Proof of $int_{mathbb{R}^d} f(x) dx = prod_{i = 1}^{d} int_{mathbb{R}} phi_i(x_i) dx_i$, where $f :=...












0















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







share|cite|improve this question















This question has an open bounty worth +50
reputation from Viktor Glombik ending in 4 days.


This question has not received enough attention.


One of my questions (star) was answered, (ast) not.
















  • I have edited my answer to cover both the questions.
    – Kavi Rama Murthy
    2 days ago
















0















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







share|cite|improve this question















This question has an open bounty worth +50
reputation from Viktor Glombik ending in 4 days.


This question has not received enough attention.


One of my questions (star) was answered, (ast) not.
















  • I have edited my answer to cover both the questions.
    – Kavi Rama Murthy
    2 days ago














0












0








0








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







share|cite|improve this question














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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 17 at 21:50









Viktor Glombik

574425




574425






This question has an open bounty worth +50
reputation from Viktor Glombik ending in 4 days.


This question has not received enough attention.


One of my questions (star) was answered, (ast) not.








This question has an open bounty worth +50
reputation from Viktor Glombik ending in 4 days.


This question has not received enough attention.


One of my questions (star) was answered, (ast) not.














  • I have edited my answer to cover both the questions.
    – Kavi Rama Murthy
    2 days ago


















  • I have edited my answer to cover both the questions.
    – Kavi Rama Murthy
    2 days ago
















I have edited my answer to cover both the questions.
– Kavi Rama Murthy
2 days ago




I have edited my answer to cover both the questions.
– Kavi Rama Murthy
2 days ago










1 Answer
1






active

oldest

votes


















1














The first step follows by a theorem on multiple integrals. Note that we can enclose the domain of integration in a set of the form $[a_1,b_1]times [a_2,b_2]times cdots times [a_n,b_n]$. In Apostol's book there is a section of 'Evaluation of multiple integral by repeated integration'. The theorem proved here for two variable case extends to any number of variables. If you apply that theorem you will get ($ast$). About ($star$): 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.






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    The first step follows by a theorem on multiple integrals. Note that we can enclose the domain of integration in a set of the form $[a_1,b_1]times [a_2,b_2]times cdots times [a_n,b_n]$. In Apostol's book there is a section of 'Evaluation of multiple integral by repeated integration'. The theorem proved here for two variable case extends to any number of variables. If you apply that theorem you will get ($ast$). About ($star$): 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.






    share|cite|improve this answer




























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      The first step follows by a theorem on multiple integrals. Note that we can enclose the domain of integration in a set of the form $[a_1,b_1]times [a_2,b_2]times cdots times [a_n,b_n]$. In Apostol's book there is a section of 'Evaluation of multiple integral by repeated integration'. The theorem proved here for two variable case extends to any number of variables. If you apply that theorem you will get ($ast$). About ($star$): 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.






      share|cite|improve this answer


























        1












        1








        1






        The first step follows by a theorem on multiple integrals. Note that we can enclose the domain of integration in a set of the form $[a_1,b_1]times [a_2,b_2]times cdots times [a_n,b_n]$. In Apostol's book there is a section of 'Evaluation of multiple integral by repeated integration'. The theorem proved here for two variable case extends to any number of variables. If you apply that theorem you will get ($ast$). About ($star$): 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.






        share|cite|improve this answer














        The first step follows by a theorem on multiple integrals. Note that we can enclose the domain of integration in a set of the form $[a_1,b_1]times [a_2,b_2]times cdots times [a_n,b_n]$. In Apostol's book there is a section of 'Evaluation of multiple integral by repeated integration'. The theorem proved here for two variable case extends to any number of variables. If you apply that theorem you will get ($ast$). About ($star$): 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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered Nov 17 at 23:37









        Kavi Rama Murthy

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