Mixing
A functional equation on concave functions [closed]
1
$begingroup$
Let $x_0,x_1in (0,1)$ such that $x_0<x_1$, and let $a,bin(0,1)$. All these quantities are fixed.
I would like to find the concave nondecreasing functions $f$ such that for all $xin[x_0,x_1]$
$$1-a + a b fleft(frac{x}{b + (1-b)x}right) = f(x).$$
real-analysis functions convex-analysis
asked Jan 8 at 10:47
BoobooBooboo
63
$endgroup$
closed as off-topic by José Carlos Santos, Chris Custer, Cesareo, Alexander Gruber♦ Jan 8 at 22:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Chris Custer, Cesareo, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
1
$begingroup$
Let $x_0,x_1in (0,1)$ such that $x_0<x_1$, and let $a,bin(0,1)$. All these quantities are fixed.
I would like to find the concave nondecreasing functions $f$ such that for all $xin[x_0,x_1]$
$$1-a + a b fleft(frac{x}{b + (1-b)x}right) = f(x).$$
real-analysis functions convex-analysis
asked Jan 8 at 10:47
BoobooBooboo
63
$endgroup$
closed as off-topic by José Carlos Santos, Chris Custer, Cesareo, Alexander Gruber♦ Jan 8 at 22:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Chris Custer, Cesareo, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
If you mean for all $a,bin(0,1)$ then there is no such function. If there were, by Froda's Theorem we could find $xin(x_0,x_1)$ at which $f$ is continuous. Take $bto 1^-$ to get $f(x)=1$, and plug that back into the original equation to get $1-a+ab=1$, which is impossible. If instead you mean that $a$ and $b$ are fixed, then it is a much harder problem, and I'm not sure there is a simple classification of all such functions.
$endgroup$
– Ben W
Jan 8 at 11:33
$begingroup$
If you mean fixed $a$ and $b$, we can still say a couple of things about $f$. First of all, it is not differentiable at 0 or 1. In fact, I believe it is not left- or right-differentiable either. There may be a theorem stating that every monotone function at each point has a left- or right-derivative, but you should check this. That would mean $0,1notin[x_0,x_1]$. Also, it is easy to see that $f(1)=(1-a)/(1-ab)$, provided $1in[x_0,x_1]$. Maybe that will help.
$endgroup$
– Ben W
Jan 8 at 11:50
add a comment |
1
1
1
$begingroup$
Let $x_0,x_1in (0,1)$ such that $x_0<x_1$, and let $a,bin(0,1)$. All these quantities are fixed.
I would like to find the concave nondecreasing functions $f$ such that for all $xin[x_0,x_1]$
$$1-a + a b fleft(frac{x}{b + (1-b)x}right) = f(x).$$
real-analysis functions convex-analysis
asked Jan 8 at 10:47
BoobooBooboo
63
$endgroup$
Let $x_0,x_1in (0,1)$ such that $x_0<x_1$, and let $a,bin(0,1)$. All these quantities are fixed.
I would like to find the concave nondecreasing functions $f$ such that for all $xin[x_0,x_1]$
$$1-a + a b fleft(frac{x}{b + (1-b)x}right) = f(x).$$
real-analysis functions convex-analysis
real-analysis functions convex-analysis
asked Jan 8 at 10:47
BoobooBooboo
63
asked Jan 8 at 10:47
BoobooBooboo
63
edited Jan 8 at 11:53
Booboo
asked Jan 8 at 10:47
BoobooBooboo
63
asked Jan 8 at 10:47
BoobooBooboo
63
asked Jan 8 at 10:47
BoobooBooboo
63
63
closed as off-topic by José Carlos Santos, Chris Custer, Cesareo, Alexander Gruber♦ Jan 8 at 22:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Chris Custer, Cesareo, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Chris Custer, Cesareo, Alexander Gruber♦ Jan 8 at 22:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Chris Custer, Cesareo, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
If you mean for all $a,bin(0,1)$ then there is no such function. If there were, by Froda's Theorem we could find $xin(x_0,x_1)$ at which $f$ is continuous. Take $bto 1^-$ to get $f(x)=1$, and plug that back into the original equation to get $1-a+ab=1$, which is impossible. If instead you mean that $a$ and $b$ are fixed, then it is a much harder problem, and I'm not sure there is a simple classification of all such functions.
