Convolution with Gaussian
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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 Nov 19 at 15:20
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 Nov 19 at 15:20
Math Learner
3109
Do you mean $Gast h$?
– Marco
Nov 19 at 16:51
@Marco: Yes. Thanks. I'll correct the typo.
– Math Learner
Nov 19 at 16:55
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
Nov 19 at 17:07
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
Nov 19 at 17:12
@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
Nov 19 at 20:03
|
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 Nov 19 at 15:20
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 Nov 19 at 15:20
Math Learner
3109
asked Nov 19 at 15:20
Math Learner
3109
edited Nov 19 at 16:55
asked Nov 19 at 15:20
Math Learner
3109
asked Nov 19 at 15:20
Math Learner
3109
asked Nov 19 at 15:20
Math Learner
3109
3109
Do you mean $Gast h$?
– Marco
Nov 19 at 16:51
@Marco: Yes. Thanks. I'll correct the typo.
– Math Learner
Nov 19 at 16:55
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
Nov 19 at 17:07
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
Nov 19 at 17:12
@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
Nov 19 at 20:03
|
show 5 more comments
Do you mean $Gast h$?
– Marco
Nov 19 at 16:51
@Marco: Yes. Thanks. I'll correct the typo.
– Math Learner
Nov 19 at 16:55
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
Nov 19 at 17:07
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
Nov 19 at 17:12
@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
Nov 19 at 20:03
Do you mean $Gast h$?
– Marco
Nov 19 at 16:51
Do you mean $Gast h$?
– Marco
Nov 19 at 16:51
@Marco: Yes. Thanks. I'll correct the typo.
– Math Learner
Nov 19 at 16:55
@Marco: Yes. Thanks. I'll correct the typo.
– Math Learner
Nov 19 at 16:55
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
Nov 19 at 17:07
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
Nov 19 at 17:07
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
Nov 19 at 17:12
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
Nov 19 at 17:12
@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
Nov 19 at 20:03
@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
Nov 19 at 20:03
|
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
3
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 Nov 19 at 17:07
Abdelmalek Abdesselam
396110
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
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 Nov 19 at 17:07
Abdelmalek Abdesselam
396110
add a comment |
up vote
3
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 Nov 19 at 17:07
Abdelmalek Abdesselam
396110
add a comment |
up vote
3
down vote
up vote
3
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 Nov 19 at 17:07
Abdelmalek Abdesselam
396110
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 Nov 19 at 17:07
Abdelmalek Abdesselam
396110
edited Nov 19 at 19:48
answered Nov 19 at 17:07
Abdelmalek Abdesselam
396110
answered Nov 19 at 17:07
Abdelmalek Abdesselam
396110
answered Nov 19 at 17:07
Abdelmalek Abdesselam
396110
396110
add a comment |
add a comment |
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Do you mean $Gast h$?
– Marco
Nov 19 at 16:51
@Marco: Yes. Thanks. I'll correct the typo.
– Math Learner
Nov 19 at 16:55
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
Nov 19 at 17:07
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
Nov 19 at 17:12
@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
Nov 19 at 20:03