# How prove this integral inequality $2\int_{-1}^{1}f(x)g(x)dx\ge\int_{-1}^{1}f(x)dx\int_{-1}^{1}g(x)dx$

Let $f:[-1,1]\longrightarrow \mathbb{R}$ be increasing on $[0,1]$ and even, i.e. $f(x)=f(-x)$ $\forall x\in [-1,1]$.

Let $g:[-1,1]\longrightarrow \mathbb{R}$ be convex, i.e. $g(tx+(1-t)y)\le tg(x)+(1-t)g(y)$ for every $x,y\in [-1,1]$ and $t\in (0,1)$,

Show that

$$2\int_{-1}^{1}f(x)g(x)dx\ge\int_{-1}^{1}f(x)dx\int_{-1}^{1}g(x)dx$$

My try: I think this can use Cauchy-Schwarz inequality

$$\int_{a}^{b}f^2(x)dx\int_{a}^{b}g^2(x)dx\ge\left(\int_{a}^{b} f(x)g(x)dx\right)^2$$

But I can't, and this problem is from China competition yesterday.

• $f$ is even function but $f$ is increasing? – Shuchang Oct 27 '13 at 5:37
• Oh,Thank you,I have edit, – user94270 Oct 27 '13 at 5:40

Accounting for the symmetry of $f(\pm x)$ we want

$$\int_{0}^{1}f(x)(g(x)+g(-x)) dx \ge \int_{0}^{1}f(x)dx\int_{0}^{1}(g(x)+g(-x)) dx$$

or

$$\int_{0}^{1}f(x)G(x) dx \ge \int_{0}^{1}f(x)dx\int_{0}^{1}G(x) dx$$

where $G(x)=g(x) + g(-x)$ for a convex function $g$. Observe that $G$ is increasing and the problem is finished (it is the rearrangement inequality for $f$ and $G$).

The increasing nature of $G$,

if $g$ is a convex function then $g(x)+g(-x)$ is increasing on any interval $[0,a]$ where it is defined

is another way to say $g(p) + g(-p) \leq g(q) + g(-q)$ when $|p| \leq |q| \leq a$, when $g$ is convex and defined on $[-a,a]$.

• Why:we only prove this $$\int_{0}^{1}f(x)(g(x)+g(-x)) dx \ge \int_{0}^{1}f(x)dx\int_{0}^{1}(g(x)+g(-x)) dx$$？ – china math Oct 27 '13 at 17:19
• Write the integrals that include $g$ as a sum of integrals on $[-1,0]$ and $[0,1]$, then change variable $x \to -x$ in the integral on $[-1,0]$. – zyx Oct 27 '13 at 17:23

I found this (written in Chinese): • you were right, my example contained an error – Ewan Delanoy Oct 27 '13 at 16:20