Proof of $int_{mathbb{R}^d} f(x) dx = prod_{i = 1}^{d} int_{mathbb{R}} phi_i(x_i) dx_i$, where $f :=...
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Let $phi_1, ldots phi_d: mathbb{R} to mathbb{R}$ be continuos and have compact support.
Show that for
$$
f := bigotimes_{i = 1}^{d} phi_i: mathbb{R}^d to mathbb{R},
(x_1, ldots, x_d) mapsto prod_{i = 1}^{d} phi_i(x_i)
$$
we have
$$
int_{mathbb{R}^d} f(x) dx = prod_{i = 1}^{d} int_{mathbb{R}} phi_i(x_i) dx_i
$$
Proof
Since $f$ is continuous and has a compact support as well, we get
begin{align*}
int_{mathbb{R}^d} f(x) dx
& = int_{text{supp}(f)} prod_{i = 1}^{d} phi_i(x_i) dx
overset{(ast)}{=} int_{text{supp}(phi_d)} left( ldots left( int_{text{supp}(phi_1)} prod_{i = 1}^{d} phi_i(x_i) dx_1 right) ldots right) d x_d \
& overset{(star)}{=} left( int_{text{supp}(phi_1)} phi_1(x_1) dx_1 right) left( int_{text{supp}(phi_d)} left( ldots left( int_{text{supp}(phi_1)} prod_{i = 2}^{d} phi_i(x_i) dx_2 right) ldots right) d x_d right) \
& = ldots
= prod_{i = 1}^{d} int_{mathbb{R}} phi_i(x_i) dx_i
end{align*}
My Question
I don't understand the steps $(ast)$ and especially $(star)$ . Can we just ''pull out'' a one-dimensional integral since it's just a constant?
The proof is from Otto Forster: Analyis 3 (in German) on page 3.
This was a homework assignment and the corrector gave full marks to this answer but suggested using induction instead of the $ldots$-step.
Could someone please show that induction in light of the question above?
Please bear in mind that this is from the very beginning of real analysis III and we have not learned measure theory yet.
real-analysis integration functional-analysis measure-theory
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up vote
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Let $phi_1, ldots phi_d: mathbb{R} to mathbb{R}$ be continuos and have compact support.
Show that for
$$
f := bigotimes_{i = 1}^{d} phi_i: mathbb{R}^d to mathbb{R},
(x_1, ldots, x_d) mapsto prod_{i = 1}^{d} phi_i(x_i)
$$
we have
$$
int_{mathbb{R}^d} f(x) dx = prod_{i = 1}^{d} int_{mathbb{R}} phi_i(x_i) dx_i
$$
Proof
Since $f$ is continuous and has a compact support as well, we get
begin{align*}
int_{mathbb{R}^d} f(x) dx
& = int_{text{supp}(f)} prod_{i = 1}^{d} phi_i(x_i) dx
overset{(ast)}{=} int_{text{supp}(phi_d)} left( ldots left( int_{text{supp}(phi_1)} prod_{i = 1}^{d} phi_i(x_i) dx_1 right) ldots right) d x_d \
& overset{(star)}{=} left( int_{text{supp}(phi_1)} phi_1(x_1) dx_1 right) left( int_{text{supp}(phi_d)} left( ldots left( int_{text{supp}(phi_1)} prod_{i = 2}^{d} phi_i(x_i) dx_2 right) ldots right) d x_d right) \
& = ldots
= prod_{i = 1}^{d} int_{mathbb{R}} phi_i(x_i) dx_i
end{align*}
My Question
I don't understand the steps $(ast)$ and especially $(star)$ . Can we just ''pull out'' a one-dimensional integral since it's just a constant?
The proof is from Otto Forster: Analyis 3 (in German) on page 3.
This was a homework assignment and the corrector gave full marks to this answer but suggested using induction instead of the $ldots$-step.
Could someone please show that induction in light of the question above?
Please bear in mind that this is from the very beginning of real analysis III and we have not learned measure theory yet.
real-analysis integration functional-analysis measure-theory
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $phi_1, ldots phi_d: mathbb{R} to mathbb{R}$ be continuos and have compact support.
Show that for
$$
f := bigotimes_{i = 1}^{d} phi_i: mathbb{R}^d to mathbb{R},
(x_1, ldots, x_d) mapsto prod_{i = 1}^{d} phi_i(x_i)
$$
we have
$$
int_{mathbb{R}^d} f(x) dx = prod_{i = 1}^{d} int_{mathbb{R}} phi_i(x_i) dx_i
$$
Proof
Since $f$ is continuous and has a compact support as well, we get
begin{align*}
int_{mathbb{R}^d} f(x) dx
& = int_{text{supp}(f)} prod_{i = 1}^{d} phi_i(x_i) dx
overset{(ast)}{=} int_{text{supp}(phi_d)} left( ldots left( int_{text{supp}(phi_1)} prod_{i = 1}^{d} phi_i(x_i) dx_1 right) ldots right) d x_d \
& overset{(star)}{=} left( int_{text{supp}(phi_1)} phi_1(x_1) dx_1 right) left( int_{text{supp}(phi_d)} left( ldots left( int_{text{supp}(phi_1)} prod_{i = 2}^{d} phi_i(x_i) dx_2 right) ldots right) d x_d right) \
& = ldots
= prod_{i = 1}^{d} int_{mathbb{R}} phi_i(x_i) dx_i
end{align*}
My Question
I don't understand the steps $(ast)$ and especially $(star)$ . Can we just ''pull out'' a one-dimensional integral since it's just a constant?
