# Spivak, Ch. 14, Prob *22: $f$ is a diff. function with $f(0)=0$ and $0<f'\leq 1$. Prove for all $x\geq 0$ we have $\int_0^x f^3\leq ( \int_0^x f )^2$

*22. Suppose that $$f$$ is a differentiable function with $$f(0)=0$$ and $$0. Prove that for all $$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 $$f(x) for $$x>0$$.

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

Therefore

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

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

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

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

• 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. Commented Jul 9, 2022 at 2:26

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).