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Is there a bijective, 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?

5

3

$begingroup$

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?

real-analysis functions recreational-mathematics real-numbers

share|cite|improve this question

edited Mar 23 at 8:30

Jack M

19.1k33983

asked Mar 22 at 17:52

cammilcammil

1436

$endgroup$

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  $f(x) = -e^-x$?
  $endgroup$
  – Daniel Schepler
  Mar 22 at 17:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielSchepler I was just about to write the same, +1.
  $endgroup$
  – Michael Hoppe
  Mar 22 at 17:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @cammil a surjection (i.e. a function whose range is equal to its codomain).
  $endgroup$
  – Jake
  Mar 22 at 18:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 22 at 20:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  A better title is "is there a bijective convex function from the reals to reals?" (I prefer convex since "convex" is simpler and more popular than "concave")
  $endgroup$
  – Apass.Jack
  Mar 22 at 22:58
  
  

  
  
  
  

 | 
show 2 more comments

5

3

$begingroup$

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?

real-analysis functions recreational-mathematics real-numbers

share|cite|improve this question

edited Mar 23 at 8:30

Jack M

19.1k33983

asked Mar 22 at 17:52

cammilcammil

1436

$endgroup$

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  $f(x) = -e^-x$?
  $endgroup$
  – Daniel Schepler
  Mar 22 at 17:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielSchepler I was just about to write the same, +1.
  $endgroup$
  – Michael Hoppe
  Mar 22 at 17:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @cammil a surjection (i.e. a function whose range is equal to its codomain).
  $endgroup$
  – Jake
  Mar 22 at 18:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 22 at 20:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  A better title is "is there a bijective convex function from the reals to reals?" (I prefer convex since "convex" is simpler and more popular than "concave")
  $endgroup$
  – Apass.Jack
  Mar 22 at 22:58
  
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

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?

real-analysis functions recreational-mathematics real-numbers

share|cite|improve this question

edited Mar 23 at 8:30

Jack M

19.1k33983

asked Mar 22 at 17:52

cammilcammil

1436

$endgroup$

share|cite|improve this question

edited Mar 23 at 8:30

Jack M

19.1k33983

asked Mar 22 at 17:52

cammilcammil

1436

share|cite|improve this question

edited Mar 23 at 8:30

Jack M

19.1k33983

asked Mar 22 at 17:52

cammilcammil

1436

edited Mar 23 at 8:30

Jack M

19.1k33983

edited Mar 23 at 8:30

Jack M

19.1k33983

asked Mar 22 at 17:52

cammilcammil

1436

asked Mar 22 at 17:52

cammilcammil

1436

3 Answers
3

active

oldest

votes

16

$begingroup$

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

EDIT: Since it was requested in the comments, here is a plot of this function and its inverse:

Note that even though the growth rate for positive $x$ is slow, the function is asymptotically linear (with slope $2-sqrt3approx 0.268$) and thus unbounded.

share|cite|improve this answer

edited Mar 23 at 16:30

answered Mar 22 at 18:38

eyeballfrogeyeballfrog

7,642634

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +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
  Mar 22 at 19:44
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^-x$), then fiddled with the parameters until both the function and its inverse came out looking nice.
  $endgroup$
  – eyeballfrog
  Mar 22 at 22:22
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
  $endgroup$
  – Apass.Jack
  Mar 22 at 22:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I could’ve never come up with this.
  $endgroup$
  – Randall
  Mar 22 at 23:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As in Ross Millikan's comment, $f$ is the lower branch of a hyperbola. Neatly, $f^-1$ is the upper branch. The lower branch passes through the points $(1,0)$, $(0,-1)$, $(4,1)$, $(-1,-4)$, $(15,4)$, $(-4,-15)$. (So the upper branch passes through the reflections of these six points in the line $x=y$.) This makes the hyperbola easy to plot (in GeoGebra, for instance). Its equation is $x^2-4xy+y^2=1$. The equation of its asymptotes is $x^2-4xy+y^2=0$.
  $endgroup$
  – Calum Gilhooley
  Mar 23 at 19:52
  
  

  
  
  
  

add a comment | 

5

$begingroup$

How about

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

share|cite|improve this answer

answered Mar 22 at 18:33

paw88789paw88789

29.6k12451

$endgroup$

add a comment | 

3

$begingroup$

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

share|cite|improve this answer

edited Mar 23 at 23:29

answered Mar 22 at 18:57

zhw.zhw.

75.6k43376

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I would have preferred to incorporate the $pi x$ term into the integral, by adding a constant to the integrand. And using $pi$ is maybe not the most obvious choice of a number larger than$~fracpi2$; the essential point is having the primitive of an everywhere decreasing function of $t$ that moreover is positively bounded away from $0$ (i.e., remains $>c$ for some constant $c>0$).
  $endgroup$
  – Marc van Leeuwen
  Mar 25 at 17:57
  

  
  
  
  

add a comment | 

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16

$begingroup$

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

EDIT: Since it was requested in the comments, here is a plot of this function and its inverse:

Note that even though the growth rate for positive $x$ is slow, the function is asymptotically linear (with slope $2-sqrt3approx 0.268$) and thus unbounded.

