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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?

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  • $\begingroup$ No proof without assuming continuity of $f'$ yet? Why not? $\endgroup$
    – user171110
    Commented Mar 6, 2016 at 15:23
  • $\begingroup$ Where is the proof without continuity of $f'$? $\endgroup$
    – user171110
    Commented Mar 24, 2016 at 10:15

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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 $x>0$.

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  • $\begingroup$ However, the exercise should be done without assuming continuity of the first derivative of f... $\endgroup$
    – user171110
    Commented Feb 11, 2016 at 21:21
  • $\begingroup$ I suspected as much. Oh well, can't help you there :) $\endgroup$
    – grand_chat
    Commented Feb 11, 2016 at 21:23
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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, $$\int_0^1 (1- 2 x^2)(e^{-0.0001x} - 1) 1000dx = 0.112762 > 0$$

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  • $\begingroup$ Maybe Wolfram Alpha has some error. Look at the same integral here: it is negative!!! $\endgroup$
    – user171110
    Commented Feb 11, 2016 at 21:16
  • $\begingroup$ very interesting. I did see the difference from the same integral. Report a bug to wolfram? $\endgroup$
    – runaround
    Commented Feb 11, 2016 at 21:42
  • $\begingroup$ Maybe you do that, because I have no time to do that... $\endgroup$
    – user171110
    Commented Feb 11, 2016 at 21:48
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    $\begingroup$ sent a report to wolfram. $\endgroup$
    – runaround
    Commented Feb 11, 2016 at 21:53
  • $\begingroup$ Maybe delete that answer? $\endgroup$
    – user171110
    Commented Mar 24, 2016 at 10:14

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