The proof is from Otto Forster: Analyis 3 (in German) on page 3.
This was a homework assignment and the corrector gave full marks to this answer but suggested using induction instead of the $ldots$-step.
Could someone please show that induction in light of the question above?
Please bear in mind that this is from the very beginning of real analysis III and we have not learned measure theory yet.
real-analysis integration functional-analysis measure-theory
Let $phi_1, ldots phi_d: mathbb{R} to mathbb{R}$ be continuos and have compact support.
Show that for
$$
f := bigotimes_{i = 1}^{d} phi_i: mathbb{R}^d to mathbb{R},
(x_1, ldots, x_d) mapsto prod_{i = 1}^{d} phi_i(x_i)
$$
we have
$$
int_{mathbb{R}^d} f(x) dx = prod_{i = 1}^{d} int_{mathbb{R}} phi_i(x_i) dx_i
$$
Proof
Since $f$ is continuous and has a compact support as well, we get
begin{align*}
int_{mathbb{R}^d} f(x) dx
& = int_{text{supp}(f)} prod_{i = 1}^{d} phi_i(x_i) dx
overset{(ast)}{=} int_{text{supp}(phi_d)} left( ldots left( int_{text{supp}(phi_1)} prod_{i = 1}^{d} phi_i(x_i) dx_1 right) ldots right) d x_d \
& overset{(star)}{=} left( int_{text{supp}(phi_1)} phi_1(x_1) dx_1 right) left( int_{text{supp}(phi_d)} left( ldots left( int_{text{supp}(phi_1)} prod_{i = 2}^{d} phi_i(x_i) dx_2 right) ldots right) d x_d right) \
& = ldots
= prod_{i = 1}^{d} int_{mathbb{R}} phi_i(x_i) dx_i
end{align*}
My Question
I don't understand the steps $(ast)$ and especially $(star)$ . Can we just ''pull out'' a one-dimensional integral since it's just a constant?
The proof is from Otto Forster: Analyis 3 (in German) on page 3.
This was a homework assignment and the corrector gave full marks to this answer but suggested using induction instead of the $ldots$-step.
Could someone please show that induction in light of the question above?
Please bear in mind that this is from the very beginning of real analysis III and we have not learned measure theory yet.
real-analysis integration functional-analysis measure-theory
real-analysis integration functional-analysis measure-theory
asked Nov 17 at 21:50
Viktor Glombik
489321
489321
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Yes, you can pull out $intphi_1(x_1), dx_1$ because it is a constant. To use induction simply note that when you pull out $intphi_1(x_1), dx_1$ you are left with a similar integral with $n-1$ variables instead of $n$. By induction hypothesis this integral w.r.t $n-1$ variables can be written as a product of $n-1$ integrals.
ok what about step $(ast)$?
– Viktor Glombik
Nov 18 at 9:15
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Yes, you can pull out $intphi_1(x_1), dx_1$ because it is a constant. To use induction simply note that when you pull out $intphi_1(x_1), dx_1$ you are left with a similar integral with $n-1$ variables instead of $n$. By induction hypothesis this integral w.r.t $n-1$ variables can be written as a product of $n-1$ integrals.
ok what about step $(ast)$?
– Viktor Glombik
Nov 18 at 9:15
add a comment |
up vote
1
down vote
Yes, you can pull out $intphi_1(x_1), dx_1$ because it is a constant. To use induction simply note that when you pull out $intphi_1(x_1), dx_1$ you are left with a similar integral with $n-1$ variables instead of $n$. By induction hypothesis this integral w.r.t $n-1$ variables can be written as a product of $n-1$ integrals.
ok what about step $(ast)$?
– Viktor Glombik
Nov 18 at 9:15
add a comment |
up vote
1
down vote
up vote
1
down vote
Yes, you can pull out $intphi_1(x_1), dx_1$ because it is a constant. To use induction simply note that when you pull out $intphi_1(x_1), dx_1$ you are left with a similar integral with $n-1$ variables instead of $n$. By induction hypothesis this integral w.r.t $n-1$ variables can be written as a product of $n-1$ integrals.
Yes, you can pull out $intphi_1(x_1), dx_1$ because it is a constant. To use induction simply note that when you pull out $intphi_1(x_1), dx_1$ you are left with a similar integral with $n-1$ variables instead of $n$. By induction hypothesis this integral w.r.t $n-1$ variables can be written as a product of $n-1$ integrals.
answered Nov 17 at 23:37
Kavi Rama Murthy
43.5k31751
43.5k31751
ok what about step $(ast)$?
– Viktor Glombik
Nov 18 at 9:15
add a comment |
ok what about step $(ast)$?
– Viktor Glombik
Nov 18 at 9:15
ok what about step $(ast)$?
– Viktor Glombik
Nov 18 at 9:15
ok what about step $(ast)$?
– Viktor Glombik
Nov 18 at 9:15
add a comment |
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