Is the product of an absolutely continuous function $f$ and a continuous function of bounded variation $g$ on $[0,1]$ for which $f(0)=0$ and $g(0)=0$, absolutely continuous?
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Why do you think it would be? – GEdgar Feb 24 '15 at 18:44
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1Take $f$ to be the constant one except near the end points (and absolutely continuous). Choose some bounded variation $g$ that is not absolutely continuous (on $[\epsilon,1-\epsilon]$) then the product will not be absolutely continuous either. – copper.hat Feb 24 '15 at 18:46
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Thank you for your answer. Would you please give me more details? – Zeinab Feb 26 '15 at 03:13
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Take $$f(x)= \begin{cases} x & x\leq \frac{1}{2}\\ \frac{1}{2} & x>\frac{1}{2} \end{cases} $$ Clearly $f$ is AC and $f(0)=0$.
Next, take $g$ to be famous deil stair function, or cantor function if prefer, then clearly $g(0)=0$, $g$ is continuous of bounded variation but not AC. However, $fg= \frac{1}{2}g$ if $x>\frac{1}{2}$ and hence it is not AC
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But $f$ is not continuous at $x=\frac{1}{2}$, so it is not absolutely continuous on $[0,1]$. – Zeinab Feb 27 '15 at 08:58
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If we let $f(x)=x$ and $g$ be the Cantor function on $[0,1]$, then $h(x)=f.g(x)=xg(x)$ is not absolutely continuous on $[0,1]$. Assume towards a contradiction that $h$ is absolutely continuous. The cantor function $g$ is differentiable almost everywhere on $[0,1]$ and $g'=0$ a.e., so $h$ is differentiable almost everywhere on $[0,1]$ and we have $$1=1g(1)-0g(0)=h(1)-h(0)=\int_0^1h'(x)dx=\int_0^1[xg'(x)+g(x)]dx=\int_0^1g(x)dx=\frac{1}{2}.$$ This contradiction completes the proof.
Zeinab
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