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Is there an injective, monotonically increasing, strictly concave function from the reals, to the reals?


Example of continuous but not absolutely continuous strictly increasing functionCan you build metric space theory without the real numbers?Proof by induction: prove that if $x_0>3$ then the following sequence is strictly increasing…A continuous function $f:Bbb Rto Bbb R$ is injective if and only if it is strictly increasing or strictly decreasingIs there a “jagged” real-valued function that is “smooth” in cardinalities greater than the reals?examples of first strictly concave then convex function?Inverse of any strictly monotonic increasing function defined over a fixed domain and range.Continuity of $argmax$ of a strictly concave functionstrictly increasing function from reals to reals which is never an algebraic numberAt which value (over $mathbbR^+$) is the gamma function strictly increasing?













2












$begingroup$


I can't come up with a single one.



The range should be the whole of the reals. The best I have is $log(x)$ but that's only on the positive real line. And there's $f(x) = x$, but this is not strictly concave. And $-e^-x$ only maps to half of the real line.



Any ideas?










share|cite|improve this question











$endgroup$







  • 5




    $begingroup$
    $f(x) = -e^-x$?
    $endgroup$
    – Daniel Schepler
    4 hours ago






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    4 hours ago










  • $begingroup$
    Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
    $endgroup$
    – cammil
    4 hours ago







  • 1




    $begingroup$
    @cammil a surjection (i.e. a function whose range is equal to its codomain).
    $endgroup$
    – Jake
    3 hours ago










  • $begingroup$
    If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
    $endgroup$
    – Ross Millikan
    2 hours ago















2












$begingroup$


I can't come up with a single one.



The range should be the whole of the reals. The best I have is $log(x)$ but that's only on the positive real line. And there's $f(x) = x$, but this is not strictly concave. And $-e^-x$ only maps to half of the real line.



Any ideas?










share|cite|improve this question











$endgroup$







  • 5




    $begingroup$
    $f(x) = -e^-x$?
    $endgroup$
    – Daniel Schepler
    4 hours ago






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    4 hours ago










  • $begingroup$
    Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
    $endgroup$
    – cammil
    4 hours ago







  • 1




    $begingroup$
    @cammil a surjection (i.e. a function whose range is equal to its codomain).
    $endgroup$
    – Jake
    3 hours ago










  • $begingroup$
    If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
    $endgroup$
    – Ross Millikan
    2 hours ago













2












2








2





$begingroup$


I can't come up with a single one.



The range should be the whole of the reals. The best I have is $log(x)$ but that's only on the positive real line. And there's $f(x) = x$, but this is not strictly concave. And $-e^-x$ only maps to half of the real line.



Any ideas?










share|cite|improve this question











$endgroup$




I can't come up with a single one.



The range should be the whole of the reals. The best I have is $log(x)$ but that's only on the positive real line. And there's $f(x) = x$, but this is not strictly concave. And $-e^-x$ only maps to half of the real line.



Any ideas?







real-analysis functions recreational-mathematics real-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago







cammil

















asked 4 hours ago









cammilcammil

1264




1264







  • 5




    $begingroup$
    $f(x) = -e^-x$?
    $endgroup$
    – Daniel Schepler
    4 hours ago






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    4 hours ago










  • $begingroup$
    Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
    $endgroup$
    – cammil
    4 hours ago







  • 1




    $begingroup$
    @cammil a surjection (i.e. a function whose range is equal to its codomain).
    $endgroup$
    – Jake
    3 hours ago










  • $begingroup$
    If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
    $endgroup$
    – Ross Millikan
    2 hours ago












  • 5




    $begingroup$
    $f(x) = -e^-x$?
    $endgroup$
    – Daniel Schepler
    4 hours ago






  • 1




    $begingroup$
    @DanielSchepler I was just about to write the same, +1.
    $endgroup$
    – Michael Hoppe
    4 hours ago










  • $begingroup$
    Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
    $endgroup$
    – cammil
    4 hours ago







  • 1




    $begingroup$
    @cammil a surjection (i.e. a function whose range is equal to its codomain).
    $endgroup$
    – Jake
    3 hours ago










  • $begingroup$
    If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
    $endgroup$
    – Ross Millikan
    2 hours ago







5




5




$begingroup$
$f(x) = -e^-x$?
$endgroup$
– Daniel Schepler
4 hours ago




$begingroup$
$f(x) = -e^-x$?
$endgroup$
– Daniel Schepler
4 hours ago




1




1




$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
4 hours ago




$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
4 hours ago












$begingroup$
Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
$endgroup$
– cammil
4 hours ago





$begingroup$
Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
$endgroup$
– cammil
4 hours ago





1




1




$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
3 hours ago




$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
3 hours ago












$begingroup$
If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
$endgroup$
– Ross Millikan
2 hours ago




$begingroup$
If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
$endgroup$
– Ross Millikan
2 hours ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

How about



$f(x)=left{beginarraycc ln(x+1)& &xge 0\1-e^-x& &x<0endarrayright.$






share|cite|improve this answer









$endgroup$




















    4












    $begingroup$

    $$
    f(x) = x-e^-x
    $$

    is such a function. Since $f''(x) = -e^-x$ is always negative, it is strictly concave, and it's not hard to show it hits every real.



