Do we have $|F|_{TV([a,b])}=lim_{epsilonto 0+}|F|_{TV([a+epsilon,b])}$ if $F:[a,b]tomathbb{R}$ is continuous?












2












$begingroup$



Given any interval ${[a,b]}$, define the total variation ${|F|_{TV([a,b])}}$ of ${F}$ on ${[a,b]}$ as
$$
displaystyle |F|_{TV([a,b])} := sup_{a leq x_0 < ldots < x_n leq b} sum_{i=1}^n |F(x_i) - F(x_{i+1})|.
$$
Let $F:[a,b]tomathbb{R}$ be a continuous function.



Can one conclude that $$|F|_{TV([a,b])}=lim_{epsilonto 0+}|F|_{TV([a+epsilon,b])}?$$





If $F$ is absolutely continuous than this can by done by noting that
$$
|F|_{TV([c,d])}=int_c^d|F'(x)| dx.
$$
What can one say in the general case?






real-analysis bounded-variation







share|cite|improve this question









$endgroup$












  • $begingroup$
    Have you tried exploring the canonical examples with infinite total variation?
    $endgroup$
    – Alex R.
    Jan 4 '17 at 0:10










  • $begingroup$
    @AlexR. In that case, I assume that one also has $lim_{epsilonto0+}|F|_{TV([a+epsilon,b])}=infty$ and thus the equality is also true?
    $endgroup$
    – Jack
    Jan 4 '17 at 0:13


















2












$begingroup$



Given any interval ${[a,b]}$, define the total variation ${|F|_{TV([a,b])}}$ of ${F}$ on ${[a,b]}$ as
$$
displaystyle |F|_{TV([a,b])} := sup_{a leq x_0 < ldots < x_n leq b} sum_{i=1}^n |F(x_i) - F(x_{i+1})|.
$$
Let $F:[a,b]tomathbb{R}$ be a continuous function.



Can one conclude that $$|F|_{TV([a,b])}=lim_{epsilonto 0+}|F|_{TV([a+epsilon,b])}?$$





If $F$ is absolutely continuous than this can by done by noting that
$$
|F|_{TV([c,d])}=int_c^d|F'(x)| dx.
$$
What can one say in the general case?






real-analysis bounded-variation







share|cite|improve this question









$endgroup$












  • $begingroup$
    Have you tried exploring the canonical examples with infinite total variation?
    $endgroup$
    – Alex R.
    Jan 4 '17 at 0:10










  • $begingroup$
    @AlexR. In that case, I assume that one also has $lim_{epsilonto0+}|F|_{TV([a+epsilon,b])}=infty$ and thus the equality is also true?
    $endgroup$
    – Jack
    Jan 4 '17 at 0:13
















2












2








2





$begingroup$



Given any interval ${[a,b]}$, define the total variation ${|F|_{TV([a,b])}}$ of ${F}$ on ${[a,b]}$ as
$$
displaystyle |F|_{TV([a,b])} := sup_{a leq x_0 < ldots < x_n leq b} sum_{i=1}^n |F(x_i) - F(x_{i+1})|.
$$
Let $F:[a,b]tomathbb{R}$ be a continuous function.



Can one conclude that $$|F|_{TV([a,b])}=lim_{epsilonto 0+}|F|_{TV([a+epsilon,b])}?$$





If $F$ is absolutely continuous than this can by done by noting that
$$
|F|_{TV([c,d])}=int_c^d|F'(x)| dx.
$$
What can one say in the general case?






real-analysis bounded-variation







share|cite|improve this question









$endgroup$





Given any interval ${[a,b]}$, define the total variation ${|F|_{TV([a,b])}}$ of ${F}$ on ${[a,b]}$ as
$$
displaystyle |F|_{TV([a,b])} := sup_{a leq x_0 < ldots < x_n leq b} sum_{i=1}^n |F(x_i) - F(x_{i+1})|.
$$
Let $F:[a,b]tomathbb{R}$ be a continuous function.



