Show that if f is differentiable and f'(x) ≥ 0 on (a, b), then f is strictly increasing












1












$begingroup$


Show that if f is differentiable and f'(x) ≥ $0$ on (a, b), then f is strictly increasing provided there is no sub interval (c, d) with с < d on which f' is identically zero.



So so far I'm trying to do this by contradiction:



Suppose not, that is suppose we have function $f$ where f(x)$geq$$0$ on (a,b) where f ' is not identically $0$ for a sub interval of (a,b) and f is not strictly increasing. Since f is not strictly increasing this implies there exists $x_1$ and $x_2$ where a <$x_1$ < $x_2$ < b and f($x_1$) = f($x_2$). Then for all y $in$ [$x_1$, $x_2$], f(y) = f($x_2$) which means that f is constant and f '(y) = 0.



Since f'(y)=$0$ for all y $in$ [$x_1$, $x_2$] this means f ' is identically $0$ which is a contradiction. Thus f is strictly increasing. $square$



I'm not sure if there is a better way to do this but any help or comments would be appreciated!






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  • $begingroup$
    I'm not sure how you conclude that $f$ is constant on a sub interval. That doesn't follow from the fact that $f$ is not strictly increasing (it may as well be decreasing or not monotonic at all).
    $endgroup$
    – freakish
    Dec 3 '18 at 11:00


















1












$begingroup$


Show that if f is differentiable and f'(x) ≥ $0$ on (a, b), then f is strictly increasing provided there is no sub interval (c, d) with с < d on which f' is identically zero.



So so far I'm trying to do this by contradiction:



Suppose not, that is suppose we have function $f$ where f(x)$geq$$0$ on (a,b) where f ' is not identically $0$ for a sub interval of (a,b) and f is not strictly increasing. Since f is not strictly increasing this implies there exists $x_1$ and $x_2$ where a <$x_1$ < $x_2$ < b and f($x_1$) = f($x_2$). Then for all y $in$ [$x_1$, $x_2$], f(y) = f($x_2$) which means that f is constant and f '(y) = 0.



Since f'(y)=$0$ for all y $in$ [$x_1$, $x_2$] this means f ' is identically $0$ which is a contradiction. Thus f is strictly increasing. $square$



I'm not sure if there is a better way to do this but any help or comments would be appreciated!






real-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    I'm not sure how you conclude that $f$ is constant on a sub interval. That doesn't follow from the fact that $f$ is not strictly increasing (it may as well be decreasing or not monotonic at all).
    $endgroup$
    – freakish
    Dec 3 '18 at 11:00
















1












1








1





$begingroup$


Show that if f is differentiable and f'(x) ≥ $0$ on (a, b), then f is strictly increasing provided there is no sub interval (c, d) with с < d on which f' is identically zero.



So so far I'm trying to do this by contradiction:



Suppose not, that is suppose we have function $f$ where f(x)$geq$$0$ on (a,b) where f ' is not identically $0$ for a sub interval of (a,b) and f is not strictly increasing. Since f is not strictly increasing this implies there exists $x_1$ and $x_2$ where a <$x_1$ < $x_2$ < b and f($x_1$) = f($x_2$). Then for all y $in$ [$x_1$, $x_2$], f(y) = f($x_2$) which means that f is constant and f '(y) = 0.



Since f'(y)=$0$ for all y $in$ [$x_1$, $x_2$] this means f ' is identically $0$ which is a contradiction. Thus f is strictly increasing. $square$



I'm not sure if there is a better way to do this but any help or comments would be appreciated!






real-analysis







share|cite|improve this question









$endgroup$




Show that if f is differentiable and f'(x) ≥ $0$ on (a, b), then f is strictly increasing provided there is no sub interval (c, d) with с < d on which f' is identically zero.



So so far I'm trying to do this by contradiction:



Suppose not, that is suppose we have function $f$ where f(x)$geq$$0$ on (a,b) where f ' is not identically $0$ for a sub interval of (a,b) and f is not strictly increasing. Since f is not strictly increasing this implies there exists $x_1$ and $x_2$ where a <$x_1$ < $x_2$ < b and f($x_1$) = f($x_2$). Then for all y $in$ [$x_1$, $x_2$], f(y) = f($x_2$) which means that f is constant and f '(y) = 0.



