# Does a non-monotone convex function go to positive infinity as $x$ goes to $\pm \infty$?

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?

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