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Minkowski functional

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Function made from a set

Inmathematics, in the field offunctional analysis, aMinkowski functional (afterHermann Minkowski) orgauge function is a function that recovers a notion of distance on a linear space.

If ${\textstyle K}$ is a subset of areal orcomplex vector space ${\textstyle X,}$ then theMinkowski functional orgauge of ${\textstyle K}$ is defined to be thefunction ${\textstyle p_{K}:X\to [0,\infty ],}$ valued in theextended real numbers, defined by $p_{K}(x):=\inf\{r\in \mathbb {R} :r>0{\text{ and }}x\in rK\}\quad {\text{ for every }}x\in X,$ where theinfimum of the empty set is defined to bepositive infinity ${\textstyle \,\infty \,}$ (which isnot a real number so that ${\textstyle p_{K}(x)}$ would thennot be real-valued).

The set ${\textstyle K}$ is often assumed/picked to have properties, such as being an absorbingdisk in ${\textstyle X}$ , that guarantee that ${\textstyle p_{K}}$ will be a real-valuedseminorm on ${\textstyle X.}$ In fact, every seminorm ${\textstyle p}$ on ${\textstyle X}$ is equal to the Minkowski functional (that is, ${\textstyle p=p_{K}}$ ) of any subset ${\textstyle K}$ of ${\textstyle X}$ satisfying

$\{x\in X:p(x)<1\}\subseteq K\subseteq \{x\in X:p(x)\leq 1\}$

(where all three of these sets are necessarily absorbing in ${\textstyle X}$ and the first and last are also disks).

Thus every seminorm (which is afunction defined by purely algebraic properties) can be associated (non-uniquely) with an absorbing disk (which is aset with certain geometric properties) and conversely, every absorbing disk can be associated with its Minkowski functional (which will necessarily be a seminorm). These relationships between seminorms, Minkowski functionals, and absorbing disks is a major reason why Minkowski functionals are studied and used in functional analysis. In particular, through these relationships, Minkowski functionals allow one to "translate" certaingeometric properties of a subset of ${\textstyle X}$ into certainalgebraic properties of a function on ${\textstyle X.}$

The Minkowski function is always non-negative (meaning ${\textstyle p_{K}\geq 0}$ ). This property of being nonnegative stands in contrast to other classes of functions, such assublinear functions and reallinear functionals, that do allow negative values. However, ${\textstyle p_{K}}$ might not be real-valued since for any given ${\textstyle x\in X,}$ the value ${\textstyle p_{K}(x)}$ is a real number if and only if ${\textstyle \{r>0:x\in rK\}}$ is notempty.Consequently, ${\textstyle K}$ is usually assumed to have properties (such as beingabsorbing in ${\textstyle X,}$ for instance) that will guarantee that ${\textstyle p_{K}}$ is real-valued.

Definition

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Let ${\textstyle K}$ be a subset of a real or complex vector space ${\textstyle X.}$ Define thegauge of ${\textstyle K}$ or theMinkowski functional associated with or induced by ${\textstyle K}$ as being the function ${\textstyle p_{K}:X\to [0,\infty ],}$ valued in theextended real numbers, defined by

$p_{K}(x):=\inf\{r>0:x\in rK\},$

(recall that theinfimum of the empty set is ${\textstyle \,\infty }$ , that is, ${\textstyle \inf \varnothing =\infty }$ ). Here, ${\textstyle \{r>0:x\in rK\}}$ is shorthand for ${\textstyle \{r\in \mathbb {R} :r>0{\text{ and }}x\in rK\}.}$

For any ${\textstyle x\in X,}$ ${\textstyle p_{K}(x)\neq \infty }$ if and only if ${\textstyle \{r>0:x\in rK\}}$ is not empty. The arithmetic operations on ${\textstyle \mathbb {R} }$ can be extended to operate on ${\textstyle \pm \infty ,}$ where ${\textstyle {\frac {r}{\pm \infty }}:=0}$ for all non-zero real ${\textstyle -\infty <r<\infty .}$ The products ${\textstyle 0\cdot \infty }$ and ${\textstyle 0\cdot -\infty }$ remain undefined.

Some conditions making a gauge real-valued

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In the field ofconvex analysis, the map ${\textstyle p_{K}}$ taking on the value of ${\textstyle \,\infty \,}$ is not necessarily an issue. However, in functional analysis ${\textstyle p_{K}}$ is almost always real-valued (that is, to never take on the value of ${\textstyle \,\infty \,}$ ), which happens if and only if the set ${\textstyle \{r>0:x\in rK\}}$ is non-empty for every ${\textstyle x\in X.}$

In order for ${\textstyle p_{K}}$ to be real-valued, it suffices for the origin of ${\textstyle X}$ to belong to thealgebraic interior orcore of ${\textstyle K}$ in ${\textstyle X.}$ ^[1] If ${\textstyle K}$ isabsorbing in ${\textstyle X,}$ where recall that this implies that ${\textstyle 0\in K,}$ then the origin belongs to thealgebraic interior of ${\textstyle K}$ in ${\textstyle X}$ and thus ${\textstyle p_{K}}$ is real-valued. Characterizations of when ${\textstyle p_{K}}$ is real-valued are given below.

Motivating examples

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Example 1

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Consider anormed vector space ${\textstyle (X,\|\,\cdot \,\|),}$ with the norm ${\textstyle \|\,\cdot \,\|}$ and let ${\textstyle U:=\{x\in X:\|x\|\leq 1\}}$ be the unit ball in ${\textstyle X.}$ Then for every ${\textstyle x\in X,}$ ${\textstyle \|x\|=p_{U}(x).}$ Thus the Minkowski functional ${\textstyle p_{U}}$ is just the norm on ${\textstyle X.}$

Example 2

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Let ${\textstyle X}$ be a vector space without topology with underlying scalar field ${\textstyle \mathbb {K} .}$ Let ${\textstyle f:X\to \mathbb {K} }$ be anylinear functional on ${\textstyle X}$ (not necessarily continuous). Fix ${\textstyle a>0.}$ Let ${\textstyle K}$ be the set $K:=\{x\in X:|f(x)|\leq a\}$ and let ${\textstyle p_{K}}$ be the Minkowski functional of ${\textstyle K.}$ Then $p_{K}(x)={\frac {1}{a}}|f(x)|\quad {\text{ for all }}x\in X.$ The function ${\textstyle p_{K}}$ has the following properties:

It issubadditive: ${\textstyle p_{K}(x+y)\leq p_{K}(x)+p_{K}(y).}$
It isabsolutely homogeneous: ${\textstyle p_{K}(sx)=|s|p_{K}(x)}$ for all scalars ${\textstyle s.}$
It isnonnegative: ${\textstyle p_{K}\geq 0.}$

Therefore, ${\textstyle p_{K}}$ is aseminorm on ${\textstyle X,}$ with an induced topology. This is characteristic of Minkowski functionals defined via "nice" sets. There is a one-to-one correspondence between seminorms and the Minkowski functional given by such sets. What is meant precisely by "nice" is discussed in the section below.

