Let \(f:[0,1]\to\mathbb R\) be continuously differentiable with \(f(0)=0\). Suppose there is a constant \(M>0\) such that \[ 0\le f'(x)\le Mf(x) \qquad(0\le x\le1). \] Prove that \(f(x)=0\) for all \(x\in[0,1]\).
Solution
1. The function \(f\) is nonnegative on \([0,1]\).
The inequality \[ f'(x)\geq0 \] shows that \(f\) is nondecreasing. Since \(f(0)=0\), \[ f(x)\geq0 \] for every \(x\in[0,1]\).
2. Define\[ h(x)=e^{-Mx}f(x). \]Then \(h\) is nonincreasing on \([0,1]\).
Differentiating gives \[ h'(x) = e^{-Mx}\bigl(f'(x)-Mf(x)\bigr). \] The hypothesis \(f'(x)\leq Mf(x)\) therefore gives \[ h'(x)\leq0. \]
3. One has\[ h(x)=0 \]for every \(x\in[0,1]\).
By step 1, \[ h(x)=e^{-Mx}f(x)\geq0. \] By step 2 and \[ h(0)=f(0)=0, \] one also has \[ h(x)\leq h(0)=0 \] for \(x\geq0\). Thus \(h(x)=0\) everywhere.
4. Consequently,\[ \boxed{f(x)=0\quad\text{for all }x\in[0,1]}. \]
Since \(e^{-Mx}>0\), step 3 implies \[ f(x)=e^{Mx}h(x)=0. \]
5. Q.E.D.
Step 4 is the required conclusion.