Since f'(y)=$0$ for all y $in$ [$x_1$, $x_2$] this means f ' is identically $0$ which is a contradiction. Thus f is strictly increasing. $square$



I'm not sure if there is a better way to do this but any help or comments would be appreciated!






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real-analysis






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asked Dec 3 '18 at 10:54









NolandoNolando

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  • $begingroup$
    I'm not sure how you conclude that $f$ is constant on a sub interval. That doesn't follow from the fact that $f$ is not strictly increasing (it may as well be decreasing or not monotonic at all).
    $endgroup$
    – freakish
    Dec 3 '18 at 11:00




















  • $begingroup$
    I'm not sure how you conclude that $f$ is constant on a sub interval. That doesn't follow from the fact that $f$ is not strictly increasing (it may as well be decreasing or not monotonic at all).
    $endgroup$
    – freakish
    Dec 3 '18 at 11:00


















$begingroup$
I'm not sure how you conclude that $f$ is constant on a sub interval. That doesn't follow from the fact that $f$ is not strictly increasing (it may as well be decreasing or not monotonic at all).
$endgroup$
– freakish
Dec 3 '18 at 11:00






$begingroup$
I'm not sure how you conclude that $f$ is constant on a sub interval. That doesn't follow from the fact that $f$ is not strictly increasing (it may as well be decreasing or not monotonic at all).
$endgroup$
– freakish
Dec 3 '18 at 11:00












2 Answers
2






active

oldest

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0












$begingroup$

Your idea of going by contradiction is a good one, but it is not performed all that well. In particular, your argument that there exist $x_1<x_2$ such that $f(x_1)=f(x_2)$ is ok, but it is not clear how to conclude from that the fact that $f$ is constant of $[x_1,x_2]$.



My advice is to either use Rolle's theorem on $x_1,x_2$, or to restart your proof, write exactly what "not strictly increasing" means, and go with the classics and employ Lagrange's mean value theorem.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I was wondering how I would define not strictly increasing, would I negate the formal definition or find the contrapositive of the definition? Or is there another way?
    $endgroup$
    – Nolando
    Dec 3 '18 at 12:02










  • $begingroup$
    @Nolando That's exactly it. A function is strictly increasing if for all $x_1,x_2$ such that $x_1<x_2$, we have $f(x_1)<f(x_2)$. Symbolically, this can be written as $$forall x_1, x_2: x_1<x_2implies f(x_1)<f(x_2)$$ By definition, a function is not strictly increasing if the negation of that holds, i.e. $$exists x_1,x_2:x_1<x_2land f(x_1)geq f(x_2)$$ meaning there exists some pair $x_1,x_2$ such that $x_1<x_2$ and $f(x_1)geq f(x_2)$.
    $endgroup$
    – 5xum
    Dec 3 '18 at 12:41










  • $begingroup$
    Thanks! I was thinking wondering if i should go along the lines of $dfrac{f(x2)-f(x1)}{x2-x1}$ $geq$ 0 since f '(x) $geq$ 0 then f(x2)-f(x1) $geq$ 0 but from the fact f is not strictly increasing and f(x2) -f(x1) $leq$ 0. So f(x2)-f(x1) = 0. I'm not sure if this is a good way to go about it either.
    $endgroup$
    – Nolando
    Dec 3 '18 at 13:27



















-1












$begingroup$

To prove that by contradiction, firstly we can prove that $f'(x)>0 implies f(x)$ strictly increasing.



Suppose indeed that exist $x_1<x_2$ such that $f(x_1)>f(x_2)$ then by MVT



$$f(x_2)-f(x_1)=(x_2-x_1)f'(c)>0$$



which is a contradiction.



Form here we can now extend the result to $f'(x)ge 0$, with $f'(x)=0$ only for a set of isolated points, $implies f(x)$ strictly increasing, indeed assuming again by contradiction that $f(x)$ is not strictly increasing then it would exist an interval with $f'(x)=0$ which is against the hypotesis.






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    2 Answers
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    2 Answers
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    0












    $begingroup$

    Your idea of going by contradiction is a good one, but it is not performed all that well. In particular, your argument that there exist $x_1<x_2$ such that $f(x_1)=f(x_2)$ is ok, but it is not clear how to conclude from that the fact that $f$ is constant of $[x_1,x_2]$.



