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Minkowski Functional differential

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2

2

$begingroup$

I have a question about Minkowski functional.

We start with a compact, convex set $C$ containing $0$ in its interior. We define the Minkowski functional of $C$ to be the function $mu: R^n to [0,infty)$ given by the formula
begineqnarray
mu _C(x) &:=& minlambda ; lambda^-1x in C\
mu_C (0) &: =& 0.
endeqnarray

Assume now that the boundary $Sigma$ of $C$ is a $C^2$ hypersurface. Then $mu_C$ is $C^2$ at every point $x neq 0$ in $R^n$.

Now my question, how can I check that
beginequation
mu_C'(x) = x|x|^-1, ;; textfor;; xneq 0.
endequation

I did some research and found that when $C$ is symmetric, $mu$ define a norm at the space. But in my case, $C$ is not symmetric.

Another thing that may help is that: Let $x in R^n$, $x neq 0$, so the segment 0x intersect the boundary $Sigma$ in exactly one point $x'$. So
beginequation
mu_C(x) = fracd(0,x)d(0,x') = frac.
endequation
But I don't know how to differentiate $x'$.

Thank you for the help.

real-analysis functional-analysis convex-analysis

share|cite|improve this question

edited Mar 12 at 18:11

Mariana

asked Mar 11 at 19:24

MarianaMariana

575

$endgroup$

add a comment | 

2

2

$begingroup$

I have a question about Minkowski functional.

We start with a compact, convex set $C$ containing $0$ in its interior. We define the Minkowski functional of $C$ to be the function $mu: R^n to [0,infty)$ given by the formula
begineqnarray
mu _C(x) &:=& minlambda ; lambda^-1x in C\
mu_C (0) &: =& 0.
endeqnarray

Assume now that the boundary $Sigma$ of $C$ is a $C^2$ hypersurface. Then $mu_C$ is $C^2$ at every point $x neq 0$ in $R^n$.

Now my question, how can I check that
beginequation
mu_C'(x) = x|x|^-1, ;; textfor;; xneq 0.
endequation

I did some research and found that when $C$ is symmetric, $mu$ define a norm at the space. But in my case, $C$ is not symmetric.

Another thing that may help is that: Let $x in R^n$, $x neq 0$, so the segment 0x intersect the boundary $Sigma$ in exactly one point $x'$. So
beginequation
mu_C(x) = fracd(0,x)d(0,x') = frac.
endequation
But I don't know how to differentiate $x'$.

Thank you for the help.

real-analysis functional-analysis convex-analysis

share|cite|improve this question

edited Mar 12 at 18:11

Mariana

asked Mar 11 at 19:24

MarianaMariana

575

$endgroup$

add a comment | 

$begingroup$

I have a question about Minkowski functional.

We start with a compact, convex set $C$ containing $0$ in its interior. We define the Minkowski functional of $C$ to be the function $mu: R^n to [0,infty)$ given by the formula
begineqnarray
mu _C(x) &:=& minlambda ; lambda^-1x in C\
mu_C (0) &: =& 0.
endeqnarray

Assume now that the boundary $Sigma$ of $C$ is a $C^2$ hypersurface. Then $mu_C$ is $C^2$ at every point $x neq 0$ in $R^n$.

Now my question, how can I check that
beginequation
mu_C'(x) = x|x|^-1, ;; textfor;; xneq 0.
endequation

I did some research and found that when $C$ is symmetric, $mu$ define a norm at the space. But in my case, $C$ is not symmetric.

Another thing that may help is that: Let $x in R^n$, $x neq 0$, so the segment 0x intersect the boundary $Sigma$ in exactly one point $x'$. So
beginequation
mu_C(x) = fracd(0,x)d(0,x') = frac.
endequation
But I don't know how to differentiate $x'$.

Thank you for the help.

real-analysis functional-analysis convex-analysis

share|cite|improve this question

edited Mar 12 at 18:11

Mariana

asked Mar 11 at 19:24

MarianaMariana

575

$endgroup$

share|cite|improve this question

edited Mar 12 at 18:11

Mariana

asked Mar 11 at 19:24

MarianaMariana

575

share|cite|improve this question

edited Mar 12 at 18:11

Mariana

asked Mar 11 at 19:24

MarianaMariana

575

asked Mar 11 at 19:24

MarianaMariana

575

asked Mar 11 at 19:24

MarianaMariana

575