1
$\begingroup$

Let $f: \mathbb R \to \mathbb R$ satisfy the following inequality:

$$ f\left(\frac{x_1 + x_2}{2}\right) < {f(x_1) + f(x_2) \over 2} $$ Show that:

$$ f\left(\frac{x_1 + x_2 + x_3}{3}\right) < {f(x_1) + f(x_2) + f(x_3) \over 3} $$

The above appeared to be fairly simple for the case of $x_1, \;x_2, \; x_3, \;x_4$. (Let's call that case $n=4$)

By the initial conditions we have: $$ f\left(\frac{x_1 + x_2}{2}\right) < {f(x_1) + f(x_2) \over 2} \\ f\left(\frac{x_3 + x_4}{2}\right) < {f(x_3) + f(x_4) \over 2} $$

Adding the inequality gives:

$$ f\left(\frac{x_1 + x_2}{2}\right) + f\left(\frac{x_3 + x_4}{2}\right) < {f(x_1) + f(x_2) + f(x_3) + f(x_4) \over 2} \iff \\ \iff {1 \over 2}\left(f\left(\frac{x_1 + x_2}{2}\right) + f\left(\frac{x_3 + x_4}{2}\right)\right) < {f(x_1) + f(x_2) + f(x_3) + f(x_4) \over 4} $$

Let $p = \frac{x_1 + x_2}{2}$, $q = \frac{x_3 + x_4}{2}$, then using the initial conditions:

$$ f\left(\frac{p + q}{2}\right) < \frac{f(p) + f(q)}{2} $$

And therefore:

$$ f\left(\frac{x_1+x_2+x_3+x_4}{4}\right) < \frac{f(x_1) + f(x_2) + f(x_3) + f(x_4)}{4} $$

I see how this may be expanded further by induction for $n = 2^k$. However I failed to prove the case of $n = 3$.

How can I prove that the inequality holds for $n = 3$?

$\endgroup$
  • 1
    $\begingroup$ Well, actually by induction you get the result for $n=2^k$. The idea for the other values of $n$ is to apply a 'down' version of induction. $\endgroup$ – Stan Tendijck Aug 7 '18 at 10:58
  • $\begingroup$ @StanTendijck What is a 'down' version of induction? $\endgroup$ – roman Aug 7 '18 at 11:00
  • $\begingroup$ With that I mean that given that it works for a general $n$ you could prove the $n-1$ case (by replacing the $n$th element with the mean of the first $n-1$) $\endgroup$ – Stan Tendijck Aug 7 '18 at 13:07
3
$\begingroup$

The inequality in the question is what defined a midpoint convex function (which is almost the same as convexity).

The inequality you are after is a special case of the proof presented in this answer. Let $\overline x=\frac{x_1+x_2+x_3}{3}$ and consider $$f(\overline x)=f\left(\frac{x_1+x_2+x_3+\overline x}{4}\right) <\frac{f(x_1)+f(x_2)+f(x_3)+f(\overline x)}{4}$$ where we have used the $n=4$ case you already have proven. Rearrange the inequality to get the desired result $$f(\overline x) <\frac{f(x_1)+f(x_2)+f(x_3)}{3}$$ We can in fact use the same technique as presented in the linked answer to prove a generalization, namely that $$f\left(\sum_{i=1}^n x_i t_i\right) < \sum_{i=1}^n t_i f(x_i)$$ for any rational $t_i$ with $\sum_{i=1}^n t_i = 1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.