Let $f,g:[0,1] \to [0, \infty)$ be two continuous functions such that $$f(0) = g(0) = 0,$$ $f$ is convex, $g$ is concave and increasing and $$\displaystyle \int_0^1f(x)dx = \int_0^1g(x)dx.$$ Prove that $$ \displaystyle \int_0^1\left(f(x) \right)^2dx \geq \int_0^1\left(g(x) \right)^2dx.$$

I don't quite know how to approach the problem.

I thought about using Chebyshev's inequality due to the fact that $g$ is increasing, $F(x) = \int_0^xf(t)dt$ is increasing (since $f$ is convex) and $G(x) = \int_0^xg(t)dt$ is increasing (since $g$ is increasing and $g(0) = 0$), but it didn't help me.

I also tried to obtain something by writing the convexity and concavity point-wise and using the fact that $\displaystyle h(x) = \frac{f(x)}{x}$ is increasing and $p(x) = \displaystyle \frac{g(x)}{x}$ is decreasing, but I got nothing.

  • 1
    $\begingroup$ Since $g$ is increasing and $g(0)=0$, shouldn't $G$ be increasing? Not sure if this will help you, but it seems you've made a mistake in the first approach you've listed, at least. $\endgroup$ Apr 9, 2018 at 13:29
  • $\begingroup$ @TheoreticalEconomist You're right, but still, I can't continue. $\endgroup$
    – C_M
    Apr 9, 2018 at 13:33

2 Answers 2


It will be easy to answer the following question first: Assume that $f(x)$ is convex, $f(0)=0$, $f(x)\geq 0$ and $\int_{0}^{1}f(x)dx=M>0$ is fixed. Which function makes the integral $$ \int_{0}^{1}f(x)^{2}dx $$ as small as possible? We will prove that linear function is the optimizer, i.e. $f(x) = kx = 2Mx$ gives the answer. To prove this, we need a lemma:

Lemma. Let $f:[0, 1]\to [0, \infty)$ be a convex continuous function with $f(0)=0$. Then we have $$ \int_{0}^{1}xf(x)dx\geq \frac{2}{3}\int_{0}^{1}f(x)dx $$

This is actually a contest problem I saw before. We will show this later.

Assume that the lemma holds. Then we have $$ 0\leq \int_{0}^{1}(f(x)-2Mx)^{2}dx = \int_{0}^{1}f(x)^{2} -4M\int_{0}^{1}xf(x)dx + \frac{4}{3}M^{2} \leq \int_{0}^{1}f(x)^{2}dx - \frac{4}{3}M^{2} $$ so $\int_{0}^{1}f(x)^{2}dx \geq \frac{4}{3}M^{2}$, where the equality holds for $f(x)=2Mx$. Similarly, the reversed inequality holds for the concave function $g(x)$ with same condition, so we get $$ \int_{0}^{1}f(x)^{2}dx \geq \frac{4}{3}M^{2} \geq \int_{0}^{1}g(x)^{2}dx. $$

Proof of Lemma. Since $f(x)$ is convex, we have $$\int_{0}^{x}f(t)dt \leq \frac{1}{2}xf(x)$$ for any $x$. (Consider the triangle with vertices $(0, 0), (x, 0), (x, f(x))$.) Then \begin{align*} \frac{1}{2}\int_{0}^{1}xf(x)dx &\geq \int_{0}^{1}\int_{0}^{x}f(t)dtdx \\ &=\int_{0}^{1}\int_{t}^{1}f(t)dxdt \,\,\quad(\text{Fubini's theorem})\\ &=\int_{0}^{1}(1-t)f(t)dt \\ &= \int_{0}^{1}f(x)dx - \int_{0}^{1}xf(x)dx \end{align*} and we get the inequality.

  • $\begingroup$ @C_M Thanks! I just edited. $\endgroup$
    – Seewoo Lee
    Apr 9, 2018 at 14:08
  • 2
    $\begingroup$ Also, your proof for the lemma is interesting, but integration by parts does the trick in the final inequation: $\int_0^1 \int_0^x f(t)dt dx = \int_0^1 x' \int_0^x f(t)dt dx = \int_0^1f(x)dx - \int_0^1xf(x)dx$. That's just a minor thing. Thanks for the answer! $\endgroup$
    – C_M
    Apr 9, 2018 at 14:14

For some $0 < \epsilon < 1$, $$ \int_\epsilon^1 (f(x)^2-g(x)^2)dx = \int_\epsilon^1 (f(x) + g(x))(f(x)-g(x)) dx $$ since $f(x)+g(x) \ge 0$, we can apply the intermediate value theorem to get a $\xi \in [\epsilon,1]$, such that $$ \int_\epsilon^1 (f(x) + g(x))(f(x)-g(x))dx = (f(\xi)+g(\xi))\int_\epsilon^1 (f(x)-g(x))dx \,. $$ Since $f(\xi)+g(\xi) \ge 0$, it remains to show that $$ \int_\epsilon^1 (f(x)-g(x))dx \ge 0 $$ $h(x) = (f(x)-g(x))$ is also convex ($-g(x)$ is convex) with ($h(0) = 0$), so $h$ can only have at most 2 roots (except $h \equiv 0$). If there is no other root $x_0 \in(0,1)$, then either $f(x) > g(x)$ or the other way around, and the integrals would not match. So there must be $x_0 \in(0,1)$ such that $h(x) \ge 0$ for all $x \ge x_0$. Therefore $f(x) \ge g(x)$ and furthermore $\int_\epsilon^1(f(x)-g(x))dx \ge 0$, with $\epsilon := x_0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.