Derivative Of A Function Defined In Terms Of An Infimum
$begingroup$
Let $A subset mathbb{R}^n$ be compact and let $Phi: mathbb{R}^n rightarrow mathbb{R}$ be defined by $displaystyle Phi(x) = inf_{p in A} ||p - x||_2^2$. I want to find the total derivative of $Phi$, but I don't know what to do with this infimum. Is it possible to interchange the infimum and a partial derivative?
EDIT: Since $A$ is compact, I suppose it should be a minimum, not an infimum. I will add then, to my question, about what happens if $A$ is bounded but not closed and we have an infimum.
real-analysis
asked Dec 8 '18 at 4:09
Frederic ChopinFrederic Chopin
321111
$endgroup$
add a comment |
$begingroup$
Let $A subset mathbb{R}^n$ be compact and let $Phi: mathbb{R}^n rightarrow mathbb{R}$ be defined by $displaystyle Phi(x) = inf_{p in A} ||p - x||_2^2$. I want to find the total derivative of $Phi$, but I don't know what to do with this infimum. Is it possible to interchange the infimum and a partial derivative?
EDIT: Since $A$ is compact, I suppose it should be a minimum, not an infimum. I will add then, to my question, about what happens if $A$ is bounded but not closed and we have an infimum.
real-analysis
asked Dec 8 '18 at 4:09
Frederic ChopinFrederic Chopin
321111
$endgroup$
1
$begingroup$
$Phi$ is not differentiable everywhere. For example, if $A$ is a square (not including the interior), then $Phi$ will not be differentiable on the diagonals of that square. The infimum/minimum and derivative are definitely NOT interchangable. To your edited question, you get the same $Phi$ from $overline A$ as from $A$ itself, so there is nothing gained from relaxing the compactness condition.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 14:55
$begingroup$
Thank you for your response! Yes, a square is a good example of what could go wrong. I'm going to post a new question asking if this could work if $A$ is a smooth, embedded submanifold of $mathbb{R}^n$ (since a square is not an example of this).
$endgroup$
– Frederic Chopin
Dec 8 '18 at 15:59
$begingroup$
Here it is for reference: math.stackexchange.com/questions/3031295/…
$endgroup$
– Frederic Chopin
Dec 8 '18 at 16:14
add a comment |
$begingroup$
Let $A subset mathbb{R}^n$ be compact and let $Phi: mathbb{R}^n rightarrow mathbb{R}$ be defined by $displaystyle Phi(x) = inf_{p in A} ||p - x||_2^2$. I want to find the total derivative of $Phi$, but I don't know what to do with this infimum. Is it possible to interchange the infimum and a partial derivative?
EDIT: Since $A$ is compact, I suppose it should be a minimum, not an infimum. I will add then, to my question, about what happens if $A$ is bounded but not closed and we have an infimum.
real-analysis
asked Dec 8 '18 at 4:09
Frederic ChopinFrederic Chopin
321111
$endgroup$
Let $A subset mathbb{R}^n$ be compact and let $Phi: mathbb{R}^n rightarrow mathbb{R}$ be defined by $displaystyle Phi(x) = inf_{p in A} ||p - x||_2^2$. I want to find the total derivative of $Phi$, but I don't know what to do with this infimum. Is it possible to interchange the infimum and a partial derivative?
EDIT: Since $A$ is compact, I suppose it should be a minimum, not an infimum. I will add then, to my question, about what happens if $A$ is bounded but not closed and we have an infimum.
real-analysis
real-analysis
asked Dec 8 '18 at 4:09
Frederic ChopinFrederic Chopin
321111
asked Dec 8 '18 at 4:09
Frederic ChopinFrederic Chopin
321111
edited Dec 8 '18 at 4:18
Frederic Chopin
asked Dec 8 '18 at 4:09
Frederic ChopinFrederic Chopin
321111
asked Dec 8 '18 at 4:09
Frederic ChopinFrederic Chopin
321111
asked Dec 8 '18 at 4:09
Frederic ChopinFrederic Chopin
321111
321111
1
$begingroup$
$Phi$ is not differentiable everywhere. For example, if $A$ is a square (not including the interior), then $Phi$ will not be differentiable on the diagonals of that square. The infimum/minimum and derivative are definitely NOT interchangable. To your edited question, you get the same $Phi$ from $overline A$ as from $A$ itself, so there is nothing gained from relaxing the compactness condition.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 14:55
$begingroup$
Thank you for your response! Yes, a square is a good example of what could go wrong. I'm going to post a new question asking if this could work if $A$ is a smooth, embedded submanifold of $mathbb{R}^n$ (since a square is not an example of this).
