Understand proof involving convolution and integration
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I'm currently trying to understand the following proof but there are some obscurities
From the second to the third line, why does $int_mathbb{R^n}|f(y)|Big(int_{mathbb{R^n}}|g(x-y)|dxBig)dy$ equal $int_mathbb{R^n}|f(y)|dyint_{mathbb{R^n}}|g(x)|dx$?
And here I don't understand why $int_{mathbb{R^n}}|g(x-y)|dy^{1/p'}=|g|_1^{1/p'}$ from the third to last line 
real-analysis analysis
asked Dec 4 '18 at 7:38
conradconrad
757
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add a comment |
$begingroup$
I'm currently trying to understand the following proof but there are some obscurities
From the second to the third line, why does $int_mathbb{R^n}|f(y)|Big(int_{mathbb{R^n}}|g(x-y)|dxBig)dy$ equal $int_mathbb{R^n}|f(y)|dyint_{mathbb{R^n}}|g(x)|dx$?
And here I don't understand why $int_{mathbb{R^n}}|g(x-y)|dy^{1/p'}=|g|_1^{1/p'}$ from the third to last line
real-analysis analysis
asked Dec 4 '18 at 7:38
conradconrad
757
$endgroup$
add a comment |
$begingroup$
I'm currently trying to understand the following proof but there are some obscurities
From the second to the third line, why does $int_mathbb{R^n}|f(y)|Big(int_{mathbb{R^n}}|g(x-y)|dxBig)dy$ equal $int_mathbb{R^n}|f(y)|dyint_{mathbb{R^n}}|g(x)|dx$?
And here I don't understand why $int_{mathbb{R^n}}|g(x-y)|dy^{1/p'}=|g|_1^{1/p'}$ from the third to last line
real-analysis analysis
asked Dec 4 '18 at 7:38
conradconrad
757
$endgroup$
I'm currently trying to understand the following proof but there are some obscurities
From the second to the third line, why does $int_mathbb{R^n}|f(y)|Big(int_{mathbb{R^n}}|g(x-y)|dxBig)dy$ equal $int_mathbb{R^n}|f(y)|dyint_{mathbb{R^n}}|g(x)|dx$?
And here I don't understand why $int_{mathbb{R^n}}|g(x-y)|dy^{1/p'}=|g|_1^{1/p'}$ from the third to last line
real-analysis analysis
real-analysis analysis
asked Dec 4 '18 at 7:38
conradconrad
757
asked Dec 4 '18 at 7:38
conradconrad
757
asked Dec 4 '18 at 7:38
conradconrad
757
asked Dec 4 '18 at 7:38
conradconrad
757
asked Dec 4 '18 at 7:38
conradconrad
757
757
add a comment |
add a comment |
3 Answers
3
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oldest
votes
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It’s actually rather easy: just notice that the integral is over $y$ and the whole space, so shifting it by $x$ won’t make any difference. (In other words, you can replace $x-y$ by $y$).
Once you have done that, it’s just a matter of definition.
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To the first question:
This is the change of variables $xmapsto x-y$, indeed,
$$
int_{-infty}^infty g(x-y)mathrm dx =int_{-infty}^infty g(x)mathrm dx
$$
For the second, this is the definition of the $L^1$ norm,
$$
||g||_1=int_{-infty}^infty |g(x)|mathrm dx=int_{-infty}^infty |g(x-y)|mathrm dx
$$
answered Dec 4 '18 at 7:43
qbertqbert
22.1k32560
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add a comment |
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The same argument for both the questions: $int h(x+a) , dx=int h(x) , dx$ for any integrable function $h$ and any $a in mathbb R^{n}$.
answered Dec 4 '18 at 7:43
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It’s actually rather easy: just notice that the integral is over $y$ and the whole space, so shifting it by $x$ won’t make any difference. (In other words, you can replace $x-y$ by $y$).
Once you have done that, it’s just a matter of definition.
$endgroup$
add a comment |
$begingroup$
It’s actually rather easy: just notice that the integral is over $y$ and the whole space, so shifting it by $x$ won’t make any difference. (In other words, you can replace $x-y$ by $y$).
