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Why the Jauge of $C$ is $infr>0mid v/rin C$ and not $supr>0mid rvin K$?
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Why the Jauge $p:mathbb R^nto mathbb R$ of a set $Csubset mathbb R^n$ is defined as $$p(v)=infr>0mid v/rin C$$ and not as $$p(v)=infr>0mid rvnotin C quad textorquad p(v)=supr>0mid rvin C ?$$
Because both explain the same concept as : the first time we enter in $C$ or go out of $C$, and it looks more easy to work with $rv$ than with $fracvr$. So I was wondering what is the motivation to use $fracvr$ instead of $rv$.
functional-analysis
asked Jul 24 at 17:33
user330587
821310
add a comment |Â
up vote
4
down vote
favorite
Why the Jauge $p:mathbb R^nto mathbb R$ of a set $Csubset mathbb R^n$ is defined as $$p(v)=infr>0mid v/rin C$$ and not as $$p(v)=infr>0mid rvnotin C quad textorquad p(v)=supr>0mid rvin C ?$$
Because both explain the same concept as : the first time we enter in $C$ or go out of $C$, and it looks more easy to work with $rv$ than with $fracvr$. So I was wondering what is the motivation to use $fracvr$ instead of $rv$.
functional-analysis
asked Jul 24 at 17:33
user330587
821310
Isn't $K$ the same as $C$?
– Anton Grudkin
Jul 24 at 18:29
@AntonGrudkin: Yes, sorry.
– user330587
Jul 24 at 20:44
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Why the Jauge $p:mathbb R^nto mathbb R$ of a set $Csubset mathbb R^n$ is defined as $$p(v)=infr>0mid v/rin C$$ and not as $$p(v)=infr>0mid rvnotin C quad textorquad p(v)=supr>0mid rvin C ?$$
Because both explain the same concept as : the first time we enter in $C$ or go out of $C$, and it looks more easy to work with $rv$ than with $fracvr$. So I was wondering what is the motivation to use $fracvr$ instead of $rv$.
functional-analysis
asked Jul 24 at 17:33
user330587
821310
Why the Jauge $p:mathbb R^nto mathbb R$ of a set $Csubset mathbb R^n$ is defined as $$p(v)=infr>0mid v/rin C$$ and not as $$p(v)=infr>0mid rvnotin C quad textorquad p(v)=supr>0mid rvin C ?$$
Because both explain the same concept as : the first time we enter in $C$ or go out of $C$, and it looks more easy to work with $rv$ than with $fracvr$. So I was wondering what is the motivation to use $fracvr$ instead of $rv$.
functional-analysis
asked Jul 24 at 17:33
user330587
821310
edited Jul 24 at 20:45
asked Jul 24 at 17:33
user330587
821310
asked Jul 24 at 17:33
user330587
821310
asked Jul 24 at 17:33
user330587
821310
821310
Isn't $K$ the same as $C$?
– Anton Grudkin
Jul 24 at 18:29
@AntonGrudkin: Yes, sorry.
– user330587
Jul 24 at 20:44
add a comment |Â
Isn't $K$ the same as $C$?
– Anton Grudkin
Jul 24 at 18:29
@AntonGrudkin: Yes, sorry.
– user330587
Jul 24 at 20:44
Isn't $K$ the same as $C$?
– Anton Grudkin
Jul 24 at 18:29
Isn't $K$ the same as $C$?
– Anton Grudkin
Jul 24 at 18:29
@AntonGrudkin: Yes, sorry.
– user330587
Jul 24 at 20:44
@AntonGrudkin: Yes, sorry.
– user330587
Jul 24 at 20:44
add a comment |Â
1 Answer
1
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For $p(v)=infr>0mid rvnotin C$, if $mathbb Rsetminus C$ contains a neighborhood of $0$, then $p(v) = 0$ for all $v$, which is not very instructive.
For fixed $v$, let $P = r>0mid v/r in C$ and $P^* = r>0mid rvin C$.
If $r>0$, we have
$$r in P^* iff rv in C iff frac v1/r in C iff frac 1r in P$$
It follows that
$$sup P^* = frac1inf P,$$
where the 'equality' $infty = 1/0$ is implicit.
Hence, with $p^*(v) = supr>0mid rvin C$, we'd have $p^*(v) = 1/p(v)$.
I guess what you should ask yourself at this point is: do you prefer to work with infinites, or with $0$s?
If you want a more prosaic answer (but probably also more correct), I'd just say that people have worked with $p(v)$ and found it to have nice properties that might not be immediately or as easily intuitive (at least notationally) when working with $1/p(v)$.
In particular, $p$ can be equivalently written as
$$p(v) = inf r>0mid v in rC,$$
which may now remind you of Minkowski functionals.
Under certain conditions $($on $C)$, these functionals have very nice properties: they are a norm for which $C$ is a ball!
answered Jul 24 at 21:26
Fimpellizieri
16.4k11735
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
For $p(v)=infr>0mid rvnotin C$, if $mathbb Rsetminus C$ contains a neighborhood of $0$, then $p(v) = 0$ for all $v$, which is not very instructive.
For fixed $v$, let $P = r>0mid v/r in C$ and $P^* = r>0mid rvin C$.
