Derivative Of Infimum - Embedded Submanifold Of $mathbb{R}^n$
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This is a follow-up question to:
Derivative Of A Function Defined In Terms Of An Infimum
Let $A$ be a compact, (smooth) embedded submanifold of $mathbb{R}^n$ and let $Phi : mathbb{R}^n rightarrow mathbb{R}$ be defined by $displaystyle Phi(x) = inf_{p in A} ||p - x||_2^2$. Does $Phi$ have a total derivative everywhere? How does one go about finding the total derivative of $Phi$? Is it possible to interchange the infimum and a partial derivative?
multivariable-calculus differential-geometry
asked Dec 8 '18 at 16:14
Frederic ChopinFrederic Chopin
333111
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add a comment |
$begingroup$
This is a follow-up question to:
Derivative Of A Function Defined In Terms Of An Infimum
Let $A$ be a compact, (smooth) embedded submanifold of $mathbb{R}^n$ and let $Phi : mathbb{R}^n rightarrow mathbb{R}$ be defined by $displaystyle Phi(x) = inf_{p in A} ||p - x||_2^2$. Does $Phi$ have a total derivative everywhere? How does one go about finding the total derivative of $Phi$? Is it possible to interchange the infimum and a partial derivative?
multivariable-calculus differential-geometry
asked Dec 8 '18 at 16:14
Frederic ChopinFrederic Chopin
333111
$endgroup$
1
$begingroup$
The reason my square counter-example in the other thread worked wasn't because it was not smooth. It worked because it wasn't convex. For any non-convex set, there will be points outside of it for which there are closest points in $A$ to it in more than one direction. $Phi$ will not be differentiable at any such point.
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– Paul Sinclair
Dec 8 '18 at 18:49
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And no, it is still NOT possible to exchange an infimum and partial derivative.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 18:50
$begingroup$
Ah, I see. Thank you.
$endgroup$
– Frederic Chopin
Dec 8 '18 at 19:45
add a comment |
$begingroup$
This is a follow-up question to:
Derivative Of A Function Defined In Terms Of An Infimum
Let $A$ be a compact, (smooth) embedded submanifold of $mathbb{R}^n$ and let $Phi : mathbb{R}^n rightarrow mathbb{R}$ be defined by $displaystyle Phi(x) = inf_{p in A} ||p - x||_2^2$. Does $Phi$ have a total derivative everywhere? How does one go about finding the total derivative of $Phi$? Is it possible to interchange the infimum and a partial derivative?
multivariable-calculus differential-geometry
asked Dec 8 '18 at 16:14
Frederic ChopinFrederic Chopin
333111
$endgroup$
This is a follow-up question to:
Derivative Of A Function Defined In Terms Of An Infimum
Let $A$ be a compact, (smooth) embedded submanifold of $mathbb{R}^n$ and let $Phi : mathbb{R}^n rightarrow mathbb{R}$ be defined by $displaystyle Phi(x) = inf_{p in A} ||p - x||_2^2$. Does $Phi$ have a total derivative everywhere? How does one go about finding the total derivative of $Phi$? Is it possible to interchange the infimum and a partial derivative?
multivariable-calculus differential-geometry
multivariable-calculus differential-geometry
asked Dec 8 '18 at 16:14
Frederic ChopinFrederic Chopin
333111
asked Dec 8 '18 at 16:14
Frederic ChopinFrederic Chopin
333111
asked Dec 8 '18 at 16:14
Frederic ChopinFrederic Chopin
333111
asked Dec 8 '18 at 16:14
Frederic ChopinFrederic Chopin
333111
asked Dec 8 '18 at 16:14
Frederic ChopinFrederic Chopin
333111
333111
1
$begingroup$
The reason my square counter-example in the other thread worked wasn't because it was not smooth. It worked because it wasn't convex. For any non-convex set, there will be points outside of it for which there are closest points in $A$ to it in more than one direction. $Phi$ will not be differentiable at any such point.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 18:49
$begingroup$
And no, it is still NOT possible to exchange an infimum and partial derivative.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 18:50
$begingroup$
Ah, I see. Thank you.
$endgroup$
– Frederic Chopin
Dec 8 '18 at 19:45
add a comment |
1
$begingroup$
The reason my square counter-example in the other thread worked wasn't because it was not smooth. It worked because it wasn't convex. For any non-convex set, there will be points outside of it for which there are closest points in $A$ to it in more than one direction. $Phi$ will not be differentiable at any such point.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 18:49
$begingroup$
And no, it is still NOT possible to exchange an infimum and partial derivative.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 18:50
$begingroup$
Ah, I see. Thank you.
$endgroup$
– Frederic Chopin
Dec 8 '18 at 19:45
1
1
$begingroup$
The reason my square counter-example in the other thread worked wasn't because it was not smooth. It worked because it wasn't convex. For any non-convex set, there will be points outside of it for which there are closest points in $A$ to it in more than one direction. $Phi$ will not be differentiable at any such point.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 18:49
$begingroup$
The reason my square counter-example in the other thread worked wasn't because it was not smooth. It worked because it wasn't convex. For any non-convex set, there will be points outside of it for which there are closest points in $A$ to it in more than one direction. $Phi$ will not be differentiable at any such point.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 18:49
$begingroup$
And no, it is still NOT possible to exchange an infimum and partial derivative.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 18:50
$begingroup$
And no, it is still NOT possible to exchange an infimum and partial derivative.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 18:50
$begingroup$
Ah, I see. Thank you.
$endgroup$
– Frederic Chopin
Dec 8 '18 at 19:45
$begingroup$
Ah, I see. Thank you.
$endgroup$
– Frederic Chopin
Dec 8 '18 at 19:45
add a comment |
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$begingroup$
The reason my square counter-example in the other thread worked wasn't because it was not smooth. It worked because it wasn't convex. For any non-convex set, there will be points outside of it for which there are closest points in $A$ to it in more than one direction. $Phi$ will not be differentiable at any such point.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 18:49
$begingroup$
And no, it is still NOT possible to exchange an infimum and partial derivative.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 18:50
$begingroup$
Ah, I see. Thank you.
$endgroup$
– Frederic Chopin
Dec 8 '18 at 19:45