$endgroup$
– Ben W
Jan 8 at 11:33
$begingroup$
If you mean fixed $a$ and $b$, we can still say a couple of things about $f$. First of all, it is not differentiable at 0 or 1. In fact, I believe it is not left- or right-differentiable either. There may be a theorem stating that every monotone function at each point has a left- or right-derivative, but you should check this. That would mean $0,1notin[x_0,x_1]$. Also, it is easy to see that $f(1)=(1-a)/(1-ab)$, provided $1in[x_0,x_1]$. Maybe that will help.
$endgroup$
– Ben W
Jan 8 at 11:50
add a comment |
1
$begingroup$
If you mean for all $a,bin(0,1)$ then there is no such function. If there were, by Froda's Theorem we could find $xin(x_0,x_1)$ at which $f$ is continuous. Take $bto 1^-$ to get $f(x)=1$, and plug that back into the original equation to get $1-a+ab=1$, which is impossible. If instead you mean that $a$ and $b$ are fixed, then it is a much harder problem, and I'm not sure there is a simple classification of all such functions.
$endgroup$
– Ben W
Jan 8 at 11:33
$begingroup$
If you mean fixed $a$ and $b$, we can still say a couple of things about $f$. First of all, it is not differentiable at 0 or 1. In fact, I believe it is not left- or right-differentiable either. There may be a theorem stating that every monotone function at each point has a left- or right-derivative, but you should check this. That would mean $0,1notin[x_0,x_1]$. Also, it is easy to see that $f(1)=(1-a)/(1-ab)$, provided $1in[x_0,x_1]$. Maybe that will help.
$endgroup$
– Ben W
Jan 8 at 11:50
1
1
$begingroup$
If you mean for all $a,bin(0,1)$ then there is no such function. If there were, by Froda's Theorem we could find $xin(x_0,x_1)$ at which $f$ is continuous. Take $bto 1^-$ to get $f(x)=1$, and plug that back into the original equation to get $1-a+ab=1$, which is impossible. If instead you mean that $a$ and $b$ are fixed, then it is a much harder problem, and I'm not sure there is a simple classification of all such functions.
$endgroup$
– Ben W
Jan 8 at 11:33
$begingroup$
If you mean for all $a,bin(0,1)$ then there is no such function. If there were, by Froda's Theorem we could find $xin(x_0,x_1)$ at which $f$ is continuous. Take $bto 1^-$ to get $f(x)=1$, and plug that back into the original equation to get $1-a+ab=1$, which is impossible. If instead you mean that $a$ and $b$ are fixed, then it is a much harder problem, and I'm not sure there is a simple classification of all such functions.
$endgroup$
– Ben W
Jan 8 at 11:33
$begingroup$
If you mean fixed $a$ and $b$, we can still say a couple of things about $f$. First of all, it is not differentiable at 0 or 1. In fact, I believe it is not left- or right-differentiable either. There may be a theorem stating that every monotone function at each point has a left- or right-derivative, but you should check this. That would mean $0,1notin[x_0,x_1]$. Also, it is easy to see that $f(1)=(1-a)/(1-ab)$, provided $1in[x_0,x_1]$. Maybe that will help.
$endgroup$
– Ben W
Jan 8 at 11:50
$begingroup$
If you mean fixed $a$ and $b$, we can still say a couple of things about $f$. First of all, it is not differentiable at 0 or 1. In fact, I believe it is not left- or right-differentiable either. There may be a theorem stating that every monotone function at each point has a left- or right-derivative, but you should check this. That would mean $0,1notin[x_0,x_1]$. Also, it is easy to see that $f(1)=(1-a)/(1-ab)$, provided $1in[x_0,x_1]$. Maybe that will help.
$endgroup$
– Ben W
Jan 8 at 11:50
add a comment |
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1
$begingroup$
If you mean for all $a,bin(0,1)$ then there is no such function. If there were, by Froda's Theorem we could find $xin(x_0,x_1)$ at which $f$ is continuous. Take $bto 1^-$ to get $f(x)=1$, and plug that back into the original equation to get $1-a+ab=1$, which is impossible. If instead you mean that $a$ and $b$ are fixed, then it is a much harder problem, and I'm not sure there is a simple classification of all such functions.
$endgroup$
– Ben W
Jan 8 at 11:33
$begingroup$
If you mean fixed $a$ and $b$, we can still say a couple of things about $f$. First of all, it is not differentiable at 0 or 1. In fact, I believe it is not left- or right-differentiable either. There may be a theorem stating that every monotone function at each point has a left- or right-derivative, but you should check this. That would mean $0,1notin[x_0,x_1]$. Also, it is easy to see that $f(1)=(1-a)/(1-ab)$, provided $1in[x_0,x_1]$. Maybe that will help.
$endgroup$
– Ben W
Jan 8 at 11:50