Solvability of integral equation $f(x)=g(A(||g||)x)$

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2
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I meet this problem in a unpublished paper. Consider integral equation
$$
f(x)=g(A(||g||)x)
$$
where $fin H^2(mathbb R^n)$ is a given function, $A:mathbb R^+rightarrow mathbb R^+$ is a continue function, and $||cdot||$ is $L^2$-norm. How to know whether there is a $gin H^2(mathbb R^n)$ meet the above equation ?

In fact, my question is harder, the integral equation is
$$
f(x)=g(A(||nabla g||)x).
$$
Since I want to deal the easy situation first, I ask the above equation. Besides, I think it is not solvability for all continue $A$. There should be some request for the $A$. But I don't know what is it.

Thanks for any hint or answer.

integration analysis

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asked Jul 28 at 15:00

lanse7pty

1,7801723

add a comment | 

up vote
2
down vote

favorite

I meet this problem in a unpublished paper. Consider integral equation
$$
f(x)=g(A(||g||)x)
$$
where $fin H^2(mathbb R^n)$ is a given function, $A:mathbb R^+rightarrow mathbb R^+$ is a continue function, and $||cdot||$ is $L^2$-norm. How to know whether there is a $gin H^2(mathbb R^n)$ meet the above equation ?

In fact, my question is harder, the integral equation is
$$
f(x)=g(A(||nabla g||)x).
$$
Since I want to deal the easy situation first, I ask the above equation. Besides, I think it is not solvability for all continue $A$. There should be some request for the $A$. But I don't know what is it.

Thanks for any hint or answer.

integration analysis

share|cite|improve this question

asked Jul 28 at 15:00

lanse7pty

1,7801723

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

I meet this problem in a unpublished paper. Consider integral equation
$$
f(x)=g(A(||g||)x)
$$
where $fin H^2(mathbb R^n)$ is a given function, $A:mathbb R^+rightarrow mathbb R^+$ is a continue function, and $||cdot||$ is $L^2$-norm. How to know whether there is a $gin H^2(mathbb R^n)$ meet the above equation ?

In fact, my question is harder, the integral equation is
$$
f(x)=g(A(||nabla g||)x).
$$
Since I want to deal the easy situation first, I ask the above equation. Besides, I think it is not solvability for all continue $A$. There should be some request for the $A$. But I don't know what is it.

Thanks for any hint or answer.

integration analysis

share|cite|improve this question

asked Jul 28 at 15:00

lanse7pty

1,7801723

I meet this problem in a unpublished paper. Consider integral equation
$$
f(x)=g(A(||g||)x)
$$
where $fin H^2(mathbb R^n)$ is a given function, $A:mathbb R^+rightarrow mathbb R^+$ is a continue function, and $||cdot||$ is $L^2$-norm. How to know whether there is a $gin H^2(mathbb R^n)$ meet the above equation ?

In fact, my question is harder, the integral equation is
$$
f(x)=g(A(||nabla g||)x).
$$
Since I want to deal the easy situation first, I ask the above equation. Besides, I think it is not solvability for all continue $A$. There should be some request for the $A$. But I don't know what is it.

Thanks for any hint or answer.

integration analysis

share|cite|improve this question

asked Jul 28 at 15:00

lanse7pty

1,7801723

share|cite|improve this question

share|cite|improve this question

asked Jul 28 at 15:00

lanse7pty

1,7801723

asked Jul 28 at 15:00

lanse7pty

1,7801723

asked Jul 28 at 15:00

lanse7pty

1,7801723

1,7801723

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
3
down vote



accepted

Let's go for the harder question. Once $g$ is fixed, $A(|nabla g|)$ is a constant. This suggests to look for $g(x)=f(x/a)$ for some $a>0$. From
$$
g(A(|nabla g|),x)=gBigl(A(|nabla g|,fracxaBigr)=f(x)
$$
we see that $a=A(|nabla g|)$. IS this possible? We have
$$
|nabla g|=a^tfracn-22,|nabla f|=A(|nabla g|)^tfracn-22,|nabla f|.
$$
We impose now the condition that the equation $b=A(b)^tfracn-22,|nabla f|$ has a solution $b>0$ and take $a=A(b)$.

share|cite|improve this answer

answered Jul 28 at 18:16

Julián Aguirre

64.4k23894

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But $g$ is not fixed, I can't understand you.
  – lanse7pty
  Jul 29 at 0:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What I mean, is that $A(nabla g|)$ is a constant (depending on $g$, but independent of $x$). I reduce the problem to find $a>0$ such that $g(x)=f(x/a)$ solves the problem.
  – Julián Aguirre
  Jul 29 at 10:17
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

Let's go for the harder question. Once $g$ is fixed, $A(|nabla g|)$ is a constant. This suggests to look for $g(x)=f(x/a)$ for some $a>0$. From
$$
g(A(|nabla g|),x)=gBigl(A(|nabla g|,fracxaBigr)=f(x)
$$
we see that $a=A(|nabla g|)$. IS this possible? We have
$$
|nabla g|=a^tfracn-22,|nabla f|=A(|nabla g|)^tfracn-22,|nabla f|.
$$
We impose now the condition that the equation $b=A(b)^tfracn-22,|nabla f|$ has a solution $b>0$ and take $a=A(b)$.