$endgroup$
– Frederic Chopin
Dec 8 '18 at 15:59
$begingroup$
Here it is for reference: math.stackexchange.com/questions/3031295/…
$endgroup$
– Frederic Chopin
Dec 8 '18 at 16:14
add a comment |
1
$begingroup$
$Phi$ is not differentiable everywhere. For example, if $A$ is a square (not including the interior), then $Phi$ will not be differentiable on the diagonals of that square. The infimum/minimum and derivative are definitely NOT interchangable. To your edited question, you get the same $Phi$ from $overline A$ as from $A$ itself, so there is nothing gained from relaxing the compactness condition.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 14:55
$begingroup$
Thank you for your response! Yes, a square is a good example of what could go wrong. I'm going to post a new question asking if this could work if $A$ is a smooth, embedded submanifold of $mathbb{R}^n$ (since a square is not an example of this).
$endgroup$
– Frederic Chopin
Dec 8 '18 at 15:59
$begingroup$
Here it is for reference: math.stackexchange.com/questions/3031295/…
$endgroup$
– Frederic Chopin
Dec 8 '18 at 16:14
1
1
$begingroup$
$Phi$ is not differentiable everywhere. For example, if $A$ is a square (not including the interior), then $Phi$ will not be differentiable on the diagonals of that square. The infimum/minimum and derivative are definitely NOT interchangable. To your edited question, you get the same $Phi$ from $overline A$ as from $A$ itself, so there is nothing gained from relaxing the compactness condition.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 14:55
$begingroup$
$Phi$ is not differentiable everywhere. For example, if $A$ is a square (not including the interior), then $Phi$ will not be differentiable on the diagonals of that square. The infimum/minimum and derivative are definitely NOT interchangable. To your edited question, you get the same $Phi$ from $overline A$ as from $A$ itself, so there is nothing gained from relaxing the compactness condition.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 14:55
$begingroup$
Thank you for your response! Yes, a square is a good example of what could go wrong. I'm going to post a new question asking if this could work if $A$ is a smooth, embedded submanifold of $mathbb{R}^n$ (since a square is not an example of this).
$endgroup$
– Frederic Chopin
Dec 8 '18 at 15:59
$begingroup$
Thank you for your response! Yes, a square is a good example of what could go wrong. I'm going to post a new question asking if this could work if $A$ is a smooth, embedded submanifold of $mathbb{R}^n$ (since a square is not an example of this).
$endgroup$
– Frederic Chopin
Dec 8 '18 at 15:59
$begingroup$
Here it is for reference: math.stackexchange.com/questions/3031295/…
$endgroup$
– Frederic Chopin
Dec 8 '18 at 16:14
$begingroup$
Here it is for reference: math.stackexchange.com/questions/3031295/…
$endgroup$
– Frederic Chopin
Dec 8 '18 at 16:14
add a comment |
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$begingroup$
$Phi$ is not differentiable everywhere. For example, if $A$ is a square (not including the interior), then $Phi$ will not be differentiable on the diagonals of that square. The infimum/minimum and derivative are definitely NOT interchangable. To your edited question, you get the same $Phi$ from $overline A$ as from $A$ itself, so there is nothing gained from relaxing the compactness condition.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 14:55
$begingroup$
Thank you for your response! Yes, a square is a good example of what could go wrong. I'm going to post a new question asking if this could work if $A$ is a smooth, embedded submanifold of $mathbb{R}^n$ (since a square is not an example of this).
$endgroup$
– Frederic Chopin
Dec 8 '18 at 15:59
$begingroup$
Here it is for reference: math.stackexchange.com/questions/3031295/…
$endgroup$
– Frederic Chopin
Dec 8 '18 at 16:14