$\int\limits_0^1 {\left( {1 - 2{x^2}} \right)f\left( x \right)dx}<0$, when $f$:convex and differentiable with $f(0)=0$

Question

Let $f:[0,1]\to\mathbb{R}$ be a differentiable function that is convex and $f(0)=0$. Prove that: $\int\limits_0^1 {\left( {1 - 2{x^2}} \right)f\left( x \right)dx}<0$.

I thought that since $f$ is convex and differentiable, $f'$ is strictly increasing, but cannot go any further. Any help?

Answer 1

Here's a proof assuming $f$ is continuously differentiable.

Use integration by parts with $u=f(x)/x$ and $v=\frac12(x^2-x^4)$ to find $$ \begin{align} \int_0^1(1-2x^2)f(x)\,dx&=\left.uv\right|_0^1-\int_0^1 vdu\\ &=\left.\frac12\frac{f(x)}x(x^2-x^4)\right|_0^1-\frac12\int_0^1(x^2-x^4)\left({xf'(x)-f(x)\over x^2}\right)\,dx\\ &=\frac12\int_0^1(x^2-1)\left(xf'(x)-f(x)\right)\,dx\,.\tag1 \end{align} $$ Since $f$ is convex and continuously differentiable, the function lies above all its tangents, i.e., for every $t$ and $a$ we have $$ f(t)\ge f(a) + f'(a)(t-a)\,.\tag2 $$ With $t=0$ and $a=x$, and using the fact that $f(0)=0$, (2) gives $$xf'(x)-f(x)\ge 0$$ for every $x$. The proof is complete after noting that the function $x\mapsto x^2-1$ is negative when $x>0$.

Answer 2

It is not true. The reason is that for any contex function close to constant 0, the integral is positive.

Let $ f(x) = (e^{-0.0001x} - 1) 1000$.

f(0) = 0;

$f" > 0$, so f is convex.

From Wolfram, $$\int_0^1 (1- 2 x^2)(e^{-0.0001x} - 1) 1000dx = 0.112762 > 0$$