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It is well known that a continuous function on a compact interval $[a,b]$ that has a bounded derivative on $(a,b)$ is of bounded variation on $[a,b]$. I am curious that whether the continuity at the endpoints $a$ and $b$ is essential for the function to be of bounded variation on $[a,b]$? In other words, is there a function on $[a,b]$ that has a bounded derivative on $(a,b)$, is NOT continuous at the endpoints $a$ and $b$, and is NOT of bounded variation on $[a,b]$?

Pavan C.
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Jason
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  • I think that you can show that being differentiable with bounded derivative on $(a,b)$ implies that the function must have one-sided limits as you approach the end points, using the nested interval theorem. Then the function would differ from a function which is continuous on a compact interval only at two points. – Joe Apr 03 '22 at 16:51
  • Great. Your hint of the existence of one-sided limits of such a function f led me to first show that f is bounded on (a,b), then use the Bolzano-Weierstrass theorem to show that the bounded sequence {f(a+1/n)} has a convergent subsequence, and finally show the existence of the one-sided limit f(a+) which is equal to the limit of the above-mentioned convergent subsequence. This approach uses the mean-value theorem for derivative and the definition of one-sided limit. I am wondering how the nested interval theorem is used to show the existence of the one-sided limits? – Jason Apr 04 '22 at 06:48
  • You can use the Nested Intervals theorem to prove the Bolzano-Weierstrass theorem (the two are equivalent), and thus prove the existence of the one-sided limits as you already have. Just because two theorems are related doesn't mean that they are equally useful for proving various things. – Paul Sinclair Apr 04 '22 at 15:30
  • I don't want to work out the algebra, but the idea I had was: if the absolute value of the derivative is bounded by $M$, then each point $(x, f(x))$ defines an allowable region for all other points on the graph (in physics, the analogous concept would be the lightcone). As $x \to a$, the sequence of half-regions are nested, hence so are their boundaries at $x=a$, which are intervals, and which have lengths that are approaching zero. To make this argument formal, I'd take an arbitrary sequence approaching $a$, and a monotonic subsequence, then I'd find a formula for the intervals. – Joe Apr 04 '22 at 15:56
  • Many thanks for the delightful comments and discussion. – Jason Apr 05 '22 at 14:59

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