We consider functions defined on an interval $[a,b]$. I have to prove that a product of functions of bounded variation is a function of bounded variation. I have to also show that this isn't true for quotient in general and tell which additional assumption guarantees that quotient IS of bounded variation.
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2So what did you do so far and what is your question? – flawr Oct 12 '14 at 13:01
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Yes flawr is right. I have given you the key step. If you tried it. You should do it now. – Ri-Li Oct 12 '14 at 13:09
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Hint:
$|(fg)(x)-(fg)(y)|\leq |f(x)||g(x)-g(y)|+|g(y)||f(x)-f(y)|$. Again $f,g$ are bounded so what will you get from here??
Ri-Li
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1If f, g is bounded then it is. You have given f,g has bounded variation right?/ – Ri-Li Oct 12 '14 at 14:11
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This came to my mind first.
$$fg = \frac{(f+g)^2 - (f-g)^2}{4}$$
Since sum of functions of bounded variation have bounded variation, it suffices to show that $f^2$ is of bounded variation when $f$ is. To show the latter, use
$$|f(x)^2-f(y)^2| = |f(x)-f(y)||f(x)+f(y)| \leq |f(x)-f(y)|(|f(x)|+|f(y)|)$$
and the fact that $f$ is bounded.
Dhruv Kohli
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Another approach: $f$ is of bounded variation iff $f=f_1-f_2$ for $f_1,f_2$ positive and increasing. Then $fg=(f_1-f_2)(g_1-g_2)=(f_1g_1+f_2g_2)-(f_1g_2+f_2g_1)$ and we are done.
Gilad Reti
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1But how do you prove the product of 2 increasing functions is increasing? (It's false in fact) – Mariana Jul 11 '21 at 02:28
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