# Properties of $\frac{f(x)}{x}$ if $f(x)$ is a convex function [Zorich's book]

Show that

a) if a convex function $$f:\mathbb{R}\to \mathbb{R}$$ is bounded, then it is constant;

b) if $$\lim \limits_{x\to -\infty}\dfrac{f(x)}{x}=\lim \limits_{x\to +\infty}\dfrac{f(x)}{x}=0,$$ for a convex function $$f:\mathbb{R}\to \mathbb{R}$$, then $$f$$ is constant.

c) for any convex function $$f$$ defined on an open interval $$a (or $$-\infty), the ratio $$\dfrac{f(x)}{x}$$ tends to a finite limit or to infinity as $$x$$ tends to infinity in the domain of definition of the function.

These problems are from Zorich's book. I have solved parts a) and b) but have some issues with part c).

I was trying to solve it by contradiction. WLOG suppose $$f(x)$$ is defined on $$(0,\infty)$$ and the $$\lim \limits_{x\to +\infty}\dfrac{f(x)}{x}$$ does not exist. Then $$\forall \delta>0$$ we can find $$x_0,y_0>\delta$$ such that $$\left|\frac{f(x_0)}{x_0}-\frac{f(y_0)}{y_0}\right|>c$$ for some $$c>0$$. Then I've tried to apply the definition of convexity to $$\delta to get contradiction but failed.

Would be very grateful if you can show how to finish the proof using my approach. Thanks in advance!

• I don't know about your method, but you can show $$\lim_{x\to\infty} \frac{f(x)}{x} = \sup_{a < x < y} \frac{f(y)-f(x)}{y-x}$$ May 31, 2021 at 2:06
• @BrianMoehring, if you can give more detailed how you obtained it and why it leads to the solution then it would be great! And I'll highly appreciate your reply!
– ZFR
May 31, 2021 at 2:37

Assume $$f$$ is convex on $$(a,+\infty)$$
For any number $$b>a$$, the convexity gives that the function $$g(x)=\frac{ f(x)-f(b)}{x-b}$$ is increasing on $$(b,+\infty)$$
Thus $$g$$ has a limit when $$x$$ converges to $$\infty$$.
On the other hand, $$\lim_{x\rightarrow +\infty} \frac{f(b)}{x-b}=0 \quad \text{and} \quad \lim_{x\rightarrow +\infty} \frac{x}{x-b}=1$$ Hence forth the conclusion.

• Sorry but I guess when you are defined the function $g(x)$ the wording is not clear. You are fixing that $b$ so it should be $g_b(x)$, right? Can you clarify your sentence please!
– ZFR
May 31, 2021 at 3:00
• Is this $b$ fixed or not?
– ZFR
May 31, 2021 at 3:19
• @ZFR It doesn't matter. What we want is to prove that $$\lim \frac{f(x)-f(b)}{x-b}$$ exists and the eventual behavior of $g$ is not related. May 31, 2021 at 3:20
• And this existence is equivalent to the existence of the very first limit. "$b$ fixed or not" doesn't matter. Anyways, you can let it fixed. May 31, 2021 at 3:21
• Thanks a lot! You really helped me!
– ZFR
May 31, 2021 at 3:24