Mixing
Artificial unity for the convolution.
Clash Royale CLAN TAG#URR8PPP
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It is not hard to show, that $(L^1(mathbb R),*)$, where $(f*g)(t)=int_mathbb R f(t-s)g(s)ds$ has no unity element.
Consider therefore the functions $phi_epsilon=I_[-epsilon,epsilon]$. If $e$ is the unit, we get for almost all $tin[-epsilon,epsilon]: 1=(e*phi_epsilon)(t)=int_[-epsilon,epsilon] e(t-s)ds$, which converges to zero for $epsilon rightarrow 0$.
My question is the following: $e$ seems to "want to be" a dirac delta but cannot. Is is possible to embed $L^1(mathbb R)$ into a bigger space, which contains something similar to a dirac delta, which gives me a unit?
convolution dirac-delta
asked Jul 24 at 11:40
Janis Papewalis
132
add a comment |Â
up vote
0
down vote
favorite
It is not hard to show, that $(L^1(mathbb R),*)$, where $(f*g)(t)=int_mathbb R f(t-s)g(s)ds$ has no unity element.
Consider therefore the functions $phi_epsilon=I_[-epsilon,epsilon]$. If $e$ is the unit, we get for almost all $tin[-epsilon,epsilon]: 1=(e*phi_epsilon)(t)=int_[-epsilon,epsilon] e(t-s)ds$, which converges to zero for $epsilon rightarrow 0$.
My question is the following: $e$ seems to "want to be" a dirac delta but cannot. Is is possible to embed $L^1(mathbb R)$ into a bigger space, which contains something similar to a dirac delta, which gives me a unit?
convolution dirac-delta
asked Jul 24 at 11:40
Janis Papewalis
132
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
It is not hard to show, that $(L^1(mathbb R),*)$, where $(f*g)(t)=int_mathbb R f(t-s)g(s)ds$ has no unity element.
Consider therefore the functions $phi_epsilon=I_[-epsilon,epsilon]$. If $e$ is the unit, we get for almost all $tin[-epsilon,epsilon]: 1=(e*phi_epsilon)(t)=int_[-epsilon,epsilon] e(t-s)ds$, which converges to zero for $epsilon rightarrow 0$.
My question is the following: $e$ seems to "want to be" a dirac delta but cannot. Is is possible to embed $L^1(mathbb R)$ into a bigger space, which contains something similar to a dirac delta, which gives me a unit?
convolution dirac-delta
asked Jul 24 at 11:40
Janis Papewalis
132
It is not hard to show, that $(L^1(mathbb R),*)$, where $(f*g)(t)=int_mathbb R f(t-s)g(s)ds$ has no unity element.
Consider therefore the functions $phi_epsilon=I_[-epsilon,epsilon]$. If $e$ is the unit, we get for almost all $tin[-epsilon,epsilon]: 1=(e*phi_epsilon)(t)=int_[-epsilon,epsilon] e(t-s)ds$, which converges to zero for $epsilon rightarrow 0$.
My question is the following: $e$ seems to "want to be" a dirac delta but cannot. Is is possible to embed $L^1(mathbb R)$ into a bigger space, which contains something similar to a dirac delta, which gives me a unit?
convolution dirac-delta
asked Jul 24 at 11:40
Janis Papewalis
132
asked Jul 24 at 11:40
Janis Papewalis
132
asked Jul 24 at 11:40
Janis Papewalis
132
asked Jul 24 at 11:40
Janis Papewalis
132
132
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
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up vote
1
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Surely. You can thin of elements of $L^1$ as (real or complex )measures and convolution of functions in $L^1$ is a special case of convolution of measures. The delta function is a unit element for measures under convolution. (To think of an $L^1$ function $f$ as mesure just consider $dmu (x) =f(x),dx$).
answered Jul 24 at 11:45
Kavi Rama Murthy
20.2k2829
Thank you, helped very much. I just don't see yet, why $int_Eint_mathbb Rf(x-y)dmu_g(y)dx=int_Eint_mathbb Rf(x-y)g(y)dydx$.
– Janis Papewalis
Jul 24 at 12:13
ah its clear, algebraic induction.
