Commutative Convolution. Problem 26 Royden 2 ed.












1












$begingroup$


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 $int f(y-x)g(x)dx=int f(z)g(y-z)(-dz)=-gast f$










share|cite|improve this question









$endgroup$








  • 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












$begingroup$


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 $int f(y-x)g(x)dx=int f(z)g(y-z)(-dz)=-gast f$










share|cite|improve this question









$endgroup$








  • 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








1


1



$begingroup$


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 $int f(y-x)g(x)dx=int f(z)g(y-z)(-dz)=-gast f$










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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
$$ (fast g)(y) = int_{-infty}^{infty} g(y-z) , f(z) , dz = (gast f)(y). $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









2












$begingroup$

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
$$ (fast g)(y) = int_{-infty}^{infty} g(y-z) , f(z) , dz = (gast f)(y). $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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
$$ (fast g)(y) = int_{-infty}^{infty} g(y-z) , f(z) , dz = (gast f)(y). $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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
$$ (fast g)(y) = int_{-infty}^{infty} g(y-z) , f(z) , dz = (gast f)(y). $$






share|cite|improve this answer









$endgroup$



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
$$ (fast g)(y) = int_{-infty}^{infty} g(y-z) , f(z) , dz = (gast f)(y). $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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