If $f$ is monotone increasing on an interval and has a jump discontinuity at $x_0$, show that the jump is...












1














If $f$ is monotone increasing on an interval and has a jump discontinuity at $x_0$ in the interior of the domain show that the jump is bounded above by $f(x_1) - f(x_2)$ for any two points $x_1$, $x_2$ of the domain surrounding $x_0$, $x_1 < x_0 < x_2$.



So I've I tried solving this here is what I have:



Let $f$ be monotone increasing on an interval $A$ that has a jump discontinuity at $x_0$ on the interior of the domain where $x_0 in A$. Let there be any $x_1, x_2 in A$ where $x_1< x_0 < x_2$. Then by definition of monotone increasing $f(x_1) leq f (x_2)$.



From here I want to say that $f(x_1) leq f(x_0) leq f(x_2) rightarrow f(x_1) - f(x_1) leq f(x_0) leq f(x_2) - f(x_1)$ and that $f(x_0) leq f(x_2) -f(x_2)$. But I'm not sure if I'm going in the right direction since I can't say for sure that $f(x_0)$ is really less than $f(x_2)-f(x_1)$. Or would I need to do something like $f(x_0)-f(x_1)leq f(x_2)-f(x_1)$ and go from there. Any help would be appreciated as I'm somewhat unsure on this problem!






real-analysis functions continuity







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    If $f$ is monotone increasing on an interval and has a jump discontinuity at $x_0$ in the interior of the domain show that the jump is bounded above by $f(x_1) - f(x_2)$ for any two points $x_1$, $x_2$ of the domain surrounding $x_0$, $x_1 < x_0 < x_2$.



    So I've I tried solving this here is what I have:



    Let $f$ be monotone increasing on an interval $A$ that has a jump discontinuity at $x_0$ on the interior of the domain where $x_0 in A$. Let there be any $x_1, x_2 in A$ where $x_1< x_0 < x_2$. Then by definition of monotone increasing $f(x_1) leq f (x_2)$.



    From here I want to say that $f(x_1) leq f(x_0) leq f(x_2) rightarrow f(x_1) - f(x_1) leq f(x_0) leq f(x_2) - f(x_1)$ and that $f(x_0) leq f(x_2) -f(x_2)$. But I'm not sure if I'm going in the right direction since I can't say for sure that $f(x_0)$ is really less than $f(x_2)-f(x_1)$. Or would I need to do something like $f(x_0)-f(x_1)leq f(x_2)-f(x_1)$ and go from there. Any help would be appreciated as I'm somewhat unsure on this problem!






    real-analysis functions continuity







    share|cite|improve this question



























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      1







      If $f$ is monotone increasing on an interval and has a jump discontinuity at $x_0$ in the interior of the domain show that the jump is bounded above by $f(x_1) - f(x_2)$ for any two points $x_1$, $x_2$ of the domain surrounding $x_0$, $x_1 < x_0 < x_2$.



      So I've I tried solving this here is what I have:



      Let $f$ be monotone increasing on an interval $A$ that has a jump discontinuity at $x_0$ on the interior of the domain where $x_0 in A$. Let there be any $x_1, x_2 in A$ where $x_1< x_0 < x_2$. Then by definition of monotone increasing $f(x_1) leq f (x_2)$.



      From here I want to say that $f(x_1) leq f(x_0) leq f(x_2) rightarrow f(x_1) - f(x_1) leq f(x_0) leq f(x_2) - f(x_1)$ and that $f(x_0) leq f(x_2) -f(x_2)$. But I'm not sure if I'm going in the right direction since I can't say for sure that $f(x_0)$ is really less than $f(x_2)-f(x_1)$. Or would I need to do something like $f(x_0)-f(x_1)leq f(x_2)-f(x_1)$ and go from there. Any help would be appreciated as I'm somewhat unsure on this problem!






      real-analysis functions continuity







      share|cite|improve this question















      If $f$ is monotone increasing on an interval and has a jump discontinuity at $x_0$ in the interior of the domain show that the jump is bounded above by $f(x_1) - f(x_2)$ for any two points $x_1$, $x_2$ of the domain surrounding $x_0$, $x_1 < x_0 < x_2$.



      So I've I tried solving this here is what I have:



      Let $f$ be monotone increasing on an interval $A$ that has a jump discontinuity at $x_0$ on the interior of the domain where $x_0 in A$. Let there be any $x_1, x_2 in A$ where $x_1< x_0 < x_2$. Then by definition of monotone increasing $f(x_1) leq f (x_2)$.



      From here I want to say that $f(x_1) leq f(x_0) leq f(x_2) rightarrow f(x_1) - f(x_1) leq f(x_0) leq f(x_2) - f(x_1)$ and that $f(x_0) leq f(x_2) -f(x_2)$. But I'm not sure if I'm going in the right direction since I can't say for sure that $f(x_0)$ is really less than $f(x_2)-f(x_1)$. Or would I need to do something like $f(x_0)-f(x_1)leq f(x_2)-f(x_1)$ and go from there. Any help would be appreciated as I'm somewhat unsure on this problem!






      real-analysis functions continuity




      real-analysis functions continuity






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      edited Nov 19 '18 at 6:24









      Taroccoesbrocco

      5,08261839




      5,08261839










      asked Nov 19 '18 at 5:05









      Nolando

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          2 Answers
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          $x_1$ and $x_2$ are supposed to be given points. You cannot let $x_2 to x_1$. If $x_0<t<x_2$ then $f(t) leq f(x_2)$. Let $t to x_0$ to get $f(x_0+) leq f(x_2)$. [ $f(x_0+)$ is the right hand limit of $f$ at $x_0$]. Similarly prove that $f(x_0-) geq f(x_1)$. From these two se get $f(x_0+)-f(x_0-) leq f(x_2)-f(x_1)$.






