3
$\begingroup$

This is the seemingly "obvious" property that the graph of a non-monotone convex function on $(-\infty,\infty )$ has a "u-shape". I wonder how I may prove this?

My approach was the following:

For $x_1 < x_2 < x_3 \in [1,\infty ) $ from the definition of convex function

$$\frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \le \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2} $$ so that $$f \left({x_3}\right) - f \left({x_2}\right) \ge \frac {(x_3 - x_2)(f \left({x_2}\right) - f \left({x_1}\right))} {x_2 - x_1} $$

So if $f \left({x_2}\right) \ge f \left({x_1}\right)$ then $f \left({x_3}\right) \ge f \left({x_2}\right)$. If there is no such $x_1, x_2$, then the function is monotonically decreasing on $[1,\infty )$.

Simarily For $x_1 < x_2 < x_3 \in (-\infty , -1] $

$$\frac {f \left({x_1}\right) - f \left({x_2}\right)} {x_1 - x_2} \le \frac {f \left({x_2}\right) - f \left({x_3}\right)} {x_2 - x_3} $$ so that $$f \left({x_1}\right) - f \left({x_2}\right) \ge \frac {(x_1-x_2)(f \left({x_2}\right) - f \left({x_3}\right))} {x_2 - x_3} $$

So if $f \left({x_2}\right) \ge f \left({x_3}\right)$ then $f \left({x_1}\right) \ge f \left({x_2}\right)$ and as above if no such $x_2, x_3$ exists then the function is monotonically decreasing on $(-\infty,-1]$

Is it possible to prove that in both cases we cannot have that the function is monotonically decreasing?

Thanks in advance!

$\endgroup$
1
$\begingroup$

If $f$ is not increasing, there exists $x_1<x_2$ such that $f(x_1)>f(x_2)$. Since $f$ is convex, the map $\tau:x\mapsto \frac{f(x_1)-f(x)}{x_1-x}$ is non-decreasing. Note that $\tau(x_2)<0$, hence there is some $\ell\in [-\infty,0)$ such that $\lim_{x\to - \infty}\tau(x) = \ell$. If $\ell$ is finite, this implies $\frac{f(x_1)-f(x)}{x_1-x} = \ell+o(1)$, that is $f(x)=\ell x + o(x)$ as $x\to -\infty$, hence $\lim_{x\to -\infty} f(x)=\infty$. If $\ell =-\infty$ the proof readily adapts.

If $f$ is not decreasing, there exists $x_1<x_2$ such that $f(x_1)<f(x_2)$. Since $f$ is convex, the map $\tau:x\mapsto \frac{f(x)-f(x_2)}{x-x_2}$ is non-decreasing. Note that $\tau(x_1)>0$, hence there is some $\ell \in (0,\infty]$ such that $\lim_{x\to \infty}\tau(x) = \ell$. If $\ell$ is finite, this implies $\frac{f(x)-f(x_2)}{x-x_2} = \ell+o(1)$, that is $f(x)=\ell x + o(x)$ as $x\to \infty$, hence $\lim_{x\to\infty} f(x)=\infty$. If $\ell =\infty$ the proof readily adapts.

$\endgroup$
  • $\begingroup$ Thanks! But what is meant with $\frac{f(x_1)-f(x)}{x_1-x} = \ell+o(1)$? $\endgroup$ – MrFranzén Sep 5 '19 at 10:34
  • 1
    $\begingroup$ @MrFranzén If you're not familiar with this notation, just note instead that $$f(x) = \frac{f(x)-f(x_1)}{x-x_1}(x-x_1) + f(x_1)$$ $\frac{f(x)-f(x_1)}{x-x_1}$ goes to $\ell<0$ and $x-x_1$ goes to $-\infty$. $\endgroup$ – Gabriel Romon Sep 5 '19 at 10:37
  • $\begingroup$ @MrFranzén. RE: $o(1)$: Do a google search for "big-O and little-o notation". $\endgroup$ – DanielWainfleet Sep 5 '19 at 11:31

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.