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!
Best Answer
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.