Question

Let \(f\) be defined for all real \(x\), and suppose that $$ |f(x)-f(y)| \leq(x-y)^{2} $$ for all real \(x\) and \(y\). Prove that \(f\) is constant.

Short answer

Question: Prove that if a function f satisfies \(|f(x) - f(y)| \leq (x - y)^2\) for all real numbers x and y, then f is constant. Answer: We have shown that the derivative \(f'(x) = 0\) for all real numbers \(x\), which implies that the function \(f\) does not change its value at any point, and therefore, it is constant.

Step-by-step solution

Write the definition of the derivative

We will use the definition of the derivative of a function, which is given by: $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$ We will need this to calculate the derivative of the function \(f\) at any point \(x\).

Use the given inequality to bound the difference quotient

We are given that \(|f(x)-f(y)| \leq (x-y)^{2}\) for all real \(x\) and \(y\). Let's replace \(y\) with \(x+h\) and analyze the inequality: $$ |f(x+h)-f(x)| \leq (h)^{2} $$ Now, we can divide both sides of the inequality by \(|h|\) to get: $$ \left|\frac{f(x+h) - f(x)}{h}\right| \leq |h| $$

Take the limit as h approaches zero

Now, we will take the limit as \(h\) approaches zero on both sides of the inequality: $$ \lim_{h \to 0} \left|\frac{f(x+h) - f(x)}{h}\right| \leq \lim_{h \to 0} |h| $$ Since \(\lim_{h \to 0} |h| = 0\), we can write: $$ \lim_{h \to 0} \left|\frac{f(x+h) - f(x)}{h}\right| \leq 0 $$ However, since the absolute value is always non-negative, we know that the left-hand side of this inequality is always greater than or equal to zero. Therefore, the only way for the inequality to hold is if: $$ \lim_{h \to 0} \left|\frac{f(x+h) - f(x)}{h}\right| = 0 $$

Conclude that the derivative is zero

We recall from Step 1 that the derivative of the function \(f\) at any point \(x\) is given by the limit: $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$ Comparing this with the result from Step 3, we can write: $$ f'(x) = \lim_{h \to 0} \left|\frac{f(x+h) - f(x)}{h}\right| = 0 $$

Prove that the function is constant

Since \(f'(x) = 0\) for all real numbers \(x\), we have shown that the function \(f\) has a derivative of zero at every point in its domain. This implies that the function \(f\) does not change its value at any point, and therefore, it is constant.

Key concepts

Derivative Concept

The concept of a derivative is fundamental in calculus, often described as the rate of change or slope of a function at a given point. Essentially, it measures how a function's output changes as its input changes. To compute a derivative, one usually employs the definition through limits: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]This formula represents how the function \(f\) changes as we make an infinitesimally small change \(h\) to the input \(x\). Calculating derivatives helps us understand and predict the behavior of functions in diverse scientific and engineering fields. In the exercise above, by proving that the derivative \(f'(x)\) is zero, it indicates that the function \(f\) does not change at all, hence it is constant.

Constant Function

A constant function is one in which the output value remains the same for any input within the function's domain. In mathematical terms, if \(f(x) = c\) for all \(x\), where \(c\) is a constant, then \(f\) is a constant function. This property is characterized by having a derivative of zero: \[ f'(x) = 0 \quad \text{for all } x \]In our exercise example, proving \(f'(x) = 0\) for all real \(x\) leads directly to the conclusion that the function \(f\) is constant. The exercise showcases one of the beautiful results in calculus where knowing properties about a function's derivative lets us conclude properties of the function itself.

Inequality Application

Inequalities are crucial tools across various fields of mathematics, including calculus. In this exercise, the inequality \(|f(x)-f(y)| \leq (x-y)^{2}\) sets a condition for the function \(f\). This inequality bounds the possible values of the function's difference based on the distance between \(x\) and \(y\). By manipulating this inequality, the exercise demonstrates the relationship between the difference quotient and the behavior of \(f\), effectively showing \( \left|\frac{f(x+h) - f(x)}{h}\right| \leq |h| \) using substitution. This crucial step allows the exercise to explore the circumstances under which the function remains unchanged, a testament to the power of employing inequalities to uncover deeper truths in calculus.

Limit Process

The limit process is a cornerstone of calculus, providing a way to understand behavior as one value approaches another. In the process of finding derivatives, limits help define the exact instantaneous rate of change of a function. Consider the limit expression used to find the derivative: \[ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]Here, taking the limit as \(h\) approaches zero measures the behavior of \(\frac{f(x+h) - f(x)}{h}\) when \(h\) becomes very small. In the exercise, by evaluating the limit \(\lim_{h \to 0} |h| = 0\), we find that the function must have a derivative of zero everywhere, leading us to conclude it's constant. Limits thus provide the precision necessary for proving such properties in functions, a technique widely applicable in mathematics.