*22. Suppose that $f$ is a differentiable function with $f(0)=0$ and $0<f'\leq 1$. Prove that for all $x\geq 0$ we have

$$\int_0^x f^3\leq \left ( \int_0^x f \right )^2$$

My proof differed substantially from the solution manual proof, and I'd like to know if it is correct.

Here is my proof

Consider the interval $[0,x]$.

Since $f'>0$, we have $f(x)>0$ for $x>0$

Since $f'\leq 1$ we have $f(x)<x$ for $x>0$.

That is $$0\leq f(x)\leq x, \text{ for } x\geq 0\tag{1}$$


$$0\leq\int_0^x f \leq \int_0^x x = \frac{x^2}{2}$$

$$\left ( \int_0^x f \right )^2 \leq \frac{x^4}{4}$$

Now from $(1)$ we have

$$0\leq [f(x)]^3 \leq x^3, \text{ for } x\geq 0$$

$$\implies 0\leq\int_0^x f^3 \leq \int_0^x x^3 = \frac{x^4}{4}=\left ( \int_0^x f \right)^2, \text{ for } x\geq 0$$

Hence, we have the desired result that for $x\geq 0$ we have

$$\int_0^x f^3 \leq \left ( \int_0^x f \right)^2$$

  • 1
    $\begingroup$ You derive two inequalities: $\int f^3 \leq x^4/4$ and $(\int f)^2 \leq x^4/4$. You can't derive a relation between $\int f^3$ and $(\int f)^2$ only from these two inequalities. $\endgroup$
    – Kolja
    Commented Jul 9, 2022 at 2:26

1 Answer 1


Unfortunately there's a mistake: you correctly derive the inequality $\displaystyle\left ( \int_0^x f \right )^2 \leq \frac{x^4}{4}$, but near the end you use an equality $\displaystyle\left ( \int_0^x f \right )^2 = \frac{x^4}{4}$ that you haven't derived (and is not true in general).


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