Mixing
Integral of average of integral equal to integral itself
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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 6 hours ago
adosdeci
276
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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 6 hours ago
adosdeci
276
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
6 hours ago
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 6 hours ago
adosdeci
276
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 6 hours ago
adosdeci
276
asked 6 hours ago
adosdeci
276
asked 6 hours ago
adosdeci
276
asked 6 hours ago
adosdeci
276
asked 6 hours ago
adosdeci
276
276
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
6 hours ago
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
6 hours ago
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
6 hours ago
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
6 hours ago
add a comment |
1 Answer
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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 6 hours ago
Kavi Rama Murthy
39.9k31750
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 6 hours ago
Kavi Rama Murthy
39.9k31750
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 6 hours ago
Kavi Rama Murthy
39.9k31750
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 6 hours ago
Kavi Rama Murthy
39.9k31750
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 6 hours ago
Kavi Rama Murthy
39.9k31750
answered 6 hours ago
Kavi Rama Murthy
39.9k31750
answered 6 hours ago
Kavi Rama Murthy
39.9k31750
answered 6 hours ago
Kavi Rama Murthy
39.9k31750
39.9k31750
add a comment |
add a comment |
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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
6 hours ago