Notice that, in contrast to a stronger requirement for a norm, ${\textstyle p_{K}(x)=0}$ need not imply ${\textstyle x=0.}$ In the above example, one can take a nonzero ${\textstyle x}$ from the kernel of ${\textstyle f.}$ Consequently, the resulting topology need not beHausdorff.

Common conditions guaranteeing gauges are seminorms

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To guarantee that ${\textstyle p_{K}(0)=0,}$ it will henceforth be assumed that ${\textstyle 0\in K.}$

In order for ${\textstyle p_{K}}$ to be a seminorm, it suffices for ${\textstyle K}$ to be adisk (that is, convex and balanced) and absorbing in ${\textstyle X,}$ which are the most common assumption placed on ${\textstyle K.}$

Theorem^[2]—If ${\textstyle K}$ is anabsorbing disk in a vector space ${\textstyle X}$ then the Minkowski functional of ${\textstyle K,}$ which is the map ${\textstyle p_{K}:X\to [0,\infty )}$ defined by $p_{K}(x):=\inf\{r>0:x\in rK\},$ is a seminorm on ${\textstyle X.}$ Moreover, $p_{K}(x)={\frac {1}{\sup\{r>0:rx\in K\}}}.$

More generally, if ${\textstyle K}$ is convex and the origin belongs to thealgebraic interior of ${\textstyle K,}$ then ${\textstyle p_{K}}$ is a nonnegativesublinear functional on ${\textstyle X,}$ which implies in particular that it issubadditive andpositive homogeneous.If ${\textstyle K}$ is absorbing in ${\textstyle X}$ then ${\textstyle p_{[0,1]K}}$ is positive homogeneous, meaning that ${\textstyle p_{[0,1]K}(sx)=sp_{[0,1]K}(x)}$ for all real ${\textstyle s\geq 0,}$ where ${\textstyle [0,1]K=\{tk:t\in [0,1],k\in K\}.}$ ^[3]If ${\textstyle q}$ is a nonnegative real-valued function on ${\textstyle X}$ that is positive homogeneous, then the sets ${\textstyle U:=\{x\in X:q(x)<1\}}$ and ${\textstyle D:=\{x\in X:q(x)\leq 1\}}$ satisfy ${\textstyle [0,1]U=U}$ and ${\textstyle [0,1]D=D;}$ if in addition ${\textstyle q}$ is absolutely homogeneous then both ${\textstyle U}$ and ${\textstyle D}$ arebalanced.^[3]

Gauges of absorbing disks

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Arguably the most common requirements placed on a set ${\textstyle K}$ to guarantee that ${\textstyle p_{K}}$ is a seminorm are that ${\textstyle K}$ be anabsorbing disk in ${\textstyle X.}$ Due to how common these assumptions are, the properties of a Minkowski functional ${\textstyle p_{K}}$ when ${\textstyle K}$ is an absorbing disk will now be investigated. Since all of the results mentioned above made few (if any) assumptions on ${\textstyle K,}$ they can be applied in this special case.

Theorem—Assume that ${\textstyle K}$ is an absorbing subset of ${\textstyle X.}$ It is shown that:

If ${\textstyle K}$ isconvex then ${\textstyle p_{K}}$ is subadditive.
If ${\textstyle K}$ isbalanced then ${\textstyle p_{K}}$ isabsolutely homogeneous; that is, ${\textstyle p_{K}(sx)=|s|p_{K}(x)}$ for all scalars ${\textstyle s.}$

Proof that the Gauge of an absorbing disk is a seminorm
Convexity and subadditivity A simple geometric argument that shows convexity of ${\textstyle K}$ implies subadditivity is as follows. Suppose for the moment that ${\textstyle p_{K}(x)=p_{K}(y)=r.}$ Then for all ${\textstyle e>0,}$ ${\textstyle x,y\in K_{e}:=(r,e)K.}$ Since ${\textstyle K}$ is convex and ${\textstyle r+e\neq 0,}$ ${\textstyle K_{e}}$ is also convex. Therefore, ${\textstyle {\frac {1}{2}}x+{\frac {1}{2}}y\in K_{e}.}$ By definition of the Minkowski functional ${\textstyle p_{K},}$ $p_{K}\left({\frac {1}{2}}x+{\frac {1}{2}}y\right)\leq r+e={\frac {1}{2}}p_{K}(x)+{\frac {1}{2}}p_{K}(y)+e.$ But the left hand side is ${\textstyle {\frac {1}{2}}p_{K}(x+y),}$ so that $p_{K}(x+y)\leq p_{K}(x)+p_{K}(y)+2e.$ Since ${\textstyle e>0}$ was arbitrary, it follows that ${\textstyle p_{K}(x+y)\leq p_{K}(x)+p_{K}(y),}$ which is the desired inequality. The general case ${\textstyle p_{K}(x)>p_{K}(y)}$ is obtained after the obvious modification. Convexity of ${\textstyle K,}$ together with the initial assumption that the set ${\textstyle \{r>0:x\in rK\}}$ is nonempty, implies that ${\textstyle K}$ isabsorbing. Balancedness and absolute homogeneity Notice that ${\textstyle K}$ being balanced implies that $\lambda x\in rK\quad {\mbox{if and only if}}\quad x\in {\frac {r}{\|\lambda \|}}K.$ Therefore $p_{K}(\lambda x)=\inf \left\{r>0:\lambda x\in rK\right\}=\inf \left\{r>0:x\in {\frac {r}{\|\lambda \|}}K\right\}=\inf \left\{\|\lambda \|{\frac {r}{\|\lambda \|}}>0:x\in {\frac {r}{\|\lambda \|}}K\right\}=\|\lambda \|p_{K}(x).$

Proof that the Gauge of an absorbing disk is a seminorm

Convexity and subadditivity

A simple geometric argument that shows convexity of ${\textstyle K}$ implies subadditivity is as follows. Suppose for the moment that ${\textstyle p_{K}(x)=p_{K}(y)=r.}$ Then for all ${\textstyle e>0,}$ ${\textstyle x,y\in K_{e}:=(r,e)K.}$ Since ${\textstyle K}$ is convex and ${\textstyle r+e\neq 0,}$ ${\textstyle K_{e}}$ is also convex. Therefore, ${\textstyle {\frac {1}{2}}x+{\frac {1}{2}}y\in K_{e}.}$ By definition of the Minkowski functional ${\textstyle p_{K},}$ $p_{K}\left({\frac {1}{2}}x+{\frac {1}{2}}y\right)\leq r+e={\frac {1}{2}}p_{K}(x)+{\frac {1}{2}}p_{K}(y)+e.$

But the left hand side is ${\textstyle {\frac {1}{2}}p_{K}(x+y),}$ so that $p_{K}(x+y)\leq p_{K}(x)+p_{K}(y)+2e.$

Since ${\textstyle e>0}$ was arbitrary, it follows that ${\textstyle p_{K}(x+y)\leq p_{K}(x)+p_{K}(y),}$ which is the desired inequality. The general case ${\textstyle p_{K}(x)>p_{K}(y)}$ is obtained after the obvious modification.

Convexity of ${\textstyle K,}$ together with the initial assumption that the set ${\textstyle \{r>0:x\in rK\}}$ is nonempty, implies that ${\textstyle K}$ isabsorbing.

