Proof that $F$ is continuous in fundamental theorem of calculus












1












$begingroup$


I am studying calculus and have arrived at the first FTC. In our book it states that:



let $f: [a,b]tomathbb{R}$ be continuous on the interval $[a,b]$ then we define the function
$F_f : [a,b]tomathbb{R} : x mapsto F_f(x) := int_a^x f(t),dt$



Then these properties hold:





  • $F_f$ is continuous on $[a,b]$


  • $F_f$ is differentiable on $]a,b[$


  • $F_f$ is a primitive for $f$


The proof of the last two properties are quite understandable and also easy to find. However I am confused about the first property (supposedly the most straightforward one) where it states:



The continuity of $F_f$ follows immediately from the fact that for every $x,y in [a,b]$ we have:



$|F_f(x) - F_f(y)| = int_a^x f(t),dt - int_a^y f(t),dt| = |int_y^x f(t),dt| leq max_{x in [a,b]} |f(x)|cdot |x-y|$



Now here is what I am not sure of:




  • Where it says $max_{x in [a,b]} |f(x)|$ are we talking about a different $x$ here? Like it could have been $max_{t in [a,b]} |f(t)|$, right?


  • This statement seems true enough and understandable (assuming the above) however: how does continuity of $F_f$ follow from it?







calculus







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$endgroup$








  • 2




    $begingroup$
    The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
    $endgroup$
    – Math1000
    Jan 4 at 22:16






  • 1




    $begingroup$
    The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
    $endgroup$
    – Math1000
    Jan 4 at 22:17










  • $begingroup$
    Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
    $endgroup$
    – Chai
    Jan 5 at 0:40


















1












$begingroup$


I am studying calculus and have arrived at the first FTC. In our book it states that:



let $f: [a,b]tomathbb{R}$ be continuous on the interval $[a,b]$ then we define the function
$F_f : [a,b]tomathbb{R} : x mapsto F_f(x) := int_a^x f(t),dt$



Then these properties hold:





  • $F_f$ is continuous on $[a,b]$


  • $F_f$ is differentiable on $]a,b[$


  • $F_f$ is a primitive for $f$


The proof of the last two properties are quite understandable and also easy to find. However I am confused about the first property (supposedly the most straightforward one) where it states:



The continuity of $F_f$ follows immediately from the fact that for every $x,y in [a,b]$ we have:



$|F_f(x) - F_f(y)| = int_a^x f(t),dt - int_a^y f(t),dt| = |int_y^x f(t),dt| leq max_{x in [a,b]} |f(x)|cdot |x-y|$



Now here is what I am not sure of:




  • Where it says $max_{x in [a,b]} |f(x)|$ are we talking about a different $x$ here? Like it could have been $max_{t in [a,b]} |f(t)|$, right?


  • This statement seems true enough and understandable (assuming the above) however: how does continuity of $F_f$ follow from it?







calculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
    $endgroup$
    – Math1000
    Jan 4 at 22:16






  • 1




    $begingroup$
    The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
    $endgroup$
    – Math1000
    Jan 4 at 22:17










  • $begingroup$
    Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
    $endgroup$
    – Chai
    Jan 5 at 0:40
















1












1








1





$begingroup$


I am studying calculus and have arrived at the first FTC. In our book it states that:



let $f: [a,b]tomathbb{R}$ be continuous on the interval $[a,b]$ then we define the function
$F_f : [a,b]tomathbb{R} : x mapsto F_f(x) := int_a^x f(t),dt$



Then these properties hold:





  • $F_f$ is continuous on $[a,b]$


  • $F_f$ is differentiable on $]a,b[$


  • $F_f$ is a primitive for $f$


The proof of the last two properties are quite understandable and also easy to find. However I am confused about the first property (supposedly the most straightforward one) where it states:



The continuity of $F_f$ follows immediately from the fact that for every $x,y in [a,b]$ we have:



$|F_f(x) - F_f(y)| = int_a^x f(t),dt - int_a^y f(t),dt| = |int_y^x f(t),dt| leq max_{x in [a,b]} |f(x)|cdot |x-y|$



Now here is what I am not sure of:




  • Where it says $max_{x in [a,b]} |f(x)|$ are we talking about a different $x$ here? Like it could have been $max_{t in [a,b]} |f(t)|$, right?


