Assume that \(f:[0,1]\to\mathbb R\) is continuous on \([0,1]\) and differentiable on \((0,1)\) with \(f(0)=f(1)=0\) and \(f(c)=1\) for some \(c\in(0,1)\). Prove that there exists some \(s\in(0,1)\) such that \(|f'(s)|>2\).
Solution
1. If \(c \ne 1/2\), use the mean value theorem directly.Proof: if \(c < 1/2\), MVT on \([0,c]\): \(f'(s) = \frac{f(c) - f(0)}{c} = \frac{1}{c} > 2\) for some \(s \in (0,c)\).If \(c > 1/2\), MVT on \([c,1]\): \(f'(s) = \frac{f(1) - f(c)}{1 - c} = -\frac{1}{1-c}\) with \(\frac{1}{1-c} > 2\), so \(|f'(s)| > 2\).
2. The critical case \(c = 1/2\): suppose, toward a contradiction, that \(|f'(x)| \le 2\) for all \(x \in (0,1)\).
3. \(f\) is the tent function on \([0,1/2]\).Proof: for \(0 \le x \le 1/2\), MVT on \([0,x]\) gives \(f(x) = f'(\xi)x\) with \(|f'(\xi)| \le 2\), so \(f(x) \le 2x\); MVT on \([x, 1/2]\) gives \(1 - f(x) = f'(\eta)(1/2 - x)\) with \(|f'(\eta)| \le 2\), so \(f(x) \ge 1 - 2(1/2 - x) = 2x\).Hence \(f(x) = 2x\) for all \(x \in [0,1/2]\).
4. \(f\) is the tent function on \([1/2,1]\) as well.Proof: for \(1/2 \le x \le 1\), the same argument on \([x,1]\) and \([1/2, x]\) gives \(f(x) = 2 - 2x\) for all \(x \in [1/2,1]\).
5. Contradiction with differentiability at \(1/2\).Proof: the left derivative of the tent at \(1/2\) is \(2\) (from \(f = 2x\)), while the right derivative is \(-2\) (from \(f = 2-2x\)); they differ, so \(f\) is not differentiable at \(1/2 \in (0,1)\) — contradicting differentiability on \((0,1)\).Hence \(|f'(x)| \le 2\) is impossible, and there is \(s \in (0,1)\) with \(|f'(s)| > 2\).
6. Q.E.D.