Let \(f : \mathbb{R} \to \mathbb{R}\) be such that \(|f(x) - f(y)| \le (x - y)^{2}\) for all \(x, y \in \mathbb{R}\). Then \(f(1) - f(0) =\) ________ (Answer in integer)
Answer
0
Explanation
For \(x \neq y\), divide the condition by \(|x - y|\):
\[ \left| \frac{f(x) - f(y)}{x - y} \right| \le |x - y| \]
As \(y \to x\) the right side tends to 0, so \(f\) is differentiable with \(f'(x) = 0\) for every \(x\). Hence \(f\) is constant and \(f(1) - f(0) = 0\).
Without derivatives: split \([0, 1]\) into \(n\) equal steps. Each step changes \(f\) by at most \(\frac{1}{n^{2}}\), so \(|f(1) - f(0)| \le n \cdot \frac{1}{n^{2}} = \frac{1}{n}\) for every \(n\), which forces the difference to be 0.
Answer: 0