Can one conclude that $$|F|_{TV([a,b])}=lim_{epsilonto 0+}|F|_{TV([a+epsilon,b])}?$$





If $F$ is absolutely continuous than this can by done by noting that
$$
|F|_{TV([c,d])}=int_c^d|F'(x)| dx.
$$
What can one say in the general case?






real-analysis bounded-variation




real-analysis bounded-variation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 3 '17 at 23:56









JackJack

27.4k1782200




27.4k1782200












  • $begingroup$
    Have you tried exploring the canonical examples with infinite total variation?
    $endgroup$
    – Alex R.
    Jan 4 '17 at 0:10










  • $begingroup$
    @AlexR. In that case, I assume that one also has $lim_{epsilonto0+}|F|_{TV([a+epsilon,b])}=infty$ and thus the equality is also true?
    $endgroup$
    – Jack
    Jan 4 '17 at 0:13




















  • $begingroup$
    Have you tried exploring the canonical examples with infinite total variation?
    $endgroup$
    – Alex R.
    Jan 4 '17 at 0:10










  • $begingroup$
    @AlexR. In that case, I assume that one also has $lim_{epsilonto0+}|F|_{TV([a+epsilon,b])}=infty$ and thus the equality is also true?
    $endgroup$
    – Jack
    Jan 4 '17 at 0:13


















$begingroup$
Have you tried exploring the canonical examples with infinite total variation?
$endgroup$
– Alex R.
Jan 4 '17 at 0:10




$begingroup$
Have you tried exploring the canonical examples with infinite total variation?
$endgroup$
– Alex R.
Jan 4 '17 at 0:10












$begingroup$
@AlexR. In that case, I assume that one also has $lim_{epsilonto0+}|F|_{TV([a+epsilon,b])}=infty$ and thus the equality is also true?
$endgroup$
– Jack
Jan 4 '17 at 0:13






$begingroup$
@AlexR. In that case, I assume that one also has $lim_{epsilonto0+}|F|_{TV([a+epsilon,b])}=infty$ and thus the equality is also true?
$endgroup$
– Jack
Jan 4 '17 at 0:13












1 Answer
1






active

oldest

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3












$begingroup$

I decided to make my comment an answer since it appears to be long.



Observe that $leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}leleftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}$ since if $P$ is a partition of $[a+epsilon,b]$



then



$$sum_{i=1}^{n}leftlvert F(x_{i})-F(x_{i-1})rightrvert leleftlvert F(a+epsilon)-F(a)rightrvert+sum_{i=1}^{n}leftlvert F(x_{i})-F(x_{i-1})rightrvertleleftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}$$



Taking a supremum over partitions gives the described result. Thus, taking a limit in $epsilon$ we obtain



$$limsup_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}leleftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}$$



Observe that through a similar proof we may obtain:



$$leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon_{1},b])}leleftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon_{2},b])}$$



when $epsilon_{1}>epsilon_{2}$. Now take a partition $P=left{a=x_{0}<ldots<x_{n}=bright}$. By continuity of $F$ at $x=a$ we can find $delta>0$ so that $leftlvert F(x)-F(a)rightrvert<eta$ if $leftlvert x-arightrvert<delta$. By perhaps refining our partition and increasing the variation we may assume $leftlvert x_{1}-arightrvert<delta$. So:



$$sum_{i=1}^{n}leftlvert F(x_{i})-F(x_{i-1})rightrvertleeta+leftlvertleftlvert Frightrvertrightrvert_{TV([x_{1},b])}leeta+liminf_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}$$



So $$leftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}leeta+liminf_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}$$



So $$leftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}leliminf_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}$$



Note that if you remove continuity of the function at $a$ then the limit is not true. For example, consider $f:[0,1]tomathbb{R}$ defined by $f(x)=chi_{left{0right}}(x)$. Then $leftlvertleftlvert frightrvertrightrvert_{TV([epsilon,1])}=0$ for $epsilon>0$ but $leftlvertleftlvert frightrvertrightvert_{TV([0,1])}=1$.






