Commutative Convolution. Problem 26 Royden 2 ed.
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Let $f$ and $g$ be functions in $L^1(—infty,infty)$, and define $fast g$ to be the
function $h$ defined by $h(y) = int f(y — x)g(x) dx$.
Why $fast g=gast f$?
I have this:
If $y-x=z$ then $int f(y-x)g(x)dx=int f(z)g(y-z)(-dz)=-gast f$
real-analysis measure-theory convolution
asked Jan 7 at 20:32
eraldcoileraldcoil
395211
$endgroup$
add a comment |
$begingroup$
Let $f$ and $g$ be functions in $L^1(—infty,infty)$, and define $fast g$ to be the
function $h$ defined by $h(y) = int f(y — x)g(x) dx$.
Why $fast g=gast f$?
I have this:
If $y-x=z$ then $int f(y-x)g(x)dx=int f(z)g(y-z)(-dz)=-gast f$
real-analysis measure-theory convolution
asked Jan 7 at 20:32
eraldcoileraldcoil
395211
$endgroup$
1
$begingroup$
See this question, or perhaps this one?
$endgroup$
– Dietrich Burde
Jan 7 at 20:35
1
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when you changed variables, you didn't change the interval of integration. $infty -> - infty$ and vice versa
$endgroup$
– Dunham
Jan 7 at 20:36
1
$begingroup$
Okay, but what happens to the limits of integration at the same time?
$endgroup$
– Doug M
Jan 7 at 20:38
add a comment |
$begingroup$
Let $f$ and $g$ be functions in $L^1(—infty,infty)$, and define $fast g$ to be the
function $h$ defined by $h(y) = int f(y — x)g(x) dx$.
Why $fast g=gast f$?
I have this:
If $y-x=z$ then $int f(y-x)g(x)dx=int f(z)g(y-z)(-dz)=-gast f$
real-analysis measure-theory convolution
asked Jan 7 at 20:32
eraldcoileraldcoil
395211
$endgroup$
Let $f$ and $g$ be functions in $L^1(—infty,infty)$, and define $fast g$ to be the
function $h$ defined by $h(y) = int f(y — x)g(x) dx$.
Why $fast g=gast f$?
I have this:
If $y-x=z$ then $int f(y-x)g(x)dx=int f(z)g(y-z)(-dz)=-gast f$
real-analysis measure-theory convolution
real-analysis measure-theory convolution
asked Jan 7 at 20:32
eraldcoileraldcoil
395211
asked Jan 7 at 20:32
eraldcoileraldcoil
395211
asked Jan 7 at 20:32
eraldcoileraldcoil
395211
asked Jan 7 at 20:32
eraldcoileraldcoil
395211
asked Jan 7 at 20:32
eraldcoileraldcoil
395211
395211
1
$begingroup$
See this question, or perhaps this one?
$endgroup$
– Dietrich Burde
Jan 7 at 20:35
1
$begingroup$
when you changed variables, you didn't change the interval of integration. $infty -> - infty$ and vice versa
$endgroup$
– Dunham
Jan 7 at 20:36
1
$begingroup$
Okay, but what happens to the limits of integration at the same time?
$endgroup$
– Doug M
Jan 7 at 20:38
add a comment |
1
$begingroup$
See this question, or perhaps this one?
$endgroup$
– Dietrich Burde
Jan 7 at 20:35
1
$begingroup$
when you changed variables, you didn't change the interval of integration. $infty -> - infty$ and vice versa
$endgroup$
– Dunham
Jan 7 at 20:36
1
$begingroup$
Okay, but what happens to the limits of integration at the same time?
$endgroup$
– Doug M
Jan 7 at 20:38
1
1
$begingroup$
See this question, or perhaps this one?
$endgroup$
– Dietrich Burde
Jan 7 at 20:35
$begingroup$
See this question, or perhaps this one?
$endgroup$
– Dietrich Burde
Jan 7 at 20:35
1
1
$begingroup$
when you changed variables, you didn't change the interval of integration. $infty -> - infty$ and vice versa
$endgroup$
– Dunham
Jan 7 at 20:36
$begingroup$
when you changed variables, you didn't change the interval of integration. $infty -> - infty$ and vice versa
$endgroup$
– Dunham
Jan 7 at 20:36
1
1
$begingroup$
Okay, but what happens to the limits of integration at the same time?
$endgroup$
– Doug M
Jan 7 at 20:38
$begingroup$
Okay, but what happens to the limits of integration at the same time?
$endgroup$
– Doug M
Jan 7 at 20:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The indefinite integral is not a value by a function (the anti-derivative of the integrand). Convolution is defined with a definite integral as follows:
$$ (fast g)(y) = int_{-infty}^{infty} f(y-x) g(x) , dx. $$
Now let's do the substitution where $z=y-x$. Then
$$ (fast g)(y) = int_{-infty}^{infty} f(z) , g(y-z) , (-dz). $$
Bringing out the negative sign and reversing the limits of integration yields
$$ (fast g)(y) = int_{-infty}^{infty} g(y-z) , f(z) , dz = (gast f)(y). $$
answered Jan 7 at 20:45
NicNic8NicNic8
4,60531123
$endgroup$
$begingroup$
The substitution of variable is valid in Lebesgue integral?
$endgroup$
– eraldcoil
Jan 9 at 3:39
$begingroup$
@eraldcoil The substitution is valid as a Riemann integral. Doesn't that make it automatically valid as a Lebesgue integral? I'm not sure.
