Decomposition of $C^{(k)}$ function
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I am stuck at the following exercise in Zorich.
Let $f,gin C^{(k)}(D;mathbb{R})$, and suppose that $f(x)=0Rightarrow g(x)=0$ in the domain $D$. Show that if grad $f neq 0$, them there is a decomposition $g=hcdot f$ in $D$, where $hin C^{(k-1)}(D;mathbb{R})$.
Could you give me a hint? Thanks in advance!
analysis
asked Dec 5 at 3:07
Jiu
38919
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show 1 more comment
up vote
0
down vote
favorite
I am stuck at the following exercise in Zorich.
Let $f,gin C^{(k)}(D;mathbb{R})$, and suppose that $f(x)=0Rightarrow g(x)=0$ in the domain $D$. Show that if grad $f neq 0$, them there is a decomposition $g=hcdot f$ in $D$, where $hin C^{(k-1)}(D;mathbb{R})$.
Could you give me a hint? Thanks in advance!
analysis
asked Dec 5 at 3:07
Jiu
38919
Dear @Jiu, is $D$ a domain in $mathbb{R}^n$? I'm confused about $h$, because the image of $f$ lies in $mathbb{R}$, not in $D$. You should require $h(0)=0$ and perhaps some additional constraints on $h$.
– user90189
Dec 5 at 4:36
@user90189 thanks for your comment! Now that you make me realize this problem, I just checked and see that in the English version, the $circ$ is replaced by $cdot$... So the $circ$ should be a typo in the German version!
– Jiu
Dec 5 at 4:53
So, I think you can try to check that $g/f$ is well-defined :)
– user90189
Dec 5 at 4:56
@user90189 yes now I know how to solve it :D
– Jiu
Dec 5 at 7:55
@user90189 In fact when I try to write down the solution I realize that it is not correct :(
– Jiu
Dec 5 at 8:57
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am stuck at the following exercise in Zorich.
Let $f,gin C^{(k)}(D;mathbb{R})$, and suppose that $f(x)=0Rightarrow g(x)=0$ in the domain $D$. Show that if grad $f neq 0$, them there is a decomposition $g=hcdot f$ in $D$, where $hin C^{(k-1)}(D;mathbb{R})$.
Could you give me a hint? Thanks in advance!
analysis
asked Dec 5 at 3:07
Jiu
38919
I am stuck at the following exercise in Zorich.
Let $f,gin C^{(k)}(D;mathbb{R})$, and suppose that $f(x)=0Rightarrow g(x)=0$ in the domain $D$. Show that if grad $f neq 0$, them there is a decomposition $g=hcdot f$ in $D$, where $hin C^{(k-1)}(D;mathbb{R})$.
Could you give me a hint? Thanks in advance!
analysis
analysis
asked Dec 5 at 3:07
Jiu
38919
asked Dec 5 at 3:07
Jiu
38919
edited Dec 5 at 4:54
asked Dec 5 at 3:07
Jiu
38919
asked Dec 5 at 3:07
Jiu
38919
asked Dec 5 at 3:07
Jiu
38919
38919
Dear @Jiu, is $D$ a domain in $mathbb{R}^n$? I'm confused about $h$, because the image of $f$ lies in $mathbb{R}$, not in $D$. You should require $h(0)=0$ and perhaps some additional constraints on $h$.
– user90189
Dec 5 at 4:36
@user90189 thanks for your comment! Now that you make me realize this problem, I just checked and see that in the English version, the $circ$ is replaced by $cdot$... So the $circ$ should be a typo in the German version!
– Jiu
Dec 5 at 4:53
So, I think you can try to check that $g/f$ is well-defined :)
– user90189
Dec 5 at 4:56
@user90189 yes now I know how to solve it :D
– Jiu
Dec 5 at 7:55
@user90189 In fact when I try to write down the solution I realize that it is not correct :(
– Jiu
Dec 5 at 8:57
|
show 1 more comment
Dear @Jiu, is $D$ a domain in $mathbb{R}^n$? I'm confused about $h$, because the image of $f$ lies in $mathbb{R}$, not in $D$. You should require $h(0)=0$ and perhaps some additional constraints on $h$.
– user90189
Dec 5 at 4:36
@user90189 thanks for your comment! Now that you make me realize this problem, I just checked and see that in the English version, the $circ$ is replaced by $cdot$... So the $circ$ should be a typo in the German version!
– Jiu
Dec 5 at 4:53
So, I think you can try to check that $g/f$ is well-defined :)
– user90189
Dec 5 at 4:56
@user90189 yes now I know how to solve it :D
– Jiu
Dec 5 at 7:55
@user90189 In fact when I try to write down the solution I realize that it is not correct :(
– Jiu
Dec 5 at 8:57
Dear @Jiu, is $D$ a domain in $mathbb{R}^n$? I'm confused about $h$, because the image of $f$ lies in $mathbb{R}$, not in $D$. You should require $h(0)=0$ and perhaps some additional constraints on $h$.
– user90189
Dec 5 at 4:36
Dear @Jiu, is $D$ a domain in $mathbb{R}^n$? I'm confused about $h$, because the image of $f$ lies in $mathbb{R}$, not in $D$. You should require $h(0)=0$ and perhaps some additional constraints on $h$.
– user90189
Dec 5 at 4:36
@user90189 thanks for your comment! Now that you make me realize this problem, I just checked and see that in the English version, the $circ$ is replaced by $cdot$... So the $circ$ should be a typo in the German version!
– Jiu
Dec 5 at 4:53
@user90189 thanks for your comment! Now that you make me realize this problem, I just checked and see that in the English version, the $circ$ is replaced by $cdot$... So the $circ$ should be a typo in the German version!
– Jiu
Dec 5 at 4:53
So, I think you can try to check that $g/f$ is well-defined :)
– user90189
Dec 5 at 4:56
So, I think you can try to check that $g/f$ is well-defined :)
– user90189
Dec 5 at 4:56
@user90189 yes now I know how to solve it :D
– Jiu
Dec 5 at 7:55
@user90189 yes now I know how to solve it :D
– Jiu
Dec 5 at 7:55
@user90189 In fact when I try to write down the solution I realize that it is not correct :(
– Jiu
Dec 5 at 8:57
@user90189 In fact when I try to write down the solution I realize that it is not correct :(
– Jiu
Dec 5 at 8:57
|
show 1 more comment
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Dear @Jiu, is $D$ a domain in $mathbb{R}^n$? I'm confused about $h$, because the image of $f$ lies in $mathbb{R}$, not in $D$. You should require $h(0)=0$ and perhaps some additional constraints on $h$.
– user90189
Dec 5 at 4:36
@user90189 thanks for your comment! Now that you make me realize this problem, I just checked and see that in the English version, the $circ$ is replaced by $cdot$... So the $circ$ should be a typo in the German version!
– Jiu
Dec 5 at 4:53
So, I think you can try to check that $g/f$ is well-defined :)
– user90189
Dec 5 at 4:56
@user90189 yes now I know how to solve it :D
– Jiu
Dec 5 at 7:55
@user90189 In fact when I try to write down the solution I realize that it is not correct :(
– Jiu
Dec 5 at 8:57