Balancedness and absolute homogeneity

Notice that ${\textstyle K}$ being balanced implies that $\lambda x\in rK\quad {\mbox{if and only if}}\quad x\in {\frac {r}{|\lambda |}}K.$

Therefore $p_{K}(\lambda x)=\inf \left\{r>0:\lambda x\in rK\right\}=\inf \left\{r>0:x\in {\frac {r}{|\lambda |}}K\right\}=\inf \left\{|\lambda |{\frac {r}{|\lambda |}}>0:x\in {\frac {r}{|\lambda |}}K\right\}=|\lambda |p_{K}(x).$

Algebraic properties

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Let ${\textstyle X}$ be a real or complex vector space and let ${\textstyle K}$ be an absorbing disk in ${\textstyle X.}$

${\textstyle p_{K}}$ is aseminorm on ${\textstyle X.}$
${\textstyle p_{K}}$ is anorm on ${\textstyle X}$ if and only if ${\textstyle K}$ does not contain a non-trivial vector subspace.^[4]
${\textstyle p_{sK}={\frac {1}{|s|}}p_{K}}$ for any scalar ${\textstyle s\neq 0.}$ ^[4]
If ${\textstyle J}$ is an absorbing disk in ${\textstyle X}$ and ${\textstyle J\subseteq K}$ then ${\textstyle p_{K}\leq p_{J}.}$
If ${\textstyle K}$ is a set satisfying ${\textstyle \{x\in X:p(x)<1\}\;\subseteq \;K\;\subseteq \;\{x\in X:p(x)\leq 1\}}$ then ${\textstyle K}$ isabsorbing in ${\textstyle X}$ and ${\textstyle p=p_{K},}$ where ${\textstyle p_{K}}$ is the Minkowski functional associated with ${\textstyle K;}$ that is, it is the gauge of ${\textstyle K.}$ ^[5]
In particular, if ${\textstyle K}$ is as above and ${\textstyle q}$ is any seminorm on ${\textstyle X,}$ then ${\textstyle q=p}$ if and only if ${\textstyle \{x\in X:q(x)<1\}\;\subseteq \;K\;\subseteq \;\{x\in X:q(x)\leq 1\}.}$ ^[5]
If ${\textstyle x\in X}$ satisfies ${\textstyle p_{K}(x)<1}$ then ${\textstyle x\in K.}$

Topological properties

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Assume that ${\textstyle X}$ is a (real or complex)topological vector space (TVS) (not necessarilyHausdorff orlocally convex) and let ${\textstyle K}$ be an absorbing disk in ${\textstyle X.}$ Then

$\operatorname {Int} _{X}K\;\subseteq \;\{x\in X:p_{K}(x)<1\}\;\subseteq \;K\;\subseteq \;\{x\in X:p_{K}(x)\leq 1\}\;\subseteq \;\operatorname {Cl} _{X}K,$

where ${\textstyle \operatorname {Int} _{X}K}$ is thetopological interior and ${\textstyle \operatorname {Cl} _{X}K}$ is thetopological closure of ${\textstyle K}$ in ${\textstyle X.}$ ^[6] Importantly, it wasnot assumed that ${\textstyle p_{K}}$ was continuous nor was it assumed that ${\textstyle K}$ had any topological properties.

Moreover, the Minkowski functional ${\textstyle p_{K}}$ is continuous if and only if ${\textstyle K}$ is a neighborhood of the origin in ${\textstyle X.}$ ^[6] If ${\textstyle p_{K}}$ is continuous then^[6] $\operatorname {Int} _{X}K=\{x\in X:p_{K}(x)<1\}\quad {\text{ and }}\quad \operatorname {Cl} _{X}K=\{x\in X:p_{K}(x)\leq 1\}.$

Minimal requirements on the set

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This section will investigate the most general case of the gauge ofany subset ${\textstyle K}$ of ${\textstyle X.}$ The more common special case where ${\textstyle K}$ is assumed to be anabsorbing disk in ${\textstyle X}$ was discussed above.

Properties

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All results in this section may be applied to the case where ${\textstyle K}$ is an absorbing disk.

Throughout, ${\textstyle K}$ is any subset of ${\textstyle X.}$

Summary—Suppose that ${\textstyle K}$ is a subset of a real or complex vector space ${\textstyle X.}$