          share|cite|improve this answer





















          • Thanks! That is very helpful
            – Nolando
            Nov 19 '18 at 5:26



















          0














          First of all you have to define " the jump at $x_0$ "



          It means $$lim _{xto x_0+}f(x) -lim _{xto x_0-}f(x) $$



          The above limits exist due to the monotonicity of $f(x)$



          The rest is not too complicated to prove and you can do it.






          share|cite|improve this answer























          • Thanks you! That definitely helps
            – Nolando
            Nov 19 '18 at 5:23











          Your Answer





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          2 Answers
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          active

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          2 Answers
          2






          active

          oldest

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          active

          oldest

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          active

          oldest

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          1














          $x_1$ and $x_2$ are supposed to be given points. You cannot let $x_2 to x_1$. If $x_0<t<x_2$ then $f(t) leq f(x_2)$. Let $t to x_0$ to get $f(x_0+) leq f(x_2)$. [ $f(x_0+)$ is the right hand limit of $f$ at $x_0$]. Similarly prove that $f(x_0-) geq f(x_1)$. From these two se get $f(x_0+)-f(x_0-) leq f(x_2)-f(x_1)$.






          share|cite|improve this answer





















          • Thanks! That is very helpful
            – Nolando
            Nov 19 '18 at 5:26
















          1














          $x_1$ and $x_2$ are supposed to be given points. You cannot let $x_2 to x_1$. If $x_0<t<x_2$ then $f(t) leq f(x_2)$. Let $t to x_0$ to get $f(x_0+) leq f(x_2)$. [ $f(x_0+)$ is the right hand limit of $f$ at $x_0$]. Similarly prove that $f(x_0-) geq f(x_1)$. From these two se get $f(x_0+)-f(x_0-) leq f(x_2)-f(x_1)$.






          share|cite|improve this answer





















          • Thanks! That is very helpful
            – Nolando
            Nov 19 '18 at 5:26














          1












          1








          1






          $x_1$ and $x_2$ are supposed to be given points. You cannot let $x_2 to x_1$. If $x_0<t<x_2$ then $f(t) leq f(x_2)$. Let $t to x_0$ to get $f(x_0+) leq f(x_2)$. [ $f(x_0+)$ is the right hand limit of $f$ at $x_0$]. Similarly prove that $f(x_0-) geq f(x_1)$. From these two se get $f(x_0+)-f(x_0-) leq f(x_2)-f(x_1)$.






          share|cite|improve this answer












          $x_1$ and $x_2$ are supposed to be given points. You cannot let $x_2 to x_1$. If $x_0<t<x_2$ then $f(t) leq f(x_2)$. Let $t to x_0$ to get $f(x_0+) leq f(x_2)$. [ $f(x_0+)$ is the right hand limit of $f$ at $x_0$]. Similarly prove that $f(x_0-) geq f(x_1)$. From these two se get $f(x_0+)-f(x_0-) leq f(x_2)-f(x_1)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 19 '18 at 5:15









          Kavi Rama Murthy

          50.3k31854




          50.3k31854












          • Thanks! That is very helpful
            – Nolando
            Nov 19 '18 at 5:26


















          • Thanks! That is very helpful
            – Nolando
            Nov 19 '18 at 5:26
















          Thanks! That is very helpful
          – Nolando
          Nov 19 '18 at 5:26




          Thanks! That is very helpful
          – Nolando
          Nov 19 '18 at 5:26











          0














          First of all you have to define " the jump at $x_0$ "



          It means $$lim _{xto x_0+}f(x) -lim _{xto x_0-}f(x) $$



          The above limits exist due to the monotonicity of $f(x)$



          The rest is not too complicated to prove and you can do it.






          share|cite|improve this answer























          • Thanks you! That definitely helps
            – Nolando
            Nov 19 '18 at 5:23
















          0














          First of all you have to define " the jump at $x_0$ "



          It means $$lim _{xto x_0+}f(x) -lim _{xto x_0-}f(x) $$



          The above limits exist due to the monotonicity of $f(x)$



          The rest is not too complicated to prove and you can do it.






          share|cite|improve this answer























          • Thanks you! That definitely helps
            – Nolando
            Nov 19 '18 at 5:23














          0












          0








          0






          First of all you have to define " the jump at $x_0$ "



          It means $$lim _{xto x_0+}f(x) -lim _{xto x_0-}f(x) $$



          The above limits exist due to the monotonicity of $f(x)$



          The rest is not too complicated to prove and you can do it.






          share|cite|improve this answer














          First of all you have to define " the jump at $x_0$ "



          It means $$lim _{xto x_0+}f(x) -lim _{xto x_0-}f(x) $$



          The above limits exist due to the monotonicity of $f(x)$



          The rest is not too complicated to prove and you can do it.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 19 '18 at 5:44

























          answered Nov 19 '18 at 5:18









          Mohammad Riazi-Kermani

          40.8k42059




          40.8k42059












          • Thanks you! That definitely helps
            – Nolando
            Nov 19 '18 at 5:23


















          • Thanks you! That definitely helps
            – Nolando
            Nov 19 '18 at 5:23
















          Thanks you! That definitely helps
          – Nolando
          Nov 19 '18 at 5:23




          Thanks you! That definitely helps
          – Nolando
          Nov 19 '18 at 5:23


















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