Proving if $f$ is strictly increasing and onto, there exists a strictly increasing and onto function $g$ such...











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Suppose $f : (a,b) → (c,d)$ is a strictly increasing onto function. Prove that there exists a $g: (a,b) → (c,d)$, which is also strictly increasing and onto, and $g(x) < f(x)$ for all $x ∈ (a,b)$.



I have tried defining $g$ as $g(x) = f(x) - k$ for some small $k$, but that does not seem like the right answer.






real-analysis







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  • I think you are proceeding in the right direction. Differentiate $g$ and you can prove that $g$ is strictly increasing function , and onto.
    – Akash Roy
    Nov 22 at 19:13










  • @AkashRoy why should $f$ be necessarily differentiable?
    – Anurag A
    Nov 22 at 19:14










  • @Anurag it is given that $f$ is strictly inreasing function so it has to be necessarily differentiable.
    – Akash Roy
    Nov 22 at 19:16










  • @AkashRoy: even if $f$ is assumed to differentiable, your approach goes wrong: e.g., if $a = c$ and $b= d$ and $f$ is the identity function?
    – Rob Arthan
    Nov 22 at 19:17










  • Differentiation is irrelevant to this question.
    – Lee Mosher
    Nov 22 at 19:17















up vote
1
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Suppose $f : (a,b) → (c,d)$ is a strictly increasing onto function. Prove that there exists a $g: (a,b) → (c,d)$, which is also strictly increasing and onto, and $g(x) < f(x)$ for all $x ∈ (a,b)$.



I have tried defining $g$ as $g(x) = f(x) - k$ for some small $k$, but that does not seem like the right answer.






real-analysis







share|cite|improve this question
























  • I think you are proceeding in the right direction. Differentiate $g$ and you can prove that $g$ is strictly increasing function , and onto.
    – Akash Roy
    Nov 22 at 19:13










  • @AkashRoy why should $f$ be necessarily differentiable?
    – Anurag A
    Nov 22 at 19:14










  • @Anurag it is given that $f$ is strictly inreasing function so it has to be necessarily differentiable.
    – Akash Roy
    Nov 22 at 19:16










  • @AkashRoy: even if $f$ is assumed to differentiable, your approach goes wrong: e.g., if $a = c$ and $b= d$ and $f$ is the identity function?
    – Rob Arthan
    Nov 22 at 19:17










  • Differentiation is irrelevant to this question.
    – Lee Mosher
    Nov 22 at 19:17













up vote
1
down vote

favorite
1









up vote
1
down vote

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Suppose $f : (a,b) → (c,d)$ is a strictly increasing onto function. Prove that there exists a $g: (a,b) → (c,d)$, which is also strictly increasing and onto, and $g(x) < f(x)$ for all $x ∈ (a,b)$.



I have tried defining $g$ as $g(x) = f(x) - k$ for some small $k$, but that does not seem like the right answer.






real-analysis







share|cite|improve this question















Suppose $f : (a,b) → (c,d)$ is a strictly increasing onto function. Prove that there exists a $g: (a,b) → (c,d)$, which is also strictly increasing and onto, and $g(x) < f(x)$ for all $x ∈ (a,b)$.



I have tried defining $g$ as $g(x) = f(x) - k$ for some small $k$, but that does not seem like the right answer.






real-analysis




real-analysis






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share|cite|improve this question













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edited Nov 23 at 3:16









user302797

19.7k92252




19.7k92252










asked Nov 22 at 19:10









Nykis

436




436












  • I think you are proceeding in the right direction. Differentiate $g$ and you can prove that $g$ is strictly increasing function , and onto.
    – Akash Roy
    Nov 22 at 19:13










  • @AkashRoy why should $f$ be necessarily differentiable?
    – Anurag A
    Nov 22 at 19:14










  • @Anurag it is given that $f$ is strictly inreasing function so it has to be necessarily differentiable.
    – Akash Roy
    Nov 22 at 19:16