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    $begingroup$

    I decided to make my comment an answer since it appears to be long.



    Observe that $leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}leleftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}$ since if $P$ is a partition of $[a+epsilon,b]$



    then



    $$sum_{i=1}^{n}leftlvert F(x_{i})-F(x_{i-1})rightrvert leleftlvert F(a+epsilon)-F(a)rightrvert+sum_{i=1}^{n}leftlvert F(x_{i})-F(x_{i-1})rightrvertleleftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}$$



    Taking a supremum over partitions gives the described result. Thus, taking a limit in $epsilon$ we obtain



    $$limsup_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}leleftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}$$



    Observe that through a similar proof we may obtain:



    $$leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon_{1},b])}leleftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon_{2},b])}$$



    when $epsilon_{1}>epsilon_{2}$. Now take a partition $P=left{a=x_{0}<ldots<x_{n}=bright}$. By continuity of $F$ at $x=a$ we can find $delta>0$ so that $leftlvert F(x)-F(a)rightrvert<eta$ if $leftlvert x-arightrvert<delta$. By perhaps refining our partition and increasing the variation we may assume $leftlvert x_{1}-arightrvert<delta$. So:



    $$sum_{i=1}^{n}leftlvert F(x_{i})-F(x_{i-1})rightrvertleeta+leftlvertleftlvert Frightrvertrightrvert_{TV([x_{1},b])}leeta+liminf_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}$$



    So $$leftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}leeta+liminf_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}$$



    So $$leftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}leliminf_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}$$



    Note that if you remove continuity of the function at $a$ then the limit is not true. For example, consider $f:[0,1]tomathbb{R}$ defined by $f(x)=chi_{left{0right}}(x)$. Then $leftlvertleftlvert frightrvertrightrvert_{TV([epsilon,1])}=0$ for $epsilon>0$ but $leftlvertleftlvert frightrvertrightvert_{TV([0,1])}=1$.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      I decided to make my comment an answer since it appears to be long.



      Observe that $leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}leleftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}$ since if $P$ is a partition of $[a+epsilon,b]$



      then



      $$sum_{i=1}^{n}leftlvert F(x_{i})-F(x_{i-1})rightrvert leleftlvert F(a+epsilon)-F(a)rightrvert+sum_{i=1}^{n}leftlvert F(x_{i})-F(x_{i-1})rightrvertleleftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}$$



      Taking a supremum over partitions gives the described result. Thus, taking a limit in $epsilon$ we obtain



      $$limsup_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}leleftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}$$



      Observe that through a similar proof we may obtain:



      $$leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon_{1},b])}leleftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon_{2},b])}$$



      when $epsilon_{1}>epsilon_{2}$. Now take a partition $P=left{a=x_{0}<ldots<x_{n}=bright}$. By continuity of $F$ at $x=a$ we can find $delta>0$ so that $leftlvert F(x)-F(a)rightrvert<eta$ if $leftlvert x-arightrvert<delta$. By perhaps refining our partition and increasing the variation we may assume $leftlvert x_{1}-arightrvert<delta$. So:



      $$sum_{i=1}^{n}leftlvert F(x_{i})-F(x_{i-1})rightrvertleeta+leftlvertleftlvert Frightrvertrightrvert_{TV([x_{1},b])}leeta+liminf_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}$$



      So $$leftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}leeta+liminf_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}$$



      So $$leftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}leliminf_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}$$



      Note that if you remove continuity of the function at $a$ then the limit is not true. For example, consider $f:[0,1]tomathbb{R}$ defined by $f(x)=chi_{left{0right}}(x)$. Then $leftlvertleftlvert frightrvertrightrvert_{TV([epsilon,1])}=0$ for $epsilon>0$ but $leftlvertleftlvert frightrvertrightvert_{TV([0,1])}=1$.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        I decided to make my comment an answer since it appears to be long.