Strict positive homogeneity: ${\textstyle p_{K}(rx)=rp_{K}(x)}$ for all ${\textstyle x\in X}$ and allpositive real ${\textstyle r>0.}$
- Positive/Nonnegative homogeneity: ${\textstyle p_{K}}$ is nonnegative homogeneous if and only if ${\textstyle p_{K}}$ is real-valued.
  - A map ${\textstyle p}$ is callednonnegative homogeneous^[7] if ${\textstyle p(rx)=rp(x)}$ for all ${\textstyle x\in X}$ and allnonnegative real ${\textstyle r\geq 0.}$ Since ${\textstyle 0\cdot \infty }$ is undefined, a map that takes infinity as a value is not nonnegative homogeneous.
Real-values: ${\textstyle (0,\infty )K}$ is the set of all points on which ${\textstyle p_{K}}$ is real valued. So ${\textstyle p_{K}}$ is real-valued if and only if ${\textstyle (0,\infty )K=X,}$ in which case ${\textstyle 0\in K.}$
- Value at ${\textstyle 0}$ : ${\textstyle p_{K}(0)\neq \infty }$ if and only if ${\textstyle 0\in K}$ if and only if ${\textstyle p_{K}(0)=0.}$
- Null space: If ${\textstyle x\in X}$ then ${\textstyle p_{K}(x)=0}$ if and only if ${\textstyle (0,\infty )x\subseteq (0,1)K}$ if and only if there exists a divergent sequence of positive real numbers ${\textstyle t_{1},t_{2},t_{3},\cdots \to \infty }$ such that ${\textstyle t_{n}x\in K}$ for all ${\textstyle n.}$ Moreover, thezero set of ${\textstyle p_{K}}$ is ${\textstyle \ker p_{K}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\left\{y\in X:p_{K}(y)=0\right\}={\textstyle \bigcap \limits _{e>0}}(0,e)K.}$
Comparison to a constant: If ${\textstyle 0\leq r\leq \infty }$ then for any ${\textstyle x\in X,}$ ${\textstyle p_{K}(x)<r}$ if and only if ${\textstyle x\in (0,r)K;}$ this can be restated as: If ${\textstyle 0\leq r\leq \infty }$ then ${\textstyle p_{K}^{-1}([0,r))=(0,r)K.}$
- It follows that if ${\textstyle 0\leq R<\infty }$ is real then ${\textstyle p_{K}^{-1}([0,R])={\textstyle \bigcap \limits _{e>0}}(0,R+e)K,}$ where the set on the right hand side denotes ${\textstyle {\textstyle \bigcap \limits _{e>0}}[(0,R+e)K]}$ and not its subset ${\textstyle \left[{\textstyle \bigcap \limits _{e>0}}(0,R+e)\right]K=(0,R]K.}$ If ${\textstyle R>0}$ then these sets are equal if and only if ${\textstyle K}$ contains ${\textstyle \left\{y\in X:p_{K}(y)=1\right\}.}$
- In particular, if ${\textstyle x\in RK}$ or ${\textstyle x\in (0,R]K}$ then ${\textstyle p_{K}(x)\leq R,}$ but importantly, the converse is not necessarily true.
Gauge comparison: For any subset ${\textstyle L\subseteq X,}$ ${\textstyle p_{K}\leq p_{L}}$ if and only if ${\textstyle (0,1)L\subseteq (0,1)K;}$ thus ${\textstyle p_{L}=p_{K}}$ if and only if ${\textstyle (0,1)L=(0,1)K.}$
- The assignment ${\textstyle L\mapsto p_{L}}$ is order-reversing in the sense that if ${\textstyle K\subseteq L}$ then ${\textstyle p_{L}\leq p_{K}.}$ ^[8]
- Because the set ${\textstyle L:=(0,1)K}$ satisfies ${\textstyle (0,1)L=(0,1)K,}$ it follows that replacing ${\textstyle K}$ with ${\textstyle p_{K}^{-1}([0,1))=(0,1)K}$ will not change the resulting Minkowski functional. The same is true of ${\textstyle L:=(0,1]K}$ and of ${\textstyle L:=p_{K}^{-1}([0,1]).}$
- If ${\textstyle D~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\left\{y\in X:p_{K}(y)=1{\text{ or }}p_{K}(y)=0\right\}}$ then ${\textstyle p_{D}=p_{K}}$ and ${\textstyle D}$ has the particularly nice property that if ${\textstyle r>0}$ is real then ${\textstyle x\in rD}$ if and only if ${\textstyle p_{D}(x)=r}$ or ${\textstyle p_{D}(x)=0.}$ ^{[note 1]} Moreover, if ${\textstyle r>0}$ is real then ${\textstyle p_{D}(x)\leq r}$ if and only if ${\textstyle x\in (0,r]D.}$
Subadditive/Triangle inequality: ${\textstyle p_{K}}$ is subadditive if and only if ${\textstyle (0,1)K}$ is convex. If ${\textstyle K}$ is convex then so are both ${\textstyle (0,1)K}$ and ${\textstyle (0,1]K}$ and moreover, ${\textstyle p_{K}}$ is subadditive.
Scaling the set: If ${\textstyle s\neq 0}$ is a scalar then ${\textstyle p_{sK}(y)=p_{K}\left({\tfrac {1}{s}}y\right)}$ for all ${\textstyle y\in X.}$ Thus if ${\textstyle 0<r<\infty }$ is real then ${\textstyle p_{rK}(y)=p_{K}\left({\tfrac {1}{r}}y\right)={\tfrac {1}{r}}p_{K}(y).}$
Symmetric: ${\textstyle p_{K}}$ is symmetric (meaning that ${\textstyle p_{K}(-y)=p_{K}(y)}$ for all ${\textstyle y\in X}$ ) if and only if ${\textstyle (0,1)K}$ is asymmetric set (meaning that ${\textstyle (0,1)K=-(0,1)K}$ ), which happens if and only if ${\textstyle p_{K}=p_{-K}.}$
Absolute homogeneity: ${\textstyle p_{K}(ux)=p_{K}(x)}$ for all ${\textstyle x\in X}$ and all unit length scalars ${\textstyle u}$ ^{[note 2]} if and only if ${\textstyle (0,1)uK\subseteq (0,1)K}$ for all unit length scalars ${\textstyle u,}$ in which case ${\textstyle p_{K}(sx)=|s|p_{K}(x)}$ for all ${\textstyle x\in X}$ and allnon-zero scalars ${\textstyle s\neq 0.}$ If in addition ${\textstyle p_{K}}$ is also real-valued then this holds forall scalars ${\textstyle s}$ (that is, ${\textstyle p_{K}}$ is absolutely homogeneous^{[note 3]}).
- ${\textstyle (0,1)uK\subseteq (0,1)K}$ for all unit length ${\textstyle u}$ if and only if ${\textstyle (0,1)uK=(0,1)K}$ for all unit length ${\textstyle u.}$
- ${\textstyle sK\subseteq K}$ for all unit scalars ${\textstyle s}$ if and only if ${\textstyle sK=K}$ for all unit scalars ${\textstyle s;}$ if this is the case then ${\textstyle (0,1)K=(0,1)sK}$ for all unit scalars ${\textstyle s.}$
- The Minkowski functional of anybalanced set is abalanced function.^[8]
Absorbing: If ${\textstyle K}$ is convexor balanced and if ${\textstyle (0,\infty )K=X}$ then ${\textstyle K}$ is absorbing in ${\textstyle X.}$
- If a set ${\textstyle A}$ is absorbing in ${\textstyle X}$ and ${\textstyle A\subseteq K}$ then ${\textstyle K}$ is absorbing in ${\textstyle X.}$
- If ${\textstyle K}$ is convex and ${\textstyle 0\in K}$ then ${\textstyle [0,1]K=K,}$ in which case ${\textstyle (0,1)K\subseteq K.}$
Restriction to a vector subspace: If ${\textstyle S}$ is a vector subspace of ${\textstyle X}$ and if ${\textstyle p_{K\cap S}:S\to [0,\infty ]}$ denotes the Minkowski functional of ${\textstyle K\cap S}$ on ${\textstyle S,}$ then ${\textstyle p_{K}{\big \vert }_{S}=p_{K\cap S},}$ where ${\textstyle p_{K}{\big \vert }_{S}}$ denotes therestriction of ${\textstyle p_{K}}$ to ${\textstyle S.}$

Proof
The proofs of these basic properties are straightforward exercises so only the proofs of the most important statements are given. The proof that a convex subset ${\textstyle A\subseteq X}$ that satisfies ${\textstyle (0,\infty )A=X}$ is necessarilyabsorbing in ${\textstyle X}$ is straightforward and can be found in the article onabsorbing sets. For any real ${\textstyle t>0,}$ $\{r>0:tx\in rK\}=\{t(r/t):x\in (r/t)K\}=t\{s>0:x\in sK\}$ so that taking the infimum of both sides shows that $p_{K}(tx)=\inf\{r>0:tx\in rK\}=t\inf\{s>0:x\in sK\}=tp_{K}(x).$ This proves that Minkowski functionals are strictly positive homogeneous. For ${\textstyle 0\cdot p_{K}(x)}$ to be well-defined, it is necessary and sufficient that ${\textstyle p_{K}(x)\neq \infty ;}$ thus ${\textstyle p_{K}(tx)=tp_{K}(x)}$ for all ${\textstyle x\in X}$ and allnon-negative real ${\textstyle t\geq 0}$ if and only if ${\textstyle p_{K}}$ is real-valued. The hypothesis of statement (7) allows us to conclude that ${\textstyle p_{K}(sx)=p_{K}(x)}$ for all ${\textstyle x\in X}$ and all scalars ${\textstyle s}$ satisfying ${\textstyle \|s\|=1.}$ Every scalar ${\textstyle s}$ is of the form ${\textstyle re^{it}}$ for some real ${\textstyle t}$ where ${\textstyle r:=\|s\|\geq 0}$ and ${\textstyle e^{it}}$ is real if and only if ${\textstyle s}$ is real. The results in the statement about absolute homogeneity follow immediately from the aforementioned conclusion, from the strict positive homogeneity of ${\textstyle p_{K},}$ and from the positive homogeneity of ${\textstyle p_{K}}$ when ${\textstyle p_{K}}$ is real-valued. ${\textstyle \blacksquare }$

Proof

The proofs of these basic properties are straightforward exercises so only the proofs of the most important statements are given.

