If $f:[a,b] \to \mathbb{R}, f(a)=0,f(b)=1$ is a convex increasing differentiable function on the interval $[a,b]$ . Prove that $$\int_a^bf^2(x)\,dx\le \frac{2}{3}\int_a^bf(x)\,dx$$

Since f is convex and increasing so $f''(x)\ge 0 $ and $f'(x)\ge 0$. Then I consider a function $g:[a,b]\to \mathbb{R}$, $g(x)=\frac{2}{3}\int_a^xf(t)\,dt-\int_a^xf^2(t)\,dt$. Now $f$ is differentiable implies $g$ is also but can't conclude $g'(x)\ge 0$.

  • $\begingroup$ Isn't $g'(x)=2/3f(x)-f^2(x)$ and so $g'(b)=2/3f(b)-f^2(b)=2/3-1=-1/3<0$? $\endgroup$ – blub Apr 1 at 12:14
  • $\begingroup$ Yea ...but from here what we can do ??? $\endgroup$ – RAM_3R Apr 1 at 12:58

In Prove $\int _0^\infty f^2 dx \leq \cdots $ for $f$ convex the following theorem was shown:

If $F$ is convex and non-negative on $[0, \infty)$ then $$ \int _0^\infty F^2(x) dx \leq \frac{2}{3}\cdot \max_{x \in \mathbb R^+} F(x) \cdot \int _0^\infty F(x) dx \, .$$

Our function $f$ is non-negative and convex on $[a, b]$ with $f(a) = 0$ and $f(b) = 1$. If we define $F$ on $[0, \infty)$ as $$ F(x) = \begin{cases} f(b-x) & \text{ for } 0 \le x \le b-a \\ 0 & \text{ for } x > b-a \end{cases} $$ then $F$ satisfies the hypotheses of the above theorem, and therefore $$ \int_a^bf^2(x)\,dx = \int _0^\infty F^2(x) dx \leq \frac{2}{3}\cdot \max_{x \in \mathbb R^+} F(x) \cdot \int _0^\infty F(x) dx = \frac{2}{3}\int_a^bf(x)\,dx \, . $$

Alternatively we can modify the proof of the above theorem for this case. Define $\varphi: [a, b] \to \Bbb R$ as $$ \varphi(x) = \frac 23 f(x) \int_a^x f(t) \, dt - \int_a^x f^2(t) \, dt \, . $$ The goal is to show that $\varphi$ is (weakly) increasing. Then the desired conclusion follows with $$ 0 = \varphi(a) \le \varphi(b) = \frac 23 \int_a^b f(t) \, dt - \int_a^b f^2(t) \, . $$ Since $f$ is assumed to be differentiable, we have $$ \varphi'(x) = \frac 23 f'(x) \int_a^x f(t) \, dt + \frac 23 f^2(x) - f^2(x) \\ = \frac 23 f'(x) \int_a^x f(t) \, dt - \frac 13 f^2(x) \, . $$ Now we distinguish two cases:

  • If $f'(x) =0$ then $f'(t) =0$ for $a \le t \le x$, so that $f(x) = f(a) = 0$ and therefore $\varphi'(x) = 0$.
  • If $f'(x) >0$ then we estimate $f(t)$ from below by the tangent at $(x, f(x))$: $$ \int_a^x f(t) \, dt \ge \int_{x-f(x)/f'(x)}^x \bigl( f(x) + (t-x)f'(x) \bigr) \, dt = \frac{f^2(x)}{2f'(x)} $$ and therefore $\varphi'(x) \ge 0$.

So $\varphi'(x) \ge 0$ for all $x \in [a, b]$, which means that $\varphi$ is increasing on the interval, and we are done.

Remark 1: The proof becomes easier if we assume that $f$ is twice differentiable. Then $$ \varphi''(x) = \frac 23 f''(x) \int_a^x f(t) \, dt \ge 0 $$ so that $\varphi'(x) \ge \varphi'(0) = 0$.

