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Suppose that a function $f$ is convex and increasing on $[0,+\infty)$ and $f(0)=0$ .Show that $$\sum _ { k = 0 } ^ { n } ( - 1 ) ^ { k } f ( a_ { k } ) \geq f \left( \sum _ { k = 0 } ^ { n } ( - 1 ) ^ { k } a _ { k } \right)$$ For any number $a _ { 0 } \geq a _ { 1 } \geq \ldots \geq a _ { n } \geq 0$

Please help me to solve this I have try using Jensen’s inequality. But I have thought it many times. I still can’t do this. Thank beforehand!

  • For even $n$ you can either apply Karamata's inequality https://en.wikipedia.org/wiki/Karamata%27s_inequality, or just induct on $n$. – richrow Jun 24 '20 at 17:19
  • Actually, it's enough to prove inequality for even $n$ (because we can set $a_n=0$). – richrow Jun 24 '20 at 17:20
  • Is this problem an exercise in a book? Are you sure it is true? – Zim Jun 24 '20 at 18:35
  • For induction step you need only to prove that for $0\leq a\leq b\leq c\leq d$ such that $a+d=b+c$ we have $f(a)+f(d)\geq f(b)+f(c)$. – richrow Jun 26 '20 at 06:51

1 Answers1

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It is enough to prove for even numbers $n=2m$, since for an odd number of $a_0,...,a_{2m-1}$ we can make it even by adding $a_{2m}=0$ to the set of numbers.

The induction start ($m=1$) with $a_0\geq a_1 \geq a_2$ follows from $$\int_{a_1-a_2}^{a_0} f'(t-a_1+a_2) \, {\rm d}t = f(a_0-a_1+a_2) - f(0) \\ =\int_{a_1-a_2}^{a_1} f'(t-a_1+a_2) \, {\rm d}t + \int_{a_1}^{a_0} f'(t-a_1+a_2) \, {\rm d}t \\ \leq f(a_2) - f(0) + \int_{a_1}^{a_0} f'(t) \, {\rm d}t \\ =f(a_0) - f(a_1) + f(a_2)$$ since $f$ is convex with $f(0)=0$.

It then follows after going from $m$ to $m+1$ $$\sum_{k=0}^{2m+2} (-1)^k f(a_k) = f(a_0) - f(a_1) + \sum_{k=2}^{2m+2} (-1)^k f(a_k) \\ \geq f(a_0) - f(a_1) + f\left(\sum_{k=2}^{2m+2} (-1)^k a_k\right)\\ \geq f\left(a_0 - a_1 + \sum_{k=2}^{2m+2} (-1)^k a_k\right) = f\left( \sum_{k=0}^{2m+2} (-1)^k a_k \right) \, $$ since $(-1)^2=1$ (allowing the shift and applying the induction hypothesis) and $\sum_{k=2}^{2m+2} (-1)^k a_k \leq a_2 \leq a_1 \leq a_0$.

Diger
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  • why $\int_a^b f'(x)dx=f(b)-f(a)$ with these assumptions? – Paul Jun 25 '20 at 08:50
  • I guess it is not too unreasonable to assume at least piecewise differentiability of $f$. In that case you can split the integral at the problematic points $t_i$ of $f'(t)$ and the extra terms in $$\int_a^b = \lim_{\epsilon \rightarrow 0} \left(\int_a^{t_i-\epsilon} + \int_{t_i+\epsilon}^b \right)$$ cancel on both sides. – Diger Jun 25 '20 at 15:07