Mixing
If $Csubsetmathbb{R}^n$ is closed, then $C=f^{-1}(0)$ for some smooth $f:mathbb{R}^ntomathbb{R}$.
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Let $Csubset mathbb{R}^n$ closed. Prove that there is a smooth function $f:mathbb{R}^ntomathbb{R}$ such that $C=f^{-1}(0)$.
I've found this solution in the internet: take a cover of balls ${B_i}_i$ of $mathbb{R}^nsetminus C$ and a partition of unity ${f_i}_i$ subordinate to ${B_i}_i$ and define $f:=sum_{i=1}^infty frac{f_i}{2^iM_i}$, where:
$$M_i:=supleft{frac{partial^kf_i}{partial x_{j_1}cdots partial x_{j_k}}(p)mid 1leq j_1<...<j_kleq n,, kleq i,,pin B_iright}$$
My question is: why do we need to define these $M_i$'s?
Why can't we just define $f=sum_{i=1}^infty frac{f_i}{2^i}$ for example?
In that case, since ${B_i}_i$ can be considered locally finite, then for $pinmathbb{R}^n$, there is some open $Usubsetmathbb{R}^nsetminus C$ containing $p$ such that $f_i|_Uequiv 0$ for all but finitely many $i$ (say $i_1,...,i_k$), so $f|_U=frac{f_{i_1}}{2^{i_1}}+...+frac{f_{i_k}}{2^{i_k}}$, which is obviously smooth.
What am I missing?
real-analysis smooth-functions
asked Feb 2 at 17:16
rmdmc89rmdmc89
2,2421923
$endgroup$
add a comment |
$begingroup$
Let $Csubset mathbb{R}^n$ closed. Prove that there is a smooth function $f:mathbb{R}^ntomathbb{R}$ such that $C=f^{-1}(0)$.
I've found this solution in the internet: take a cover of balls ${B_i}_i$ of $mathbb{R}^nsetminus C$ and a partition of unity ${f_i}_i$ subordinate to ${B_i}_i$ and define $f:=sum_{i=1}^infty frac{f_i}{2^iM_i}$, where:
$$M_i:=supleft{frac{partial^kf_i}{partial x_{j_1}cdots partial x_{j_k}}(p)mid 1leq j_1<...<j_kleq n,, kleq i,,pin B_iright}$$
My question is: why do we need to define these $M_i$'s?
Why can't we just define $f=sum_{i=1}^infty frac{f_i}{2^i}$ for example?
In that case, since ${B_i}_i$ can be considered locally finite, then for $pinmathbb{R}^n$, there is some open $Usubsetmathbb{R}^nsetminus C$ containing $p$ such that $f_i|_Uequiv 0$ for all but finitely many $i$ (say $i_1,...,i_k$), so $f|_U=frac{f_{i_1}}{2^{i_1}}+...+frac{f_{i_k}}{2^{i_k}}$, which is obviously smooth.
What am I missing?
real-analysis smooth-functions
asked Feb 2 at 17:16
rmdmc89rmdmc89
2,2421923
$endgroup$
1
$begingroup$
i'm not certain, but what if $pin C$ in your last paragraph? the $B_i$ are only locally finite at points outside of C, right?
$endgroup$
– Tim kinsella
Feb 2 at 19:05
$begingroup$
You're right @Timkinsella, I should have said $Usubset mathbb{R}^nsetminus C$. I've corrected it
$endgroup$
– rmdmc89
Feb 2 at 19:38
$begingroup$
if $pin C$ then there is no such $U$
$endgroup$
– Tim kinsella
Feb 2 at 22:53
1
$begingroup$
my point (of which, as i say, i'm not certain) is that the $M_i$ are there to guarantee smoothness at points of $C$ somehow.
