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Size of neighborhood given by implicit function theorem
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Let $f_i:mathbbR^n+1to mathbbR^n$ for $iin mathbbN$ be collection of smooth maps such that,
- $f_i(0)=0$ for all $i$;
- The Jacobian $Df_i$ is invertible at $0$ for all $i$;
- There exists $M>0$ such that $|f_i|^2leq M$ for all $i$ (i.e. $M$ is a uniform (independent of $i$) bound on the size of all $f_i$).
By (1) and (2), we can apply the implicit function theorem to all the $f_i$ at $0$: as a consequence, there exist neighborhoods $U_isubset mathbbR^n$ of $0in mathbbR^n$ and smooth maps $g_i:U_ito mathbbR$ such that $g_i(0)=0$ and $f_i(x,g_i(x))=0$ for all $xin U_i$.
Let $B_r_i(0)subset g_i(U_i)$ be the largest ball, centered at $0in mathbbR$, which is contained in the image $g_i(U_i)$. Note that the implicit function theorem ensures that $r_i>0$ for each $i$.
Question 1: Does there exist $R>0$ such that $r_igeq R$ for all $i$? In other words, can the size of $g_i(U_i)$ go to zero as $ito infty$?
Question 2: More generally, what controls the size of the neighborhoods $U_i$ (or the size of the images $g_i(U_i)$) whose existence is given by the implicit function theorem?
real-analysis analysis
asked Jul 21 at 22:47
rpf
929512
add a comment |Â
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Let $f_i:mathbbR^n+1to mathbbR^n$ for $iin mathbbN$ be collection of smooth maps such that,
- $f_i(0)=0$ for all $i$;
- The Jacobian $Df_i$ is invertible at $0$ for all $i$;
- There exists $M>0$ such that $|f_i|^2leq M$ for all $i$ (i.e. $M$ is a uniform (independent of $i$) bound on the size of all $f_i$).
By (1) and (2), we can apply the implicit function theorem to all the $f_i$ at $0$: as a consequence, there exist neighborhoods $U_isubset mathbbR^n$ of $0in mathbbR^n$ and smooth maps $g_i:U_ito mathbbR$ such that $g_i(0)=0$ and $f_i(x,g_i(x))=0$ for all $xin U_i$.
Let $B_r_i(0)subset g_i(U_i)$ be the largest ball, centered at $0in mathbbR$, which is contained in the image $g_i(U_i)$. Note that the implicit function theorem ensures that $r_i>0$ for each $i$.
Question 1: Does there exist $R>0$ such that $r_igeq R$ for all $i$? In other words, can the size of $g_i(U_i)$ go to zero as $ito infty$?
Question 2: More generally, what controls the size of the neighborhoods $U_i$ (or the size of the images $g_i(U_i)$) whose existence is given by the implicit function theorem?
real-analysis analysis
asked Jul 21 at 22:47
rpf
929512
1
Let $g:mathbb Rtomathbb R$ be a bounded increasing function with $g(0)=0$ (e.g. $g(x)=arctan(x)$). A counterexample to Q.1 is $n=1$ and $f_i(x,y)=g(y-fracg(x)i)$. I don't know about Q.2, but a lower bound on $|D_xf_i|$ might work.
– user254433
Jul 22 at 0:18
Thanks for the counterexample. By $|D_xf_i|$ do you mean the operator norm? Or determinant? Why would such a lower bound help?
– rpf
Jul 22 at 15:24
I was thinking operator norm, but you probably need more, like a bound on $D_yf_i(0)$ as well as an upper bound on $y''_i(x)$... In the $n=1$ case, differentiating $f_i(x,y_i(x))=0$ gives $D_xf_i+D_yf_i y'_i(x)=0$, so $y'_i(x)=(D_yf_i)^-1D_xf_i$. If $|D_xf_i(0)|>epsilon$ and $|D_yf_i(0)|<delta$, then $|y'_i(0)|>epsilon/delta$ uniformly in $i$. The problem is if $y''_i(x)$ gets large as $itoinfty$, then this isn't enough.
– user254433
Jul 22 at 19:18
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f_i:mathbbR^n+1to mathbbR^n$ for $iin mathbbN$ be collection of smooth maps such that,
- $f_i(0)=0$ for all $i$;
- The Jacobian $Df_i$ is invertible at $0$ for all $i$;
- There exists $M>0$ such that $|f_i|^2leq M$ for all $i$ (i.e. $M$ is a uniform (independent of $i$) bound on the size of all $f_i$).
