Suppose $f$ and $g$ are two functions can be composed. If $f$ is continuous and bounded variation and $g$ is bounded variation, is that true $f\circ g$ is bounded variation?
As we know, if $f$ is Lipschitz, the statement is true, but I don't know if it is true for the above situation.

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    $\begingroup$ Consider $g(x)=\sqrt x$ and $f(x)=x^2\sin (1/x)$ $\endgroup$ Nov 3, 2016 at 2:34
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    $\begingroup$ For $f\circ g(x)=x\sin(1/\sqrt{x})$, I tried to pick $x_{2k}=\frac{1}{(2k\pi)^2}$ and $x_{2k-1}=\frac{1}{(2k\pi-\pi/2)^2}$, but it seems the variation will be a sum of series of the form $\sum\frac{1}{n^2}$, which is not divergent. $\endgroup$
    – Connor
    Nov 3, 2016 at 2:53
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    $\begingroup$ yes you are right. Take $f(x)=\sqrt x$ and $g(x)=x^2\sin ^2(1/x)$ $\endgroup$ Nov 3, 2016 at 2:59

1 Answer 1


As already mentioned in the comments, there are continuous functions $f$ and functions of bounded variation $g$ such that $f\circ g$ is not of bounded variation. But I think it is worth to point out that there is more which is already known on the topic of composition of functions with functions of bounded variation.

Theorem (Josephy (1981), p. 355, theorem 4) For a given function $f:[0,1]\to[0,1]$, the composition $f\circ g$ is of bounded variation for all functions $g:[0,1]\to[0,1]$ of bounded variation if and only if $f$ satisfies a Lipschitz condition on $[0,1]$.


  • The fact that the domain and codomain of the functions in the statement of theorem is $[0,1]$ does not reduce its generality: every finite interval will do.
  • In the same paper, theorem 3 states a necessary and sufficient condition for a function $g$ such that $f\circ g$ is of bounded variation for each $f$ of bounded variation, and identifies the class for such functions.

[1] Josephy, Michael (1981), "Composing Functions of Bounded Variation", Proceedings of the American Mathematical Society Vol. 83, No. 2, pp. 354-356, MR 624930, Zbl 0475.26005.


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