Mixing
Understanding a function space
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I was reading a paper on Homogenization theory, where the author uses the spaces of vector valued functions. Let us consider such a space $D[Omega; C^infty_P(Y)]$, consisting of all the compactly supported smooth functions $psi : Omega rightarrow C^infty_P(Y)$. $Y=[0,1]^n$, $Omega$ is an open set in $R^n$ and $C^infty_P(Y)$ is the space of all smooth functions on $R^n$ that are Y-periodic. A function $f$ is called Y-periodic if $f(x+ke_i)=f(x)$ for all $x$ in $R^n$ and integers $k$ and for all $i$ $in$ {1,2,...,n} ; $e_i$'s being standard basis vectors in $R^n$. Note that, the smoothness (differentiablity) is defined on $ psi$ as Fréchet derivative between the normed spaces $R^n$ and $C^infty_P(Y)$. The author did not mention which norm to consider on $C^infty_P(Y)$. Let us take the sup norm on this space. Clearly $psi(x,.)$ is an element of $C^infty _P(Y)$ for each $x$ in $Omega$. Also, $psi (x,y) $ is a real number for each $x$ in $ Omega$ and $y$ in $R^n$. So, we can also think of such a $psi$ as a real valued function defined on $Omega times R^n$.
It seems that the author is treating such a $psi(x,y)$ as a real valued SMOOTH function defined on $ Omega times R^n$. But how to argue that $psi(x,y)$ is smooth on $ Omega times R^n$? Thank you.
calculus functional-analysis measure-theory pde applications
asked Jan 23 at 1:13
Apratim BhattacharyaApratim Bhattacharya
213
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add a comment |
$begingroup$
I was reading a paper on Homogenization theory, where the author uses the spaces of vector valued functions. Let us consider such a space $D[Omega; C^infty_P(Y)]$, consisting of all the compactly supported smooth functions $psi : Omega rightarrow C^infty_P(Y)$. $Y=[0,1]^n$, $Omega$ is an open set in $R^n$ and $C^infty_P(Y)$ is the space of all smooth functions on $R^n$ that are Y-periodic. A function $f$ is called Y-periodic if $f(x+ke_i)=f(x)$ for all $x$ in $R^n$ and integers $k$ and for all $i$ $in$ {1,2,...,n} ; $e_i$'s being standard basis vectors in $R^n$. Note that, the smoothness (differentiablity) is defined on $ psi$ as Fréchet derivative between the normed spaces $R^n$ and $C^infty_P(Y)$. The author did not mention which norm to consider on $C^infty_P(Y)$. Let us take the sup norm on this space. Clearly $psi(x,.)$ is an element of $C^infty _P(Y)$ for each $x$ in $Omega$. Also, $psi (x,y) $ is a real number for each $x$ in $ Omega$ and $y$ in $R^n$. So, we can also think of such a $psi$ as a real valued function defined on $Omega times R^n$.
It seems that the author is treating such a $psi(x,y)$ as a real valued SMOOTH function defined on $ Omega times R^n$. But how to argue that $psi(x,y)$ is smooth on $ Omega times R^n$? Thank you.
calculus functional-analysis measure-theory pde applications
asked Jan 23 at 1:13
Apratim BhattacharyaApratim Bhattacharya
213
$endgroup$
$begingroup$
There is a confusion in your notation, if $x in Omega$, $psi(x) in C^{infty}(Omega)$. Moreover, what is the relation between $psi(x)$ and $psi(x,y)$?
$endgroup$
– Evan William Chandra
Jan 23 at 1:18
$begingroup$
$psi (x)$ is a function, whose domain is $R^n$, as you change x, you get another function; basically $psi(x,.)$ is a real number, where $.$ runs over $R^n$.
$endgroup$
– Apratim Bhattacharya
Jan 23 at 1:24
add a comment |
$begingroup$
I was reading a paper on Homogenization theory, where the author uses the spaces of vector valued functions. Let us consider such a space $D[Omega; C^infty_P(Y)]$, consisting of all the compactly supported smooth functions $psi : Omega rightarrow C^infty_P(Y)$. $Y=[0,1]^n$, $Omega$ is an open set in $R^n$ and $C^infty_P(Y)$ is the space of all smooth functions on $R^n$ that are Y-periodic. A function $f$ is called Y-periodic if $f(x+ke_i)=f(x)$ for all $x$ in $R^n$ and integers $k$ and for all $i$ $in$ {1,2,...,n} ; $e_i$'s being standard basis vectors in $R^n$. Note that, the smoothness (differentiablity) is defined on $ psi$ as Fréchet derivative between the normed spaces $R^n$ and $C^infty_P(Y)$. The author did not mention which norm to consider on $C^infty_P(Y)$. Let us take the sup norm on this space. Clearly $psi(x,.)$ is an element of $C^infty _P(Y)$ for each $x$ in $Omega$. Also, $psi (x,y) $ is a real number for each $x$ in $ Omega$ and $y$ in $R^n$. So, we can also think of such a $psi$ as a real valued function defined on $Omega times R^n$.
