Regarding absolute continuity of some function
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$f (y) $ is continuous function of y.
$int_-infty^infty |f(y)||(x-y)|^2dy$ is finite for all x
Given $h(x)= int_-infty^infty f(y)(x-y)^2dy=int_-infty^infty f(y+x)(y)^2dy$
is $h(x)$ always absolutely continuous?
If no. What conditions on $f(y)$ makes $h(x)$ absolutely continuous?
calculus real-analysis sequences-and-series functional-analysis functions
asked May 18 '16 at 5:48
Solvera
1
add a comment |Â
up vote
0
down vote
favorite
$f (y) $ is continuous function of y.
$int_-infty^infty |f(y)||(x-y)|^2dy$ is finite for all x
Given $h(x)= int_-infty^infty f(y)(x-y)^2dy=int_-infty^infty f(y+x)(y)^2dy$
is $h(x)$ always absolutely continuous?
If no. What conditions on $f(y)$ makes $h(x)$ absolutely continuous?
calculus real-analysis sequences-and-series functional-analysis functions
asked May 18 '16 at 5:48
Solvera
1
i couldn't construct a counter example
– Solvera
May 18 '16 at 5:49
why do you say is finite if you take $f(y)=c$ then you are basically integrating $y^2$ over $mathbb R$ which is not finite
– Spotty
May 18 '16 at 7:18
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$f (y) $ is continuous function of y.
$int_-infty^infty |f(y)||(x-y)|^2dy$ is finite for all x
Given $h(x)= int_-infty^infty f(y)(x-y)^2dy=int_-infty^infty f(y+x)(y)^2dy$
is $h(x)$ always absolutely continuous?
If no. What conditions on $f(y)$ makes $h(x)$ absolutely continuous?
calculus real-analysis sequences-and-series functional-analysis functions
asked May 18 '16 at 5:48
Solvera
1
$f (y) $ is continuous function of y.
$int_-infty^infty |f(y)||(x-y)|^2dy$ is finite for all x
Given $h(x)= int_-infty^infty f(y)(x-y)^2dy=int_-infty^infty f(y+x)(y)^2dy$
is $h(x)$ always absolutely continuous?
If no. What conditions on $f(y)$ makes $h(x)$ absolutely continuous?
calculus real-analysis sequences-and-series functional-analysis functions
asked May 18 '16 at 5:48
Solvera
1
edited May 18 '16 at 6:01
asked May 18 '16 at 5:48
Solvera
1
asked May 18 '16 at 5:48
Solvera
1
asked May 18 '16 at 5:48
Solvera
1
1
i couldn't construct a counter example
– Solvera
May 18 '16 at 5:49
why do you say is finite if you take $f(y)=c$ then you are basically integrating $y^2$ over $mathbb R$ which is not finite
– Spotty
May 18 '16 at 7:18
add a comment |Â
i couldn't construct a counter example
– Solvera
May 18 '16 at 5:49
why do you say is finite if you take $f(y)=c$ then you are basically integrating $y^2$ over $mathbb R$ which is not finite
– Spotty
May 18 '16 at 7:18
i couldn't construct a counter example
– Solvera
May 18 '16 at 5:49
i couldn't construct a counter example
– Solvera
May 18 '16 at 5:49
why do you say is finite if you take $f(y)=c$ then you are basically integrating $y^2$ over $mathbb R$ which is not finite
– Spotty
May 18 '16 at 7:18
why do you say is finite if you take $f(y)=c$ then you are basically integrating $y^2$ over $mathbb R$ which is not finite
– Spotty
May 18 '16 at 7:18
add a comment |Â
1 Answer
1
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votes
up vote
0
down vote
accepted
You can write
$$
int_-infty^inftyf(y)(x-y)^2dy=x^2int_-infty^inftyf(y)dy-2xint_-infty^inftyf(y)ydy+int_-infty^inftyf(y)y^2dy.
$$
Because of your condition, its not difficult to show that the integrals on the right are absolutely convergent.
answered May 18 '16 at 11:16
DisintegratingByParts
56.2k42373
Please could you show $sum_k |x_k+1-x_k| le epsilon$ implies $sum_k |h(x_k+1)-h(x_k)| le delta$ ?