    Even better,
    $$
    f(x) = 2x -sqrt1+3x^2
    $$

    has $f''(x) = -3(1+3x^2)^-3/2 < 0$ everywhere and the explicit inverse $f^-1(x) = 2x+sqrt1+3x^2$, clearly defined for all $x$.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      +1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
      $endgroup$
      – Calum Gilhooley
      2 hours ago



















    0












    $begingroup$

    $F(x) = pi x+ int_0^x arctan (-t),dt$ is an example. Many more examples like this one can be constructed.






    share|cite|improve this answer









    $endgroup$












      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      How about



      $f(x)=left{beginarraycc ln(x+1)& &xge 0\1-e^-x& &x<0endarrayright.$






      share|cite|improve this answer









      $endgroup$

















        4












        $begingroup$

        How about



        $f(x)=left{beginarraycc ln(x+1)& &xge 0\1-e^-x& &x<0endarrayright.$






        share|cite|improve this answer









        $endgroup$















          4












          4








          4





          $begingroup$

          How about



          $f(x)=left{beginarraycc ln(x+1)& &xge 0\1-e^-x& &x<0endarrayright.$






          share|cite|improve this answer









          $endgroup$



          How about



          $f(x)=left{beginarraycc ln(x+1)& &xge 0\1-e^-x& &x<0endarrayright.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          paw88789paw88789

          29.4k12349




          29.4k12349





















              4












              $begingroup$

              $$
              f(x) = x-e^-x
              $$

              is such a function. Since $f''(x) = -e^-x$ is always negative, it is strictly concave, and it's not hard to show it hits every real.



              Even better,
              $$
              f(x) = 2x -sqrt1+3x^2
              $$

              has $f''(x) = -3(1+3x^2)^-3/2 < 0$ everywhere and the explicit inverse $f^-1(x) = 2x+sqrt1+3x^2$, clearly defined for all $x$.






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                +1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
                $endgroup$
                – Calum Gilhooley
                2 hours ago
















              4












              $begingroup$

              $$
              f(x) = x-e^-x
              $$

              is such a function. Since $f''(x) = -e^-x$ is always negative, it is strictly concave, and it's not hard to show it hits every real.



              Even better,
              $$
              f(x) = 2x -sqrt1+3x^2
              $$

              has $f''(x) = -3(1+3x^2)^-3/2 < 0$ everywhere and the explicit inverse $f^-1(x) = 2x+sqrt1+3x^2$, clearly defined for all $x$.






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                +1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
                $endgroup$
                – Calum Gilhooley
                2 hours ago














              4












              4








              4





              $begingroup$

              $$
              f(x) = x-e^-x
              $$

              is such a function. Since $f''(x) = -e^-x$ is always negative, it is strictly concave, and it's not hard to show it hits every real.



              Even better,
              $$
              f(x) = 2x -sqrt1+3x^2
              $$

              has $f''(x) = -3(1+3x^2)^-3/2 < 0$ everywhere and the explicit inverse $f^-1(x) = 2x+sqrt1+3x^2$, clearly defined for all $x$.






              share|cite|improve this answer











              $endgroup$



              $$
              f(x) = x-e^-x
              $$

              is such a function. Since $f''(x) = -e^-x$ is always negative, it is strictly concave, and it's not hard to show it hits every real.



              Even better,
              $$
              f(x) = 2x -sqrt1+3x^2
              $$

              has $f''(x) = -3(1+3x^2)^-3/2 < 0$ everywhere and the explicit inverse $f^-1(x) = 2x+sqrt1+3x^2$, clearly defined for all $x$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 3 hours ago

























              answered 3 hours ago









              eyeballfrogeyeballfrog

              6,664630




              6,664630











              • $begingroup$
                +1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
                $endgroup$
                – Calum Gilhooley
                2 hours ago

















              • $begingroup$
                +1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
                $endgroup$
                – Calum Gilhooley
                2 hours ago
















              $begingroup$
              +1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
              $endgroup$
              – Calum Gilhooley
              2 hours ago





              $begingroup$
              +1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
              $endgroup$
              – Calum Gilhooley
              2 hours ago












              0












              $begingroup$

              $F(x) = pi x+ int_0^x arctan (-t),dt$ is an example. Many more examples like this one can be constructed.






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                $F(x) = pi x+ int_0^x arctan (-t),dt$ is an example. Many more examples like this one can be constructed.






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  $F(x) = pi x+ int_0^x arctan (-t),dt$ is an example. Many more examples like this one can be constructed.






                  share|cite|improve this answer









                  $endgroup$



                  $F(x) = pi x+ int_0^x arctan (-t),dt$ is an example. Many more examples like this one can be constructed.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  zhw.zhw.

                  74.5k43175




                  74.5k43175



























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