Mixing
If $Lf=bf'+frac12σ^2f''$, $L^ast$ is the $L^2$-adjoint of $L$, $ϱ$ is a solution of $L^astϱ=0$ and...
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Let $bin C^1(mathbb R)$, $sigmain C^2(mathbb R)$, $$Lf:=bf'+frac12sigma^2f'';;;text{for }fin C^2(mathbb R)$$ and $$L^ast g:=frac12(sigma^2g)''-(bg)';;;text{for }gin C^2(mathbb R).$$ Moreover, let $varrhoin C^2(mathbb R)$ with $$intvarrho(x):{rm d}x=1tag1$$ and $$L^astvarrho=0tag2$$ Now, let $mu$ denote the measure with density $varrho$ with respect to the Lebesgue measure $lambda$.
Are we able to show that $$int f(Lg):{rm d}mu=int g(Lf):{rm d}mutag3$$ for all $f,gin C_c^2(mathbb R)$?
Clearly, $$int(Lf)g:{rm d}lambda=int f(L^ast g):{rm d}mutag4$$ for all $f,gin C^2(mathbb R)$ such that $f$ or $g$ is compactly supported. So, if $fin C^2(mathbb R)$ and $gin C_c^2(mathbb R)$, then $varrho gin C_c^2(mathbb R)$ and $$int(Lf)g:{rm d}mu=int fL^ast(varrho g):{rm d}lambdatag5.$$ Now, $$L^ast(varrho g)=underbrace{(L^astvarrho)}_{=:0}g+(sigma^2varrho)'g'+varrhounderbrace{left(frac12sigma^2g''-bg'right)}_{=:L^ast g:+:b'g}tag6.$$ However, I don't know how we need to proceed from here.
real-analysis integration functional-analysis operator-theory
asked Jan 28 at 21:43
0xbadf00d0xbadf00d
1,79441533
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add a comment |
$begingroup$
Let $bin C^1(mathbb R)$, $sigmain C^2(mathbb R)$, $$Lf:=bf'+frac12sigma^2f'';;;text{for }fin C^2(mathbb R)$$ and $$L^ast g:=frac12(sigma^2g)''-(bg)';;;text{for }gin C^2(mathbb R).$$ Moreover, let $varrhoin C^2(mathbb R)$ with $$intvarrho(x):{rm d}x=1tag1$$ and $$L^astvarrho=0tag2$$ Now, let $mu$ denote the measure with density $varrho$ with respect to the Lebesgue measure $lambda$.
Are we able to show that $$int f(Lg):{rm d}mu=int g(Lf):{rm d}mutag3$$ for all $f,gin C_c^2(mathbb R)$?
Clearly, $$int(Lf)g:{rm d}lambda=int f(L^ast g):{rm d}mutag4$$ for all $f,gin C^2(mathbb R)$ such that $f$ or $g$ is compactly supported. So, if $fin C^2(mathbb R)$ and $gin C_c^2(mathbb R)$, then $varrho gin C_c^2(mathbb R)$ and $$int(Lf)g:{rm d}mu=int fL^ast(varrho g):{rm d}lambdatag5.$$ Now, $$L^ast(varrho g)=underbrace{(L^astvarrho)}_{=:0}g+(sigma^2varrho)'g'+varrhounderbrace{left(frac12sigma^2g''-bg'right)}_{=:L^ast g:+:b'g}tag6.$$ However, I don't know how we need to proceed from here.
real-analysis integration functional-analysis operator-theory
asked Jan 28 at 21:43
0xbadf00d0xbadf00d
1,79441533
$endgroup$
$begingroup$
Do you want to have $rho$ instead of $g$ in (2) (and in the line afterwards)?
$endgroup$
– gerw
Jan 29 at 12:54
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@gerw Yes, of course.
$endgroup$
– 0xbadf00d
Jan 29 at 12:55
$begingroup$
Why do you think that (3) should hold?
$endgroup$
– gerw
Jan 29 at 13:22
$begingroup$
@gerw Meanwhile, I think it doesn't hold, but $L$ can be decomposed into a symmetric and an anti-symmetric part.
$endgroup$
– 0xbadf00d
Jan 29 at 13:45
add a comment |
$begingroup$
Let $bin C^1(mathbb R)$, $sigmain C^2(mathbb R)$, $$Lf:=bf'+frac12sigma^2f'';;;text{for }fin C^2(mathbb R)$$ and $$L^ast g:=frac12(sigma^2g)''-(bg)';;;text{for }gin C^2(mathbb R).$$ Moreover, let $varrhoin C^2(mathbb R)$ with $$intvarrho(x):{rm d}x=1tag1$$ and $$L^astvarrho=0tag2$$ Now, let $mu$ denote the measure with density $varrho$ with respect to the Lebesgue measure $lambda$.
Are we able to show that $$int f(Lg):{rm d}mu=int g(Lf):{rm d}mutag3$$ for all $f,gin C_c^2(mathbb R)$?