By (1) and (2), we can apply the implicit function theorem to all the $f_i$ at $0$: as a consequence, there exist neighborhoods $U_isubset mathbbR^n$ of $0in mathbbR^n$ and smooth maps $g_i:U_ito mathbbR$ such that $g_i(0)=0$ and $f_i(x,g_i(x))=0$ for all $xin U_i$.
Let $B_r_i(0)subset g_i(U_i)$ be the largest ball, centered at $0in mathbbR$, which is contained in the image $g_i(U_i)$. Note that the implicit function theorem ensures that $r_i>0$ for each $i$.
Question 1: Does there exist $R>0$ such that $r_igeq R$ for all $i$? In other words, can the size of $g_i(U_i)$ go to zero as $ito infty$?
Question 2: More generally, what controls the size of the neighborhoods $U_i$ (or the size of the images $g_i(U_i)$) whose existence is given by the implicit function theorem?
real-analysis analysis
asked Jul 21 at 22:47
rpf
929512
Let $f_i:mathbbR^n+1to mathbbR^n$ for $iin mathbbN$ be collection of smooth maps such that,
- $f_i(0)=0$ for all $i$;
- The Jacobian $Df_i$ is invertible at $0$ for all $i$;
- There exists $M>0$ such that $|f_i|^2leq M$ for all $i$ (i.e. $M$ is a uniform (independent of $i$) bound on the size of all $f_i$).
By (1) and (2), we can apply the implicit function theorem to all the $f_i$ at $0$: as a consequence, there exist neighborhoods $U_isubset mathbbR^n$ of $0in mathbbR^n$ and smooth maps $g_i:U_ito mathbbR$ such that $g_i(0)=0$ and $f_i(x,g_i(x))=0$ for all $xin U_i$.
Let $B_r_i(0)subset g_i(U_i)$ be the largest ball, centered at $0in mathbbR$, which is contained in the image $g_i(U_i)$. Note that the implicit function theorem ensures that $r_i>0$ for each $i$.
Question 1: Does there exist $R>0$ such that $r_igeq R$ for all $i$? In other words, can the size of $g_i(U_i)$ go to zero as $ito infty$?
Question 2: More generally, what controls the size of the neighborhoods $U_i$ (or the size of the images $g_i(U_i)$) whose existence is given by the implicit function theorem?
real-analysis analysis
asked Jul 21 at 22:47
rpf
929512
asked Jul 21 at 22:47
rpf
929512
asked Jul 21 at 22:47
rpf
929512
asked Jul 21 at 22:47
rpf
929512
929512
1
Let $g:mathbb Rtomathbb R$ be a bounded increasing function with $g(0)=0$ (e.g. $g(x)=arctan(x)$). A counterexample to Q.1 is $n=1$ and $f_i(x,y)=g(y-fracg(x)i)$. I don't know about Q.2, but a lower bound on $|D_xf_i|$ might work.
– user254433
Jul 22 at 0:18
Thanks for the counterexample. By $|D_xf_i|$ do you mean the operator norm? Or determinant? Why would such a lower bound help?
– rpf
Jul 22 at 15:24
I was thinking operator norm, but you probably need more, like a bound on $D_yf_i(0)$ as well as an upper bound on $y''_i(x)$... In the $n=1$ case, differentiating $f_i(x,y_i(x))=0$ gives $D_xf_i+D_yf_i y'_i(x)=0$, so $y'_i(x)=(D_yf_i)^-1D_xf_i$. If $|D_xf_i(0)|>epsilon$ and $|D_yf_i(0)|<delta$, then $|y'_i(0)|>epsilon/delta$ uniformly in $i$. The problem is if $y''_i(x)$ gets large as $itoinfty$, then this isn't enough.
– user254433
Jul 22 at 19:18
add a comment |Â
1
Let $g:mathbb Rtomathbb R$ be a bounded increasing function with $g(0)=0$ (e.g. $g(x)=arctan(x)$). A counterexample to Q.1 is $n=1$ and $f_i(x,y)=g(y-fracg(x)i)$. I don't know about Q.2, but a lower bound on $|D_xf_i|$ might work.
– user254433
Jul 22 at 0:18
Thanks for the counterexample. By $|D_xf_i|$ do you mean the operator norm? Or determinant? Why would such a lower bound help?