$endgroup$
– NicNic8
Jan 9 at 4:59
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The indefinite integral is not a value by a function (the anti-derivative of the integrand). Convolution is defined with a definite integral as follows:
$$ (fast g)(y) = int_{-infty}^{infty} f(y-x) g(x) , dx. $$
Now let's do the substitution where $z=y-x$. Then
$$ (fast g)(y) = int_{-infty}^{infty} f(z) , g(y-z) , (-dz). $$
Bringing out the negative sign and reversing the limits of integration yields
$$ (fast g)(y) = int_{-infty}^{infty} g(y-z) , f(z) , dz = (gast f)(y). $$
answered Jan 7 at 20:45
NicNic8NicNic8
4,60531123
$endgroup$
$begingroup$
The substitution of variable is valid in Lebesgue integral?
$endgroup$
– eraldcoil
Jan 9 at 3:39
$begingroup$
@eraldcoil The substitution is valid as a Riemann integral. Doesn't that make it automatically valid as a Lebesgue integral? I'm not sure.
$endgroup$
– NicNic8
Jan 9 at 4:59
add a comment |
$begingroup$
The indefinite integral is not a value by a function (the anti-derivative of the integrand). Convolution is defined with a definite integral as follows:
$$ (fast g)(y) = int_{-infty}^{infty} f(y-x) g(x) , dx. $$
Now let's do the substitution where $z=y-x$. Then
$$ (fast g)(y) = int_{-infty}^{infty} f(z) , g(y-z) , (-dz). $$
Bringing out the negative sign and reversing the limits of integration yields
$$ (fast g)(y) = int_{-infty}^{infty} g(y-z) , f(z) , dz = (gast f)(y). $$
answered Jan 7 at 20:45
NicNic8NicNic8
4,60531123
$endgroup$
$begingroup$
The substitution of variable is valid in Lebesgue integral?
$endgroup$
– eraldcoil
Jan 9 at 3:39
$begingroup$
@eraldcoil The substitution is valid as a Riemann integral. Doesn't that make it automatically valid as a Lebesgue integral? I'm not sure.
$endgroup$
– NicNic8
Jan 9 at 4:59
add a comment |
$begingroup$
The indefinite integral is not a value by a function (the anti-derivative of the integrand). Convolution is defined with a definite integral as follows:
$$ (fast g)(y) = int_{-infty}^{infty} f(y-x) g(x) , dx. $$
Now let's do the substitution where $z=y-x$. Then
$$ (fast g)(y) = int_{-infty}^{infty} f(z) , g(y-z) , (-dz). $$
Bringing out the negative sign and reversing the limits of integration yields
$$ (fast g)(y) = int_{-infty}^{infty} g(y-z) , f(z) , dz = (gast f)(y). $$
answered Jan 7 at 20:45
NicNic8NicNic8
4,60531123
$endgroup$
The indefinite integral is not a value by a function (the anti-derivative of the integrand). Convolution is defined with a definite integral as follows:
$$ (fast g)(y) = int_{-infty}^{infty} f(y-x) g(x) , dx. $$
Now let's do the substitution where $z=y-x$. Then
$$ (fast g)(y) = int_{-infty}^{infty} f(z) , g(y-z) , (-dz). $$
Bringing out the negative sign and reversing the limits of integration yields
$$ (fast g)(y) = int_{-infty}^{infty} g(y-z) , f(z) , dz = (gast f)(y). $$
answered Jan 7 at 20:45
NicNic8NicNic8
4,60531123
answered Jan 7 at 20:45
NicNic8NicNic8
4,60531123
answered Jan 7 at 20:45
NicNic8NicNic8
4,60531123
answered Jan 7 at 20:45
NicNic8NicNic8
4,60531123
4,60531123
$begingroup$
The substitution of variable is valid in Lebesgue integral?
$endgroup$
– eraldcoil
Jan 9 at 3:39
$begingroup$
@eraldcoil The substitution is valid as a Riemann integral. Doesn't that make it automatically valid as a Lebesgue integral? I'm not sure.
$endgroup$
– NicNic8
Jan 9 at 4:59
add a comment |
$begingroup$
The substitution of variable is valid in Lebesgue integral?
$endgroup$
– eraldcoil
Jan 9 at 3:39
$begingroup$
@eraldcoil The substitution is valid as a Riemann integral. Doesn't that make it automatically valid as a Lebesgue integral? I'm not sure.
$endgroup$
– NicNic8
Jan 9 at 4:59
$begingroup$
The substitution of variable is valid in Lebesgue integral?
$endgroup$
– eraldcoil
Jan 9 at 3:39
$begingroup$
The substitution of variable is valid in Lebesgue integral?
$endgroup$
– eraldcoil
Jan 9 at 3:39
$begingroup$
@eraldcoil The substitution is valid as a Riemann integral. Doesn't that make it automatically valid as a Lebesgue integral? I'm not sure.
$endgroup$
– NicNic8
Jan 9 at 4:59
$begingroup$
@eraldcoil The substitution is valid as a Riemann integral. Doesn't that make it automatically valid as a Lebesgue integral? I'm not sure.
$endgroup$
– NicNic8
Jan 9 at 4:59
add a comment |
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1
$begingroup$
See this question, or perhaps this one?
$endgroup$
– Dietrich Burde
Jan 7 at 20:35
1
$begingroup$
when you changed variables, you didn't change the interval of integration. $infty -> - infty$ and vice versa
$endgroup$
– Dunham
Jan 7 at 20:36
1
$begingroup$
Okay, but what happens to the limits of integration at the same time?
$endgroup$
– Doug M
Jan 7 at 20:38