Once you have done that, it’s just a matter of definition.
$endgroup$
add a comment |
$begingroup$
It’s actually rather easy: just notice that the integral is over $y$ and the whole space, so shifting it by $x$ won’t make any difference. (In other words, you can replace $x-y$ by $y$).
Once you have done that, it’s just a matter of definition.
$endgroup$
It’s actually rather easy: just notice that the integral is over $y$ and the whole space, so shifting it by $x$ won’t make any difference. (In other words, you can replace $x-y$ by $y$).
Once you have done that, it’s just a matter of definition.
answered Dec 4 '18 at 7:42
community wiki
b00n heT
add a comment |
add a comment |
$begingroup$
To the first question:
This is the change of variables $xmapsto x-y$, indeed,
$$
int_{-infty}^infty g(x-y)mathrm dx =int_{-infty}^infty g(x)mathrm dx
$$
For the second, this is the definition of the $L^1$ norm,
$$
||g||_1=int_{-infty}^infty |g(x)|mathrm dx=int_{-infty}^infty |g(x-y)|mathrm dx
$$
answered Dec 4 '18 at 7:43
qbertqbert
22.1k32560
$endgroup$
add a comment |
$begingroup$
To the first question:
This is the change of variables $xmapsto x-y$, indeed,
$$
int_{-infty}^infty g(x-y)mathrm dx =int_{-infty}^infty g(x)mathrm dx
$$
For the second, this is the definition of the $L^1$ norm,
$$
||g||_1=int_{-infty}^infty |g(x)|mathrm dx=int_{-infty}^infty |g(x-y)|mathrm dx
$$
answered Dec 4 '18 at 7:43
qbertqbert
22.1k32560
$endgroup$
add a comment |
$begingroup$
To the first question:
This is the change of variables $xmapsto x-y$, indeed,
$$
int_{-infty}^infty g(x-y)mathrm dx =int_{-infty}^infty g(x)mathrm dx
$$
For the second, this is the definition of the $L^1$ norm,
$$
||g||_1=int_{-infty}^infty |g(x)|mathrm dx=int_{-infty}^infty |g(x-y)|mathrm dx
$$
answered Dec 4 '18 at 7:43
qbertqbert
22.1k32560
$endgroup$
To the first question:
This is the change of variables $xmapsto x-y$, indeed,
$$
int_{-infty}^infty g(x-y)mathrm dx =int_{-infty}^infty g(x)mathrm dx
$$
For the second, this is the definition of the $L^1$ norm,
$$
||g||_1=int_{-infty}^infty |g(x)|mathrm dx=int_{-infty}^infty |g(x-y)|mathrm dx
$$
answered Dec 4 '18 at 7:43
qbertqbert
22.1k32560
answered Dec 4 '18 at 7:43
qbertqbert
22.1k32560
answered Dec 4 '18 at 7:43
qbertqbert
22.1k32560
answered Dec 4 '18 at 7:43
qbertqbert
22.1k32560
22.1k32560
add a comment |
add a comment |
$begingroup$
The same argument for both the questions: $int h(x+a) , dx=int h(x) , dx$ for any integrable function $h$ and any $a in mathbb R^{n}$.
answered Dec 4 '18 at 7:43
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
$endgroup$
add a comment |
$begingroup$
The same argument for both the questions: $int h(x+a) , dx=int h(x) , dx$ for any integrable function $h$ and any $a in mathbb R^{n}$.
answered Dec 4 '18 at 7:43
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
$endgroup$
add a comment |
$begingroup$
The same argument for both the questions: $int h(x+a) , dx=int h(x) , dx$ for any integrable function $h$ and any $a in mathbb R^{n}$.
answered Dec 4 '18 at 7:43
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
$endgroup$
The same argument for both the questions: $int h(x+a) , dx=int h(x) , dx$ for any integrable function $h$ and any $a in mathbb R^{n}$.
answered Dec 4 '18 at 7:43
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
answered Dec 4 '18 at 7:43
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
answered Dec 4 '18 at 7:43
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
answered Dec 4 '18 at 7:43
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
55.9k42158
add a comment |
add a comment |
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