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Claim

$f, g : [a, b] → R$ be of bounded variation. Then show that $fg$ is of bounded variation.

To prove above claim, I would like to derive the fact such as $\mid f(x_i) g(x_i)-f(x_{i-1})g(x_{i-1})\mid\le\mid f(x_i) -f(x_{i-1})\mid \mid g(x_i)-g(x_{i-1})\mid$ (*)

any advice to handle this absolute inequality so that I could prove (*)?

Daschin
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    With intuition based on the product rule $ d(fg) = fdg + dgf$, it is more reasonable to expect an equality of the form $$ |(fg)(x_i) - (fg)(x_{i-1})| \leq |f|{\sup}|g(x_i) - g(x{i-1})| + |g|{\sup}|f(x_i) - f(x{i-1})| $$ which you can indeed prove. – Sangchul Lee May 27 '17 at 09:36
  • @Sangchul Lee upon my understanding, there's no derivation in the concept of BV any reason you come up with product rule first? – Daschin May 27 '17 at 09:38
  • @SangchulLee anyway, even though this place is not for social activity, good to see the person of Korean nationality same with me. – Daschin May 27 '17 at 09:41
  • Although BV functions are often non-differentiable, it is possible to formalize differential $df$ in terms of Riemann-Stieltjes integral. But I am not telling that the precise nature of this object is important. The point here is that $f \mapsto f(x_i) - f(x_{i-1})$ can be thought as a discretized version of differential, so you will expect some analogy. (p.s. Glad to see you too! :)) – Sangchul Lee May 27 '17 at 09:41

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Try this: $∣f(x_i)g(x_i)−f(x_{i−1})g(x_{i−1})∣ = |f(x_i)g(x_i)−f(x_{i−1})g(x_{i−1}) +f(x_i)g(x_{i-1})-f(x_i)g(x_{i-1})| = |f(x_i)(g(x_i)-g(x_{i-1})) + (f(x_i) −f(x_{i−1}))g(x_{i−1}) |$

duncan
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  • I had used what you proposed, but having a problem with dealing with $f(x_i)$ and $g(x_{i-1})$. The terms consists of two - $g(x_i)-g(x_{i-1})$ and $f(x_i) −f(x_{i−1})$ are easy two handle since their bounded property are derived from their of Bounded Variation property, but how could I be sure of boundedness of summation of $f(x_i)$ and $g(x_{i-1})$? – Daschin May 27 '17 at 10:32
  • if $f, g$ are of bounded variation, they are bounded. $|f(x)| < M, |g(x)| < L$. the fact is described here https://math.stackexchange.com/questions/2298594/f-is-of-bounded-variation-rightarrow-f-is-bounded/2298608#2298608 – duncan May 27 '17 at 11:00