The proof that a convex subset ${\textstyle A\subseteq X}$ that satisfies ${\textstyle (0,\infty )A=X}$ is necessarilyabsorbing in ${\textstyle X}$ is straightforward and can be found in the article onabsorbing sets.

For any real ${\textstyle t>0,}$

$\{r>0:tx\in rK\}=\{t(r/t):x\in (r/t)K\}=t\{s>0:x\in sK\}$

so that taking the infimum of both sides shows that

$p_{K}(tx)=\inf\{r>0:tx\in rK\}=t\inf\{s>0:x\in sK\}=tp_{K}(x).$

This proves that Minkowski functionals are strictly positive homogeneous. For ${\textstyle 0\cdot p_{K}(x)}$ to be well-defined, it is necessary and sufficient that ${\textstyle p_{K}(x)\neq \infty ;}$ thus ${\textstyle p_{K}(tx)=tp_{K}(x)}$ for all ${\textstyle x\in X}$ and allnon-negative real ${\textstyle t\geq 0}$ if and only if ${\textstyle p_{K}}$ is real-valued.

The hypothesis of statement (7) allows us to conclude that ${\textstyle p_{K}(sx)=p_{K}(x)}$ for all ${\textstyle x\in X}$ and all scalars ${\textstyle s}$ satisfying ${\textstyle |s|=1.}$ Every scalar ${\textstyle s}$ is of the form ${\textstyle re^{it}}$ for some real ${\textstyle t}$ where ${\textstyle r:=|s|\geq 0}$ and ${\textstyle e^{it}}$ is real if and only if ${\textstyle s}$ is real. The results in the statement about absolute homogeneity follow immediately from the aforementioned conclusion, from the strict positive homogeneity of ${\textstyle p_{K},}$ and from the positive homogeneity of ${\textstyle p_{K}}$ when ${\textstyle p_{K}}$ is real-valued. ${\textstyle \blacksquare }$

Examples

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If ${\textstyle {\mathcal {L}}}$ is a non-empty collection of subsets of ${\textstyle X}$ then ${\textstyle p_{\cup {\mathcal {L}}}(x)=\inf \left\{p_{L}(x):L\in {\mathcal {L}}\right\}}$ for all ${\textstyle x\in X,}$ where ${\textstyle \cup {\mathcal {L}}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\textstyle \bigcup \limits _{L\in {\mathcal {L}}}}L.}$
- Thus ${\textstyle p_{K\cup L}(x)=\min \left\{p_{K}(x),p_{L}(x)\right\}}$ for all ${\textstyle x\in X.}$
If ${\textstyle {\mathcal {L}}}$ is a non-empty collection of subsets of ${\textstyle X}$ and ${\textstyle I\subseteq X}$ satisfies

$\left\{x\in X:p_{L}(x)<1{\text{ for all }}L\in {\mathcal {L}}\right\}\quad \subseteq \quad I\quad \subseteq \quad \left\{x\in X:p_{L}(x)\leq 1{\text{ for all }}L\in {\mathcal {L}}\right\}$ then ${\textstyle p_{I}(x)=\sup \left\{p_{L}(x):L\in {\mathcal {L}}\right\}}$ for all ${\textstyle x\in X.}$

The following examples show that the containment ${\textstyle (0,R]K\;\subseteq \;{\textstyle \bigcap \limits _{e>0}}(0,R+e)K}$ could be proper.

Example: If ${\textstyle R=0}$ and ${\textstyle K=X}$ then ${\textstyle (0,R]K=(0,0]X=\varnothing X=\varnothing }$ but ${\textstyle {\textstyle \bigcap \limits _{e>0}}(0,e)K={\textstyle \bigcap \limits _{e>0}}X=X,}$ which shows that its possible for ${\textstyle (0,R]K}$ to be a proper subset of ${\textstyle {\textstyle \bigcap \limits _{e>0}}(0,R+e)K}$ when ${\textstyle R=0.}$ ${\textstyle \blacksquare }$

The next example shows that the containment can be proper when ${\textstyle R=1;}$ the example may be generalized to any real ${\textstyle R>0.}$ Assuming that ${\textstyle [0,1]K\subseteq K,}$ the following example is representative of how it happens that ${\textstyle x\in X}$ satisfies ${\textstyle p_{K}(x)=1}$ but ${\textstyle x\not \in (0,1]K.}$

Example: Let ${\textstyle x\in X}$ be non-zero and let ${\textstyle K=[0,1)x}$ so that ${\textstyle [0,1]K=K}$ and ${\textstyle x\not \in K.}$ From ${\textstyle x\not \in (0,1)K=K}$ it follows that ${\textstyle p_{K}(x)\geq 1.}$ That ${\textstyle p_{K}(x)\leq 1}$ follows from observing that for every ${\textstyle e>0,}$ ${\textstyle (0,1+e)K=[0,1+e)([0,1)x)=[0,1+e)x,}$ which contains ${\textstyle x.}$ Thus ${\textstyle p_{K}(x)=1}$ and ${\textstyle x\in {\textstyle \bigcap \limits _{e>0}}(0,1+e)K.}$ However, ${\textstyle (0,1]K=(0,1]([0,1)x)=[0,1)x=K}$ so that ${\textstyle x\not \in (0,1]K,}$ as desired. ${\textstyle \blacksquare }$

Positive homogeneity characterizes Minkowski functionals

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The next theorem shows that Minkowski functionals areexactly those functions ${\textstyle f:X\to [0,\infty ]}$ that have a certain purely algebraic property that is commonly encountered.

Theorem—Let ${\textstyle f:X\to [0,\infty ]}$ be any function. The following statements are equivalent:

Strict positive homogeneity: ${\textstyle \;f(tx)=tf(x)}$ for all ${\textstyle x\in X}$ and allpositive real ${\textstyle t>0.}$
- This statement is equivalent to: ${\textstyle f(tx)\leq tf(x)}$ for all ${\textstyle x\in X}$ and all positive real ${\textstyle t>0.}$
${\textstyle f}$ is a Minkowski functional: meaning that there exists a subset ${\textstyle S\subseteq X}$ such that ${\textstyle f=p_{S}.}$
${\textstyle f=p_{K}}$ where ${\textstyle K:=\{x\in X:f(x)\leq 1\}.}$
${\textstyle f=p_{V}\,}$ where ${\textstyle V\,:=\{x\in X:f(x)<1\}.}$

Moreover, if ${\textstyle f}$ never takes on the value ${\textstyle \,\infty \,}$ (so that the product ${\textstyle 0\cdot f(x)}$ is always well-defined) then this list may be extended to include:

Positive/Nonnegative homogeneity: ${\textstyle f(tx)=tf(x)}$ for all ${\textstyle x\in X}$ and allnonnegative real ${\textstyle t\geq 0}$ .

