(January 2005 #6) Suppose that \(f:[0,1]\to\mathbb R\) is a differentiable function with \(f(0)=0\) and that there exists some \(K>0\) so that \(|f'(x)|\le K|f(x)|\) for all \(x\in[0,1]\). Prove that \(f(x)=0\) on \([0,1]\).
Note: See [Rud76, Chap 5, #26] for a significant hint.
Solution
1. \(f\) is bounded.Proof: let \(M = \sup_{[0,1]}|f(x)| < \infty\) (\(f\) differentiable on a compact interval is continuous, hence bounded).
2. Iterated Gronwall bound.Proof: for \(x \in [0,1]\), \(f\) is differentiable with \(|f'| \le K|f| \le KM\).Integrating: \[|f(x)| = \left|\int_0^x f'(t)\,dt\right| \le \int_0^x |f'(t)|\,dt \le K\int_0^x |f(t)|\,dt \le K\int_0^x KM\,dt = MK^2 x.\] Iterating: substitute the improved bound \(|f(t)| \le MK^2 t\) into the integral: \[|f(x)| \le K\int_0^x MK^2 t\,dt = MK^3\frac{x^2}{2},\] and by induction \(|f(x)| \le MK^{n+1}\frac{x^n}{n!}\) for every \(n \ge 0\).
3. Conclude.Proof: for fixed \(x \in [0,1]\), \(|f(x)| \le MK^{n+1}\frac{x^n}{n!} \le M K \frac{(Kx)^n}{n!} \to 0\) as \(n \to \infty\) (for any fixed \(Kx\), the factorial beats the exponential).Hence \(f(x) = 0\) for every \(x \in [0,1]\).
4. Q.E.D.