Mixing
Is any $C_c^infty(mathbb R)$-function the product of two $C_c^infty(mathbb R)$-functions?
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Let $fin C_c^infty(mathbb R)$. Are we able to write $$f=gh$$ for some $g,hin C_c^infty(mathbb R)$? Unfortunately, I've no idea how to I could prove or disprove this.
functional-analysis analysis smooth-functions
asked Jan 29 at 12:54
0xbadf00d0xbadf00d
1,80941534
$endgroup$
add a comment |
$begingroup$
Let $fin C_c^infty(mathbb R)$. Are we able to write $$f=gh$$ for some $g,hin C_c^infty(mathbb R)$? Unfortunately, I've no idea how to I could prove or disprove this.
functional-analysis analysis smooth-functions
asked Jan 29 at 12:54
0xbadf00d0xbadf00d
1,80941534
$endgroup$
1
$begingroup$
We can find $g in C_c(Bbb R)$ such that $g=1 $on $text{supp} f$ by Urysohn lemma.
$endgroup$
– Song
Jan 29 at 12:58
$begingroup$
@Song I didn't intend to write "for some $C_c^infty(mathbb R)$" (which is not even a meaningful sentence), but "for some $g,hin C_c^infty(mathbb R)$".
$endgroup$
– 0xbadf00d
Jan 29 at 13:04
1
$begingroup$
I didn't recognize it was $C^infty_c(Bbb R)$, sorry. But we can take care of it by convoluting $g$ with smooth function $eta$ with support in $[-epsilon,epsilon]$ and $int eta =1$. You can see this Wikipedia page on mollifier and smooth cut-off function.
$endgroup$
– Song
Jan 29 at 13:11
add a comment |
$begingroup$
Let $fin C_c^infty(mathbb R)$. Are we able to write $$f=gh$$ for some $g,hin C_c^infty(mathbb R)$? Unfortunately, I've no idea how to I could prove or disprove this.
functional-analysis analysis smooth-functions
asked Jan 29 at 12:54
0xbadf00d0xbadf00d
1,80941534
$endgroup$
Let $fin C_c^infty(mathbb R)$. Are we able to write $$f=gh$$ for some $g,hin C_c^infty(mathbb R)$? Unfortunately, I've no idea how to I could prove or disprove this.
functional-analysis analysis smooth-functions
functional-analysis analysis smooth-functions
asked Jan 29 at 12:54
0xbadf00d0xbadf00d
1,80941534
asked Jan 29 at 12:54
0xbadf00d0xbadf00d
1,80941534
edited Jan 29 at 13:03
0xbadf00d
asked Jan 29 at 12:54
0xbadf00d0xbadf00d
1,80941534
asked Jan 29 at 12:54
0xbadf00d0xbadf00d
1,80941534
asked Jan 29 at 12:54
0xbadf00d0xbadf00d
1,80941534
1,80941534
1
$begingroup$
We can find $g in C_c(Bbb R)$ such that $g=1 $on $text{supp} f$ by Urysohn lemma.
$endgroup$
– Song
Jan 29 at 12:58
$begingroup$
@Song I didn't intend to write "for some $C_c^infty(mathbb R)$" (which is not even a meaningful sentence), but "for some $g,hin C_c^infty(mathbb R)$".
$endgroup$
– 0xbadf00d
Jan 29 at 13:04
1
$begingroup$
I didn't recognize it was $C^infty_c(Bbb R)$, sorry. But we can take care of it by convoluting $g$ with smooth function $eta$ with support in $[-epsilon,epsilon]$ and $int eta =1$. You can see this Wikipedia page on mollifier and smooth cut-off function.
$endgroup$
– Song
Jan 29 at 13:11
add a comment |
1
$begingroup$
We can find $g in C_c(Bbb R)$ such that $g=1 $on $text{supp} f$ by Urysohn lemma.
$endgroup$
– Song
Jan 29 at 12:58
$begingroup$
@Song I didn't intend to write "for some $C_c^infty(mathbb R)$" (which is not even a meaningful sentence), but "for some $g,hin C_c^infty(mathbb R)$".
$endgroup$
– 0xbadf00d
Jan 29 at 13:04
1
$begingroup$
I didn't recognize it was $C^infty_c(Bbb R)$, sorry. But we can take care of it by convoluting $g$ with smooth function $eta$ with support in $[-epsilon,epsilon]$ and $int eta =1$. You can see this Wikipedia page on mollifier and smooth cut-off function.
$endgroup$
– Song
Jan 29 at 13:11
1
1
$begingroup$
We can find $g in C_c(Bbb R)$ such that $g=1 $on $text{supp} f$ by Urysohn lemma.
$endgroup$
– Song
Jan 29 at 12:58
$begingroup$
We can find $g in C_c(Bbb R)$ such that $g=1 $on $text{supp} f$ by Urysohn lemma.
$endgroup$
– Song
Jan 29 at 12:58
$begingroup$
@Song I didn't intend to write "for some $C_c^infty(mathbb R)$" (which is not even a meaningful sentence), but "for some $g,hin C_c^infty(mathbb R)$".
$endgroup$
– 0xbadf00d
Jan 29 at 13:04
$begingroup$
@Song I didn't intend to write "for some $C_c^infty(mathbb R)$" (which is not even a meaningful sentence), but "for some $g,hin C_c^infty(mathbb R)$".
