How do I prove that if $f$ is continuous, then $F_{0}'(x) = f(x)$?












1












$begingroup$


If $f$ is continuous on the interval $I = [a,b]$, then $f$ has an antiderivative on $I$.



Define a function $F_0(x) = int_a^x f$ for $a le x le b$



If $x$ and $x+h$ both lie in $I$, then



$F_{0}(x+h) -F_{0}(x) = int_a^{x+h} f - int_a^x f$



= $int_x^{x+h} f = f(bar{x})h$



where we use the mean value theorem for integrals and $bar{x}$ lies between $x$ and $x+h$. divide by $h$, and let $h$ approach $0$; since $bar{x}$ must then approach $x$, and since $f$ is continuous, we find $F'_{0}(x) = f(x)$, and $F_{0}$ is an antiderivative of $f$. (When $x$ is an endpoint of $I$, the argument shows the appropriate one-sided derivative of $F_{0}$ has the correct value.)



Looking at the part that says that "since $f$ is continuous, we find $F'_{0}(x) = f(x)$." I wanted to know if there's a way to show more precisely that if $f$ is continuous, then $F_{0}'(x) = f(x)$?






calculus real-analysis







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$endgroup$

















    1












    $begingroup$


    If $f$ is continuous on the interval $I = [a,b]$, then $f$ has an antiderivative on $I$.



    Define a function $F_0(x) = int_a^x f$ for $a le x le b$



    If $x$ and $x+h$ both lie in $I$, then



    $F_{0}(x+h) -F_{0}(x) = int_a^{x+h} f - int_a^x f$



    = $int_x^{x+h} f = f(bar{x})h$



    where we use the mean value theorem for integrals and $bar{x}$ lies between $x$ and $x+h$. divide by $h$, and let $h$ approach $0$; since $bar{x}$ must then approach $x$, and since $f$ is continuous, we find $F'_{0}(x) = f(x)$, and $F_{0}$ is an antiderivative of $f$. (When $x$ is an endpoint of $I$, the argument shows the appropriate one-sided derivative of $F_{0}$ has the correct value.)



    Looking at the part that says that "since $f$ is continuous, we find $F'_{0}(x) = f(x)$." I wanted to know if there's a way to show more precisely that if $f$ is continuous, then $F_{0}'(x) = f(x)$?






    calculus real-analysis







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      If $f$ is continuous on the interval $I = [a,b]$, then $f$ has an antiderivative on $I$.



      Define a function $F_0(x) = int_a^x f$ for $a le x le b$



      If $x$ and $x+h$ both lie in $I$, then



      $F_{0}(x+h) -F_{0}(x) = int_a^{x+h} f - int_a^x f$



      = $int_x^{x+h} f = f(bar{x})h$



      where we use the mean value theorem for integrals and $bar{x}$ lies between $x$ and $x+h$. divide by $h$, and let $h$ approach $0$; since $bar{x}$ must then approach $x$, and since $f$ is continuous, we find $F'_{0}(x) = f(x)$, and $F_{0}$ is an antiderivative of $f$. (When $x$ is an endpoint of $I$, the argument shows the appropriate one-sided derivative of $F_{0}$ has the correct value.)



      Looking at the part that says that "since $f$ is continuous, we find $F'_{0}(x) = f(x)$." I wanted to know if there's a way to show more precisely that if $f$ is continuous, then $F_{0}'(x) = f(x)$?






      calculus real-analysis







      share|cite|improve this question









      $endgroup$




      If $f$ is continuous on the interval $I = [a,b]$, then $f$ has an antiderivative on $I$.



      Define a function $F_0(x) = int_a^x f$ for $a le x le b$



      If $x$ and $x+h$ both lie in $I$, then



      $F_{0}(x+h) -F_{0}(x) = int_a^{x+h} f - int_a^x f$



      = $int_x^{x+h} f = f(bar{x})h$



      where we use the mean value theorem for integrals and $bar{x}$ lies between $x$ and $x+h$. divide by $h$, and let $h$ approach $0$; since $bar{x}$ must then approach $x$, and since $f$ is continuous, we find $F'_{0}(x) = f(x)$, and $F_{0}$ is an antiderivative of $f$. (When $x$ is an endpoint of $I$, the argument shows the appropriate one-sided derivative of $F_{0}$ has the correct value.)