Clearly, $$int(Lf)g:{rm d}lambda=int f(L^ast g):{rm d}mutag4$$ for all $f,gin C^2(mathbb R)$ such that $f$ or $g$ is compactly supported. So, if $fin C^2(mathbb R)$ and $gin C_c^2(mathbb R)$, then $varrho gin C_c^2(mathbb R)$ and $$int(Lf)g:{rm d}mu=int fL^ast(varrho g):{rm d}lambdatag5.$$ Now, $$L^ast(varrho g)=underbrace{(L^astvarrho)}_{=:0}g+(sigma^2varrho)'g'+varrhounderbrace{left(frac12sigma^2g''-bg'right)}_{=:L^ast g:+:b'g}tag6.$$ However, I don't know how we need to proceed from here.
real-analysis integration functional-analysis operator-theory
asked Jan 28 at 21:43
0xbadf00d0xbadf00d
1,79441533
$endgroup$
Let $bin C^1(mathbb R)$, $sigmain C^2(mathbb R)$, $$Lf:=bf'+frac12sigma^2f'';;;text{for }fin C^2(mathbb R)$$ and $$L^ast g:=frac12(sigma^2g)''-(bg)';;;text{for }gin C^2(mathbb R).$$ Moreover, let $varrhoin C^2(mathbb R)$ with $$intvarrho(x):{rm d}x=1tag1$$ and $$L^astvarrho=0tag2$$ Now, let $mu$ denote the measure with density $varrho$ with respect to the Lebesgue measure $lambda$.
Are we able to show that $$int f(Lg):{rm d}mu=int g(Lf):{rm d}mutag3$$ for all $f,gin C_c^2(mathbb R)$?
Clearly, $$int(Lf)g:{rm d}lambda=int f(L^ast g):{rm d}mutag4$$ for all $f,gin C^2(mathbb R)$ such that $f$ or $g$ is compactly supported. So, if $fin C^2(mathbb R)$ and $gin C_c^2(mathbb R)$, then $varrho gin C_c^2(mathbb R)$ and $$int(Lf)g:{rm d}mu=int fL^ast(varrho g):{rm d}lambdatag5.$$ Now, $$L^ast(varrho g)=underbrace{(L^astvarrho)}_{=:0}g+(sigma^2varrho)'g'+varrhounderbrace{left(frac12sigma^2g''-bg'right)}_{=:L^ast g:+:b'g}tag6.$$ However, I don't know how we need to proceed from here.
real-analysis integration functional-analysis operator-theory
real-analysis integration functional-analysis operator-theory
asked Jan 28 at 21:43
0xbadf00d0xbadf00d
1,79441533
asked Jan 28 at 21:43
0xbadf00d0xbadf00d
1,79441533
edited Jan 29 at 12:55
0xbadf00d
asked Jan 28 at 21:43
0xbadf00d0xbadf00d
1,79441533
asked Jan 28 at 21:43
0xbadf00d0xbadf00d
1,79441533
asked Jan 28 at 21:43
0xbadf00d0xbadf00d
1,79441533
1,79441533
$begingroup$
Do you want to have $rho$ instead of $g$ in (2) (and in the line afterwards)?
$endgroup$
– gerw
Jan 29 at 12:54
$begingroup$
@gerw Yes, of course.
$endgroup$
– 0xbadf00d
Jan 29 at 12:55
$begingroup$
Why do you think that (3) should hold?
$endgroup$
– gerw
Jan 29 at 13:22
$begingroup$
@gerw Meanwhile, I think it doesn't hold, but $L$ can be decomposed into a symmetric and an anti-symmetric part.
$endgroup$
– 0xbadf00d
Jan 29 at 13:45
add a comment |
$begingroup$
Do you want to have $rho$ instead of $g$ in (2) (and in the line afterwards)?
$endgroup$
– gerw
Jan 29 at 12:54
$begingroup$
@gerw Yes, of course.
$endgroup$
– 0xbadf00d
Jan 29 at 12:55
$begingroup$
Why do you think that (3) should hold?
$endgroup$
– gerw
Jan 29 at 13:22
$begingroup$
@gerw Meanwhile, I think it doesn't hold, but $L$ can be decomposed into a symmetric and an anti-symmetric part.
$endgroup$
– 0xbadf00d
Jan 29 at 13:45
$begingroup$
Do you want to have $rho$ instead of $g$ in (2) (and in the line afterwards)?
$endgroup$
– gerw
Jan 29 at 12:54
$begingroup$
Do you want to have $rho$ instead of $g$ in (2) (and in the line afterwards)?
$endgroup$
– gerw
Jan 29 at 12:54
$begingroup$
@gerw Yes, of course.
$endgroup$
– 0xbadf00d
Jan 29 at 12:55
$begingroup$
@gerw Yes, of course.
$endgroup$
– 0xbadf00d
Jan 29 at 12:55
$begingroup$
Why do you think that (3) should hold?
$endgroup$
– gerw
Jan 29 at 13:22
$begingroup$
Why do you think that (3) should hold?
$endgroup$
– gerw
Jan 29 at 13:22
$begingroup$
@gerw Meanwhile, I think it doesn't hold, but $L$ can be decomposed into a symmetric and an anti-symmetric part.
$endgroup$
– 0xbadf00d
Jan 29 at 13:45
$begingroup$
@gerw Meanwhile, I think it doesn't hold, but $L$ can be decomposed into a symmetric and an anti-symmetric part.
$endgroup$
– 0xbadf00d
Jan 29 at 13:45
add a comment |
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$begingroup$
Do you want to have $rho$ instead of $g$ in (2) (and in the line afterwards)?
$endgroup$
– gerw
Jan 29 at 12:54
$begingroup$
@gerw Yes, of course.
$endgroup$
– 0xbadf00d
Jan 29 at 12:55
$begingroup$
Why do you think that (3) should hold?
$endgroup$
– gerw
Jan 29 at 13:22
$begingroup$
@gerw Meanwhile, I think it doesn't hold, but $L$ can be decomposed into a symmetric and an anti-symmetric part.
$endgroup$
– 0xbadf00d
Jan 29 at 13:45