Mixing
Extension of a integration by parts formula for a linear operator on $C_0(mathbb R)$
$begingroup$
Let
$(mathcal D(A),A)$ be a linear operator on $C_0(mathbb R)$ (the space of continuous functions vanishing at infinity equipped with the supremum norm $left|;cdot;right|_infty$) such that $C_c^infty(mathbb R)$ is a core of $(mathcal D(A),A)$
$mu$ be a probability measure on $(mathbb R,mathcal B(mathbb R))$, $$Gamma(f,g):=frac12left(A(fg)-fAg-gAfright);;;text{for }f,gin C_c^infty(mathbb R)$$ and $$mathcal E(f,g):=intGamma(f,g):{rm d}mu;;;text{for }f,gin C_c^infty(mathbb R)$$
Assuming $$int Af:{rm d}mu=0;;;text{for all }fin C_c^infty(mathbb R)tag1$$ and $$int fAg:{rm d}mu=int gAf:{rm d}mu;;;text{for all }f,gin C_c^infty(mathbb R),tag2$$ it's easy to see that $$mathcal E(f,g)=-int fAg:{rm d}mu=-int gAf:{rm d}mu;;;text{for all }f,gin C_c^infty(mathbb R)tag3.$$
Are we able to extend the definition of $mathcal E$, preserving $(3)$, for $finmathcal D(A)$ and $gin C_c^infty(mathbb R)$?
Since $finmathcal D(A)$ and $C_c^infty(mathbb R)$ is a core of $(mathcal D(A),A)$, we should find a $(f_n)_{ninmathbb N}subseteq C_c^infty(mathbb R)$ with $$left|f_n-fright|_infty+left|Af_n-Afright|_inftyxrightarrow{ntoinfty}0.tag4$$ By submultiplicativity of $left|;cdot;right|_infty$, we should have $$left|mathcal E(f_n,g)-mathcal E(f,g)right|lefrac12left(left|A(f_ng)-A(fg)right|_infty+left|f_n-fright|_inftyleft|Agright|_infty+left|gright|_inftyleft|Af_n-Afright|_inftyright)tag5$$ and by $(4)$ the last two terms on the right-hand side tend to $0$ as $ntoinfty$.
Do we have $left|A(f_ng)-A(fg)right|_inftyxrightarrow{ntoinfty}0$ too?
On the other hand, $mathcal E(f_n,g)=-mu(f_nAg)=-mu(gAf_n)$ by $(3)$ for all $ninmathbb N$ and we should easily obtain $$left|mu(f_nAg)-mu(fAg)right|+left|mu(gAf_n)-mu(gAf)right|xrightarrow{ntoinfty}0$$ by $(4)$ and the dominated convergence theorem. Thus, we would be able to conclude the desired claim.
It might be useful to observe that $f_ng,fgin C_c^infty(mathbb R)$ (since $gin C_c^infty(mathbb R)$) for all $ninmathbb N$.
integration functional-analysis operator-theory
asked Jan 30 at 13:54
0xbadf00d0xbadf00d
1,80641534
$endgroup$
add a comment |
$begingroup$
Let
$(mathcal D(A),A)$ be a linear operator on $C_0(mathbb R)$ (the space of continuous functions vanishing at infinity equipped with the supremum norm $left|;cdot;right|_infty$) such that $C_c^infty(mathbb R)$ is a core of $(mathcal D(A),A)$
$mu$ be a probability measure on $(mathbb R,mathcal B(mathbb R))$, $$Gamma(f,g):=frac12left(A(fg)-fAg-gAfright);;;text{for }f,gin C_c^infty(mathbb R)$$ and $$mathcal E(f,g):=intGamma(f,g):{rm d}mu;;;text{for }f,gin C_c^infty(mathbb R)$$
Assuming $$int Af:{rm d}mu=0;;;text{for all }fin C_c^infty(mathbb R)tag1$$ and $$int fAg:{rm d}mu=int gAf:{rm d}mu;;;text{for all }f,gin C_c^infty(mathbb R),tag2$$ it's easy to see that $$mathcal E(f,g)=-int fAg:{rm d}mu=-int gAf:{rm d}mu;;;text{for all }f,gin C_c^infty(mathbb R)tag3.$$
Are we able to extend the definition of $mathcal E$, preserving $(3)$, for $finmathcal D(A)$ and $gin C_c^infty(mathbb R)$?
Since $finmathcal D(A)$ and $C_c^infty(mathbb R)$ is a core of $(mathcal D(A),A)$, we should find a $(f_n)_{ninmathbb N}subseteq C_c^infty(mathbb R)$ with $$left|f_n-fright|_infty+left|Af_n-Afright|_inftyxrightarrow{ntoinfty}0.tag4$$ By submultiplicativity of $left|;cdot;right|_infty$, we should have $$left|mathcal E(f_n,g)-mathcal E(f,g)right|lefrac12left(left|A(f_ng)-A(fg)right|_infty+left|f_n-fright|_inftyleft|Agright|_infty+left|gright|_inftyleft|Af_n-Afright|_inftyright)tag5$$ and by $(4)$ the last two terms on the right-hand side tend to $0$ as $ntoinfty$.
