Mixing
Convolution with Gaussian
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2
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Let $f, gin mathcal{S}(mathbb R)$ (Schwartz class function), $delta_0$ (dirac delta distribution).
Consider distribution as follows:
$$G(x, y)= f(x)g(x)delta_0(y)-f(y)g(y)delta_0(x), (x, yin mathbb R)$$
Let $h(x,y)= e^{-(x^2+y^2)}.$
My Question is:
Can we expect that $Gast h in L^{1}(mathbb R^2)$?
where $ast$ denotes the convolution.
functional-analysis distribution-theory convolution
asked yesterday
Math Learner
3109
|
show 5 more comments
up vote
2
down vote
favorite
Let $f, gin mathcal{S}(mathbb R)$ (Schwartz class function), $delta_0$ (dirac delta distribution).
Consider distribution as follows:
$$G(x, y)= f(x)g(x)delta_0(y)-f(y)g(y)delta_0(x), (x, yin mathbb R)$$
Let $h(x,y)= e^{-(x^2+y^2)}.$
My Question is:
Can we expect that $Gast h in L^{1}(mathbb R^2)$?
where $ast$ denotes the convolution.
functional-analysis distribution-theory convolution
asked yesterday
Math Learner
3109
Do you mean $Gast h$?
– Marco
yesterday
@Marco: Yes. Thanks. I'll correct the typo.
– Math Learner
yesterday
Also, could you please be more precise about how the distribution $G$ acts on Schwarz functions? For example for a writing like $f(x)delta_0(y)$ you mean that $leftlangle f(x)delta_0(y), phi(x,y)rightrangle=?$
– Marco
yesterday
BTW it is a bit strange to formulate the question as you did instead of replacing $fg$ by a single Schwartz function. This is because every Schwartz function can be written as $fg$ for some Schwartz functions $f,g$.
– Abdelmalek Abdesselam
yesterday
@AbdelmalekAbdesselam I'd say if $T in S', varphi in S$ then $f mapsto f ast (varphi T)$ is a continuous map $S to S$. Not sure by which argument. When replacing $S$ by $C^infty_c$ it is obvious.
– reuns
yesterday
|
show 5 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f, gin mathcal{S}(mathbb R)$ (Schwartz class function), $delta_0$ (dirac delta distribution).
Consider distribution as follows:
$$G(x, y)= f(x)g(x)delta_0(y)-f(y)g(y)delta_0(x), (x, yin mathbb R)$$
Let $h(x,y)= e^{-(x^2+y^2)}.$
My Question is:
Can we expect that $Gast h in L^{1}(mathbb R^2)$?
where $ast$ denotes the convolution.
functional-analysis distribution-theory convolution
asked yesterday
Math Learner
3109
Let $f, gin mathcal{S}(mathbb R)$ (Schwartz class function), $delta_0$ (dirac delta distribution).
Consider distribution as follows:
$$G(x, y)= f(x)g(x)delta_0(y)-f(y)g(y)delta_0(x), (x, yin mathbb R)$$
Let $h(x,y)= e^{-(x^2+y^2)}.$
My Question is:
Can we expect that $Gast h in L^{1}(mathbb R^2)$?
where $ast$ denotes the convolution.
functional-analysis distribution-theory convolution
functional-analysis distribution-theory convolution
asked yesterday
Math Learner
3109
asked yesterday
Math Learner
3109
edited yesterday
asked yesterday
Math Learner
3109
asked yesterday
Math Learner
3109
asked yesterday
Math Learner
3109
3109
Do you mean $Gast h$?
– Marco
yesterday
@Marco: Yes. Thanks. I'll correct the typo.
– Math Learner
yesterday
Also, could you please be more precise about how the distribution $G$ acts on Schwarz functions? For example for a writing like $f(x)delta_0(y)$ you mean that $leftlangle f(x)delta_0(y), phi(x,y)rightrangle=?$
– Marco
yesterday
BTW it is a bit strange to formulate the question as you did instead of replacing $fg$ by a single Schwartz function. This is because every Schwartz function can be written as $fg$ for some Schwartz functions $f,g$.
– Abdelmalek Abdesselam
yesterday
@AbdelmalekAbdesselam I'd say if $T in S', varphi in S$ then $f mapsto f ast (varphi T)$ is a continuous map $S to S$. Not sure by which argument. When replacing $S$ by $C^infty_c$ it is obvious.