    My advice is to either use Rolle's theorem on $x_1,x_2$, or to restart your proof, write exactly what "not strictly increasing" means, and go with the classics and employ Lagrange's mean value theorem.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I was wondering how I would define not strictly increasing, would I negate the formal definition or find the contrapositive of the definition? Or is there another way?
      $endgroup$
      – Nolando
      Dec 3 '18 at 12:02










    • $begingroup$
      @Nolando That's exactly it. A function is strictly increasing if for all $x_1,x_2$ such that $x_1<x_2$, we have $f(x_1)<f(x_2)$. Symbolically, this can be written as $$forall x_1, x_2: x_1<x_2implies f(x_1)<f(x_2)$$ By definition, a function is not strictly increasing if the negation of that holds, i.e. $$exists x_1,x_2:x_1<x_2land f(x_1)geq f(x_2)$$ meaning there exists some pair $x_1,x_2$ such that $x_1<x_2$ and $f(x_1)geq f(x_2)$.
      $endgroup$
      – 5xum
      Dec 3 '18 at 12:41










    • $begingroup$
      Thanks! I was thinking wondering if i should go along the lines of $dfrac{f(x2)-f(x1)}{x2-x1}$ $geq$ 0 since f '(x) $geq$ 0 then f(x2)-f(x1) $geq$ 0 but from the fact f is not strictly increasing and f(x2) -f(x1) $leq$ 0. So f(x2)-f(x1) = 0. I'm not sure if this is a good way to go about it either.
      $endgroup$
      – Nolando
      Dec 3 '18 at 13:27
















    0












    $begingroup$

    Your idea of going by contradiction is a good one, but it is not performed all that well. In particular, your argument that there exist $x_1<x_2$ such that $f(x_1)=f(x_2)$ is ok, but it is not clear how to conclude from that the fact that $f$ is constant of $[x_1,x_2]$.



    My advice is to either use Rolle's theorem on $x_1,x_2$, or to restart your proof, write exactly what "not strictly increasing" means, and go with the classics and employ Lagrange's mean value theorem.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I was wondering how I would define not strictly increasing, would I negate the formal definition or find the contrapositive of the definition? Or is there another way?
      $endgroup$
      – Nolando
      Dec 3 '18 at 12:02










    • $begingroup$
      @Nolando That's exactly it. A function is strictly increasing if for all $x_1,x_2$ such that $x_1<x_2$, we have $f(x_1)<f(x_2)$. Symbolically, this can be written as $$forall x_1, x_2: x_1<x_2implies f(x_1)<f(x_2)$$ By definition, a function is not strictly increasing if the negation of that holds, i.e. $$exists x_1,x_2:x_1<x_2land f(x_1)geq f(x_2)$$ meaning there exists some pair $x_1,x_2$ such that $x_1<x_2$ and $f(x_1)geq f(x_2)$.
      $endgroup$
      – 5xum
      Dec 3 '18 at 12:41










    • $begingroup$
      Thanks! I was thinking wondering if i should go along the lines of $dfrac{f(x2)-f(x1)}{x2-x1}$ $geq$ 0 since f '(x) $geq$ 0 then f(x2)-f(x1) $geq$ 0 but from the fact f is not strictly increasing and f(x2) -f(x1) $leq$ 0. So f(x2)-f(x1) = 0. I'm not sure if this is a good way to go about it either.
      $endgroup$
      – Nolando
      Dec 3 '18 at 13:27














    0












    0








    0





    $begingroup$

    Your idea of going by contradiction is a good one, but it is not performed all that well. In particular, your argument that there exist $x_1<x_2$ such that $f(x_1)=f(x_2)$ is ok, but it is not clear how to conclude from that the fact that $f$ is constant of $[x_1,x_2]$.



    My advice is to either use Rolle's theorem on $x_1,x_2$, or to restart your proof, write exactly what "not strictly increasing" means, and go with the classics and employ Lagrange's mean value theorem.






    share|cite|improve this answer









    $endgroup$



    Your idea of going by contradiction is a good one, but it is not performed all that well. In particular, your argument that there exist $x_1<x_2$ such that $f(x_1)=f(x_2)$ is ok, but it is not clear how to conclude from that the fact that $f$ is constant of $[x_1,x_2]$.