If $r>0$, we have
$$r in P^* iff rv in C iff frac v1/r in C iff frac 1r in P$$
It follows that
$$sup P^* = frac1inf P,$$
where the 'equality' $infty = 1/0$ is implicit.
Hence, with $p^*(v) = supr>0mid rvin C$, we'd have $p^*(v) = 1/p(v)$.
I guess what you should ask yourself at this point is: do you prefer to work with infinites, or with $0$s?
If you want a more prosaic answer (but probably also more correct), I'd just say that people have worked with $p(v)$ and found it to have nice properties that might not be immediately or as easily intuitive (at least notationally) when working with $1/p(v)$.
In particular, $p$ can be equivalently written as
$$p(v) = inf r>0mid v in rC,$$
which may now remind you of Minkowski functionals.
Under certain conditions $($on $C)$, these functionals have very nice properties: they are a norm for which $C$ is a ball!
answered Jul 24 at 21:26
Fimpellizieri
16.4k11735
add a comment |Â
up vote
2
down vote
accepted
For $p(v)=infr>0mid rvnotin C$, if $mathbb Rsetminus C$ contains a neighborhood of $0$, then $p(v) = 0$ for all $v$, which is not very instructive.
For fixed $v$, let $P = r>0mid v/r in C$ and $P^* = r>0mid rvin C$.
If $r>0$, we have
$$r in P^* iff rv in C iff frac v1/r in C iff frac 1r in P$$
It follows that
$$sup P^* = frac1inf P,$$
where the 'equality' $infty = 1/0$ is implicit.
Hence, with $p^*(v) = supr>0mid rvin C$, we'd have $p^*(v) = 1/p(v)$.
I guess what you should ask yourself at this point is: do you prefer to work with infinites, or with $0$s?
If you want a more prosaic answer (but probably also more correct), I'd just say that people have worked with $p(v)$ and found it to have nice properties that might not be immediately or as easily intuitive (at least notationally) when working with $1/p(v)$.
In particular, $p$ can be equivalently written as
$$p(v) = inf r>0mid v in rC,$$
which may now remind you of Minkowski functionals.
Under certain conditions $($on $C)$, these functionals have very nice properties: they are a norm for which $C$ is a ball!
answered Jul 24 at 21:26
Fimpellizieri
16.4k11735
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For $p(v)=infr>0mid rvnotin C$, if $mathbb Rsetminus C$ contains a neighborhood of $0$, then $p(v) = 0$ for all $v$, which is not very instructive.
For fixed $v$, let $P = r>0mid v/r in C$ and $P^* = r>0mid rvin C$.
If $r>0$, we have
$$r in P^* iff rv in C iff frac v1/r in C iff frac 1r in P$$
It follows that
$$sup P^* = frac1inf P,$$
where the 'equality' $infty = 1/0$ is implicit.
Hence, with $p^*(v) = supr>0mid rvin C$, we'd have $p^*(v) = 1/p(v)$.
I guess what you should ask yourself at this point is: do you prefer to work with infinites, or with $0$s?
If you want a more prosaic answer (but probably also more correct), I'd just say that people have worked with $p(v)$ and found it to have nice properties that might not be immediately or as easily intuitive (at least notationally) when working with $1/p(v)$.
In particular, $p$ can be equivalently written as
$$p(v) = inf r>0mid v in rC,$$
which may now remind you of Minkowski functionals.
Under certain conditions $($on $C)$, these functionals have very nice properties: they are a norm for which $C$ is a ball!
answered Jul 24 at 21:26
Fimpellizieri
16.4k11735
For $p(v)=infr>0mid rvnotin C$, if $mathbb Rsetminus C$ contains a neighborhood of $0$, then $p(v) = 0$ for all $v$, which is not very instructive.
For fixed $v$, let $P = r>0mid v/r in C$ and $P^* = r>0mid rvin C$.
If $r>0$, we have
$$r in P^* iff rv in C iff frac v1/r in C iff frac 1r in P$$
It follows that
$$sup P^* = frac1inf P,$$
where the 'equality' $infty = 1/0$ is implicit.
Hence, with $p^*(v) = supr>0mid rvin C$, we'd have $p^*(v) = 1/p(v)$.
I guess what you should ask yourself at this point is: do you prefer to work with infinites, or with $0$s?
If you want a more prosaic answer (but probably also more correct), I'd just say that people have worked with $p(v)$ and found it to have nice properties that might not be immediately or as easily intuitive (at least notationally) when working with $1/p(v)$.
In particular, $p$ can be equivalently written as
$$p(v) = inf r>0mid v in rC,$$
which may now remind you of Minkowski functionals.
Under certain conditions $($on $C)$, these functionals have very nice properties: they are a norm for which $C$ is a ball!
answered Jul 24 at 21:26
Fimpellizieri
16.4k11735
answered Jul 24 at 21:26
Fimpellizieri
16.4k11735
answered Jul 24 at 21:26
Fimpellizieri
16.4k11735
answered Jul 24 at 21:26
Fimpellizieri
16.4k11735
16.4k11735
add a comment |Â
add a comment |Â
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Isn't $K$ the same as $C$?
– Anton Grudkin
Jul 24 at 18:29
@AntonGrudkin: Yes, sorry.
– user330587
Jul 24 at 20:44