It seems that the author is treating such a $psi(x,y)$ as a real valued SMOOTH function defined on $ Omega times R^n$. But how to argue that $psi(x,y)$ is smooth on $ Omega times R^n$? Thank you.
calculus functional-analysis measure-theory pde applications
asked Jan 23 at 1:13
Apratim BhattacharyaApratim Bhattacharya
213
$endgroup$
I was reading a paper on Homogenization theory, where the author uses the spaces of vector valued functions. Let us consider such a space $D[Omega; C^infty_P(Y)]$, consisting of all the compactly supported smooth functions $psi : Omega rightarrow C^infty_P(Y)$. $Y=[0,1]^n$, $Omega$ is an open set in $R^n$ and $C^infty_P(Y)$ is the space of all smooth functions on $R^n$ that are Y-periodic. A function $f$ is called Y-periodic if $f(x+ke_i)=f(x)$ for all $x$ in $R^n$ and integers $k$ and for all $i$ $in$ {1,2,...,n} ; $e_i$'s being standard basis vectors in $R^n$. Note that, the smoothness (differentiablity) is defined on $ psi$ as Fréchet derivative between the normed spaces $R^n$ and $C^infty_P(Y)$. The author did not mention which norm to consider on $C^infty_P(Y)$. Let us take the sup norm on this space. Clearly $psi(x,.)$ is an element of $C^infty _P(Y)$ for each $x$ in $Omega$. Also, $psi (x,y) $ is a real number for each $x$ in $ Omega$ and $y$ in $R^n$. So, we can also think of such a $psi$ as a real valued function defined on $Omega times R^n$.
It seems that the author is treating such a $psi(x,y)$ as a real valued SMOOTH function defined on $ Omega times R^n$. But how to argue that $psi(x,y)$ is smooth on $ Omega times R^n$? Thank you.
calculus functional-analysis measure-theory pde applications
calculus functional-analysis measure-theory pde applications
asked Jan 23 at 1:13
Apratim BhattacharyaApratim Bhattacharya
213
asked Jan 23 at 1:13
Apratim BhattacharyaApratim Bhattacharya
213
edited Feb 1 at 22:52
Apratim Bhattacharya
asked Jan 23 at 1:13
Apratim BhattacharyaApratim Bhattacharya
213
asked Jan 23 at 1:13
Apratim BhattacharyaApratim Bhattacharya
213
asked Jan 23 at 1:13
Apratim BhattacharyaApratim Bhattacharya
213
213
$begingroup$
There is a confusion in your notation, if $x in Omega$, $psi(x) in C^{infty}(Omega)$. Moreover, what is the relation between $psi(x)$ and $psi(x,y)$?
$endgroup$
– Evan William Chandra
Jan 23 at 1:18
$begingroup$
$psi (x)$ is a function, whose domain is $R^n$, as you change x, you get another function; basically $psi(x,.)$ is a real number, where $.$ runs over $R^n$.
$endgroup$
– Apratim Bhattacharya
Jan 23 at 1:24
add a comment |
$begingroup$
There is a confusion in your notation, if $x in Omega$, $psi(x) in C^{infty}(Omega)$. Moreover, what is the relation between $psi(x)$ and $psi(x,y)$?
$endgroup$
– Evan William Chandra
Jan 23 at 1:18
$begingroup$
$psi (x)$ is a function, whose domain is $R^n$, as you change x, you get another function; basically $psi(x,.)$ is a real number, where $.$ runs over $R^n$.
$endgroup$
– Apratim Bhattacharya
Jan 23 at 1:24
$begingroup$
There is a confusion in your notation, if $x in Omega$, $psi(x) in C^{infty}(Omega)$. Moreover, what is the relation between $psi(x)$ and $psi(x,y)$?
$endgroup$
– Evan William Chandra
Jan 23 at 1:18
$begingroup$
There is a confusion in your notation, if $x in Omega$, $psi(x) in C^{infty}(Omega)$. Moreover, what is the relation between $psi(x)$ and $psi(x,y)$?
$endgroup$
– Evan William Chandra
Jan 23 at 1:18
$begingroup$
$psi (x)$ is a function, whose domain is $R^n$, as you change x, you get another function; basically $psi(x,.)$ is a real number, where $.$ runs over $R^n$.
$endgroup$
– Apratim Bhattacharya
Jan 23 at 1:24
$begingroup$
$psi (x)$ is a function, whose domain is $R^n$, as you change x, you get another function; basically $psi(x,.)$ is a real number, where $.$ runs over $R^n$.
$endgroup$
– Apratim Bhattacharya
Jan 23 at 1:24
add a comment |
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$begingroup$
There is a confusion in your notation, if $x in Omega$, $psi(x) in C^{infty}(Omega)$. Moreover, what is the relation between $psi(x)$ and $psi(x,y)$?
$endgroup$
– Evan William Chandra
Jan 23 at 1:18
$begingroup$
$psi (x)$ is a function, whose domain is $R^n$, as you change x, you get another function; basically $psi(x,.)$ is a real number, where $.$ runs over $R^n$.
$endgroup$
– Apratim Bhattacharya
Jan 23 at 1:24