– Solvera
May 18 '16 at 11:47
1
I'll bet that you can show how that works for the second order polynomial $h(x)=Ax^2+Bx+C$. That's what you have, with constants $A,B,C$.
– DisintegratingByParts
May 18 '16 at 11:50
1
ok thanks. the last integral contains $y^2$ i think
– Solvera
May 18 '16 at 11:53
@stalker2133 : Thanks, I fixed that. Any continuously differentiable function is absolutely continuous.
– DisintegratingByParts
May 18 '16 at 11:56
@TrialAndErrror. thanks for the answer. I was wondering if you looked at math.stackexchange.com/questions/1782377/…
– Solvera
May 18 '16 at 11:58
 |Â
show 16 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You can write
$$
int_-infty^inftyf(y)(x-y)^2dy=x^2int_-infty^inftyf(y)dy-2xint_-infty^inftyf(y)ydy+int_-infty^inftyf(y)y^2dy.
$$
Because of your condition, its not difficult to show that the integrals on the right are absolutely convergent.
answered May 18 '16 at 11:16
DisintegratingByParts
56.2k42373
Please could you show $sum_k |x_k+1-x_k| le epsilon$ implies $sum_k |h(x_k+1)-h(x_k)| le delta$ ?
– Solvera
May 18 '16 at 11:47
1
I'll bet that you can show how that works for the second order polynomial $h(x)=Ax^2+Bx+C$. That's what you have, with constants $A,B,C$.
– DisintegratingByParts
May 18 '16 at 11:50
1
ok thanks. the last integral contains $y^2$ i think
– Solvera
May 18 '16 at 11:53
@stalker2133 : Thanks, I fixed that. Any continuously differentiable function is absolutely continuous.
– DisintegratingByParts
May 18 '16 at 11:56
@TrialAndErrror. thanks for the answer. I was wondering if you looked at math.stackexchange.com/questions/1782377/…
– Solvera
May 18 '16 at 11:58
 |Â
show 16 more comments
up vote
0
down vote
accepted
You can write
$$
int_-infty^inftyf(y)(x-y)^2dy=x^2int_-infty^inftyf(y)dy-2xint_-infty^inftyf(y)ydy+int_-infty^inftyf(y)y^2dy.
$$
Because of your condition, its not difficult to show that the integrals on the right are absolutely convergent.
answered May 18 '16 at 11:16
DisintegratingByParts
56.2k42373
Please could you show $sum_k |x_k+1-x_k| le epsilon$ implies $sum_k |h(x_k+1)-h(x_k)| le delta$ ?
– Solvera
May 18 '16 at 11:47
1
I'll bet that you can show how that works for the second order polynomial $h(x)=Ax^2+Bx+C$. That's what you have, with constants $A,B,C$.
– DisintegratingByParts
May 18 '16 at 11:50
1
ok thanks. the last integral contains $y^2$ i think
– Solvera
May 18 '16 at 11:53
@stalker2133 : Thanks, I fixed that. Any continuously differentiable function is absolutely continuous.
– DisintegratingByParts
May 18 '16 at 11:56
@TrialAndErrror. thanks for the answer. I was wondering if you looked at math.stackexchange.com/questions/1782377/…
– Solvera
May 18 '16 at 11:58
 |Â
show 16 more comments
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You can write
$$
int_-infty^inftyf(y)(x-y)^2dy=x^2int_-infty^inftyf(y)dy-2xint_-infty^inftyf(y)ydy+int_-infty^inftyf(y)y^2dy.
$$
Because of your condition, its not difficult to show that the integrals on the right are absolutely convergent.
answered May 18 '16 at 11:16
DisintegratingByParts
56.2k42373
You can write
$$
int_-infty^inftyf(y)(x-y)^2dy=x^2int_-infty^inftyf(y)dy-2xint_-infty^inftyf(y)ydy+int_-infty^inftyf(y)y^2dy.
$$
Because of your condition, its not difficult to show that the integrals on the right are absolutely convergent.
answered May 18 '16 at 11:16
DisintegratingByParts
56.2k42373
edited May 18 '16 at 11:55
answered May 18 '16 at 11:16
DisintegratingByParts
56.2k42373
answered May 18 '16 at 11:16
DisintegratingByParts
56.2k42373
answered May 18 '16 at 11:16
DisintegratingByParts
56.2k42373
56.2k42373
Please could you show $sum_k |x_k+1-x_k| le epsilon$ implies $sum_k |h(x_k+1)-h(x_k)| le delta$ ?