Proof
If ${\textstyle f(tx)\leq tf(x)}$ holds for all ${\textstyle x\in X}$ and real ${\textstyle t>0}$ then ${\textstyle tf(x)=tf\left({\tfrac {1}{t}}(tx)\right)\leq t{\tfrac {1}{t}}f(tx)=f(tx)\leq tf(x)}$ so that ${\textstyle tf(x)=f(tx).}$ Only (1) implies (3) will be proven because afterwards, the rest of the theorem follows immediately from the basic properties of Minkowski functionals described earlier; properties that will henceforth be used without comment. So assume that ${\textstyle f:X\to [0,\infty ]}$ is a function such that ${\textstyle f(tx)=tf(x)}$ for all ${\textstyle x\in X}$ and all real ${\textstyle t>0}$ and let ${\textstyle K:=\{y\in X:f(y)\leq 1\}.}$ For all real ${\textstyle t>0,}$ ${\textstyle f(0)=f(t0)=tf(0)}$ so by taking ${\textstyle t=2}$ for instance, it follows that either ${\textstyle f(0)=0}$ or ${\textstyle f(0)=\infty .}$ Let ${\textstyle x\in X.}$ It remains to show that ${\textstyle f(x)=p_{K}(x).}$ It will now be shown that if ${\textstyle f(x)=0}$ or ${\textstyle f(x)=\infty }$ then ${\textstyle f(x)=p_{K}(x),}$ so that in particular, it will follow that ${\textstyle f(0)=p_{K}(0).}$ So suppose that ${\textstyle f(x)=0}$ or ${\textstyle f(x)=\infty ;}$ in either case ${\textstyle f(tx)=tf(x)=f(x)}$ for all real ${\textstyle t>0.}$ Now if ${\textstyle f(x)=0}$ then this implies that that ${\textstyle tx\in K}$ for all real ${\textstyle t>0}$ (since ${\textstyle f(tx)=0\leq 1}$ ), which implies that ${\textstyle p_{K}(x)=0,}$ as desired. Similarly, if ${\textstyle f(x)=\infty }$ then ${\textstyle tx\not \in K}$ for all real ${\textstyle t>0,}$ which implies that ${\textstyle p_{K}(x)=\infty ,}$ as desired. Thus, it will henceforth be assumed that ${\textstyle R:=f(x)}$ a positive real number and that ${\textstyle x\neq 0}$ (importantly, however, the possibility that ${\textstyle p_{K}(x)}$ is ${\textstyle 0}$ or ${\textstyle \,\infty \,}$ has not yet been ruled out). Recall that just like ${\textstyle f,}$ the function ${\textstyle p_{K}}$ satisfies ${\textstyle p_{K}(tx)=tp_{K}(x)}$ for all real ${\textstyle t>0.}$ Since ${\textstyle 0<{\tfrac {1}{R}}<\infty ,}$ ${\textstyle p_{K}(x)=R=f(x)}$ if and only if ${\textstyle p_{K}\left({\tfrac {1}{R}}x\right)=1=f\left({\tfrac {1}{R}}x\right)}$ so assume without loss of generality that ${\textstyle R=1}$ and it remains to show that ${\textstyle p_{K}\left({\tfrac {1}{R}}x\right)=1.}$ Since ${\textstyle f(x)=1,}$ ${\textstyle x\in K\subseteq (0,1]K,}$ which implies that ${\textstyle p_{K}(x)\leq 1}$ (so in particular, ${\textstyle p_{K}(x)\neq \infty }$ is guaranteed). It remains to show that ${\textstyle p_{K}(x)\geq 1,}$ which recall happens if and only if ${\textstyle x\not \in (0,1)K.}$ So assume for the sake of contradiction that ${\textstyle x\in (0,1)K}$ and let ${\textstyle 0<r<1}$ and ${\textstyle k\in K}$ be such that ${\textstyle x=rk,}$ where note that ${\textstyle k\in K}$ implies that ${\textstyle f(k)\leq 1.}$ Then ${\textstyle 1=f(x)=f(rk)=rf(k)\leq r<1.}$ ${\textstyle \blacksquare }$

Proof

If ${\textstyle f(tx)\leq tf(x)}$ holds for all ${\textstyle x\in X}$ and real ${\textstyle t>0}$ then ${\textstyle tf(x)=tf\left({\tfrac {1}{t}}(tx)\right)\leq t{\tfrac {1}{t}}f(tx)=f(tx)\leq tf(x)}$ so that ${\textstyle tf(x)=f(tx).}$

Only (1) implies (3) will be proven because afterwards, the rest of the theorem follows immediately from the basic properties of Minkowski functionals described earlier; properties that will henceforth be used without comment. So assume that ${\textstyle f:X\to [0,\infty ]}$ is a function such that ${\textstyle f(tx)=tf(x)}$ for all ${\textstyle x\in X}$ and all real ${\textstyle t>0}$ and let ${\textstyle K:=\{y\in X:f(y)\leq 1\}.}$

For all real ${\textstyle t>0,}$ ${\textstyle f(0)=f(t0)=tf(0)}$ so by taking ${\textstyle t=2}$ for instance, it follows that either ${\textstyle f(0)=0}$ or ${\textstyle f(0)=\infty .}$ Let ${\textstyle x\in X.}$ It remains to show that ${\textstyle f(x)=p_{K}(x).}$

It will now be shown that if ${\textstyle f(x)=0}$ or ${\textstyle f(x)=\infty }$ then ${\textstyle f(x)=p_{K}(x),}$ so that in particular, it will follow that ${\textstyle f(0)=p_{K}(0).}$ So suppose that ${\textstyle f(x)=0}$ or ${\textstyle f(x)=\infty ;}$ in either case ${\textstyle f(tx)=tf(x)=f(x)}$ for all real ${\textstyle t>0.}$ Now if ${\textstyle f(x)=0}$ then this implies that that ${\textstyle tx\in K}$ for all real ${\textstyle t>0}$ (since ${\textstyle f(tx)=0\leq 1}$ ), which implies that ${\textstyle p_{K}(x)=0,}$ as desired. Similarly, if ${\textstyle f(x)=\infty }$ then ${\textstyle tx\not \in K}$ for all real ${\textstyle t>0,}$ which implies that ${\textstyle p_{K}(x)=\infty ,}$ as desired. Thus, it will henceforth be assumed that ${\textstyle R:=f(x)}$ a positive real number and that ${\textstyle x\neq 0}$ (importantly, however, the possibility that ${\textstyle p_{K}(x)}$ is ${\textstyle 0}$ or ${\textstyle \,\infty \,}$ has not yet been ruled out).