$endgroup$
– Tim kinsella
Feb 2 at 22:55
$begingroup$
Oh, I see what you mean. I'll think about it
$endgroup$
– rmdmc89
Feb 3 at 12:04
add a comment |
$begingroup$
Let $Csubset mathbb{R}^n$ closed. Prove that there is a smooth function $f:mathbb{R}^ntomathbb{R}$ such that $C=f^{-1}(0)$.
I've found this solution in the internet: take a cover of balls ${B_i}_i$ of $mathbb{R}^nsetminus C$ and a partition of unity ${f_i}_i$ subordinate to ${B_i}_i$ and define $f:=sum_{i=1}^infty frac{f_i}{2^iM_i}$, where:
$$M_i:=supleft{frac{partial^kf_i}{partial x_{j_1}cdots partial x_{j_k}}(p)mid 1leq j_1<...<j_kleq n,, kleq i,,pin B_iright}$$
My question is: why do we need to define these $M_i$'s?
Why can't we just define $f=sum_{i=1}^infty frac{f_i}{2^i}$ for example?
In that case, since ${B_i}_i$ can be considered locally finite, then for $pinmathbb{R}^n$, there is some open $Usubsetmathbb{R}^nsetminus C$ containing $p$ such that $f_i|_Uequiv 0$ for all but finitely many $i$ (say $i_1,...,i_k$), so $f|_U=frac{f_{i_1}}{2^{i_1}}+...+frac{f_{i_k}}{2^{i_k}}$, which is obviously smooth.
What am I missing?
real-analysis smooth-functions
asked Feb 2 at 17:16
rmdmc89rmdmc89
2,2421923
$endgroup$
Let $Csubset mathbb{R}^n$ closed. Prove that there is a smooth function $f:mathbb{R}^ntomathbb{R}$ such that $C=f^{-1}(0)$.
I've found this solution in the internet: take a cover of balls ${B_i}_i$ of $mathbb{R}^nsetminus C$ and a partition of unity ${f_i}_i$ subordinate to ${B_i}_i$ and define $f:=sum_{i=1}^infty frac{f_i}{2^iM_i}$, where:
$$M_i:=supleft{frac{partial^kf_i}{partial x_{j_1}cdots partial x_{j_k}}(p)mid 1leq j_1<...<j_kleq n,, kleq i,,pin B_iright}$$
My question is: why do we need to define these $M_i$'s?
Why can't we just define $f=sum_{i=1}^infty frac{f_i}{2^i}$ for example?
In that case, since ${B_i}_i$ can be considered locally finite, then for $pinmathbb{R}^n$, there is some open $Usubsetmathbb{R}^nsetminus C$ containing $p$ such that $f_i|_Uequiv 0$ for all but finitely many $i$ (say $i_1,...,i_k$), so $f|_U=frac{f_{i_1}}{2^{i_1}}+...+frac{f_{i_k}}{2^{i_k}}$, which is obviously smooth.
What am I missing?
real-analysis smooth-functions
real-analysis smooth-functions
asked Feb 2 at 17:16
rmdmc89rmdmc89
2,2421923
asked Feb 2 at 17:16
rmdmc89rmdmc89
2,2421923
edited Feb 2 at 19:36
rmdmc89
asked Feb 2 at 17:16
rmdmc89rmdmc89
2,2421923
asked Feb 2 at 17:16
rmdmc89rmdmc89
2,2421923
asked Feb 2 at 17:16
rmdmc89rmdmc89
2,2421923
2,2421923
1
$begingroup$
i'm not certain, but what if $pin C$ in your last paragraph? the $B_i$ are only locally finite at points outside of C, right?
$endgroup$
– Tim kinsella
Feb 2 at 19:05
$begingroup$
You're right @Timkinsella, I should have said $Usubset mathbb{R}^nsetminus C$. I've corrected it
$endgroup$
– rmdmc89
Feb 2 at 19:38
$begingroup$
if $pin C$ then there is no such $U$
$endgroup$
– Tim kinsella
Feb 2 at 22:53
1
$begingroup$
my point (of which, as i say, i'm not certain) is that the $M_i$ are there to guarantee smoothness at points of $C$ somehow.