  • This statement seems true enough and understandable (assuming the above) however: how does continuity of $F_f$ follow from it?







calculus







share|cite|improve this question











$endgroup$




I am studying calculus and have arrived at the first FTC. In our book it states that:



let $f: [a,b]tomathbb{R}$ be continuous on the interval $[a,b]$ then we define the function
$F_f : [a,b]tomathbb{R} : x mapsto F_f(x) := int_a^x f(t),dt$



Then these properties hold:





  • $F_f$ is continuous on $[a,b]$


  • $F_f$ is differentiable on $]a,b[$


  • $F_f$ is a primitive for $f$


The proof of the last two properties are quite understandable and also easy to find. However I am confused about the first property (supposedly the most straightforward one) where it states:



The continuity of $F_f$ follows immediately from the fact that for every $x,y in [a,b]$ we have:



$|F_f(x) - F_f(y)| = int_a^x f(t),dt - int_a^y f(t),dt| = |int_y^x f(t),dt| leq max_{x in [a,b]} |f(x)|cdot |x-y|$



Now here is what I am not sure of:




  • Where it says $max_{x in [a,b]} |f(x)|$ are we talking about a different $x$ here? Like it could have been $max_{t in [a,b]} |f(t)|$, right?


  • This statement seems true enough and understandable (assuming the above) however: how does continuity of $F_f$ follow from it?







calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 0:26







Chai

















asked Jan 4 at 22:03









ChaiChai

1276




1276








  • 2




    $begingroup$
    The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
    $endgroup$
    – Math1000
    Jan 4 at 22:16






  • 1




    $begingroup$
    The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
    $endgroup$
    – Math1000
    Jan 4 at 22:17










  • $begingroup$
    Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
    $endgroup$
    – Chai
    Jan 5 at 0:40
















  • 2




    $begingroup$
    The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
    $endgroup$
    – Math1000
    Jan 4 at 22:16






  • 1




    $begingroup$
    The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
    $endgroup$
    – Math1000
    Jan 4 at 22:17










  • $begingroup$
    Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
    $endgroup$
    – Chai
    Jan 5 at 0:40










2




2




$begingroup$
The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
$endgroup$
– Math1000
Jan 4 at 22:16




$begingroup$
The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
$endgroup$
– Math1000
Jan 4 at 22:16




1




1




$begingroup$
The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
$endgroup$
– Math1000
Jan 4 at 22:17




$begingroup$
The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
$endgroup$
– Math1000
Jan 4 at 22:17












$begingroup$
Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
$endgroup$
– Chai
Jan 5 at 0:40






$begingroup$
Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
$endgroup$
– Chai
Jan 5 at 0:40












2 Answers
2






active

oldest

votes


















4












$begingroup$


  • Right.

  • Call it $M$. That is, let $M=sup_{xin[a,b]}bigllvert f(x)bigrrvert$. Then$$(forall x,yin[a,b]):bigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert.$$So, for any $varepsilon>0$, if you take $delta=fracvarepsilon M$, then$$lvert x-yrvert<delta=fracvarepsilon Mimpliesbigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert<Mfracvarepsilon M=varepsilon.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
    $endgroup$
    – Chai
    Jan 5 at 0:56






  • 2




    $begingroup$
    Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
    $endgroup$
    – José Carlos Santos
    Jan 5 at 9:17



















1












$begingroup$

$F$ is continuous if



$lim_limits{xto y} F(x) = F(y)$ for all $y$ in $[a,b]$



$forallepsilon>0,exists delta >0,forall x,y in [a,b]: |x-y|<delta implies |F(x) - F(y)|<epsilon$



$F(x) - F(y) = int_a^x f(x) dx - int_a^y f(x) dx = int_y^x f(x) dx$



$f(x)$ is bounded over the interval, and has a least upper bound.
Let $M$ be the least upper bound of $|f(x)|$



$M = sup {|f(x)|: xin[x,y]}$



$|int_y^x f(x) dx| le |x-y|M$



$delta = frac {epsilon}{M}$



$F(x)$ is continuous.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$


    • Right.

    • Call it $M$. That is, let $M=sup_{xin[a,b]}bigllvert f(x)bigrrvert$. Then$$(forall x,yin[a,b]):bigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert.$$So, for any $varepsilon>0$, if you take $delta=fracvarepsilon M$, then$$lvert x-yrvert<delta=fracvarepsilon Mimpliesbigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert<Mfracvarepsilon M=varepsilon.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
      $endgroup$
      – Chai
      Jan 5 at 0:56






    • 2




      $begingroup$
      Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
      $endgroup$
      – José Carlos Santos
      Jan 5 at 9:17
















    4












    $begingroup$


    • Right.

    • Call it $M$. That is, let $M=sup_{xin[a,b]}bigllvert f(x)bigrrvert$. Then$$(forall x,yin[a,b]):bigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert.$$So, for any $varepsilon>0$, if you take $delta=fracvarepsilon M$, then$$lvert x-yrvert<delta=fracvarepsilon Mimpliesbigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert<Mfracvarepsilon M=varepsilon.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
      $endgroup$
      – Chai
      Jan 5 at 0:56






    • 2




      $begingroup$
      Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
      $endgroup$
      – José Carlos Santos
      Jan 5 at 9:17














    4












    4








    4





    $begingroup$


    • Right.

