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 $a<b<c \in I$ such that $f(a)<f(b)$ and $f(c)\leq f(b)$ (draw a picture). So if $a_0<a_1<\ldots <a_n$ 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 $f(a_k)<f(a_{k+1})$ we must have $f(a_i)<f(a_{i+1})$ 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 $I$.

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    $\begingroup$ Shouldn't the last expression be $f(a_0)-f(a_k)+f(a_n)-f(a_k) = f(a_0) - 2f(a_k) + f(a_n)$? $\endgroup$ – Martin R Nov 27 '20 at 12:43
  • $\begingroup$ Yep, of course, thanks $\endgroup$ – Jonathan Hole Nov 27 '20 at 12:45
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    $\begingroup$ 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. $\endgroup$ – Martin R Nov 28 '20 at 11:48
  • $\begingroup$ Yeah, I forgot that convexity only implies continuity on the interior. $\endgroup$ – Jonathan Hole Nov 28 '20 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.

  • $\begingroup$ 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.. $\endgroup$ – JacobsonRadical Nov 27 '20 at 22:30
  • $\begingroup$ @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 $(0, 1)$. $\endgroup$ – Martin R Nov 27 '20 at 22:34
  • $\begingroup$ 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? $\endgroup$ – JacobsonRadical Nov 27 '20 at 22:38
  • $\begingroup$ @JacobsonRadical: No, continuity at the boundary is not needed. $\endgroup$ – Martin R Nov 27 '20 at 22:52
  • $\begingroup$ 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? $\endgroup$ – JacobsonRadical Nov 27 '20 at 22:53

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