Do we have $left|A(f_ng)-A(fg)right|_inftyxrightarrow{ntoinfty}0$ too?
On the other hand, $mathcal E(f_n,g)=-mu(f_nAg)=-mu(gAf_n)$ by $(3)$ for all $ninmathbb N$ and we should easily obtain $$left|mu(f_nAg)-mu(fAg)right|+left|mu(gAf_n)-mu(gAf)right|xrightarrow{ntoinfty}0$$ by $(4)$ and the dominated convergence theorem. Thus, we would be able to conclude the desired claim.
It might be useful to observe that $f_ng,fgin C_c^infty(mathbb R)$ (since $gin C_c^infty(mathbb R)$) for all $ninmathbb N$.
integration functional-analysis operator-theory
asked Jan 30 at 13:54
0xbadf00d0xbadf00d
1,80641534
$endgroup$
$begingroup$
I'm not sure if I understand your question correctly. To me it seems like you want to define $mathcal{E}(f,g)$ for $fin D(A)$ and $gin C_c^infty(mathbb{R})$ in such a way that (3) holds and $mathcal{E}(f,g)$ depends continuously on $f$ w.r.t the graph norm. If this is the case, why don't you just take the right side of (3) as definition?
$endgroup$
– MaoWao
Jan 31 at 9:45
add a comment |
$begingroup$
Let
$(mathcal D(A),A)$ be a linear operator on $C_0(mathbb R)$ (the space of continuous functions vanishing at infinity equipped with the supremum norm $left|;cdot;right|_infty$) such that $C_c^infty(mathbb R)$ is a core of $(mathcal D(A),A)$
$mu$ be a probability measure on $(mathbb R,mathcal B(mathbb R))$, $$Gamma(f,g):=frac12left(A(fg)-fAg-gAfright);;;text{for }f,gin C_c^infty(mathbb R)$$ and $$mathcal E(f,g):=intGamma(f,g):{rm d}mu;;;text{for }f,gin C_c^infty(mathbb R)$$
Assuming $$int Af:{rm d}mu=0;;;text{for all }fin C_c^infty(mathbb R)tag1$$ and $$int fAg:{rm d}mu=int gAf:{rm d}mu;;;text{for all }f,gin C_c^infty(mathbb R),tag2$$ it's easy to see that $$mathcal E(f,g)=-int fAg:{rm d}mu=-int gAf:{rm d}mu;;;text{for all }f,gin C_c^infty(mathbb R)tag3.$$
Are we able to extend the definition of $mathcal E$, preserving $(3)$, for $finmathcal D(A)$ and $gin C_c^infty(mathbb R)$?
Since $finmathcal D(A)$ and $C_c^infty(mathbb R)$ is a core of $(mathcal D(A),A)$, we should find a $(f_n)_{ninmathbb N}subseteq C_c^infty(mathbb R)$ with $$left|f_n-fright|_infty+left|Af_n-Afright|_inftyxrightarrow{ntoinfty}0.tag4$$ By submultiplicativity of $left|;cdot;right|_infty$, we should have $$left|mathcal E(f_n,g)-mathcal E(f,g)right|lefrac12left(left|A(f_ng)-A(fg)right|_infty+left|f_n-fright|_inftyleft|Agright|_infty+left|gright|_inftyleft|Af_n-Afright|_inftyright)tag5$$ and by $(4)$ the last two terms on the right-hand side tend to $0$ as $ntoinfty$.
Do we have $left|A(f_ng)-A(fg)right|_inftyxrightarrow{ntoinfty}0$ too?
On the other hand, $mathcal E(f_n,g)=-mu(f_nAg)=-mu(gAf_n)$ by $(3)$ for all $ninmathbb N$ and we should easily obtain $$left|mu(f_nAg)-mu(fAg)right|+left|mu(gAf_n)-mu(gAf)right|xrightarrow{ntoinfty}0$$ by $(4)$ and the dominated convergence theorem. Thus, we would be able to conclude the desired claim.
It might be useful to observe that $f_ng,fgin C_c^infty(mathbb R)$ (since $gin C_c^infty(mathbb R)$) for all $ninmathbb N$.
integration functional-analysis operator-theory
asked Jan 30 at 13:54
0xbadf00d0xbadf00d
1,80641534
$endgroup$
Let
$(mathcal D(A),A)$ be a linear operator on $C_0(mathbb R)$ (the space of continuous functions vanishing at infinity equipped with the supremum norm $left|;cdot;right|_infty$) such that $C_c^infty(mathbb R)$ is a core of $(mathcal D(A),A)$
$mu$ be a probability measure on $(mathbb R,mathcal B(mathbb R))$, $$Gamma(f,g):=frac12left(A(fg)-fAg-gAfright);;;text{for }f,gin C_c^infty(mathbb R)$$ and $$mathcal E(f,g):=intGamma(f,g):{rm d}mu;;;text{for }f,gin C_c^infty(mathbb R)$$
Assuming $$int Af:{rm d}mu=0;;;text{for all }fin C_c^infty(mathbb R)tag1$$ and $$int fAg:{rm d}mu=int gAf:{rm d}mu;;;text{for all }f,gin C_c^infty(mathbb R),tag2$$ it's easy to see that $$mathcal E(f,g)=-int fAg:{rm d}mu=-int gAf:{rm d}mu;;;text{for all }f,gin C_c^infty(mathbb R)tag3.$$
Are we able to extend the definition of $mathcal E$, preserving $(3)$, for $finmathcal D(A)$ and $gin C_c^infty(mathbb R)$?
