Minkowski functional

In mathematics, in the field of functional analysis, a Minkowski functional (after Hermann Minkowski) or gauge function is a function that recovers a notion of distance on a linear space.

If $K$ is a subset of a real or complex vector space $X,$ then the Minkowski functional or gauge of $K$ is defined to be the function $p_{K}:X\to [0,\infty ],$ valued in the extended 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 the infimum of the empty set is defined to be positive infinity $\,\infty \,$ (which is not a real number so that $p_{K}(x)$ would then not be real-valued).

The set $K$ is often assumed/picked to have properties, such as being an absorbing disk in $X,$ that guarantee that $p_{K}$ will be a real-valued seminorm on $X.$ In fact, every seminorm $p$ on $X$ is equal to the Minkowski functional (that is, $p=p_{K}$ ) of any subset $K$ of $X$ 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 $X$ and the first and last are also disks).

Thus every seminorm (which is a function defined by purely algebraic properties) can be associated (non-uniquely) with an absorbing disk (which is a set 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" certain geometric properties of a subset of $X$ into certain algebraic properties of a function on $X.$

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

Definition

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

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

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

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

Some conditions making a gauge real-valued

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

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

Motivating examples

Example 1

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

Example 2

Let $X$ be a vector space without topology with underlying scalar field $\mathbb {K} .$ Let $f:X\to \mathbb {K}$ be any linear functional on $X$ (not necessarily continuous). Fix $a>0.$ Let $K$ be the set

K:=\{x\in X:|f(x)|\leq a\}

and let $p_{K}$ be the Minkowski functional of $K.$ Then

p_{K}(x)={\frac {1}{a}}|f(x)|\quad {\text{ for all }}x\in X.

The function $p_{K}$ has the following properties:

It is subadditive: $p_{K}(x+y)\leq p_{K}(x)+p_{K}(y).$
It is absolutely homogeneous: $p_{K}(sx)=|s|p_{K}(x)$ for all scalars $s.$
It is nonnegative: $p_{K}\geq 0.$

Therefore, $p_{K}$ is a seminorm on $X,$ 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, $p_{K}(x)=0$ need not imply $x=0.$ In the above example, one can take a nonzero $x$ from the kernel of $f.$ Consequently, the resulting topology need not be Hausdorff.

Common conditions guaranteeing gauges are seminorms

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

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

Theorem[2] — If $K$ is an absorbing disk in a vector space $X$ then the Minkowski functional of $K,$ which is the map $p_{K}:X\to [0,\infty )$ defined by

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

is a seminorm on $X.$ Moreover,

p_{K}(x)={\frac {1}{\sup\{r>0:rx\in K\}}}.

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

Gauges of absorbing disks

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

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

If $K$ is convex then $p_{K}$ is subadditive.
If $K$ is balanced then $p_{K}$ is absolutely homogeneous; that is, $p_{K}(sx)=|s|p_{K}(x)$ for all scalars $s.$

Proof that the Gauge of an absorbing disk is a seminorm

Convexity and subadditivity

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

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 ${\frac {1}{2}}p_{K}(x+y),$ so that

p_{K}(x+y)\leq p_{K}(x)+p_{K}(y)+2e.

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

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

Balancedness and absolute homogeneity

Notice that $K$ 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

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

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

Topological properties

Assume that $X$ is a (real or complex) topological vector space (TVS) (not necessarily Hausdorff or locally convex) and let $K$ be an absorbing disk in $X.$ 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 $\operatorname {Int} _{X}K$ is the topological interior and $\operatorname {Cl} _{X}K$ is the topological closure of $K$ in $X.$ [6] Importantly, it was not assumed that $p_{K}$ was continuous nor was it assumed that $K$ had any topological properties.

Moreover, the Minkowski functional $p_{K}$ is continuous if and only if $K$ is a neighborhood of the origin in $X.$ [6] If $p_{K}$ 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

This section will investigate the most general case of the gauge of any subset $K$ of $X.$ The more common special case where $K$ is assumed to be an absorbing disk in $X$ was discussed above.

Properties

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

Throughout, $K$ is any subset of $X.$

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

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

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 $A\subseteq X$ that satisfies $(0,\infty )A=X$ is necessarily absorbing in $X$ is straightforward and can be found in the article on absorbing sets.

For any real $t>0,$

\{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 $0\cdot p_{K}(x)$ to be well-defined, it is necessary and sufficient that ${\displaystyle p_{K}(x)\neq \infty$ thus $p_{K}(tx)=tp_{K}(x)$ for all $x\in X$ and all non-negative real $t\geq 0$ if and only if $p_{K}$ is real-valued.

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

Examples

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

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

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

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

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

Positive homogeneity characterizes Minkowski functionals

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

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

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

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

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

Proof

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

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 $f:X\to [0,\infty ]$ is a function such that $f(tx)=tf(x)$ for all $x\in X$ and all real $t>0$ and let $K:=\{y\in X:f(y)\leq 1\}.$

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

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

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

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

Characterizing Minkowski functionals on star sets

Proposition[9] — Let $f:X\to [0,\infty ]$ be any function and $K\subseteq X$ be any subset. The following statements are equivalent:

$f$ is (strictly) positive homogeneous, $f(0)=0,$ and $\{x\in X:f(x)<1\}\;\subseteq \;K\;\subseteq \;\{x\in X:f(x)\leq 1\}.$
$f$ $f$ is the Minkowski functional of $K$ $K$ (that is, $f=p_{K}$ $f=p_{K}$ ), $K$ $K$ contains the origin, and $K$ $K$ is star-shaped at the origin.
- The set $K$ is star-shaped at the origin if and only if $tk\in K$ whenever $k\in K$ and $0\leq t\leq 1.$ A set that is star-shaped at the origin is sometimes called a star set.[10]

Characterizing Minkowski functionals that are seminorms

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

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

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

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

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

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

Positive sublinear functions and Minkowski functionals

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

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

Correspondence between open convex sets and positive continuous sublinear functions

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

Proof

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

Notes

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

References

Narici & Beckenstein 2011, p. 109.
Narici & Beckenstein 2011, p. 119.
Jarchow 1981, pp. 104–108.
Narici & Beckenstein 2011, pp. 115–154.
Schaefer 1999, p. 40.
Narici & Beckenstein 2011, p. 119-120.
Kubrusly 2011, p. 200.
Schechter 1996, p. 316.
Schechter 1996, pp. 313–317.
Schechter 1996, p. 303.
Narici & Beckenstein 2011, pp. 192–193.

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Minkowski functional

Definition

Motivating examples

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

See also

Notes

References

Further reading