share|cite|improve this answer

edited Mar 23 at 16:30

answered Mar 22 at 18:38

eyeballfrogeyeballfrog

7,642634

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +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
  Mar 22 at 19:44
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^-x$), then fiddled with the parameters until both the function and its inverse came out looking nice.
  $endgroup$
  – eyeballfrog
  Mar 22 at 22:22
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
  $endgroup$
  – Apass.Jack
  Mar 22 at 22:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I could’ve never come up with this.
  $endgroup$
  – Randall
  Mar 22 at 23:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As in Ross Millikan's comment, $f$ is the lower branch of a hyperbola. Neatly, $f^-1$ is the upper branch. The lower branch passes through the points $(1,0)$, $(0,-1)$, $(4,1)$, $(-1,-4)$, $(15,4)$, $(-4,-15)$. (So the upper branch passes through the reflections of these six points in the line $x=y$.) This makes the hyperbola easy to plot (in GeoGebra, for instance). Its equation is $x^2-4xy+y^2=1$. The equation of its asymptotes is $x^2-4xy+y^2=0$.
  $endgroup$
  – Calum Gilhooley
  Mar 23 at 19:52
  
  

  
  
  
  

add a comment | 

16

$begingroup$

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

EDIT: Since it was requested in the comments, here is a plot of this function and its inverse:

Note that even though the growth rate for positive $x$ is slow, the function is asymptotically linear (with slope $2-sqrt3approx 0.268$) and thus unbounded.

share|cite|improve this answer

edited Mar 23 at 16:30

answered Mar 22 at 18:38

eyeballfrogeyeballfrog

7,642634

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +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
  Mar 22 at 19:44
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^-x$), then fiddled with the parameters until both the function and its inverse came out looking nice.
  $endgroup$
  – eyeballfrog
  Mar 22 at 22:22
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
  $endgroup$
  – Apass.Jack
  Mar 22 at 22:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I could’ve never come up with this.
  $endgroup$
  – Randall
  Mar 22 at 23:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As in Ross Millikan's comment, $f$ is the lower branch of a hyperbola. Neatly, $f^-1$ is the upper branch. The lower branch passes through the points $(1,0)$, $(0,-1)$, $(4,1)$, $(-1,-4)$, $(15,4)$, $(-4,-15)$. (So the upper branch passes through the reflections of these six points in the line $x=y$.) This makes the hyperbola easy to plot (in GeoGebra, for instance). Its equation is $x^2-4xy+y^2=1$. The equation of its asymptotes is $x^2-4xy+y^2=0$.
  $endgroup$
  – Calum Gilhooley
  Mar 23 at 19:52
  
  

  
  
  
  

add a comment | 

$begingroup$

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

EDIT: Since it was requested in the comments, here is a plot of this function and its inverse:

Note that even though the growth rate for positive $x$ is slow, the function is asymptotically linear (with slope $2-sqrt3approx 0.268$) and thus unbounded.

share|cite|improve this answer

edited Mar 23 at 16:30

answered Mar 22 at 18:38

eyeballfrogeyeballfrog

7,642634

$endgroup$

share|cite|improve this answer

edited Mar 23 at 16:30

answered Mar 22 at 18:38

eyeballfrogeyeballfrog

7,642634

answered Mar 22 at 18:38

eyeballfrogeyeballfrog

7,642634

answered Mar 22 at 18:38

eyeballfrogeyeballfrog

7,642634

5

$begingroup$

How about

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

share|cite|improve this answer

answered Mar 22 at 18:33

paw88789paw88789

29.6k12451

$endgroup$

add a comment | 

5

$begingroup$

How about

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

share|cite|improve this answer

answered Mar 22 at 18:33

paw88789paw88789

29.6k12451

$endgroup$

add a comment | 

$begingroup$

How about

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

share|cite|improve this answer

answered Mar 22 at 18:33

paw88789paw88789

29.6k12451

$endgroup$

share|cite|improve this answer

answered Mar 22 at 18:33

paw88789paw88789

29.6k12451

answered Mar 22 at 18:33

paw88789paw88789

29.6k12451

answered Mar 22 at 18:33

paw88789paw88789

29.6k12451

3

$begingroup$

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

share|cite|improve this answer

edited Mar 23 at 23:29

answered Mar 22 at 18:57

zhw.zhw.

75.6k43376

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I would have preferred to incorporate the $pi x$ term into the integral, by adding a constant to the integrand. And using $pi$ is maybe not the most obvious choice of a number larger than$~fracpi2$; the essential point is having the primitive of an everywhere decreasing function of $t$ that moreover is positively bounded away from $0$ (i.e., remains $>c$ for some constant $c>0$).
  $endgroup$
  – Marc van Leeuwen
  Mar 25 at 17:57
  

  
  
  
  

add a comment | 

3

$begingroup$

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

share|cite|improve this answer

edited Mar 23 at 23:29

answered Mar 22 at 18:57

zhw.zhw.

75.6k43376

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I would have preferred to incorporate the $pi x$ term into the integral, by adding a constant to the integrand. And using $pi$ is maybe not the most obvious choice of a number larger than$~fracpi2$; the essential point is having the primitive of an everywhere decreasing function of $t$ that moreover is positively bounded away from $0$ (i.e., remains $>c$ for some constant $c>0$).
  $endgroup$
  – Marc van Leeuwen
  Mar 25 at 17:57
  

  
  
  
  

add a comment | 

$begingroup$

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

share|cite|improve this answer

edited Mar 23 at 23:29

answered Mar 22 at 18:57

zhw.zhw.

75.6k43376

$endgroup$

share|cite|improve this answer

edited Mar 23 at 23:29

answered Mar 22 at 18:57

zhw.zhw.

75.6k43376

answered Mar 22 at 18:57

zhw.zhw.

75.6k43376

answered Mar 22 at 18:57

zhw.zhw.

75.6k43376