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Show that if $f,g \in BV([a,b]) $ then $$fg \in BV([a,b]) $$ My idea was to write $f = f_1 - f_2$ and $g = g_1 - g_2$ (which are non decreasing functions) and write $fg$ as: $$ fg = (f_1 g_1 + f_2 g_2 ) - (f_1 g_2 + f_2 g_1)$$

which are two non decreasing function hence $fg$ is a bounded variation function. Is this correct?

Julian Vené
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    Please show your attempt. – Kavi Rama Murthy Jun 16 '21 at 08:54
  • I uploaded the question – Julian Vené Jun 16 '21 at 08:59
  • You are very confused. You want to prove that the difference of two BV functions is BV. Why do you take the product $fg$? – Kavi Rama Murthy Jun 16 '21 at 09:02
  • nono sorry i explained very badly. I want to prove that $f,g \in BV$ implies $fg \in BV$, but I proved this (as you can see) assuming that any difference of two non decreasing function "creates" a bounded variation function. My question is about this last fact: is this statement true (any difference of two non decreasing function "creates" a bounded variation function)? – Julian Vené Jun 16 '21 at 09:06
  • Proving the difference of two BV functions is BV is his lemma to proving that the product is. This is a good question communicated poorly. – aristotlefromgreece Jun 16 '21 at 09:11
  • Do you know that a monotonic function on $[a,b]$ is of bouned variation? (A fact which is immediate from the definition). If so the question is trivial and I think you are not attempting to prove it yourself. – Kavi Rama Murthy Jun 16 '21 at 09:18
  • No. Let $f_1(x) = x = g_1(x)$, and $f_2 = -1 = g_2$. Each of those four functions is non-decreasing, but $f_1 g_2 + f_2 g_1 = -2x$ is decreasing. – user0 Aug 02 '21 at 18:31

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Proposition: The difference of two increasing functions on a compact set is BV

Let $f_1,f_2$ be increasing and $g=f_1-f_2$. Now consider any partition of of $[a,b]$, $a=x_0<x_1<...<x_n=b$.

$$\sum^n_{j=1}|g(x_j)-g(x_{j-1})| \\=\sum|f_1(x_j)-f_2(x_j)-f_1(x_{j-1})+f_2(x_{j-1})|\\=\sum |(f_1(x_j)-f_1(x_{j-1}))+(-f_2(x_j)+f_2(x_{j-1}))|\\ \le \sum |f_1(x_j)-f_1(x_{j-1})|+|f_2(x_j)-f_2(x_{j-1})|\\=\sum f_1(x_j)-f_1(x_{j-1})+f_2(x_j)-f_2(x_{j-1})\\=f_1(x_n)-f_1(x_0)+f_2(x_n)-f_2(x_0)\\\square$$

EDIT: The question has changed since I looked, but yes, what you have now is correct and the above proves it.

aristotlefromgreece
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