share|cite|improve this answer

answered Jul 28 at 18:16

Julián Aguirre

64.4k23894

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But $g$ is not fixed, I can't understand you.
  – lanse7pty
  Jul 29 at 0:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What I mean, is that $A(nabla g|)$ is a constant (depending on $g$, but independent of $x$). I reduce the problem to find $a>0$ such that $g(x)=f(x/a)$ solves the problem.
  – Julián Aguirre
  Jul 29 at 10:17
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

Let's go for the harder question. Once $g$ is fixed, $A(|nabla g|)$ is a constant. This suggests to look for $g(x)=f(x/a)$ for some $a>0$. From
$$
g(A(|nabla g|),x)=gBigl(A(|nabla g|,fracxaBigr)=f(x)
$$
we see that $a=A(|nabla g|)$. IS this possible? We have
$$
|nabla g|=a^tfracn-22,|nabla f|=A(|nabla g|)^tfracn-22,|nabla f|.
$$
We impose now the condition that the equation $b=A(b)^tfracn-22,|nabla f|$ has a solution $b>0$ and take $a=A(b)$.

share|cite|improve this answer

answered Jul 28 at 18:16

Julián Aguirre

64.4k23894

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But $g$ is not fixed, I can't understand you.
  – lanse7pty
  Jul 29 at 0:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What I mean, is that $A(nabla g|)$ is a constant (depending on $g$, but independent of $x$). I reduce the problem to find $a>0$ such that $g(x)=f(x/a)$ solves the problem.
  – Julián Aguirre
  Jul 29 at 10:17
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

Let's go for the harder question. Once $g$ is fixed, $A(|nabla g|)$ is a constant. This suggests to look for $g(x)=f(x/a)$ for some $a>0$. From
$$
g(A(|nabla g|),x)=gBigl(A(|nabla g|,fracxaBigr)=f(x)
$$
we see that $a=A(|nabla g|)$. IS this possible? We have
$$
|nabla g|=a^tfracn-22,|nabla f|=A(|nabla g|)^tfracn-22,|nabla f|.
$$
We impose now the condition that the equation $b=A(b)^tfracn-22,|nabla f|$ has a solution $b>0$ and take $a=A(b)$.

share|cite|improve this answer

answered Jul 28 at 18:16

Julián Aguirre

64.4k23894

Let's go for the harder question. Once $g$ is fixed, $A(|nabla g|)$ is a constant. This suggests to look for $g(x)=f(x/a)$ for some $a>0$. From
$$
g(A(|nabla g|),x)=gBigl(A(|nabla g|,fracxaBigr)=f(x)
$$
we see that $a=A(|nabla g|)$. IS this possible? We have
$$
|nabla g|=a^tfracn-22,|nabla f|=A(|nabla g|)^tfracn-22,|nabla f|.
$$
We impose now the condition that the equation $b=A(b)^tfracn-22,|nabla f|$ has a solution $b>0$ and take $a=A(b)$.

share|cite|improve this answer

answered Jul 28 at 18:16

Julián Aguirre

64.4k23894

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 18:16

Julián Aguirre

64.4k23894

answered Jul 28 at 18:16

Julián Aguirre

64.4k23894

answered Jul 28 at 18:16

Julián Aguirre

64.4k23894

64.4k23894

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But $g$ is not fixed, I can't understand you.
  – lanse7pty
  Jul 29 at 0:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What I mean, is that $A(nabla g|)$ is a constant (depending on $g$, but independent of $x$). I reduce the problem to find $a>0$ such that $g(x)=f(x/a)$ solves the problem.
  – Julián Aguirre
  Jul 29 at 10:17
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But $g$ is not fixed, I can't understand you.
  – lanse7pty
  Jul 29 at 0:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What I mean, is that $A(nabla g|)$ is a constant (depending on $g$, but independent of $x$). I reduce the problem to find $a>0$ such that $g(x)=f(x/a)$ solves the problem.
  – Julián Aguirre
  Jul 29 at 10:17
  

  
  
  
  

But $g$ is not fixed, I can't understand you.
– lanse7pty
Jul 29 at 0:01

But $g$ is not fixed, I can't understand you.
– lanse7pty
Jul 29 at 0:01

What I mean, is that $A(nabla g|)$ is a constant (depending on $g$, but independent of $x$). I reduce the problem to find $a>0$ such that $g(x)=f(x/a)$ solves the problem.
– Julián Aguirre
Jul 29 at 10:17

What I mean, is that $A(nabla g|)$ is a constant (depending on $g$, but independent of $x$). I reduce the problem to find $a>0$ such that $g(x)=f(x/a)$ solves the problem.
– Julián Aguirre
Jul 29 at 10:17

add a comment | 

 

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