Suppose that \(f:[0,1]\to\mathbb R\) is differentiable, \(f(0)=0\), and there is \(k>0\) such that \[ |f'(x)|\le k|f(x)| \qquad(x\in[0,1]). \] Prove that \(f(x)=0\) for all \(x\in[0,1]\).
Solution
1. Obtain an integral inequality.
Since \(f\) is continuous on \([0,1]\), let \[ M:=\max_{[0,1]}|f|<\infty. \] The derivative bound gives \(|f'|\le kM\), so \(f\) is Lipschitz and hence absolutely continuous. Therefore, using \(f(0)=0\), \[ |f(x)| =\left|\int_0^x f'(t)\,dt\right| \le k\int_0^x |f(t)|\,dt \qquad(0\le x\le1). \]
2. Iterate the inequality.
We claim that for every integer \(n\ge1\), \[ |f(x)|\le M\frac{(kx)^n}{n!} \qquad(0\le x\le1). \] For \(n=1\), Step 1 and \(|f(t)|\le M\) give \[ |f(x)|\le kMx. \] If the estimate holds for \(n\), then Step 1 gives \[ |f(x)| \le k\int_0^x M\frac{(kt)^n}{n!}\,dt =M\frac{(kx)^{n+1}}{(n+1)!}. \] Thus the claim follows by induction.
3. Let the iteration order tend to infinity.
For each fixed \(x\in[0,1]\), \[ 0\le |f(x)|\le M\frac{(kx)^n}{n!}\longrightarrow0 \qquad(n\to\infty). \] Hence \(f(x)=0\). Since \(x\) was arbitrary, \[ \boxed{f\equiv0\text{ on }[0,1].} \]