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$f,g$ be of bounded variation $\Rightarrow$ $f \circ g$ is of bounded variation?

if not, any counter example?

delog
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1 Answers1

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Let $f$ be the sign function, $f(x) = -1$ if $x\leq 0$, $f(x) = 1$ if $x>0$, and let $$ g(x) = \sum_{n=1}^\infty \frac{(-1)^n}{n^2} \chi_{[n, n+1)} (x), $$ where $\chi_A$ denotes the characteristic function of the set $A$.

It is easily seen that $f, g\in BV(\mathbb{R})$, but $$ (f\circ g)(x) = \sum_{n=1}^\infty (-1)^n \chi_{[n, n+1)}(x) $$ has not finite variation on $\mathbb{R}$.

If you prefer an example with continuous functions: let us consider the functions $f,g \in C([0,1])$ defined by $$ f(x) = \sqrt{x}, \qquad g(x) = \begin{cases} x^2 \sin^2(1/x), &\text{if}\ x\in (0,1],\\ 0, &\text{if}\ x = 0. \end{cases} $$ You can check that $f,g\in BV([0,1])$, but $f\circ g\not\in BV([0,1])$.

Rigel
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  • It also works with $g(x) = e^{-|x|}\sin(x)$, which seems easier to read and write. But good example nonetheless. It would also be interesting to explore what happens if we also require $f$ and $g$ to both be continuous. Can you make something continuous that still looks enough like $f$ to make the composition not have bounded variation? – Arthur May 28 '17 at 09:26
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    Sure, it's enough to change sign countably many times. I have added an example with continuous functions. – Rigel May 28 '17 at 10:34
  • how could I understand $g(x)$ is $BV(\Bbb R)$? and $TV$ of $f$ is $2$ right? – delog May 28 '17 at 10:38
  • @delog The total variation of $g$ is $2\cdot\frac11+2\cdot\frac14+2\cdot\frac19+\cdots=\frac{\pi^2}3$. – Arthur May 28 '17 at 10:42
  • @Arthur would you elaborate more the process to get the number 2 in front of each term? – delog May 28 '17 at 10:50
  • It goes from $0$ to $\frac11$. Then it goes from $\frac11$ to $-\frac14$. Then it goes from $-\frac14$ to $\frac19$, and so on. – Arthur May 28 '17 at 10:53