of \(\mathbb{R}^{n}\) to be convex. a. Define what it means for a subset \(X\) of \(C([a, b], \mathbb{R})\) to be convex. b. Prove that convex subsets of \(C([a, b], \mathbb{R})\) are pathwise- connected. c. Prove that open balls in \(C([a, b], \mathbb{R})\) are convex. d. Prove that open balls in \(C([a, b], \mathbb{R})\) are connected.

Suppose that the function \(f:[a, b] \rightarrow \mathbb{R}\) is continuous. Prove that \(f:[a, b] \rightarrow \mathbb{R}\) is Lipschitz if and only if \(f:(a, b) \rightarrow \mathbb{R}\) is Lipschitz.

For each positive integer \(k,\) define \(f_{k}(x)=e^{x / k}\) for \(0 \leq x \leq 1 .\) Is the sequence \(\left\\{f_{k}:[0,1] \rightarrow \mathbb{R}\right\\}\) a Cauchy sequence in the metric space \(C([0,1], \mathbb{R}) ?\)
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