    • Call it $M$. That is, let $M=sup_{xin[a,b]}bigllvert f(x)bigrrvert$. Then$$(forall x,yin[a,b]):bigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert.$$So, for any $varepsilon>0$, if you take $delta=fracvarepsilon M$, then$$lvert x-yrvert<delta=fracvarepsilon Mimpliesbigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert<Mfracvarepsilon M=varepsilon.$$






    share|cite|improve this answer











    $endgroup$




    • Right.

    • Call it $M$. That is, let $M=sup_{xin[a,b]}bigllvert f(x)bigrrvert$. Then$$(forall x,yin[a,b]):bigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert.$$So, for any $varepsilon>0$, if you take $delta=fracvarepsilon M$, then$$lvert x-yrvert<delta=fracvarepsilon Mimpliesbigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert<Mfracvarepsilon M=varepsilon.$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 5 at 9:13

























    answered Jan 4 at 22:18









    José Carlos SantosJosé Carlos Santos

    173k23133241




    173k23133241












    • $begingroup$
      Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
      $endgroup$
      – Chai
      Jan 5 at 0:56






    • 2




      $begingroup$
      Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
      $endgroup$
      – José Carlos Santos
      Jan 5 at 9:17


















    • $begingroup$
      Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
      $endgroup$
      – Chai
      Jan 5 at 0:56






    • 2




      $begingroup$
      Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
      $endgroup$
      – José Carlos Santos
      Jan 5 at 9:17
















    $begingroup$
    Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
    $endgroup$
    – Chai
    Jan 5 at 0:56




    $begingroup$
    Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
    $endgroup$
    – Chai
    Jan 5 at 0:56




    2




    2




    $begingroup$
    Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
    $endgroup$
    – José Carlos Santos
    Jan 5 at 9:17




    $begingroup$
    Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
    $endgroup$
    – José Carlos Santos
    Jan 5 at 9:17











    1












    $begingroup$

    $F$ is continuous if



    $lim_limits{xto y} F(x) = F(y)$ for all $y$ in $[a,b]$



    $forallepsilon>0,exists delta >0,forall x,y in [a,b]: |x-y|<delta implies |F(x) - F(y)|<epsilon$



    $F(x) - F(y) = int_a^x f(x) dx - int_a^y f(x) dx = int_y^x f(x) dx$



    $f(x)$ is bounded over the interval, and has a least upper bound.
    Let $M$ be the least upper bound of $|f(x)|$



    $M = sup {|f(x)|: xin[x,y]}$



    $|int_y^x f(x) dx| le |x-y|M$



    $delta = frac {epsilon}{M}$



    $F(x)$ is continuous.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $F$ is continuous if



      $lim_limits{xto y} F(x) = F(y)$ for all $y$ in $[a,b]$



      $forallepsilon>0,exists delta >0,forall x,y in [a,b]: |x-y|<delta implies |F(x) - F(y)|<epsilon$



      $F(x) - F(y) = int_a^x f(x) dx - int_a^y f(x) dx = int_y^x f(x) dx$



      $f(x)$ is bounded over the interval, and has a least upper bound.
      Let $M$ be the least upper bound of $|f(x)|$



      $M = sup {|f(x)|: xin[x,y]}$



      $|int_y^x f(x) dx| le |x-y|M$



      $delta = frac {epsilon}{M}$



      $F(x)$ is continuous.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $F$ is continuous if



        $lim_limits{xto y} F(x) = F(y)$ for all $y$ in $[a,b]$



        $forallepsilon>0,exists delta >0,forall x,y in [a,b]: |x-y|<delta implies |F(x) - F(y)|<epsilon$



        $F(x) - F(y) = int_a^x f(x) dx - int_a^y f(x) dx = int_y^x f(x) dx$



        $f(x)$ is bounded over the interval, and has a least upper bound.
        Let $M$ be the least upper bound of $|f(x)|$



        $M = sup {|f(x)|: xin[x,y]}$



        $|int_y^x f(x) dx| le |x-y|M$



        $delta = frac {epsilon}{M}$



        $F(x)$ is continuous.






        share|cite|improve this answer









        $endgroup$



        $F$ is continuous if



        $lim_limits{xto y} F(x) = F(y)$ for all $y$ in $[a,b]$



        $forallepsilon>0,exists delta >0,forall x,y in [a,b]: |x-y|<delta implies |F(x) - F(y)|<epsilon$



        $F(x) - F(y) = int_a^x f(x) dx - int_a^y f(x) dx = int_y^x f(x) dx$



        $f(x)$ is bounded over the interval, and has a least upper bound.
        Let $M$ be the least upper bound of $|f(x)|$



        $M = sup {|f(x)|: xin[x,y]}$



        $|int_y^x f(x) dx| le |x-y|M$



        $delta = frac {epsilon}{M}$



        $F(x)$ is continuous.







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        answered Jan 4 at 22:37









        Doug MDoug M

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