– reuns
yesterday
|
show 5 more comments
Do you mean $Gast h$?
– Marco
yesterday
@Marco: Yes. Thanks. I'll correct the typo.
– Math Learner
yesterday
Also, could you please be more precise about how the distribution $G$ acts on Schwarz functions? For example for a writing like $f(x)delta_0(y)$ you mean that $leftlangle f(x)delta_0(y), phi(x,y)rightrangle=?$
– Marco
yesterday
BTW it is a bit strange to formulate the question as you did instead of replacing $fg$ by a single Schwartz function. This is because every Schwartz function can be written as $fg$ for some Schwartz functions $f,g$.
– Abdelmalek Abdesselam
yesterday
@AbdelmalekAbdesselam I'd say if $T in S', varphi in S$ then $f mapsto f ast (varphi T)$ is a continuous map $S to S$. Not sure by which argument. When replacing $S$ by $C^infty_c$ it is obvious.
– reuns
yesterday
Do you mean $Gast h$?
– Marco
yesterday
Do you mean $Gast h$?
– Marco
yesterday
@Marco: Yes. Thanks. I'll correct the typo.
– Math Learner
yesterday
@Marco: Yes. Thanks. I'll correct the typo.
– Math Learner
yesterday
Also, could you please be more precise about how the distribution $G$ acts on Schwarz functions? For example for a writing like $f(x)delta_0(y)$ you mean that $leftlangle f(x)delta_0(y), phi(x,y)rightrangle=?$
– Marco
yesterday
Also, could you please be more precise about how the distribution $G$ acts on Schwarz functions? For example for a writing like $f(x)delta_0(y)$ you mean that $leftlangle f(x)delta_0(y), phi(x,y)rightrangle=?$
– Marco
yesterday
BTW it is a bit strange to formulate the question as you did instead of replacing $fg$ by a single Schwartz function. This is because every Schwartz function can be written as $fg$ for some Schwartz functions $f,g$.
– Abdelmalek Abdesselam
yesterday
BTW it is a bit strange to formulate the question as you did instead of replacing $fg$ by a single Schwartz function. This is because every Schwartz function can be written as $fg$ for some Schwartz functions $f,g$.
– Abdelmalek Abdesselam
yesterday
@AbdelmalekAbdesselam I'd say if $T in S', varphi in S$ then $f mapsto f ast (varphi T)$ is a continuous map $S to S$. Not sure by which argument. When replacing $S$ by $C^infty_c$ it is obvious.
– reuns
yesterday
@AbdelmalekAbdesselam I'd say if $T in S', varphi in S$ then $f mapsto f ast (varphi T)$ is a continuous map $S to S$. Not sure by which argument. When replacing $S$ by $C^infty_c$ it is obvious.
– reuns
yesterday
|
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
You can just do the explicit computation:
$$
(Gast h)(x,y)=intint G(u,v)h(x-u,y-v)dudv
$$
$$
=intint f(u)g(u)delta(v)h(x-u,y-v)dudv-
intint f(v)g(v)delta(u)h(x-u,y-v)dudv
$$
$$
=int f(u)g(u)h(x-u,y)du-int f(v)g(v)h(x,y-v)dv=gamma(y)((fg)astgamma)(x)-
gamma(x)((fg)astgamma)(y)
$$
where $gamma(z)=e^{-z^2}$ which is in $mathcal{S}(mathbb{R})$. The latter is stable by product and convolution. Moreover, the product of a Schwartz function in $x$ and a Schwartz function in $y$ is in $mathcal{S}(mathbb{R}^2)$. So the result is not only in $L^1(mathbb{R}^2)$ but in fact in $mathcal{S}(mathbb{R}^2)$.
answered yesterday
Abdelmalek Abdesselam
386110
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You can just do the explicit computation:
$$
(Gast h)(x,y)=intint G(u,v)h(x-u,y-v)dudv
$$
$$
=intint f(u)g(u)delta(v)h(x-u,y-v)dudv-
intint f(v)g(v)delta(u)h(x-u,y-v)dudv
$$
$$
=int f(u)g(u)h(x-u,y)du-int f(v)g(v)h(x,y-v)dv=gamma(y)((fg)astgamma)(x)-
gamma(x)((fg)astgamma)(y)
$$
where $gamma(z)=e^{-z^2}$ which is in $mathcal{S}(mathbb{R})$. The latter is stable by product and convolution. Moreover, the product of a Schwartz function in $x$ and a Schwartz function in $y$ is in $mathcal{S}(mathbb{R}^2)$. So the result is not only in $L^1(mathbb{R}^2)$ but in fact in $mathcal{S}(mathbb{R}^2)$.