– Solvera
May 18 '16 at 11:47
1
I'll bet that you can show how that works for the second order polynomial $h(x)=Ax^2+Bx+C$. That's what you have, with constants $A,B,C$.
– DisintegratingByParts
May 18 '16 at 11:50
1
ok thanks. the last integral contains $y^2$ i think
– Solvera
May 18 '16 at 11:53
@stalker2133 : Thanks, I fixed that. Any continuously differentiable function is absolutely continuous.
– DisintegratingByParts
May 18 '16 at 11:56
@TrialAndErrror. thanks for the answer. I was wondering if you looked at math.stackexchange.com/questions/1782377/…
– Solvera
May 18 '16 at 11:58
 |Â
show 16 more comments
Please could you show $sum_k |x_k+1-x_k| le epsilon$ implies $sum_k |h(x_k+1)-h(x_k)| le delta$ ?
– Solvera
May 18 '16 at 11:47
1
I'll bet that you can show how that works for the second order polynomial $h(x)=Ax^2+Bx+C$. That's what you have, with constants $A,B,C$.
– DisintegratingByParts
May 18 '16 at 11:50
1
ok thanks. the last integral contains $y^2$ i think
– Solvera
May 18 '16 at 11:53
@stalker2133 : Thanks, I fixed that. Any continuously differentiable function is absolutely continuous.
– DisintegratingByParts
May 18 '16 at 11:56
@TrialAndErrror. thanks for the answer. I was wondering if you looked at math.stackexchange.com/questions/1782377/…
– Solvera
May 18 '16 at 11:58
Please could you show $sum_k |x_k+1-x_k| le epsilon$ implies $sum_k |h(x_k+1)-h(x_k)| le delta$ ?
– Solvera
May 18 '16 at 11:47
Please could you show $sum_k |x_k+1-x_k| le epsilon$ implies $sum_k |h(x_k+1)-h(x_k)| le delta$ ?
– Solvera
May 18 '16 at 11:47
1
1
I'll bet that you can show how that works for the second order polynomial $h(x)=Ax^2+Bx+C$. That's what you have, with constants $A,B,C$.
– DisintegratingByParts
May 18 '16 at 11:50
I'll bet that you can show how that works for the second order polynomial $h(x)=Ax^2+Bx+C$. That's what you have, with constants $A,B,C$.
– DisintegratingByParts
May 18 '16 at 11:50
1
1
ok thanks. the last integral contains $y^2$ i think
– Solvera
May 18 '16 at 11:53
ok thanks. the last integral contains $y^2$ i think
– Solvera
May 18 '16 at 11:53
@stalker2133 : Thanks, I fixed that. Any continuously differentiable function is absolutely continuous.
– DisintegratingByParts
May 18 '16 at 11:56
@stalker2133 : Thanks, I fixed that. Any continuously differentiable function is absolutely continuous.
– DisintegratingByParts
May 18 '16 at 11:56
@TrialAndErrror. thanks for the answer. I was wondering if you looked at math.stackexchange.com/questions/1782377/…
– Solvera
May 18 '16 at 11:58
@TrialAndErrror. thanks for the answer. I was wondering if you looked at math.stackexchange.com/questions/1782377/…
– Solvera
May 18 '16 at 11:58
 |Â
show 16 more comments
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i couldn't construct a counter example
– Solvera
May 18 '16 at 5:49
why do you say is finite if you take $f(y)=c$ then you are basically integrating $y^2$ over $mathbb R$ which is not finite
– Spotty
May 18 '16 at 7:18