Remark 2: The proof works even without the assumption that $f$ is differentiable: As a convex function, $f$ has a right derivative $$ f_+'(x) = \lim_{\substack{h \to 0\\ h > 0}} \frac{f(x+h)-f(x)}{h} $$ everywhere in $[a, b)$, and we can replace $f'$ by $f_+'$ and $\varphi'$ by $\varphi_+'$ in the above argument.


Without any loss of generality, we shift and scale to set $a=0, b=1$. And now we consider the integrals, $ \int_0^1{f(x)dx} $ and $ \int_0^1{f^2(x) dx} $.

Convexity of $ f(x) $ assures that $f(x)\leq x$. (1)

Now, we write the integrals as limits of Riemann sums.$ \int_0^1{f^2(x) dx} = \lim_{h \to 0, N \to \infty}{\sum_{r=0}^N(f^2(rh) \times h)}$ $\int_0^1{f(x) dx} = \lim_{h \to 0, N \to \infty}{\sum_{r=0}^N(f(rh) \times h)} $

$ f^2(rh) / f(rh) =f(rh) \leq rh$ (from (1)). For this ratio to be maximum (that is when the ratio $ \frac{ \int_0^1{f^2(x) dx}}{\int_0^1{f(x) dx}}$ is maximum), $ f(rh)=rh $ for all $r$. $\Rightarrow f(x)=x $. (2)

This means $ \frac{ \int_0^1{f^2(x) dx}}{\int_0^1{f(x) dx}} \leq \frac{ \int_0^1{x^2 dx}}{\int_0^1{x dx}} = \frac{2}{3}$

Edit: The maximisation holds if a unique maximum function exists in every interval of (2). This holds if $f(x)$ is convex. Otherwise as correctly pointed out by Martin, this ratio can be more than $\frac{2}{3}$. For example, if $ f(x)= sin^2(\frac{\pi x}{2})$, this ratio is 3/4.

  • $\begingroup$ The ratio of Riemann sums is $\dfrac{\sum_{r=1}^N f^2(\frac rN)}{\sum_{r=1}^N f(\frac rN)}$. Why is that maximal for $f(x) = x$? – If your argument is correct then the integral inequality would hold for all functions on $[0, 1]$ with $0 \le f(x) \le x$. $\endgroup$ – Martin R Apr 2 at 11:24
  • $\begingroup$ Because in every small interval $ (rh, (r+1)h) $ the ratio $ 𝑓^2(𝑟ℎ)/𝑓(𝑟ℎ)=𝑓(𝑟ℎ)$. As a convex function will lie below (or coincide with) the chord $y=x$, we must have $f(rh) \leq rh$. And yes, the integral inequality would hold if $0 \leq f(x) \leq x$. The convexity assures that only. $\endgroup$ – Anirban Apr 2 at 14:19
  • $\begingroup$ Here is a counter-example (unless I made some error): $f(x) = \sqrt{2x-1}$ for $0.5 \le x \le 1$, $f(x) = 0$ otherwise. Then $\int_0^1 f^2(x)dx = \frac 14$ and $\int_0^1 f(x) dx = \frac 13$. $\endgroup$ – Martin R Apr 2 at 14:34
  • $\begingroup$ @ Martin R, But then $f(x)$ is not increasing always. $f(x) >0$ is implied by your condition that $ f(x) $ is increasing. That the ratio, $ f^2(rh)/f(rh)$ is defined is assumed in my calculations. $\endgroup$ – Anirban Apr 2 at 15:36
  • $\begingroup$ As they didn't allow me to edit the comment and I am new in this site to know this, here is a more helpful comment. your counterexample function, $f(x)=0 $ for $ 0<x<1/2 $ violates your premise in that it is not increasing always. $f(x) >0$ is implied by your condition that $ f(x) $ is increasing. That the ratio, $ f^2(rh)/f(rh)$ is defined is assumed in my calculations. $\endgroup$ – Anirban Apr 2 at 15:46

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