        Observe that $leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}leleftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}$ since if $P$ is a partition of $[a+epsilon,b]$



        then



        $$sum_{i=1}^{n}leftlvert F(x_{i})-F(x_{i-1})rightrvert leleftlvert F(a+epsilon)-F(a)rightrvert+sum_{i=1}^{n}leftlvert F(x_{i})-F(x_{i-1})rightrvertleleftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}$$



        Taking a supremum over partitions gives the described result. Thus, taking a limit in $epsilon$ we obtain



        $$limsup_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}leleftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}$$



        Observe that through a similar proof we may obtain:



        $$leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon_{1},b])}leleftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon_{2},b])}$$



        when $epsilon_{1}>epsilon_{2}$. Now take a partition $P=left{a=x_{0}<ldots<x_{n}=bright}$. By continuity of $F$ at $x=a$ we can find $delta>0$ so that $leftlvert F(x)-F(a)rightrvert<eta$ if $leftlvert x-arightrvert<delta$. By perhaps refining our partition and increasing the variation we may assume $leftlvert x_{1}-arightrvert<delta$. So:



        $$sum_{i=1}^{n}leftlvert F(x_{i})-F(x_{i-1})rightrvertleeta+leftlvertleftlvert Frightrvertrightrvert_{TV([x_{1},b])}leeta+liminf_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}$$



        So $$leftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}leeta+liminf_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}$$



        So $$leftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}leliminf_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}$$



        Note that if you remove continuity of the function at $a$ then the limit is not true. For example, consider $f:[0,1]tomathbb{R}$ defined by $f(x)=chi_{left{0right}}(x)$. Then $leftlvertleftlvert frightrvertrightrvert_{TV([epsilon,1])}=0$ for $epsilon>0$ but $leftlvertleftlvert frightrvertrightvert_{TV([0,1])}=1$.






        share|cite|improve this answer











        $endgroup$



        I decided to make my comment an answer since it appears to be long.



        Observe that $leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}leleftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}$ since if $P$ is a partition of $[a+epsilon,b]$



        then



        $$sum_{i=1}^{n}leftlvert F(x_{i})-F(x_{i-1})rightrvert leleftlvert F(a+epsilon)-F(a)rightrvert+sum_{i=1}^{n}leftlvert F(x_{i})-F(x_{i-1})rightrvertleleftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}$$



        Taking a supremum over partitions gives the described result. Thus, taking a limit in $epsilon$ we obtain



        $$limsup_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}leleftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}$$



        Observe that through a similar proof we may obtain:



        $$leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon_{1},b])}leleftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon_{2},b])}$$



        when $epsilon_{1}>epsilon_{2}$. Now take a partition $P=left{a=x_{0}<ldots<x_{n}=bright}$. By continuity of $F$ at $x=a$ we can find $delta>0$ so that $leftlvert F(x)-F(a)rightrvert<eta$ if $leftlvert x-arightrvert<delta$. By perhaps refining our partition and increasing the variation we may assume $leftlvert x_{1}-arightrvert<delta$. So:



        $$sum_{i=1}^{n}leftlvert F(x_{i})-F(x_{i-1})rightrvertleeta+leftlvertleftlvert Frightrvertrightrvert_{TV([x_{1},b])}leeta+liminf_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}$$



        So $$leftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}leeta+liminf_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}$$



        So $$leftlvertleftlvert Frightrvertrightrvert_{TV([a,b])}leliminf_{epsilonto0^{+}}leftlvertleftlvert Frightrvertrightrvert_{TV([a+epsilon,b])}$$



        Note that if you remove continuity of the function at $a$ then the limit is not true. For example, consider $f:[0,1]tomathbb{R}$ defined by $f(x)=chi_{left{0right}}(x)$. Then $leftlvertleftlvert frightrvertrightrvert_{TV([epsilon,1])}=0$ for $epsilon>0$ but $leftlvertleftlvert frightrvertrightvert_{TV([0,1])}=1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 22 '18 at 16:50


























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