Let \(f\) be continuous and nonnegative on \([0,1]\), and suppose \[ f(t)^2\le 1+2\int_0^t f(s)\,ds. \] Prove that \[ f(t)\le1+t \qquad(0\le t\le1). \]
Solution
Define \[ F(t) \coloneqq \sqrt{ 1+2\int_0^t f(s)\,ds }. \]
1. For every \(t\in[0,1]\),\[ F(t)\geq1 \]and\[ f(t)\leq F(t). \]
Since \(f\) is nonnegative, \[ \int_0^t f(s)\,ds\geq0, \] so the expression under the square root is at least \(1\), giving \(F(t)\geq1\).
The hypothesis says \[ f(t)^2 \leq F(t)^2. \] Both \(f(t)\) and \(F(t)\) are nonnegative, hence \[ f(t)\leq F(t). \]
2. The function \(F\) is continuously differentiable and satisfies\[ F'(t) = \frac{f(t)}{F(t)}. \]
The function \[ t\longmapsto 1+2\int_0^t f(s)\,ds \] is continuously differentiable with derivative \(2f(t)\), and by step 1 it is strictly positive. The chain rule therefore gives \[ F'(t) = \frac{1}{2F(t)}\,2f(t) = \frac{f(t)}{F(t)}. \]
3. For every \(t\in[0,1]\),\[ F'(t)\leq1. \]
By step 1, \[ 0\leq f(t)\leq F(t). \] Divide by the positive number \(F(t)\) and use step 2.
4. For every \(t\in[0,1]\),\[ F(t)\leq1+t. \]
One has \[ F(0)=1. \] Integrating the estimate from step 3, \[ F(t)-F(0) = \int_0^tF'(s)\,ds \leq \int_0^t1\,ds =t. \] Thus \(F(t)\leq1+t\).
5. For every \(t\in[0,1]\),\[ \boxed{f(t)\leq1+t}. \]
Step 1 gives \(f(t)\leq F(t)\), and step 4 gives \(F(t)\leq1+t\).
6. Q.E.D.
Step 5 is the required estimate.