      Looking at the part that says that "since $f$ is continuous, we find $F'_{0}(x) = f(x)$." I wanted to know if there's a way to show more precisely that if $f$ is continuous, then $F_{0}'(x) = f(x)$?






      calculus real-analysis




      calculus real-analysis






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      asked Nov 28 '18 at 20:43









      K.MK.M

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          $begingroup$

          Following your argument, you have that $$frac{F_0(x_0+h)-F_0(x_0)}{h}=f(bar x),$$ but the mean value theorem aserts that $bar xin[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $bar x$ depends on $h$.



          So if $hto0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $bar xto x_0$ as $hto 0$. So, if you put all together, you'll have $$F_0'(x_0)=lim_{hto0}frac{F_0(x_0+h)-F_0(x_0)}{h}=lim_{hto0}f(bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.






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            1












            $begingroup$

            If $f$ is continuous at $x$ then, for any $varepsilon > 0$ there is an $1 geq r > 0$ such that the relation $|h| < r$ implies $|f(x+h)-f(x)|<varepsilon.$ By noticing, $F_0(x+h)-F_0(x)=intlimits_x^{x+h}(f-f(x))+f(x)h$ we reach at once $left| dfrac{F_0(x+h)-F_0(x) - f(x)}{h}right| leq varepsilon |h| leq varepsilon.$ (The same proof applies whatever the normed space $F$ and $f$ were defined.) Q.E.D






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              where does the $(f -f(x)) + f(x)h$ come from?
              $endgroup$
              – K.M
              Nov 28 '18 at 21:06










            • $begingroup$
              Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
              $endgroup$
              – Will M.
              Nov 28 '18 at 21:09













            Your Answer





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            $begingroup$

            Following your argument, you have that $$frac{F_0(x_0+h)-F_0(x_0)}{h}=f(bar x),$$ but the mean value theorem aserts that $bar xin[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $bar x$ depends on $h$.



            So if $hto0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $bar xto x_0$ as $hto 0$. So, if you put all together, you'll have $$F_0'(x_0)=lim_{hto0}frac{F_0(x_0+h)-F_0(x_0)}{h}=lim_{hto0}f(bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Following your argument, you have that $$frac{F_0(x_0+h)-F_0(x_0)}{h}=f(bar x),$$ but the mean value theorem aserts that $bar xin[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $bar x$ depends on $h$.



              So if $hto0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $bar xto x_0$ as $hto 0$. So, if you put all together, you'll have $$F_0'(x_0)=lim_{hto0}frac{F_0(x_0+h)-F_0(x_0)}{h}=lim_{hto0}f(bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Following your argument, you have that $$frac{F_0(x_0+h)-F_0(x_0)}{h}=f(bar x),$$ but the mean value theorem aserts that $bar xin[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $bar x$ depends on $h$.



                So if $hto0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $bar xto x_0$ as $hto 0$. So, if you put all together, you'll have $$F_0'(x_0)=lim_{hto0}frac{F_0(x_0+h)-F_0(x_0)}{h}=lim_{hto0}f(bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.






                share|cite|improve this answer









                $endgroup$



                Following your argument, you have that $$frac{F_0(x_0+h)-F_0(x_0)}{h}=f(bar x),$$ but the mean value theorem aserts that $bar xin[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $bar x$ depends on $h$.



