Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be such that \( |f(x) - f(y)| \leq (x - y)^2 \) for all \( x, y \in \mathbb{R} \). Then \[ f(1) - f(0) = \ \underline{\hspace{2cm}} \quad \text{(Answer in integer)} \]

Hint:

When solving functional equations involving bounds like \( |f(x) - f(y)| \), examine the behavior as \( x \) and \( y \) get closer. Such conditions often suggest continuity or constancy of the function.

Answer 1

Step 1: Analyze the given condition.
We are given that \( |f(x) - f(y)| \leq (x - y)^2 \) for all \( x, y \in \mathbb{R} \). Let’s use this condition with \( x = 1 \) and \( y = 0 \): \[ |f(1) - f(0)| \leq (1 - 0)^2 = 1 \] Thus, we have: \[ |f(1) - f(0)| \leq 1 \] This implies: \[ -1 \leq f(1) - f(0) \leq 1 \] Step 2: Consider the behavior of \( f \).
Notice that the condition \( |f(x) - f(y)| \leq (x - y)^2 \) suggests that \( f(x) \) must be a very smooth function, and in fact, it implies that \( f \) is a constant function because the difference \( |f(x) - f(y)| \) is very tightly bound by \( (x - y)^2 \), which tends to 0 as \( x \) approaches \( y \). This implies that: \[ f(x) = f(0) \quad \text{for all} \ x \in \mathbb{R}. \] Step 3: Conclusion. Since \( f(x) = f(0) \) for all \( x \), we have: \[ f(1) - f(0) = 0 \]