– Janis Papewalis
Jul 24 at 12:22
One can also embed $L^1$ in the space of distributions $mathscr D'.$
– md2perpe
Jul 24 at 13:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Surely. You can thin of elements of $L^1$ as (real or complex )measures and convolution of functions in $L^1$ is a special case of convolution of measures. The delta function is a unit element for measures under convolution. (To think of an $L^1$ function $f$ as mesure just consider $dmu (x) =f(x),dx$).
answered Jul 24 at 11:45
Kavi Rama Murthy
20.2k2829
Thank you, helped very much. I just don't see yet, why $int_Eint_mathbb Rf(x-y)dmu_g(y)dx=int_Eint_mathbb Rf(x-y)g(y)dydx$.
– Janis Papewalis
Jul 24 at 12:13
ah its clear, algebraic induction.
– Janis Papewalis
Jul 24 at 12:22
One can also embed $L^1$ in the space of distributions $mathscr D'.$
– md2perpe
Jul 24 at 13:07
add a comment |Â
up vote
1
down vote
accepted
Surely. You can thin of elements of $L^1$ as (real or complex )measures and convolution of functions in $L^1$ is a special case of convolution of measures. The delta function is a unit element for measures under convolution. (To think of an $L^1$ function $f$ as mesure just consider $dmu (x) =f(x),dx$).
answered Jul 24 at 11:45
Kavi Rama Murthy
20.2k2829
Thank you, helped very much. I just don't see yet, why $int_Eint_mathbb Rf(x-y)dmu_g(y)dx=int_Eint_mathbb Rf(x-y)g(y)dydx$.
– Janis Papewalis
Jul 24 at 12:13
ah its clear, algebraic induction.
– Janis Papewalis
Jul 24 at 12:22
One can also embed $L^1$ in the space of distributions $mathscr D'.$
– md2perpe
Jul 24 at 13:07
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Surely. You can thin of elements of $L^1$ as (real or complex )measures and convolution of functions in $L^1$ is a special case of convolution of measures. The delta function is a unit element for measures under convolution. (To think of an $L^1$ function $f$ as mesure just consider $dmu (x) =f(x),dx$).
answered Jul 24 at 11:45
Kavi Rama Murthy
20.2k2829
Surely. You can thin of elements of $L^1$ as (real or complex )measures and convolution of functions in $L^1$ is a special case of convolution of measures. The delta function is a unit element for measures under convolution. (To think of an $L^1$ function $f$ as mesure just consider $dmu (x) =f(x),dx$).
answered Jul 24 at 11:45
Kavi Rama Murthy
20.2k2829
answered Jul 24 at 11:45
Kavi Rama Murthy
20.2k2829
answered Jul 24 at 11:45
Kavi Rama Murthy
20.2k2829
answered Jul 24 at 11:45
Kavi Rama Murthy
20.2k2829
20.2k2829
Thank you, helped very much. I just don't see yet, why $int_Eint_mathbb Rf(x-y)dmu_g(y)dx=int_Eint_mathbb Rf(x-y)g(y)dydx$.
– Janis Papewalis
Jul 24 at 12:13
ah its clear, algebraic induction.
– Janis Papewalis
Jul 24 at 12:22
One can also embed $L^1$ in the space of distributions $mathscr D'.$
– md2perpe
Jul 24 at 13:07
add a comment |Â
Thank you, helped very much. I just don't see yet, why $int_Eint_mathbb Rf(x-y)dmu_g(y)dx=int_Eint_mathbb Rf(x-y)g(y)dydx$.
– Janis Papewalis
Jul 24 at 12:13
ah its clear, algebraic induction.
– Janis Papewalis
Jul 24 at 12:22
One can also embed $L^1$ in the space of distributions $mathscr D'.$
– md2perpe
Jul 24 at 13:07
Thank you, helped very much. I just don't see yet, why $int_Eint_mathbb Rf(x-y)dmu_g(y)dx=int_Eint_mathbb Rf(x-y)g(y)dydx$.
– Janis Papewalis
Jul 24 at 12:13
Thank you, helped very much. I just don't see yet, why $int_Eint_mathbb Rf(x-y)dmu_g(y)dx=int_Eint_mathbb Rf(x-y)g(y)dydx$.
– Janis Papewalis
Jul 24 at 12:13
ah its clear, algebraic induction.
– Janis Papewalis
Jul 24 at 12:22
ah its clear, algebraic induction.
– Janis Papewalis
Jul 24 at 12:22
One can also embed $L^1$ in the space of distributions $mathscr D'.$
– md2perpe
Jul 24 at 13:07
One can also embed $L^1$ in the space of distributions $mathscr D'.$
– md2perpe
Jul 24 at 13:07
add a comment |Â
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