    My advice is to either use Rolle's theorem on $x_1,x_2$, or to restart your proof, write exactly what "not strictly increasing" means, and go with the classics and employ Lagrange's mean value theorem.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 3 '18 at 10:58









    5xum5xum

    90.2k393161




    90.2k393161












    • $begingroup$
      I was wondering how I would define not strictly increasing, would I negate the formal definition or find the contrapositive of the definition? Or is there another way?
      $endgroup$
      – Nolando
      Dec 3 '18 at 12:02










    • $begingroup$
      @Nolando That's exactly it. A function is strictly increasing if for all $x_1,x_2$ such that $x_1<x_2$, we have $f(x_1)<f(x_2)$. Symbolically, this can be written as $$forall x_1, x_2: x_1<x_2implies f(x_1)<f(x_2)$$ By definition, a function is not strictly increasing if the negation of that holds, i.e. $$exists x_1,x_2:x_1<x_2land f(x_1)geq f(x_2)$$ meaning there exists some pair $x_1,x_2$ such that $x_1<x_2$ and $f(x_1)geq f(x_2)$.
      $endgroup$
      – 5xum
      Dec 3 '18 at 12:41










    • $begingroup$
      Thanks! I was thinking wondering if i should go along the lines of $dfrac{f(x2)-f(x1)}{x2-x1}$ $geq$ 0 since f '(x) $geq$ 0 then f(x2)-f(x1) $geq$ 0 but from the fact f is not strictly increasing and f(x2) -f(x1) $leq$ 0. So f(x2)-f(x1) = 0. I'm not sure if this is a good way to go about it either.
      $endgroup$
      – Nolando
      Dec 3 '18 at 13:27


















    • $begingroup$
      I was wondering how I would define not strictly increasing, would I negate the formal definition or find the contrapositive of the definition? Or is there another way?
      $endgroup$
      – Nolando
      Dec 3 '18 at 12:02










    • $begingroup$
      @Nolando That's exactly it. A function is strictly increasing if for all $x_1,x_2$ such that $x_1<x_2$, we have $f(x_1)<f(x_2)$. Symbolically, this can be written as $$forall x_1, x_2: x_1<x_2implies f(x_1)<f(x_2)$$ By definition, a function is not strictly increasing if the negation of that holds, i.e. $$exists x_1,x_2:x_1<x_2land f(x_1)geq f(x_2)$$ meaning there exists some pair $x_1,x_2$ such that $x_1<x_2$ and $f(x_1)geq f(x_2)$.
      $endgroup$
      – 5xum
      Dec 3 '18 at 12:41










    • $begingroup$
      Thanks! I was thinking wondering if i should go along the lines of $dfrac{f(x2)-f(x1)}{x2-x1}$ $geq$ 0 since f '(x) $geq$ 0 then f(x2)-f(x1) $geq$ 0 but from the fact f is not strictly increasing and f(x2) -f(x1) $leq$ 0. So f(x2)-f(x1) = 0. I'm not sure if this is a good way to go about it either.
      $endgroup$
      – Nolando
      Dec 3 '18 at 13:27
















    $begingroup$
    I was wondering how I would define not strictly increasing, would I negate the formal definition or find the contrapositive of the definition? Or is there another way?
    $endgroup$
    – Nolando
    Dec 3 '18 at 12:02




    $begingroup$
    I was wondering how I would define not strictly increasing, would I negate the formal definition or find the contrapositive of the definition? Or is there another way?
    $endgroup$
    – Nolando
    Dec 3 '18 at 12:02












    $begingroup$
    @Nolando That's exactly it. A function is strictly increasing if for all $x_1,x_2$ such that $x_1<x_2$, we have $f(x_1)<f(x_2)$. Symbolically, this can be written as $$forall x_1, x_2: x_1<x_2implies f(x_1)<f(x_2)$$ By definition, a function is not strictly increasing if the negation of that holds, i.e. $$exists x_1,x_2:x_1<x_2land f(x_1)geq f(x_2)$$ meaning there exists some pair $x_1,x_2$ such that $x_1<x_2$ and $f(x_1)geq f(x_2)$.
    $endgroup$
    – 5xum
    Dec 3 '18 at 12:41