$endgroup$
– Tim kinsella
Feb 2 at 22:55
$begingroup$
Oh, I see what you mean. I'll think about it
$endgroup$
– rmdmc89
Feb 3 at 12:04
add a comment |
1
$begingroup$
i'm not certain, but what if $pin C$ in your last paragraph? the $B_i$ are only locally finite at points outside of C, right?
$endgroup$
– Tim kinsella
Feb 2 at 19:05
$begingroup$
You're right @Timkinsella, I should have said $Usubset mathbb{R}^nsetminus C$. I've corrected it
$endgroup$
– rmdmc89
Feb 2 at 19:38
$begingroup$
if $pin C$ then there is no such $U$
$endgroup$
– Tim kinsella
Feb 2 at 22:53
1
$begingroup$
my point (of which, as i say, i'm not certain) is that the $M_i$ are there to guarantee smoothness at points of $C$ somehow.
$endgroup$
– Tim kinsella
Feb 2 at 22:55
$begingroup$
Oh, I see what you mean. I'll think about it
$endgroup$
– rmdmc89
Feb 3 at 12:04
1
1
$begingroup$
i'm not certain, but what if $pin C$ in your last paragraph? the $B_i$ are only locally finite at points outside of C, right?
$endgroup$
– Tim kinsella
Feb 2 at 19:05
$begingroup$
i'm not certain, but what if $pin C$ in your last paragraph? the $B_i$ are only locally finite at points outside of C, right?
$endgroup$
– Tim kinsella
Feb 2 at 19:05
$begingroup$
You're right @Timkinsella, I should have said $Usubset mathbb{R}^nsetminus C$. I've corrected it
$endgroup$
– rmdmc89
Feb 2 at 19:38
$begingroup$
You're right @Timkinsella, I should have said $Usubset mathbb{R}^nsetminus C$. I've corrected it
$endgroup$
– rmdmc89
Feb 2 at 19:38
$begingroup$
if $pin C$ then there is no such $U$
$endgroup$
– Tim kinsella
Feb 2 at 22:53
$begingroup$
if $pin C$ then there is no such $U$
$endgroup$
– Tim kinsella
Feb 2 at 22:53
1
1
$begingroup$
my point (of which, as i say, i'm not certain) is that the $M_i$ are there to guarantee smoothness at points of $C$ somehow.
$endgroup$
– Tim kinsella
Feb 2 at 22:55
$begingroup$
my point (of which, as i say, i'm not certain) is that the $M_i$ are there to guarantee smoothness at points of $C$ somehow.
$endgroup$
– Tim kinsella
Feb 2 at 22:55
$begingroup$
Oh, I see what you mean. I'll think about it
$endgroup$
– rmdmc89
Feb 3 at 12:04
$begingroup$
Oh, I see what you mean. I'll think about it
$endgroup$
– rmdmc89
Feb 3 at 12:04
add a comment |
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1
$begingroup$
i'm not certain, but what if $pin C$ in your last paragraph? the $B_i$ are only locally finite at points outside of C, right?
$endgroup$
– Tim kinsella
Feb 2 at 19:05
$begingroup$
You're right @Timkinsella, I should have said $Usubset mathbb{R}^nsetminus C$. I've corrected it
$endgroup$
– rmdmc89
Feb 2 at 19:38
$begingroup$
if $pin C$ then there is no such $U$
$endgroup$
– Tim kinsella
Feb 2 at 22:53
1
$begingroup$
my point (of which, as i say, i'm not certain) is that the $M_i$ are there to guarantee smoothness at points of $C$ somehow.
$endgroup$
– Tim kinsella
Feb 2 at 22:55
$begingroup$
Oh, I see what you mean. I'll think about it
$endgroup$
– rmdmc89
Feb 3 at 12:04