$endgroup$
– 0xbadf00d
Jan 29 at 13:04
1
1
$begingroup$
I didn't recognize it was $C^infty_c(Bbb R)$, sorry. But we can take care of it by convoluting $g$ with smooth function $eta$ with support in $[-epsilon,epsilon]$ and $int eta =1$. You can see this Wikipedia page on mollifier and smooth cut-off function.
$endgroup$
– Song
Jan 29 at 13:11
$begingroup$
I didn't recognize it was $C^infty_c(Bbb R)$, sorry. But we can take care of it by convoluting $g$ with smooth function $eta$ with support in $[-epsilon,epsilon]$ and $int eta =1$. You can see this Wikipedia page on mollifier and smooth cut-off function.
$endgroup$
– Song
Jan 29 at 13:11
add a comment |
1 Answer
1
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$begingroup$
There exists a smooth compactly supported function $g$ that is $1$ on an open ball containing the support of $f$. Then $f=fg$.
For the definition of $g$: let $h(r)=0$ if $r leq 0$ and $h(r)=e^{-1/r}$ if $r > 0$. It is well-known that $h$ is smooth. Let $h_1(r)=frac{h(r)}{h(r)+h(1-r)}$: $h_1$ is smooth, zero on $(-infty,0]$ and $1$ on $[1,infty)$.
Take now $g(x)=h_1left(2-frac{|x|^2}{R^2+1}right)$ where ${f neq 0} subset mathcal{B}(0,R)$.
answered Jan 29 at 12:59
MindlackMindlack
4,910211
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1 Answer
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1 Answer
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$begingroup$
There exists a smooth compactly supported function $g$ that is $1$ on an open ball containing the support of $f$. Then $f=fg$.
For the definition of $g$: let $h(r)=0$ if $r leq 0$ and $h(r)=e^{-1/r}$ if $r > 0$. It is well-known that $h$ is smooth. Let $h_1(r)=frac{h(r)}{h(r)+h(1-r)}$: $h_1$ is smooth, zero on $(-infty,0]$ and $1$ on $[1,infty)$.
Take now $g(x)=h_1left(2-frac{|x|^2}{R^2+1}right)$ where ${f neq 0} subset mathcal{B}(0,R)$.
answered Jan 29 at 12:59
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
There exists a smooth compactly supported function $g$ that is $1$ on an open ball containing the support of $f$. Then $f=fg$.
For the definition of $g$: let $h(r)=0$ if $r leq 0$ and $h(r)=e^{-1/r}$ if $r > 0$. It is well-known that $h$ is smooth. Let $h_1(r)=frac{h(r)}{h(r)+h(1-r)}$: $h_1$ is smooth, zero on $(-infty,0]$ and $1$ on $[1,infty)$.
Take now $g(x)=h_1left(2-frac{|x|^2}{R^2+1}right)$ where ${f neq 0} subset mathcal{B}(0,R)$.
answered Jan 29 at 12:59
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
There exists a smooth compactly supported function $g$ that is $1$ on an open ball containing the support of $f$. Then $f=fg$.
For the definition of $g$: let $h(r)=0$ if $r leq 0$ and $h(r)=e^{-1/r}$ if $r > 0$. It is well-known that $h$ is smooth. Let $h_1(r)=frac{h(r)}{h(r)+h(1-r)}$: $h_1$ is smooth, zero on $(-infty,0]$ and $1$ on $[1,infty)$.
Take now $g(x)=h_1left(2-frac{|x|^2}{R^2+1}right)$ where ${f neq 0} subset mathcal{B}(0,R)$.
answered Jan 29 at 12:59
MindlackMindlack
4,910211
$endgroup$
There exists a smooth compactly supported function $g$ that is $1$ on an open ball containing the support of $f$. Then $f=fg$.
For the definition of $g$: let $h(r)=0$ if $r leq 0$ and $h(r)=e^{-1/r}$ if $r > 0$. It is well-known that $h$ is smooth. Let $h_1(r)=frac{h(r)}{h(r)+h(1-r)}$: $h_1$ is smooth, zero on $(-infty,0]$ and $1$ on $[1,infty)$.
Take now $g(x)=h_1left(2-frac{|x|^2}{R^2+1}right)$ where ${f neq 0} subset mathcal{B}(0,R)$.
answered Jan 29 at 12:59
MindlackMindlack
4,910211
edited Jan 29 at 14:06
answered Jan 29 at 12:59
MindlackMindlack
4,910211
answered Jan 29 at 12:59
MindlackMindlack
4,910211
answered Jan 29 at 12:59
MindlackMindlack
4,910211
4,910211
add a comment |
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$begingroup$
We can find $g in C_c(Bbb R)$ such that $g=1 $on $text{supp} f$ by Urysohn lemma.
$endgroup$
– Song
Jan 29 at 12:58
$begingroup$
@Song I didn't intend to write "for some $C_c^infty(mathbb R)$" (which is not even a meaningful sentence), but "for some $g,hin C_c^infty(mathbb R)$".
$endgroup$
– 0xbadf00d
Jan 29 at 13:04
1
$begingroup$
I didn't recognize it was $C^infty_c(Bbb R)$, sorry. But we can take care of it by convoluting $g$ with smooth function $eta$ with support in $[-epsilon,epsilon]$ and $int eta =1$. You can see this Wikipedia page on mollifier and smooth cut-off function.
$endgroup$
– Song
Jan 29 at 13:11