                So if $hto0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $bar xto x_0$ as $hto 0$. So, if you put all together, you'll have $$F_0'(x_0)=lim_{hto0}frac{F_0(x_0+h)-F_0(x_0)}{h}=lim_{hto0}f(bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 28 '18 at 20:53









                Tito EliatronTito Eliatron

                1,513622




                1,513622























                    1












                    $begingroup$

                    If $f$ is continuous at $x$ then, for any $varepsilon > 0$ there is an $1 geq r > 0$ such that the relation $|h| < r$ implies $|f(x+h)-f(x)|<varepsilon.$ By noticing, $F_0(x+h)-F_0(x)=intlimits_x^{x+h}(f-f(x))+f(x)h$ we reach at once $left| dfrac{F_0(x+h)-F_0(x) - f(x)}{h}right| leq varepsilon |h| leq varepsilon.$ (The same proof applies whatever the normed space $F$ and $f$ were defined.) Q.E.D






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      where does the $(f -f(x)) + f(x)h$ come from?
                      $endgroup$
                      – K.M
                      Nov 28 '18 at 21:06










                    • $begingroup$
                      Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
                      $endgroup$
                      – Will M.
                      Nov 28 '18 at 21:09


















                    1












                    $begingroup$

                    If $f$ is continuous at $x$ then, for any $varepsilon > 0$ there is an $1 geq r > 0$ such that the relation $|h| < r$ implies $|f(x+h)-f(x)|<varepsilon.$ By noticing, $F_0(x+h)-F_0(x)=intlimits_x^{x+h}(f-f(x))+f(x)h$ we reach at once $left| dfrac{F_0(x+h)-F_0(x) - f(x)}{h}right| leq varepsilon |h| leq varepsilon.$ (The same proof applies whatever the normed space $F$ and $f$ were defined.) Q.E.D






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      where does the $(f -f(x)) + f(x)h$ come from?
                      $endgroup$
                      – K.M
                      Nov 28 '18 at 21:06










                    • $begingroup$
                      Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
                      $endgroup$
                      – Will M.
                      Nov 28 '18 at 21:09
















                    1












                    1








                    1





                    $begingroup$

                    If $f$ is continuous at $x$ then, for any $varepsilon > 0$ there is an $1 geq r > 0$ such that the relation $|h| < r$ implies $|f(x+h)-f(x)|<varepsilon.$ By noticing, $F_0(x+h)-F_0(x)=intlimits_x^{x+h}(f-f(x))+f(x)h$ we reach at once $left| dfrac{F_0(x+h)-F_0(x) - f(x)}{h}right| leq varepsilon |h| leq varepsilon.$ (The same proof applies whatever the normed space $F$ and $f$ were defined.) Q.E.D






                    share|cite|improve this answer











                    $endgroup$



                    If $f$ is continuous at $x$ then, for any $varepsilon > 0$ there is an $1 geq r > 0$ such that the relation $|h| < r$ implies $|f(x+h)-f(x)|<varepsilon.$ By noticing, $F_0(x+h)-F_0(x)=intlimits_x^{x+h}(f-f(x))+f(x)h$ we reach at once $left| dfrac{F_0(x+h)-F_0(x) - f(x)}{h}right| leq varepsilon |h| leq varepsilon.$ (The same proof applies whatever the normed space $F$ and $f$ were defined.) Q.E.D







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 28 '18 at 21:09

























                    answered Nov 28 '18 at 20:58









                    Will M.Will M.

                    2,630315




                    2,630315












                    • $begingroup$
                      where does the $(f -f(x)) + f(x)h$ come from?
                      $endgroup$
                      – K.M
                      Nov 28 '18 at 21:06










                    • $begingroup$
                      Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
                      $endgroup$
                      – Will M.
                      Nov 28 '18 at 21:09




















                    • $begingroup$
                      where does the $(f -f(x)) + f(x)h$ come from?
                      $endgroup$
                      – K.M
                      Nov 28 '18 at 21:06










                    • $begingroup$
                      Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
                      $endgroup$
                      – Will M.
                      Nov 28 '18 at 21:09


















                    $begingroup$
                    where does the $(f -f(x)) + f(x)h$ come from?
                    $endgroup$
                    – K.M
                    Nov 28 '18 at 21:06




                    $begingroup$
                    where does the $(f -f(x)) + f(x)h$ come from?
                    $endgroup$
                    – K.M
                    Nov 28 '18 at 21:06












                    $begingroup$
                    Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
                    $endgroup$
                    – Will M.
                    Nov 28 '18 at 21:09






                    $begingroup$
                    Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
                    $endgroup$
                    – Will M.
                    Nov 28 '18 at 21:09




















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