Since $finmathcal D(A)$ and $C_c^infty(mathbb R)$ is a core of $(mathcal D(A),A)$, we should find a $(f_n)_{ninmathbb N}subseteq C_c^infty(mathbb R)$ with $$left|f_n-fright|_infty+left|Af_n-Afright|_inftyxrightarrow{ntoinfty}0.tag4$$ By submultiplicativity of $left|;cdot;right|_infty$, we should have $$left|mathcal E(f_n,g)-mathcal E(f,g)right|lefrac12left(left|A(f_ng)-A(fg)right|_infty+left|f_n-fright|_inftyleft|Agright|_infty+left|gright|_inftyleft|Af_n-Afright|_inftyright)tag5$$ and by $(4)$ the last two terms on the right-hand side tend to $0$ as $ntoinfty$.
Do we have $left|A(f_ng)-A(fg)right|_inftyxrightarrow{ntoinfty}0$ too?
On the other hand, $mathcal E(f_n,g)=-mu(f_nAg)=-mu(gAf_n)$ by $(3)$ for all $ninmathbb N$ and we should easily obtain $$left|mu(f_nAg)-mu(fAg)right|+left|mu(gAf_n)-mu(gAf)right|xrightarrow{ntoinfty}0$$ by $(4)$ and the dominated convergence theorem. Thus, we would be able to conclude the desired claim.
It might be useful to observe that $f_ng,fgin C_c^infty(mathbb R)$ (since $gin C_c^infty(mathbb R)$) for all $ninmathbb N$.
integration functional-analysis operator-theory
integration functional-analysis operator-theory
asked Jan 30 at 13:54
0xbadf00d0xbadf00d
1,80641534
asked Jan 30 at 13:54
0xbadf00d0xbadf00d
1,80641534
edited Jan 30 at 19:59
0xbadf00d
asked Jan 30 at 13:54
0xbadf00d0xbadf00d
1,80641534
asked Jan 30 at 13:54
0xbadf00d0xbadf00d
1,80641534
asked Jan 30 at 13:54
0xbadf00d0xbadf00d
1,80641534
1,80641534
$begingroup$
I'm not sure if I understand your question correctly. To me it seems like you want to define $mathcal{E}(f,g)$ for $fin D(A)$ and $gin C_c^infty(mathbb{R})$ in such a way that (3) holds and $mathcal{E}(f,g)$ depends continuously on $f$ w.r.t the graph norm. If this is the case, why don't you just take the right side of (3) as definition?
$endgroup$
– MaoWao
Jan 31 at 9:45
add a comment |
$begingroup$
I'm not sure if I understand your question correctly. To me it seems like you want to define $mathcal{E}(f,g)$ for $fin D(A)$ and $gin C_c^infty(mathbb{R})$ in such a way that (3) holds and $mathcal{E}(f,g)$ depends continuously on $f$ w.r.t the graph norm. If this is the case, why don't you just take the right side of (3) as definition?
$endgroup$
– MaoWao
Jan 31 at 9:45
$begingroup$
I'm not sure if I understand your question correctly. To me it seems like you want to define $mathcal{E}(f,g)$ for $fin D(A)$ and $gin C_c^infty(mathbb{R})$ in such a way that (3) holds and $mathcal{E}(f,g)$ depends continuously on $f$ w.r.t the graph norm. If this is the case, why don't you just take the right side of (3) as definition?
$endgroup$
– MaoWao
Jan 31 at 9:45
$begingroup$
I'm not sure if I understand your question correctly. To me it seems like you want to define $mathcal{E}(f,g)$ for $fin D(A)$ and $gin C_c^infty(mathbb{R})$ in such a way that (3) holds and $mathcal{E}(f,g)$ depends continuously on $f$ w.r.t the graph norm. If this is the case, why don't you just take the right side of (3) as definition?
$endgroup$
– MaoWao
Jan 31 at 9:45
add a comment |
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$begingroup$
I'm not sure if I understand your question correctly. To me it seems like you want to define $mathcal{E}(f,g)$ for $fin D(A)$ and $gin C_c^infty(mathbb{R})$ in such a way that (3) holds and $mathcal{E}(f,g)$ depends continuously on $f$ w.r.t the graph norm. If this is the case, why don't you just take the right side of (3) as definition?
$endgroup$
– MaoWao
Jan 31 at 9:45