$\int_a^bf^2(x)\,dx\le \frac{2}{3}\int_a^bf(x)\,dx$ for a convex differentiable function

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 $$g'(x)\ge 0$$.

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

• 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 $0 \le f(x) \le x$. – Martin R Apr 2 at 11:24
• 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. – Anirban Apr 2 at 14:19
• 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 $\int_0^1 f(x) dx = \frac 13$. – Martin R Apr 2 at 14:34
• @ 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. – Anirban Apr 2 at 15:36
• 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. – Anirban Apr 2 at 15:46