$x sin(1 over x)$ can be decomposed?

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Let $f(x)=xsin(1/x)$ for $0<xle 1$, and $f(0)=0$. I was told that every continuous function on $[0,1]$ can be written as $a(x)+s(x)$, where $a(x)$ is absolute continuous and $s(x)$ is singular. But how to write $f(x)$ this way?



Note: I was able to verify that $f(x)$ is indeed continuous. Further, I took the derivative and it turns out it is not Lebesgue integrable on $[0,1]$, so $f(x)$ is not abs. continuous. But I am stuck to find $a(x)$ and $s(x)$.




real-analysis lebesgue-integral absolute-continuity




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    The ones who downvote, instead of downvoting and continue on please explain what made you downvote.
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Let $f(x)=xsin(1/x)$ for $0<xle 1$, and $f(0)=0$. I was told that every continuous function on $[0,1]$ can be written as $a(x)+s(x)$, where $a(x)$ is absolute continuous and $s(x)$ is singular. But how to write $f(x)$ this way?



Note: I was able to verify that $f(x)$ is indeed continuous. Further, I took the derivative and it turns out it is not Lebesgue integrable on $[0,1]$, so $f(x)$ is not abs. continuous. But I am stuck to find $a(x)$ and $s(x)$.




real-analysis lebesgue-integral absolute-continuity




share|cite|improve this question

















  • 4




    The ones who downvote, instead of downvoting and continue on please explain what made you downvote.
    – Holo
    Jul 28 at 20:30












up vote
5
down vote

favorite
2









up vote
5
down vote

favorite
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2





Let $f(x)=xsin(1/x)$ for $0<xle 1$, and $f(0)=0$. I was told that every continuous function on $[0,1]$ can be written as $a(x)+s(x)$, where $a(x)$ is absolute continuous and $s(x)$ is singular. But how to write $f(x)$ this way?



Note: I was able to verify that $f(x)$ is indeed continuous. Further, I took the derivative and it turns out it is not Lebesgue integrable on $[0,1]$, so $f(x)$ is not abs. continuous. But I am stuck to find $a(x)$ and $s(x)$.




real-analysis lebesgue-integral absolute-continuity




share|cite|improve this question













Let $f(x)=xsin(1/x)$ for $0<xle 1$, and $f(0)=0$. I was told that every continuous function on $[0,1]$ can be written as $a(x)+s(x)$, where $a(x)$ is absolute continuous and $s(x)$ is singular. But how to write $f(x)$ this way?



Note: I was able to verify that $f(x)$ is indeed continuous. Further, I took the derivative and it turns out it is not Lebesgue integrable on $[0,1]$, so $f(x)$ is not abs. continuous. But I am stuck to find $a(x)$ and $s(x)$.





real-analysis lebesgue-integral absolute-continuity





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edited Jul 28 at 20:11









Davide Morgante

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asked Jul 28 at 20:07









Miranda

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  • 4




    The ones who downvote, instead of downvoting and continue on please explain what made you downvote.
    – Holo
    Jul 28 at 20:30












  • 4




    The ones who downvote, instead of downvoting and continue on please explain what made you downvote.
    – Holo
    Jul 28 at 20:30







4




4




The ones who downvote, instead of downvoting and continue on please explain what made you downvote.
– Holo
Jul 28 at 20:30




The ones who downvote, instead of downvoting and continue on please explain what made you downvote.
– Holo
Jul 28 at 20:30










1 Answer
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A continuous function of bounded variation has that property. In general, a function of bounded variation can be decomposed into the sum of an absolutely continuous function, a singular function, and a jump function.



The function $x mapsto x sin(1/x)$ is not of bounded variation on $[0,1].$






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  • You are the man! Many thanks.
    – Miranda
    Jul 30 at 19:24










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1 Answer
1






active

oldest

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active

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active

oldest

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up vote
2
down vote













A continuous function of bounded variation has that property. In general, a function of bounded variation can be decomposed into the sum of an absolutely continuous function, a singular function, and a jump function.



The function $x mapsto x sin(1/x)$ is not of bounded variation on $[0,1].$






share|cite|improve this answer























  • You are the man! Many thanks.
    – Miranda
    Jul 30 at 19:24














up vote
2
down vote













A continuous function of bounded variation has that property. In general, a function of bounded variation can be decomposed into the sum of an absolutely continuous function, a singular function, and a jump function.



The function $x mapsto x sin(1/x)$ is not of bounded variation on $[0,1].$






share|cite|improve this answer























  • You are the man! Many thanks.
    – Miranda
    Jul 30 at 19:24












up vote
2
down vote










up vote
2
down vote









A continuous function of bounded variation has that property. In general, a function of bounded variation can be decomposed into the sum of an absolutely continuous function, a singular function, and a jump function.



The function $x mapsto x sin(1/x)$ is not of bounded variation on $[0,1].$






share|cite|improve this answer















A continuous function of bounded variation has that property. In general, a function of bounded variation can be decomposed into the sum of an absolutely continuous function, a singular function, and a jump function.



The function $x mapsto x sin(1/x)$ is not of bounded variation on $[0,1].$







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 28 at 21:45


























answered Jul 28 at 21:36









RRL

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  • You are the man! Many thanks.
    – Miranda
    Jul 30 at 19:24
















  • You are the man! Many thanks.
    – Miranda
    Jul 30 at 19:24















You are the man! Many thanks.
– Miranda
Jul 30 at 19:24




You are the man! Many thanks.
– Miranda
Jul 30 at 19:24












 

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