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 $a<x<+\infty$ (or $-\infty<x<a$), 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 $\delta<x_0<y_0$ to get contradiction but failed.

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

  • $\begingroup$ I don't know about your method, but you can show $$\lim_{x\to\infty} \frac{f(x)}{x} = \sup_{a < x < y} \frac{f(y)-f(x)}{y-x}$$ $\endgroup$ May 31, 2021 at 2:06
  • $\begingroup$ @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! $\endgroup$
    – RFZ
    May 31, 2021 at 2:37

1 Answer 1


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.

  • $\begingroup$ 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! $\endgroup$
    – RFZ
    May 31, 2021 at 3:00
  • $\begingroup$ Is this $b$ fixed or not? $\endgroup$
    – RFZ
    May 31, 2021 at 3:19
  • $\begingroup$ @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. $\endgroup$ May 31, 2021 at 3:20
  • $\begingroup$ 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. $\endgroup$ May 31, 2021 at 3:21
  • 1
    $\begingroup$ Thanks a lot! You really helped me! $\endgroup$
    – RFZ
    May 31, 2021 at 3:24

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