Problem 49 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove that if \(|f(y)-f(x)| \leq M(y-x)^{2}\) for all \(x\) and \(y\) then \(f\) is a constant function.

Step-by-Step Solution

Answer

If \(|f(y)-f(x)| \leq M(y-x)^2\) for all \(x\) and \(y\), then \(f\) is constant.

Step 1: Understanding the Problem

We need to prove that if the absolute difference between values of a function at two points is bound by a quadratic function of the distance between the two points, then the function is constant. In mathematical terms, if \(|f(y) - f(x)| \leq M(y-x)^2\) for some constant \(M\) and for all \(x\) and \(y\), then \(f(x) = f(y)\) for any \(x\) and \(y\).

Step 1: Assume x and y are Close

Assume \(x\) and \(y\) are two arbitrary points such that \(x < y\). According to the problem, \(|f(y) - f(x)| \leq M(y-x)^2\). Given that this inequality holds for arbitrarily small distances \((y-x)\), it suggests a very rigid behavior for the function \(f\).

Step 2: Limit the Inequality

Consider the limit as \(y\) approaches \(x\). As \(y\) gets closer to \(x\), \((y-x)^2\) approaches zero. The inequality \(|f(y) - f(x)| \leq M(y-x)^2\) implies that \( |f(y) - f(x)| \leq 0 \). This means \(f(y) = f(x)\) as \((y-x)\to 0\).

Step 3: Conclusion from Continuity

Since the relationship \(|f(y) - f(x)| \leq M(y-x)^2\) holds for all points \(x\) and \(y\), taking the limit shows that \(f(y)\) and \(f(x)\) are equal when \(y\) is arbitrarily close to \(x\). Thus, \(f\) must be the same for all \(x\) and \(y\), proving that \(f\) is constant.

Key Concepts

Understanding Function Inequality

In mathematics, understanding function inequality is crucial for evaluating how functions behave under certain conditions. The problem given involves an inequality:

• \(|f(y) - f(x)| \leq M(y-x)^2\), ensuring that the difference in function values is restricted by a quadratic function of the distance between \(x\) and \(y\).

This is significant because quadratic functions increase quickly as the distance \((y-x)\) increases.

When the inequality

• \(|f(y) - f(x)| \leq M(y-x)^2\)

is satisfied for all \(x\) and \(y\), especially when they are very close, it implies that the function changes very slowly or not at all as you move from \(x\) to \(y\).

This is an essential aspect of understanding why the function might be constant. If the function's change is always less than or equal to some very small value when measured over small intervals, this means it might not change at all.

Exploring Continuity

Continuity in functions relates to how a function behaves without sudden changes or interruptions. In our problem, continuity comes into play when we consider

• the behavior of \(f\) as \(y\) becomes very close to \(x\).

The inequality \(|f(y) - f(x)| \leq M(y-x)^2\) must hold for all possible pairs of points.

As \(y\) approaches \(x\), the term \((y-x)^2\) becomes extremely small, suggesting that the difference in the function's values also becomes near zero.

This scenario is key to understanding continuity:

• It implies that small changes in input lead to small changes in output.
• And in this case, essentially no change as \(y\) approaches \(x\), reinforcing the notion of the function being constant over its entire domain.

Knowing that a function remains smoothly consistent, without erratic behavior, helps confirm that indeed, the function is constant.

Harnessing Mathematical Proof

Mathematical proof is a logical argument that verifies the truth of a statement using reasoning and evidence. In proving that \(f\) is a constant function,

• we rely on the given inequality \(|f(y) - f(x)| \leq M(y-x)^2\).

By systematically analyzing this inequality, we explore two core ideas:

• the function's behavior as \(y\) approaches \(x\)
• and the broader implications across all possible \(x\) and \(y\).

Taking the limit as \(y\) tends towards \(x\) reveals

• that \(|f(y) - f(x)|\) converges on zero,
• leading to \(f(y)=f(x)\).

Each step in the logical sequence rigorously demonstrates that if the inequality condition is true everywhere, then no matter where on the number line you choose \(x\) and \(y\),

• \(f\) must always equate to the same value.

Thus, by connecting these deductions, the proof effectively demonstrates that \(f\) is a constant function, which stays unchanging across its entire domain.