Integral of average of integral equal to integral itself
up vote
0
down vote
favorite
I was trying to prove the following and had totally no idea how to start.
Let $f$ be integrable over $(-infty, infty)$, and let $h> 0$ be fixed. Prove that $$int_{-infty}^{infty} (frac{1}{2h} int_{x-h}^{x+h} f(y) mathrm{d}y) mathrm{d}x = int_{-infty}^{infty}f(x) mathrm{d}x$$
When $f$ is given integrability, I can use linearity but not so sure if that is the right thing to do here. Also I was considering double integral as the left hand side of the equality looks like that. That is my intention, for example with the help of Fubini's Theorem. However $f$ is defined on one dimension, I struggled with constructing a bi-variate function.
real-analysis lebesgue-integral
asked Nov 19 at 7:32
adosdeci
326
add a comment |
up vote
0
down vote
favorite
I was trying to prove the following and had totally no idea how to start.
Let $f$ be integrable over $(-infty, infty)$, and let $h> 0$ be fixed. Prove that $$int_{-infty}^{infty} (frac{1}{2h} int_{x-h}^{x+h} f(y) mathrm{d}y) mathrm{d}x = int_{-infty}^{infty}f(x) mathrm{d}x$$
When $f$ is given integrability, I can use linearity but not so sure if that is the right thing to do here. Also I was considering double integral as the left hand side of the equality looks like that. That is my intention, for example with the help of Fubini's Theorem. However $f$ is defined on one dimension, I struggled with constructing a bi-variate function.
real-analysis lebesgue-integral
asked Nov 19 at 7:32
adosdeci
326
Try writing the LHS as an integral from $-R$ to $R$ and then take the limit, usually with these kinds of problems you just have to do the math head on and suddenly the magic happens.
– tommy1996q
Nov 19 at 7:35
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was trying to prove the following and had totally no idea how to start.
Let $f$ be integrable over $(-infty, infty)$, and let $h> 0$ be fixed. Prove that $$int_{-infty}^{infty} (frac{1}{2h} int_{x-h}^{x+h} f(y) mathrm{d}y) mathrm{d}x = int_{-infty}^{infty}f(x) mathrm{d}x$$
When $f$ is given integrability, I can use linearity but not so sure if that is the right thing to do here. Also I was considering double integral as the left hand side of the equality looks like that. That is my intention, for example with the help of Fubini's Theorem. However $f$ is defined on one dimension, I struggled with constructing a bi-variate function.
real-analysis lebesgue-integral
asked Nov 19 at 7:32
adosdeci
326
I was trying to prove the following and had totally no idea how to start.
Let $f$ be integrable over $(-infty, infty)$, and let $h> 0$ be fixed. Prove that $$int_{-infty}^{infty} (frac{1}{2h} int_{x-h}^{x+h} f(y) mathrm{d}y) mathrm{d}x = int_{-infty}^{infty}f(x) mathrm{d}x$$
When $f$ is given integrability, I can use linearity but not so sure if that is the right thing to do here. Also I was considering double integral as the left hand side of the equality looks like that. That is my intention, for example with the help of Fubini's Theorem. However $f$ is defined on one dimension, I struggled with constructing a bi-variate function.
real-analysis lebesgue-integral
real-analysis lebesgue-integral
asked Nov 19 at 7:32
adosdeci
326
asked Nov 19 at 7:32
adosdeci
326
asked Nov 19 at 7:32
adosdeci
326
asked Nov 19 at 7:32
adosdeci
326
asked Nov 19 at 7:32
adosdeci
326
326
Try writing the LHS as an integral from $-R$ to $R$ and then take the limit, usually with these kinds of problems you just have to do the math head on and suddenly the magic happens.
– tommy1996q
Nov 19 at 7:35
add a comment |
Try writing the LHS as an integral from $-R$ to $R$ and then take the limit, usually with these kinds of problems you just have to do the math head on and suddenly the magic happens.
– tommy1996q
Nov 19 at 7:35
Try writing the LHS as an integral from $-R$ to $R$ and then take the limit, usually with these kinds of problems you just have to do the math head on and suddenly the magic happens.
– tommy1996q
Nov 19 at 7:35
Try writing the LHS as an integral from $-R$ to $R$ and then take the limit, usually with these kinds of problems you just have to do the math head on and suddenly the magic happens.
– tommy1996q
Nov 19 at 7:35
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
LHS $=frac 1 {2h}int_{-infty}^{infty} int_{y-h}^{y+h} dx f(y) dy$ by Fubini's Theorem because $x-h<y<x+h$ is equivalent to $ y-h<x<y+h$. Rest is obvious.
answered Nov 19 at 7:36
Kavi Rama Murthy
45.9k31854
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
LHS $=frac 1 {2h}int_{-infty}^{infty} int_{y-h}^{y+h} dx f(y) dy$ by Fubini's Theorem because $x-h<y<x+h$ is equivalent to $ y-h<x<y+h$. Rest is obvious.
answered Nov 19 at 7:36
Kavi Rama Murthy
45.9k31854
add a comment |
up vote
1
down vote
LHS $=frac 1 {2h}int_{-infty}^{infty} int_{y-h}^{y+h} dx f(y) dy$ by Fubini's Theorem because $x-h<y<x+h$ is equivalent to $ y-h<x<y+h$. Rest is obvious.
answered Nov 19 at 7:36
Kavi Rama Murthy
45.9k31854
add a comment |
up vote
1
down vote
up vote
1
down vote
LHS $=frac 1 {2h}int_{-infty}^{infty} int_{y-h}^{y+h} dx f(y) dy$ by Fubini's Theorem because $x-h<y<x+h$ is equivalent to $ y-h<x<y+h$. Rest is obvious.
answered Nov 19 at 7:36
Kavi Rama Murthy
45.9k31854
LHS $=frac 1 {2h}int_{-infty}^{infty} int_{y-h}^{y+h} dx f(y) dy$ by Fubini's Theorem because $x-h<y<x+h$ is equivalent to $ y-h<x<y+h$. Rest is obvious.
answered Nov 19 at 7:36
Kavi Rama Murthy
45.9k31854
answered Nov 19 at 7:36
Kavi Rama Murthy
45.9k31854
answered Nov 19 at 7:36
Kavi Rama Murthy
45.9k31854
answered Nov 19 at 7:36
Kavi Rama Murthy
45.9k31854
45.9k31854
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004620%2fintegral-of-average-of-integral-equal-to-integral-itself%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Try writing the LHS as an integral from $-R$ to $R$ and then take the limit, usually with these kinds of problems you just have to do the math head on and suddenly the magic happens.
– tommy1996q
Nov 19 at 7:35