  • @AkashRoy: even if $f$ is assumed to differentiable, your approach goes wrong: e.g., if $a = c$ and $b= d$ and $f$ is the identity function?
    – Rob Arthan
    Nov 22 at 19:17










  • Differentiation is irrelevant to this question.
    – Lee Mosher
    Nov 22 at 19:17


















  • I think you are proceeding in the right direction. Differentiate $g$ and you can prove that $g$ is strictly increasing function , and onto.
    – Akash Roy
    Nov 22 at 19:13










  • @AkashRoy why should $f$ be necessarily differentiable?
    – Anurag A
    Nov 22 at 19:14










  • @Anurag it is given that $f$ is strictly inreasing function so it has to be necessarily differentiable.
    – Akash Roy
    Nov 22 at 19:16










  • @AkashRoy: even if $f$ is assumed to differentiable, your approach goes wrong: e.g., if $a = c$ and $b= d$ and $f$ is the identity function?
    – Rob Arthan
    Nov 22 at 19:17










  • Differentiation is irrelevant to this question.
    – Lee Mosher
    Nov 22 at 19:17
















I think you are proceeding in the right direction. Differentiate $g$ and you can prove that $g$ is strictly increasing function , and onto.
– Akash Roy
Nov 22 at 19:13




I think you are proceeding in the right direction. Differentiate $g$ and you can prove that $g$ is strictly increasing function , and onto.
– Akash Roy
Nov 22 at 19:13












@AkashRoy why should $f$ be necessarily differentiable?
– Anurag A
Nov 22 at 19:14




@AkashRoy why should $f$ be necessarily differentiable?
– Anurag A
Nov 22 at 19:14












@Anurag it is given that $f$ is strictly inreasing function so it has to be necessarily differentiable.
– Akash Roy
Nov 22 at 19:16




@Anurag it is given that $f$ is strictly inreasing function so it has to be necessarily differentiable.
– Akash Roy
Nov 22 at 19:16












@AkashRoy: even if $f$ is assumed to differentiable, your approach goes wrong: e.g., if $a = c$ and $b= d$ and $f$ is the identity function?
– Rob Arthan
Nov 22 at 19:17




@AkashRoy: even if $f$ is assumed to differentiable, your approach goes wrong: e.g., if $a = c$ and $b= d$ and $f$ is the identity function?
– Rob Arthan
Nov 22 at 19:17












Differentiation is irrelevant to this question.
– Lee Mosher
Nov 22 at 19:17




Differentiation is irrelevant to this question.
– Lee Mosher
Nov 22 at 19:17










2 Answers
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You can't use $g(x) = f(x)-k$ because that won't have image $(c,d)$ if $k ne 0$ (because $f(x)$ does have image $(c,d)$).



But you can do something similar. First change coordinates on $(c,d)$, using the coordinate change function
$$h : (c,d) to mathbb R
$$
given by
$$h(x) = tanleft(pileft(frac{x-c}{d-c}right)-frac{pi}{2}right)
$$
This function $h : (c,d) to mathbb R$ is strictly increasing and onto, hence its inverse $h^{-1} : mathbb R to (c,d)$ is also strictly increasing and onto.



Now define
$$g(x) = h^{-1}bigl(h(f(x))-k)bigr)
$$






share|cite|improve this answer




























    up vote
    1
    down vote













    Hint: this is easy if $a = c = -infty$ and $b = d = +infty$. You can transform the problem for finite $a$, $b$, $c$ and $d$ into the easy case using the functions $tan$ and $arctan$.






    share|cite|improve this answer





















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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      5
      down vote



      accepted










      You can't use $g(x) = f(x)-k$ because that won't have image $(c,d)$ if $k ne 0$ (because $f(x)$ does have image $(c,d)$).



