Show that if $f neq 0$ and $x < y Rightarrow f(x) le frac {f(y) - f(x)}{y - x } le f(y)$ then $f$ is...











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I have the following problem :




$f neq 0$ satisfies $x < y Rightarrow f(x) le frac {f(y) - f(x)}{y - x } le f(y)$ then $f$ is differentiable





Try



For $h > 0, f(x) le frac {f(x +h) - f(x)}{h} le f(x + h)$.



By squeeze theorem, $frac {f(x +h) - f(x)}{h} le f(x + h) to f(x)$ as $h to 0^+$, but this assumes continuity of $f$. I'm stuck at showing $f$ is continuous.






calculus real-analysis







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    How exactly did you make this so unreadable?
    – Lord Shark the Unknown
    Nov 17 at 8:22










  • @LordSharktheUnknown Edited.
    – Moreblue
    Nov 17 at 8:24

















up vote
3
down vote

favorite
1












I have the following problem :




$f neq 0$ satisfies $x < y Rightarrow f(x) le frac {f(y) - f(x)}{y - x } le f(y)$ then $f$ is differentiable





Try



For $h > 0, f(x) le frac {f(x +h) - f(x)}{h} le f(x + h)$.



By squeeze theorem, $frac {f(x +h) - f(x)}{h} le f(x + h) to f(x)$ as $h to 0^+$, but this assumes continuity of $f$. I'm stuck at showing $f$ is continuous.






calculus real-analysis







share|cite|improve this question




















  • 1




    How exactly did you make this so unreadable?
    – Lord Shark the Unknown
    Nov 17 at 8:22










  • @LordSharktheUnknown Edited.
    – Moreblue
    Nov 17 at 8:24















up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





I have the following problem :




$f neq 0$ satisfies $x < y Rightarrow f(x) le frac {f(y) - f(x)}{y - x } le f(y)$ then $f$ is differentiable





Try



For $h > 0, f(x) le frac {f(x +h) - f(x)}{h} le f(x + h)$.



By squeeze theorem, $frac {f(x +h) - f(x)}{h} le f(x + h) to f(x)$ as $h to 0^+$, but this assumes continuity of $f$. I'm stuck at showing $f$ is continuous.






calculus real-analysis







share|cite|improve this question















I have the following problem :




$f neq 0$ satisfies $x < y Rightarrow f(x) le frac {f(y) - f(x)}{y - x } le f(y)$ then $f$ is differentiable





Try



For $h > 0, f(x) le frac {f(x +h) - f(x)}{h} le f(x + h)$.



By squeeze theorem, $frac {f(x +h) - f(x)}{h} le f(x + h) to f(x)$ as $h to 0^+$, but this assumes continuity of $f$. I'm stuck at showing $f$ is continuous.






calculus real-analysis




calculus real-analysis






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edited Nov 17 at 8:54









DanielV

17.7k42753




17.7k42753










asked Nov 17 at 8:21









Moreblue

770216




770216








  • 1




    How exactly did you make this so unreadable?
    – Lord Shark the Unknown
    Nov 17 at 8:22










  • @LordSharktheUnknown Edited.
    – Moreblue
    Nov 17 at 8:24
















  • 1




    How exactly did you make this so unreadable?
    – Lord Shark the Unknown
    Nov 17 at 8:22










  • @LordSharktheUnknown Edited.
    – Moreblue
    Nov 17 at 8:24










1




1




How exactly did you make this so unreadable?
– Lord Shark the Unknown
Nov 17 at 8:22




How exactly did you make this so unreadable?
– Lord Shark the Unknown
Nov 17 at 8:22












@LordSharktheUnknown Edited.
– Moreblue
Nov 17 at 8:24






@LordSharktheUnknown Edited.
– Moreblue
Nov 17 at 8:24












1 Answer
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$f(x)(y - x) le f(y) - f(x) le f(y)(y - x)$.

Thus as $x to y, f(x) to f(y).$



From such considerations and that $f$ is ascending,

continuity should be forth coming.






share|cite|improve this answer























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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    $f(x)(y - x) le f(y) - f(x) le f(y)(y - x)$.

    Thus as $x to y, f(x) to f(y).$



    From such considerations and that $f$ is ascending,

    continuity should be forth coming.






    share|cite|improve this answer



























      up vote
      2
      down vote



      accepted










      $f(x)(y - x) le f(y) - f(x) le f(y)(y - x)$.

      Thus as $x to y, f(x) to f(y).$



      From such considerations and that $f$ is ascending,

      continuity should be forth coming.






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        $f(x)(y - x) le f(y) - f(x) le f(y)(y - x)$.

        Thus as $x to y, f(x) to f(y).$



        From such considerations and that $f$ is ascending,

        continuity should be forth coming.






        share|cite|improve this answer














        $f(x)(y - x) le f(y) - f(x) le f(y)(y - x)$.

        Thus as $x to y, f(x) to f(y).$



        From such considerations and that $f$ is ascending,

        continuity should be forth coming.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 17 at 9:39









        Moreblue

        770216




        770216










        answered Nov 17 at 8:47









        William Elliot

        6,8702518




        6,8702518






























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