Let \(f:\mathbb R\to\mathbb R\) be infinitely differentiable. Suppose that, for some positive integer \(n\), \[ f(1)=f(0)=f'(0)=f''(0)=\cdots=f^{(n)}(0)=0. \] Prove that \(f^{(n+1)}(x)=0\) for some \(x\in(0,1)\).
Solution
1. There exists \(x_1\in(0,1)\) such that\[ f'(x_1)=0. \]
The hypotheses give \(f(0)=f(1)=0\). Rolle’s theorem applied to \(f\) on \([0,1]\) gives such a point \(x_1\).
2. Suppose \(1\le k\le n\) and there exists \(x_k\in(0,1)\) such that\[ f^{(k)}(x_k)=0. \]Then there exists \(x_{k+1}\in(0,x_k)\) such that\[ f^{(k+1)}(x_{k+1})=0. \]
By hypothesis, \[ f^{(k)}(0)=0, \] and by assumption \(f^{(k)}(x_k)=0\). Rolle’s theorem applied to \(f^{(k)}\) on \([0,x_k]\) gives a point \(x_{k+1}\in(0,x_k)\) with \[ (f^{(k)})'(x_{k+1})=f^{(k+1)}(x_{k+1})=0. \]
3. There exists \(x_{n+1}\in(0,1)\) such that\[ f^{(n+1)}(x_{n+1})=0. \]
Step 1 supplies \(x_1\). Apply step 2 successively for \[ k=1,2,\ldots,n. \] This produces \[ 0<x_{n+1}<x_n<\cdots<x_1<1 \] and \(f^{(n+1)}(x_{n+1})=0\).
4. Q.E.D.
Step 3 gives the required point in \((0,1)\).