– rpf
Jul 22 at 15:24
I was thinking operator norm, but you probably need more, like a bound on $D_yf_i(0)$ as well as an upper bound on $y''_i(x)$... In the $n=1$ case, differentiating $f_i(x,y_i(x))=0$ gives $D_xf_i+D_yf_i y'_i(x)=0$, so $y'_i(x)=(D_yf_i)^-1D_xf_i$. If $|D_xf_i(0)|>epsilon$ and $|D_yf_i(0)|<delta$, then $|y'_i(0)|>epsilon/delta$ uniformly in $i$. The problem is if $y''_i(x)$ gets large as $itoinfty$, then this isn't enough.
– user254433
Jul 22 at 19:18
1
1
Let $g:mathbb Rtomathbb R$ be a bounded increasing function with $g(0)=0$ (e.g. $g(x)=arctan(x)$). A counterexample to Q.1 is $n=1$ and $f_i(x,y)=g(y-fracg(x)i)$. I don't know about Q.2, but a lower bound on $|D_xf_i|$ might work.
– user254433
Jul 22 at 0:18
Let $g:mathbb Rtomathbb R$ be a bounded increasing function with $g(0)=0$ (e.g. $g(x)=arctan(x)$). A counterexample to Q.1 is $n=1$ and $f_i(x,y)=g(y-fracg(x)i)$. I don't know about Q.2, but a lower bound on $|D_xf_i|$ might work.
– user254433
Jul 22 at 0:18
Thanks for the counterexample. By $|D_xf_i|$ do you mean the operator norm? Or determinant? Why would such a lower bound help?
– rpf
Jul 22 at 15:24
Thanks for the counterexample. By $|D_xf_i|$ do you mean the operator norm? Or determinant? Why would such a lower bound help?
– rpf
Jul 22 at 15:24
I was thinking operator norm, but you probably need more, like a bound on $D_yf_i(0)$ as well as an upper bound on $y''_i(x)$... In the $n=1$ case, differentiating $f_i(x,y_i(x))=0$ gives $D_xf_i+D_yf_i y'_i(x)=0$, so $y'_i(x)=(D_yf_i)^-1D_xf_i$. If $|D_xf_i(0)|>epsilon$ and $|D_yf_i(0)|<delta$, then $|y'_i(0)|>epsilon/delta$ uniformly in $i$. The problem is if $y''_i(x)$ gets large as $itoinfty$, then this isn't enough.
– user254433
Jul 22 at 19:18
I was thinking operator norm, but you probably need more, like a bound on $D_yf_i(0)$ as well as an upper bound on $y''_i(x)$... In the $n=1$ case, differentiating $f_i(x,y_i(x))=0$ gives $D_xf_i+D_yf_i y'_i(x)=0$, so $y'_i(x)=(D_yf_i)^-1D_xf_i$. If $|D_xf_i(0)|>epsilon$ and $|D_yf_i(0)|<delta$, then $|y'_i(0)|>epsilon/delta$ uniformly in $i$. The problem is if $y''_i(x)$ gets large as $itoinfty$, then this isn't enough.
– user254433
Jul 22 at 19:18
add a comment |Â
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1
Let $g:mathbb Rtomathbb R$ be a bounded increasing function with $g(0)=0$ (e.g. $g(x)=arctan(x)$). A counterexample to Q.1 is $n=1$ and $f_i(x,y)=g(y-fracg(x)i)$. I don't know about Q.2, but a lower bound on $|D_xf_i|$ might work.
– user254433
Jul 22 at 0:18
Thanks for the counterexample. By $|D_xf_i|$ do you mean the operator norm? Or determinant? Why would such a lower bound help?
– rpf
Jul 22 at 15:24
I was thinking operator norm, but you probably need more, like a bound on $D_yf_i(0)$ as well as an upper bound on $y''_i(x)$... In the $n=1$ case, differentiating $f_i(x,y_i(x))=0$ gives $D_xf_i+D_yf_i y'_i(x)=0$, so $y'_i(x)=(D_yf_i)^-1D_xf_i$. If $|D_xf_i(0)|>epsilon$ and $|D_yf_i(0)|<delta$, then $|y'_i(0)|>epsilon/delta$ uniformly in $i$. The problem is if $y''_i(x)$ gets large as $itoinfty$, then this isn't enough.
– user254433
Jul 22 at 19:18