      But you can do something similar. First change coordinates on $(c,d)$, using the coordinate change function
      $$h : (c,d) to mathbb R
      $$
      given by
      $$h(x) = tanleft(pileft(frac{x-c}{d-c}right)-frac{pi}{2}right)
      $$
      This function $h : (c,d) to mathbb R$ is strictly increasing and onto, hence its inverse $h^{-1} : mathbb R to (c,d)$ is also strictly increasing and onto.



      Now define
      $$g(x) = h^{-1}bigl(h(f(x))-k)bigr)
      $$






      share|cite|improve this answer

























        up vote
        5
        down vote



        accepted










        You can't use $g(x) = f(x)-k$ because that won't have image $(c,d)$ if $k ne 0$ (because $f(x)$ does have image $(c,d)$).



        But you can do something similar. First change coordinates on $(c,d)$, using the coordinate change function
        $$h : (c,d) to mathbb R
        $$
        given by
        $$h(x) = tanleft(pileft(frac{x-c}{d-c}right)-frac{pi}{2}right)
        $$
        This function $h : (c,d) to mathbb R$ is strictly increasing and onto, hence its inverse $h^{-1} : mathbb R to (c,d)$ is also strictly increasing and onto.



        Now define
        $$g(x) = h^{-1}bigl(h(f(x))-k)bigr)
        $$






        share|cite|improve this answer























          up vote
          5
          down vote



          accepted







          up vote
          5
          down vote



          accepted






          You can't use $g(x) = f(x)-k$ because that won't have image $(c,d)$ if $k ne 0$ (because $f(x)$ does have image $(c,d)$).



          But you can do something similar. First change coordinates on $(c,d)$, using the coordinate change function
          $$h : (c,d) to mathbb R
          $$
          given by
          $$h(x) = tanleft(pileft(frac{x-c}{d-c}right)-frac{pi}{2}right)
          $$
          This function $h : (c,d) to mathbb R$ is strictly increasing and onto, hence its inverse $h^{-1} : mathbb R to (c,d)$ is also strictly increasing and onto.



          Now define
          $$g(x) = h^{-1}bigl(h(f(x))-k)bigr)
          $$






          share|cite|improve this answer












          You can't use $g(x) = f(x)-k$ because that won't have image $(c,d)$ if $k ne 0$ (because $f(x)$ does have image $(c,d)$).



          But you can do something similar. First change coordinates on $(c,d)$, using the coordinate change function
          $$h : (c,d) to mathbb R
          $$
          given by
          $$h(x) = tanleft(pileft(frac{x-c}{d-c}right)-frac{pi}{2}right)
          $$
          This function $h : (c,d) to mathbb R$ is strictly increasing and onto, hence its inverse $h^{-1} : mathbb R to (c,d)$ is also strictly increasing and onto.



          Now define
          $$g(x) = h^{-1}bigl(h(f(x))-k)bigr)
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 19:28









          Lee Mosher

          47.7k33681




          47.7k33681






















              up vote
              1
              down vote













              Hint: this is easy if $a = c = -infty$ and $b = d = +infty$. You can transform the problem for finite $a$, $b$, $c$ and $d$ into the easy case using the functions $tan$ and $arctan$.






              share|cite|improve this answer

























                up vote
                1
                down vote













                Hint: this is easy if $a = c = -infty$ and $b = d = +infty$. You can transform the problem for finite $a$, $b$, $c$ and $d$ into the easy case using the functions $tan$ and $arctan$.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Hint: this is easy if $a = c = -infty$ and $b = d = +infty$. You can transform the problem for finite $a$, $b$, $c$ and $d$ into the easy case using the functions $tan$ and $arctan$.






                  share|cite|improve this answer












                  Hint: this is easy if $a = c = -infty$ and $b = d = +infty$. You can transform the problem for finite $a$, $b$, $c$ and $d$ into the easy case using the functions $tan$ and $arctan$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 19:13









                  Rob Arthan

                  28.7k42865




                  28.7k42865






























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