Recall that just like ${\textstyle f,}$ the function ${\textstyle p_{K}}$ satisfies ${\textstyle p_{K}(tx)=tp_{K}(x)}$ for all real ${\textstyle t>0.}$ Since ${\textstyle 0<{\tfrac {1}{R}}<\infty ,}$ ${\textstyle p_{K}(x)=R=f(x)}$ if and only if ${\textstyle p_{K}\left({\tfrac {1}{R}}x\right)=1=f\left({\tfrac {1}{R}}x\right)}$ so assume without loss of generality that ${\textstyle R=1}$ and it remains to show that ${\textstyle p_{K}\left({\tfrac {1}{R}}x\right)=1.}$ Since ${\textstyle f(x)=1,}$ ${\textstyle x\in K\subseteq (0,1]K,}$ which implies that ${\textstyle p_{K}(x)\leq 1}$ (so in particular, ${\textstyle p_{K}(x)\neq \infty }$ is guaranteed). It remains to show that ${\textstyle p_{K}(x)\geq 1,}$ which recall happens if and only if ${\textstyle x\not \in (0,1)K.}$ So assume for the sake of contradiction that ${\textstyle x\in (0,1)K}$ and let ${\textstyle 0<r<1}$ and ${\textstyle k\in K}$ be such that ${\textstyle x=rk,}$ where note that ${\textstyle k\in K}$ implies that ${\textstyle f(k)\leq 1.}$ Then ${\textstyle 1=f(x)=f(rk)=rf(k)\leq r<1.}$ ${\textstyle \blacksquare }$

This theorem can be extended to characterize certain classes of ${\textstyle [-\infty ,\infty ]}$ -valued maps (for example, real-valuedsublinear functions) in terms of Minkowski functionals. For instance, it can be used to describe how every real homogeneous function ${\textstyle f:X\to \mathbb {R} }$ (such as linear functionals) can be written in terms of a unique Minkowski functional having a certain property.

Characterizing Minkowski functionals on star sets

[edit]

Proposition^[10]—Let ${\textstyle f:X\to [0,\infty ]}$ be any function and ${\textstyle K\subseteq X}$ be any subset. The following statements are equivalent:

${\textstyle f}$ is (strictly) positive homogeneous, ${\textstyle f(0)=0,}$ and
$\{x\in X:f(x)<1\}\;\subseteq \;K\;\subseteq \;\{x\in X:f(x)\leq 1\}.$
${\textstyle f}$ is the Minkowski functional of ${\textstyle K}$ (that is, ${\textstyle f=p_{K}}$ ), ${\textstyle K}$ contains the origin, and ${\textstyle K}$ isstar-shaped at the origin.
- The set ${\textstyle K}$ is star-shaped at the origin if and only if ${\textstyle tk\in K}$ whenever ${\textstyle k\in K}$ and ${\textstyle 0\leq t\leq 1.}$ A set that is star-shaped at the origin is sometimes called astar set.^[9]

Characterizing Minkowski functionals that are seminorms

[edit]

In this next theorem, which follows immediately from the statements above, ${\textstyle K}$ isnot assumed to be absorbing in ${\textstyle X}$ and instead, it is deduced that ${\textstyle (0,1)K}$ is absorbing when ${\textstyle p_{K}}$ is a seminorm. It is also not assumed that ${\textstyle K}$ isbalanced (which is a property that ${\textstyle K}$ is often required to have); in its place is the weaker condition that ${\textstyle (0,1)sK\subseteq (0,1)K}$ for all scalars ${\textstyle s}$ satisfying ${\textstyle |s|=1.}$ The common requirement that ${\textstyle K}$ be convex is also weakened to only requiring that ${\textstyle (0,1)K}$ be convex.

Theorem—Let ${\textstyle K}$ be a subset of a real or complex vector space ${\textstyle X.}$ Then ${\textstyle p_{K}}$ is aseminorm on ${\textstyle X}$ if and only if all of the following conditions hold:

${\textstyle (0,\infty )K=X}$ (or equivalently, ${\textstyle p_{K}}$ is real-valued).
${\textstyle (0,1)K}$ is convex (or equivalently, ${\textstyle p_{K}}$ issubadditive).
- It suffices (but is not necessary) for ${\textstyle K}$ to be convex.
${\textstyle (0,1)uK\subseteq (0,1)K}$ for all unit scalars ${\textstyle u.}$
- This condition is satisfied if ${\textstyle K}$ isbalanced or more generally if ${\textstyle uK\subseteq K}$ for all unit scalars ${\textstyle u.}$

in which case ${\textstyle 0\in K}$ and both ${\textstyle (0,1)K=\{x\in X:p(x)<1\}}$ and ${\textstyle \bigcap _{e>0}(0,1+e)K=\left\{x\in X:p_{K}(x)\leq 1\right\}}$ will be convex, balanced, andabsorbing subsets of ${\textstyle X.}$

Conversely, if ${\textstyle f}$ is a seminorm on ${\textstyle X}$ then the set ${\textstyle V:=\{x\in X:f(x)<1\}}$ satisfies all three of the above conditions (and thus also the conclusions) and also ${\textstyle f=p_{V};}$ moreover, ${\textstyle V}$ is necessarily convex, balanced, absorbing, and satisfies ${\textstyle (0,1)V=V=[0,1]V.}$

Corollary—If ${\textstyle K}$ is a convex, balanced, and absorbing subset of a real or complex vector space ${\textstyle X,}$ then ${\textstyle p_{K}}$ is aseminorm on ${\textstyle X.}$

Positive sublinear functions and Minkowski functionals

[edit]

It may be shown that a real-valuedsubadditive function ${\textstyle f:X\to \mathbb {R} }$ on an arbitrarytopological vector space ${\textstyle X}$ is continuous at the origin if and only if it is uniformly continuous, where if in addition ${\textstyle f}$ is nonnegative, then ${\textstyle f}$ is continuous if and only if ${\textstyle V:=\{x\in X:f(x)<1\}}$ is an open neighborhood in ${\textstyle X.}$ ^[11] If ${\textstyle f:X\to \mathbb {R} }$ is subadditive and satisfies ${\textstyle f(0)=0,}$ then ${\textstyle f}$ is continuous if and only if its absolute value ${\textstyle |f|:X\to [0,\infty )}$ is continuous.

Anonnegativesublinear function is anonnegative homogeneous function ${\textstyle f:X\to [0,\infty )}$ that satisfies the triangle inequality. It follows immediately from the results below that for such a function ${\textstyle f,}$ if ${\textstyle V:=\{x\in X:f(x)<1\}}$ then ${\textstyle f=p_{V}.}$ Given ${\textstyle K\subseteq X,}$ the Minkowski functional ${\textstyle p_{K}}$ is a sublinear function if and only if it is real-valued and subadditive, which is happens if and only if ${\textstyle (0,\infty )K=X}$ and ${\textstyle (0,1)K}$ is convex.

