Proving a convex function is of bounded variation

Let $$f: I \rightarrow \mathbb{R}$$ be a convex function, where $$I$$ denotes a closed interval. If $$f$$ is convex, how would I prove that $$f$$ is of bounded variation?

By convexity there cannot exist $$a such that $$f(a) and $$f(c)\leq f(b)$$ (draw a picture). So if $$a_0 is a partition of $$I$$ we might start off by having $$f(a_0)>f(a_1)$$, but the moment we have a $$k$$ such that $$f(a_k) we must have $$f(a_i) for all $$i\geq k$$. Then $$\sum\limits_{i=0}^{n-1} |f(a_{i+1})-f(a_i)|=\sum\limits_{i=0}^{k-1} f(a_i)-f(a_{i+1})+ \sum\limits_{i=k}^{n-1} f(a_{i+1})-f(a_i)=f(a_0)-2f(a_k)+f(a_n),$$

and so we only have to show that $$f$$ is bounded on the closed interval $$I$$ and we will be done. For this, write $$I=[x, z]$$ and let $$y$$ be any point strictly in between $$x$$ and $$z$$. Let $$f_1$$ be the linear interpolation of the points $$(x,f(x))$$ and $$(y, f(y))$$, and let $$f_2$$ be the linear interpolation of the points $$(y, f(y))$$ and $$(z, f(z))$$. We then see that for $$a\in [x,y]$$, $$f_2(a)\leq f(a) \leq f_1(a)$$ and for $$a\in [y,z]$$, $$f_1(a)\leq f(a) \leq f_2(a)$$, so $$f$$ is bounded on $$I$$.

• Shouldn't the last expression be $f(a_0)-f(a_k)+f(a_n)-f(a_k) = f(a_0) - 2f(a_k) + f(a_n)$? Nov 27, 2020 at 12:43
• Yep, of course, thanks Nov 27, 2020 at 12:45
• There is a minor technical problem: A convex function is continuous on an open interval, but not necessarily at the boundary points. An example would be $f(0) = 1$ and $f(x) = x$ for $0 < x \le 1$. – Your solution still works because a convex function on a compact interval is bounded above and below. Nov 28, 2020 at 11:48
• Yeah, I forgot that convexity only implies continuity on the interior. Nov 28, 2020 at 13:22

A small variation (!) of Jonathan's answer: Let $$f: [a, b] \to \Bbb R$$ be convex. Let $$c = \sup \{ x \in [a, b] \mid \text{f is decreasing on [a, x]} \} \, .$$ Then $$f$$ is decreasing on $$[a, c]$$ and increasing on $$[c, b]$$, and therefore $$V_a^b(f) = V_a^c(f) + V_c^b(f) = (f(a)-f(c)) + (f(b) - f(c)) = f(a) - 2f(c) + f(b) < \infty \, .$$

In other words: a convex function is piecewise monotone and therefore of bounded variation on a compact interval.

• Sorry for reopening an old post. In this proof, do we need $f$ to be continuous on the whole $[a,b]$? for example, if $f$ is only continuous on $(a,b)$, perhaps it will blow up at $a$ or $b$? so perhaps the conclusion should be "a continuous convex function is piecewise monotone and therefore of bounded variation on a compact interval"? I am not sure, I have recently been confused about the subtlety of convex function on compact intervals.. Nov 27, 2020 at 22:30
• @JacobsonRadical: If $f$ is convex on an open interval $(a, b)$ then it has (finite or infinite) limits at the boundary points. If either limit is infinite then the function is not of bounded variation. An example would be $f(x) = 1/x$ on $(0, 1)$. Nov 27, 2020 at 22:34
• Yes. This is exactly the reason of my question. The function in this post is convex on the compact interval $[a,b]$, do we need the function to be continuous at the boundary points to make the argument work? or we don't? Nov 27, 2020 at 22:38
• @JacobsonRadical: No, continuity at the boundary is not needed. Nov 27, 2020 at 22:52
• I am sorry, but why..? What if $f(b)$ is infinite? I am sorry for the confusion. so on compact interval, $f$ will not be infinite at boundary points? Nov 27, 2020 at 22:53