$M={(x,y)mapsto sum_{j=1}^{n}f_j(x)g_j(y)}$ Then sum and product of such functions is again in $M$
$begingroup$
Let $M$ be the set of all functions of the type $(x,y)mapsto sum_{j=1}^{n}f_j(x)g_j(y)$ with $f_j in C(X, mathbb{R})$ and $g_j in C(Y, mathbb{R})$ where $X,Y$ are compact topological spaces.
I need to show that the sum and product of two of such functions $a:=sum_{j=1}^{n}f^a_j(x)g^a_j(y)$ and $b:=sum_{i=1}^{m}f^b_i(x)g^b_i(y)$ is again in $M$. Note that $^a$ and $^b$ it just for notational reasons, so $f^a$ doens't mean $f$ multiplied by itself $a$ times
For the product:
$ab=sum_{j=1}^{n}f^a_j(x)g^a_j(y)sum_{i=1}^{m}f^b_i(x)g^b_i(y)=sum_{(j,i) in {1ldots n }times{1ldots m}}f^a_jf_i^b(x)g_j^ag_i^b(y)$
Does this already show that $ab$ is in $M$?
For the sum:
$a+b=sum_{j=1}^{n}f^a_j(x)g^a_j(y)+sum_{i=1}^{m}f^b_i(x)g^b_i(y)$. But how can I write this as one sum?
real-analysis calculus
$endgroup$
add a comment |
$begingroup$
Let $M$ be the set of all functions of the type $(x,y)mapsto sum_{j=1}^{n}f_j(x)g_j(y)$ with $f_j in C(X, mathbb{R})$ and $g_j in C(Y, mathbb{R})$ where $X,Y$ are compact topological spaces.
I need to show that the sum and product of two of such functions $a:=sum_{j=1}^{n}f^a_j(x)g^a_j(y)$ and $b:=sum_{i=1}^{m}f^b_i(x)g^b_i(y)$ is again in $M$. Note that $^a$ and $^b$ it just for notational reasons, so $f^a$ doens't mean $f$ multiplied by itself $a$ times
For the product:
$ab=sum_{j=1}^{n}f^a_j(x)g^a_j(y)sum_{i=1}^{m}f^b_i(x)g^b_i(y)=sum_{(j,i) in {1ldots n }times{1ldots m}}f^a_jf_i^b(x)g_j^ag_i^b(y)$
Does this already show that $ab$ is in $M$?
For the sum:
$a+b=sum_{j=1}^{n}f^a_j(x)g^a_j(y)+sum_{i=1}^{m}f^b_i(x)g^b_i(y)$. But how can I write this as one sum?
real-analysis calculus
$endgroup$
1
$begingroup$
it already is written as a sum, from 1 to (m+n) of products of functions of x times functions of y
$endgroup$
– Sorin Tirc
Dec 20 '18 at 14:49
add a comment |
$begingroup$
Let $M$ be the set of all functions of the type $(x,y)mapsto sum_{j=1}^{n}f_j(x)g_j(y)$ with $f_j in C(X, mathbb{R})$ and $g_j in C(Y, mathbb{R})$ where $X,Y$ are compact topological spaces.
I need to show that the sum and product of two of such functions $a:=sum_{j=1}^{n}f^a_j(x)g^a_j(y)$ and $b:=sum_{i=1}^{m}f^b_i(x)g^b_i(y)$ is again in $M$. Note that $^a$ and $^b$ it just for notational reasons, so $f^a$ doens't mean $f$ multiplied by itself $a$ times
For the product:
$ab=sum_{j=1}^{n}f^a_j(x)g^a_j(y)sum_{i=1}^{m}f^b_i(x)g^b_i(y)=sum_{(j,i) in {1ldots n }times{1ldots m}}f^a_jf_i^b(x)g_j^ag_i^b(y)$
Does this already show that $ab$ is in $M$?
For the sum:
$a+b=sum_{j=1}^{n}f^a_j(x)g^a_j(y)+sum_{i=1}^{m}f^b_i(x)g^b_i(y)$. But how can I write this as one sum?
real-analysis calculus
$endgroup$
Let $M$ be the set of all functions of the type $(x,y)mapsto sum_{j=1}^{n}f_j(x)g_j(y)$ with $f_j in C(X, mathbb{R})$ and $g_j in C(Y, mathbb{R})$ where $X,Y$ are compact topological spaces.
I need to show that the sum and product of two of such functions $a:=sum_{j=1}^{n}f^a_j(x)g^a_j(y)$ and $b:=sum_{i=1}^{m}f^b_i(x)g^b_i(y)$ is again in $M$. Note that $^a$ and $^b$ it just for notational reasons, so $f^a$ doens't mean $f$ multiplied by itself $a$ times
For the product:
$ab=sum_{j=1}^{n}f^a_j(x)g^a_j(y)sum_{i=1}^{m}f^b_i(x)g^b_i(y)=sum_{(j,i) in {1ldots n }times{1ldots m}}f^a_jf_i^b(x)g_j^ag_i^b(y)$
Does this already show that $ab$ is in $M$?
For the sum:
$a+b=sum_{j=1}^{n}f^a_j(x)g^a_j(y)+sum_{i=1}^{m}f^b_i(x)g^b_i(y)$. But how can I write this as one sum?
real-analysis calculus
real-analysis calculus
asked Dec 20 '18 at 13:42
user626880user626880
204
204
1
$begingroup$
it already is written as a sum, from 1 to (m+n) of products of functions of x times functions of y
$endgroup$
– Sorin Tirc
Dec 20 '18 at 14:49
add a comment |
1
$begingroup$
it already is written as a sum, from 1 to (m+n) of products of functions of x times functions of y
$endgroup$
– Sorin Tirc
Dec 20 '18 at 14:49
1
1
$begingroup$
it already is written as a sum, from 1 to (m+n) of products of functions of x times functions of y
$endgroup$
– Sorin Tirc
Dec 20 '18 at 14:49
$begingroup$
it already is written as a sum, from 1 to (m+n) of products of functions of x times functions of y
$endgroup$
– Sorin Tirc
Dec 20 '18 at 14:49
add a comment |
0
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$begingroup$
it already is written as a sum, from 1 to (m+n) of products of functions of x times functions of y
$endgroup$
– Sorin Tirc
Dec 20 '18 at 14:49