Correspondence between open convex sets and positive continuous sublinear functions

[edit]

Theorem^[11]—Suppose that ${\textstyle X}$ is atopological vector space (not necessarilylocally convex or Hausdorff) over the real or complex numbers. Then the non-empty open convex subsets of ${\textstyle X}$ are exactly those sets that are of the form ${\textstyle z+\{x\in X:p(x)<1\}=\{x\in X:p(x-z)<1\}}$ for some ${\textstyle z\in X}$ and some positive continuoussublinear function ${\textstyle p}$ on ${\textstyle X.}$

Proof
Let ${\textstyle V\neq \varnothing }$ be an open convex subset of ${\textstyle X.}$ If ${\textstyle 0\in V}$ then let ${\textstyle z:=0}$ and otherwise let ${\textstyle z\in V}$ be arbitrary. Let ${\textstyle p=p_{K}:X\to [0,\infty )}$ be the Minkowski functional of ${\textstyle K:=V-z}$ where this convex open neighborhood of the origin satisfies ${\textstyle (0,1)K=K.}$ Then ${\textstyle p}$ is a continuous sublinear function on ${\textstyle X}$ since ${\textstyle V-z}$ is convex, absorbing, and open (however, ${\textstyle p}$ is not necessarily a seminorm since it is not necessarily absolutely homogeneous). From the properties of Minkowski functionals, we have ${\textstyle p_{K}^{-1}([0,1))=(0,1)K,}$ from which it follows that ${\textstyle V-z=\{x\in X:p(x)<1\}}$ and so ${\textstyle V=z+\{x\in X:p(x)<1\}.}$ Since ${\textstyle z+\{x\in X:p(x)<1\}=\{x\in X:p(x-z)<1\},}$ this completes the proof. ${\textstyle \blacksquare }$

Proof

Let ${\textstyle V\neq \varnothing }$ be an open convex subset of ${\textstyle X.}$ If ${\textstyle 0\in V}$ then let ${\textstyle z:=0}$ and otherwise let ${\textstyle z\in V}$ be arbitrary. Let ${\textstyle p=p_{K}:X\to [0,\infty )}$ be the Minkowski functional of ${\textstyle K:=V-z}$ where this convex open neighborhood of the origin satisfies ${\textstyle (0,1)K=K.}$ Then ${\textstyle p}$ is a continuous sublinear function on ${\textstyle X}$ since ${\textstyle V-z}$ is convex, absorbing, and open (however, ${\textstyle p}$ is not necessarily a seminorm since it is not necessarily absolutely homogeneous). From the properties of Minkowski functionals, we have ${\textstyle p_{K}^{-1}([0,1))=(0,1)K,}$ from which it follows that ${\textstyle V-z=\{x\in X:p(x)<1\}}$ and so ${\textstyle V=z+\{x\in X:p(x)<1\}.}$ Since ${\textstyle z+\{x\in X:p(x)<1\}=\{x\in X:p(x-z)<1\},}$ this completes the proof. ${\textstyle \blacksquare }$

Notes

[edit]

^It is in generalfalse that ${\textstyle x\in rD}$ if and only if ${\textstyle p_{D}(x)=r}$ (for example, consider when ${\textstyle p_{K}}$ is anorm or a seminorm). The correct statement is: If ${\textstyle 0<r<\infty }$ then ${\textstyle x\in rD}$ if and only if ${\textstyle p_{D}(x)=r}$ or ${\textstyle p_{D}(x)=0.}$
^ ${\textstyle u}$ is having unit length means that ${\textstyle |u|=1.}$
^The map ${\textstyle p_{K}}$ is calledabsolutely homogeneous if ${\textstyle |s|p_{K}(x)}$ is well-defined and ${\textstyle p_{K}(sx)=|s|p_{K}(x)}$ for all ${\textstyle x\in X}$ and all scalars ${\textstyle s}$ (not just non-zero scalars).

References

[edit]

^Narici & Beckenstein 2011, p. 109.
^Narici & Beckenstein 2011, p. 119.
^^a ^bJarchow 1981, pp. 104–108.
^^a ^bNarici & Beckenstein 2011, pp. 115–154.
^^a ^bSchaefer 1999, p. 40.
^^a ^b ^cNarici & Beckenstein 2011, p. 119-120.
^Kubrusly 2011, p. 200.
^^a ^bSchechter 1996, p. 316.
^Schechter 1996, p. 303.
^Schechter 1996, pp. 313–317.
^^a ^bNarici & Beckenstein 2011, pp. 192–193.

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v t e Topological vector spaces (TVSs)
Basic concepts	Banach space Completeness Continuous linear operator Linear functional Fréchet space Linear map Locally convex space Metrizability Operator topologies Topological vector space Vector space
Main results	Anderson–Kadec Banach–Alaoglu Closed graph theorem F. Riesz's Hahn–Banach (hyperplane separation Vector-valued Hahn–Banach) Open mapping (Banach–Schauder) Bounded inverse Uniform boundedness (Banach–Steinhaus)
Maps	Bilinear operator form Linear map Almost open Bounded Continuous Closed Compact Densely defined Discontinuous Topological homomorphism Functional Linear Bilinear Sesquilinear Norm Seminorm Sublinear function Transpose
Types of sets	Absolutely convex/disk Absorbing/Radial Affine Balanced/Circled Banach disks Bounding points Bounded Complemented subspace Convex Convex cone(subset) Linear cone(subset) Extreme point Pre-compact/Totally bounded Prevalent/Shy Radial Radially convex/Star-shaped Symmetric
Set operations	Affine hull (Relative) Algebraic interior (core) Convex hull Linear span Minkowski addition Polar (Quasi) Relative interior
Types of TVSs	Asplund B-complete/Ptak Banach (Countably) Barrelled BK-space (Ultra-) Bornological Brauner Complete Convenient (DF)-space Distinguished F-space FK-AK space FK-space Fréchet tame Fréchet Grothendieck Hilbert Infrabarreled Interpolation space K-space LB-space LF-space Locally convex space Mackey (Pseudo)Metrizable Montel Quasibarrelled Quasi-complete Quasinormed (Polynomially Semi-) Reflexive Riesz Schwartz Semi-complete Smith Stereotype (B Strictly Uniformly) convex (Quasi-) Ultrabarrelled Uniformly smooth Webbed With the approximation property
Category

v t e Convex analysis andvariational analysis
Basic concepts	Convex combination Convex function Convex set
Topics (list)	Choquet theory Convex geometry Convex metric space Convex optimization Duality Lagrange multiplier Legendre transformation Locally convex topological vector space Simplex
Maps	Convex conjugate Concave (Closed K- Logarithmically Proper Pseudo- Quasi-)Convex function Invex function Legendre transformation Semi-continuity Subderivative
Mainresults (list)	Carathéodory's theorem Ekeland's variational principle Fenchel–Moreau theorem Fenchel-Young inequality Jensen's inequality Hermite–Hadamard inequality Krein–Milman theorem Mazur's lemma Shapley–Folkman lemma Robinson–Ursescu Simons Ursescu
Sets	Convex hull (Orthogonally,Pseudo-)Convex set Effective domain Epigraph Hypograph John ellipsoid Lens Radial set/Algebraic interior Zonotope
Series	Convex series related ((cs, lcs)-closed,(cs, bcs)-complete,(lower) ideally convex,(Hx), and(Hwx))
Duality	Dual system Duality gap Strong duality Weak duality
Applications and related	Convexity in economics

Movatterモバイル変換

Minkowski functional

Definition

Some conditions making a gauge real-valued

Motivating examples

Example 1

Example 2

Common conditions guaranteeing gauges are seminorms

Gauges of absorbing disks

Algebraic properties

Topological properties

Minimal requirements on the set

Properties

Examples

Positive homogeneity characterizes Minkowski functionals

Characterizing Minkowski functionals on star sets

Characterizing Minkowski functionals that are seminorms

Positive sublinear functions and Minkowski functionals

Correspondence between open convex sets and positive continuous sublinear functions

See also

Notes

References

Further reading