    $begingroup$
    @Nolando That's exactly it. A function is strictly increasing if for all $x_1,x_2$ such that $x_1<x_2$, we have $f(x_1)<f(x_2)$. Symbolically, this can be written as $$forall x_1, x_2: x_1<x_2implies f(x_1)<f(x_2)$$ By definition, a function is not strictly increasing if the negation of that holds, i.e. $$exists x_1,x_2:x_1<x_2land f(x_1)geq f(x_2)$$ meaning there exists some pair $x_1,x_2$ such that $x_1<x_2$ and $f(x_1)geq f(x_2)$.
    $endgroup$
    – 5xum
    Dec 3 '18 at 12:41












    $begingroup$
    Thanks! I was thinking wondering if i should go along the lines of $dfrac{f(x2)-f(x1)}{x2-x1}$ $geq$ 0 since f '(x) $geq$ 0 then f(x2)-f(x1) $geq$ 0 but from the fact f is not strictly increasing and f(x2) -f(x1) $leq$ 0. So f(x2)-f(x1) = 0. I'm not sure if this is a good way to go about it either.
    $endgroup$
    – Nolando
    Dec 3 '18 at 13:27




    $begingroup$
    Thanks! I was thinking wondering if i should go along the lines of $dfrac{f(x2)-f(x1)}{x2-x1}$ $geq$ 0 since f '(x) $geq$ 0 then f(x2)-f(x1) $geq$ 0 but from the fact f is not strictly increasing and f(x2) -f(x1) $leq$ 0. So f(x2)-f(x1) = 0. I'm not sure if this is a good way to go about it either.
    $endgroup$
    – Nolando
    Dec 3 '18 at 13:27











    -1












    $begingroup$

    To prove that by contradiction, firstly we can prove that $f'(x)>0 implies f(x)$ strictly increasing.



    Suppose indeed that exist $x_1<x_2$ such that $f(x_1)>f(x_2)$ then by MVT



    $$f(x_2)-f(x_1)=(x_2-x_1)f'(c)>0$$



    which is a contradiction.



    Form here we can now extend the result to $f'(x)ge 0$, with $f'(x)=0$ only for a set of isolated points, $implies f(x)$ strictly increasing, indeed assuming again by contradiction that $f(x)$ is not strictly increasing then it would exist an interval with $f'(x)=0$ which is against the hypotesis.






    share|cite|improve this answer











    $endgroup$


















      -1












      $begingroup$

      To prove that by contradiction, firstly we can prove that $f'(x)>0 implies f(x)$ strictly increasing.



      Suppose indeed that exist $x_1<x_2$ such that $f(x_1)>f(x_2)$ then by MVT



      $$f(x_2)-f(x_1)=(x_2-x_1)f'(c)>0$$



      which is a contradiction.



      Form here we can now extend the result to $f'(x)ge 0$, with $f'(x)=0$ only for a set of isolated points, $implies f(x)$ strictly increasing, indeed assuming again by contradiction that $f(x)$ is not strictly increasing then it would exist an interval with $f'(x)=0$ which is against the hypotesis.






      share|cite|improve this answer











      $endgroup$
















        -1












        -1








        -1





        $begingroup$

        To prove that by contradiction, firstly we can prove that $f'(x)>0 implies f(x)$ strictly increasing.



        Suppose indeed that exist $x_1<x_2$ such that $f(x_1)>f(x_2)$ then by MVT



        $$f(x_2)-f(x_1)=(x_2-x_1)f'(c)>0$$



        which is a contradiction.



        Form here we can now extend the result to $f'(x)ge 0$, with $f'(x)=0$ only for a set of isolated points, $implies f(x)$ strictly increasing, indeed assuming again by contradiction that $f(x)$ is not strictly increasing then it would exist an interval with $f'(x)=0$ which is against the hypotesis.






        share|cite|improve this answer











        $endgroup$



        To prove that by contradiction, firstly we can prove that $f'(x)>0 implies f(x)$ strictly increasing.



        Suppose indeed that exist $x_1<x_2$ such that $f(x_1)>f(x_2)$ then by MVT



        $$f(x_2)-f(x_1)=(x_2-x_1)f'(c)>0$$



        which is a contradiction.



        Form here we can now extend the result to $f'(x)ge 0$, with $f'(x)=0$ only for a set of isolated points, $implies f(x)$ strictly increasing, indeed assuming again by contradiction that $f(x)$ is not strictly increasing then it would exist an interval with $f'(x)=0$ which is against the hypotesis.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 11:34

























        answered Dec 3 '18 at 11:24









        gimusigimusi

        92.9k94494




        92.9k94494






























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