answered yesterday
Abdelmalek Abdesselam
386110
add a comment |
up vote
2
down vote
You can just do the explicit computation:
$$
(Gast h)(x,y)=intint G(u,v)h(x-u,y-v)dudv
$$
$$
=intint f(u)g(u)delta(v)h(x-u,y-v)dudv-
intint f(v)g(v)delta(u)h(x-u,y-v)dudv
$$
$$
=int f(u)g(u)h(x-u,y)du-int f(v)g(v)h(x,y-v)dv=gamma(y)((fg)astgamma)(x)-
gamma(x)((fg)astgamma)(y)
$$
where $gamma(z)=e^{-z^2}$ which is in $mathcal{S}(mathbb{R})$. The latter is stable by product and convolution. Moreover, the product of a Schwartz function in $x$ and a Schwartz function in $y$ is in $mathcal{S}(mathbb{R}^2)$. So the result is not only in $L^1(mathbb{R}^2)$ but in fact in $mathcal{S}(mathbb{R}^2)$.
answered yesterday
Abdelmalek Abdesselam
386110
add a comment |
up vote
2
down vote
up vote
2
down vote
You can just do the explicit computation:
$$
(Gast h)(x,y)=intint G(u,v)h(x-u,y-v)dudv
$$
$$
=intint f(u)g(u)delta(v)h(x-u,y-v)dudv-
intint f(v)g(v)delta(u)h(x-u,y-v)dudv
$$
$$
=int f(u)g(u)h(x-u,y)du-int f(v)g(v)h(x,y-v)dv=gamma(y)((fg)astgamma)(x)-
gamma(x)((fg)astgamma)(y)
$$
where $gamma(z)=e^{-z^2}$ which is in $mathcal{S}(mathbb{R})$. The latter is stable by product and convolution. Moreover, the product of a Schwartz function in $x$ and a Schwartz function in $y$ is in $mathcal{S}(mathbb{R}^2)$. So the result is not only in $L^1(mathbb{R}^2)$ but in fact in $mathcal{S}(mathbb{R}^2)$.
answered yesterday
Abdelmalek Abdesselam
386110
You can just do the explicit computation:
$$
(Gast h)(x,y)=intint G(u,v)h(x-u,y-v)dudv
$$
$$
=intint f(u)g(u)delta(v)h(x-u,y-v)dudv-
intint f(v)g(v)delta(u)h(x-u,y-v)dudv
$$
$$
=int f(u)g(u)h(x-u,y)du-int f(v)g(v)h(x,y-v)dv=gamma(y)((fg)astgamma)(x)-
gamma(x)((fg)astgamma)(y)
$$
where $gamma(z)=e^{-z^2}$ which is in $mathcal{S}(mathbb{R})$. The latter is stable by product and convolution. Moreover, the product of a Schwartz function in $x$ and a Schwartz function in $y$ is in $mathcal{S}(mathbb{R}^2)$. So the result is not only in $L^1(mathbb{R}^2)$ but in fact in $mathcal{S}(mathbb{R}^2)$.
answered yesterday
Abdelmalek Abdesselam
386110
edited yesterday
answered yesterday
Abdelmalek Abdesselam
386110
answered yesterday
Abdelmalek Abdesselam
386110
answered yesterday
Abdelmalek Abdesselam
386110
386110
add a comment |
add a comment |
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Do you mean $Gast h$?
– Marco
yesterday
@Marco: Yes. Thanks. I'll correct the typo.
– Math Learner
yesterday
Also, could you please be more precise about how the distribution $G$ acts on Schwarz functions? For example for a writing like $f(x)delta_0(y)$ you mean that $leftlangle f(x)delta_0(y), phi(x,y)rightrangle=?$
– Marco
yesterday
BTW it is a bit strange to formulate the question as you did instead of replacing $fg$ by a single Schwartz function. This is because every Schwartz function can be written as $fg$ for some Schwartz functions $f,g$.
– Abdelmalek Abdesselam
yesterday
@AbdelmalekAbdesselam I'd say if $T in S', varphi in S$ then $f mapsto f ast (varphi T)$ is a continuous map $S to S$. Not sure by which argument. When replacing $S$ by $C^infty_c$ it is obvious.
– reuns
yesterday