# Minkowski functional

Short description: Function made from a set

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 $\displaystyle{ K }$ is a subset of a real or complex vector space $\displaystyle{ X, }$ then the Minkowski functional or gauge of $\displaystyle{ K }$ is defined to be the function $\displaystyle{ p_K : X \to [0, \infty], }$ valued in the extended real numbers, defined by $\displaystyle{ p_K(x) := \inf \{r \in \R : r \gt 0 \text{ and } x \in r K\} \quad \text{ for every } x \in X, }$ where the infimum of the empty set is defined to be positive infinity $\displaystyle{ \,\infty\, }$ (which is not a real number so that $\displaystyle{ p_K(x) }$ would then not be real-valued).

The set $\displaystyle{ K }$ is often assumed/picked to have properties, such as being an absorbing disk in $\displaystyle{ X, }$ that guarantee that $\displaystyle{ p_K }$ will be a real-valued seminorm on $\displaystyle{ X. }$ In fact, every seminorm $\displaystyle{ p }$ on $\displaystyle{ X }$ is equal to the Minkowski functional (that is, $\displaystyle{ p = p_K }$) of any subset $\displaystyle{ K }$ of $\displaystyle{ X }$ satisfying $\displaystyle{ \{x \in X : p(x) \lt 1\} \subseteq K \subseteq \{x \in X : p(x) \leq 1\} }$ (where all three of these sets are necessarily absorbing in $\displaystyle{ X }$ 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 $\displaystyle{ X }$ into certain algebraic properties of a function on $\displaystyle{ X. }$

The Minkowski function is always non-negative (meaning $\displaystyle{ p_K \geq 0 }$). 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, $\displaystyle{ p_K }$ might not be real-valued since for any given $\displaystyle{ x \in X, }$ the value $\displaystyle{ p_K(x) }$ is a real number if and only if $\displaystyle{ \{r \gt 0 : x \in r K\} }$ is not empty. Consequently, $\displaystyle{ K }$ is usually assumed to have properties (such as being absorbing in $\displaystyle{ X, }$ for instance) that will guarantee that $\displaystyle{ p_K }$ is real-valued.

## Definition

Let $\displaystyle{ K }$ be a subset of a real or complex vector space $\displaystyle{ X. }$ Define the gauge of $\displaystyle{ K }$ or the Minkowski functional associated with or induced by $\displaystyle{ K }$ as being the function $\displaystyle{ p_K : X \to [0, \infty], }$ valued in the extended real numbers, defined by $\displaystyle{ p_K(x) := \inf \{r \gt 0 : x \in r K\}, }$ where recall that the infimum of the empty set is $\displaystyle{ \,\infty\, }$ (that is, $\displaystyle{ \inf \varnothing = \infty }$). Here, $\displaystyle{ \{r \gt 0 : x \in r K\} }$ is shorthand for $\displaystyle{ \{r \in \R : r \gt 0 \text{ and } x \in r K\}. }$

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

Some conditions making a gauge real-valued

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

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

## Motivating examples

Example 1

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

Example 2

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

1. It is subadditive: $\displaystyle{ p_K(x + y) \leq p_K(x) + p_K(y). }$
2. It is absolutely homogeneous: $\displaystyle{ p_K(s x) = |s| p_K(x) }$ for all scalars $\displaystyle{ s. }$
3. It is nonnegative: $\displaystyle{ p_K \geq 0. }$

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

## Common conditions guaranteeing gauges are seminorms

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

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

Theorem[2] — If $\displaystyle{ K }$ is an absorbing disk in a vector space $\displaystyle{ X }$ then the Minkowski functional of $\displaystyle{ K, }$ which is the map $\displaystyle{ p_K : X \to [0, \infty) }$ defined by $\displaystyle{ p_K(x) := \inf \{r \gt 0 : x \in r K\}, }$ is a seminorm on $\displaystyle{ X. }$ Moreover, $\displaystyle{ p_K(x) = \frac{1}{\sup \{r \gt 0 : r x \in K\}}. }$

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

### Gauges of absorbing disks

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

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

1. If $\displaystyle{ K }$ is convex then $\displaystyle{ p_K }$ is subadditive.
2. If $\displaystyle{ K }$ is balanced then $\displaystyle{ p_K }$ is absolutely homogeneous; that is, $\displaystyle{ p_K(s x) = |s| p_K(x) }$ for all scalars $\displaystyle{ s. }$
Proof that the Gauge of an absorbing disk is a seminorm

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

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

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

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

Balancedness and absolute homogeneity

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

Therefore $\displaystyle{ p_K (\lambda x) = \inf \left\{r \gt 0 : \lambda x \in r K \right\} = \inf \left\{r \gt 0 : x \in \frac{r}{|\lambda|} K \right\} = \inf \left\{|\lambda|\frac{r}{|\lambda|} \gt 0 : x \in \frac{r}{|\lambda|} K \right\} = |\lambda| p_K(x). }$

### Algebraic properties

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

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

### Topological properties

Assume that $\displaystyle{ X }$ is a (real or complex) topological vector space (TVS) (not necessarily Hausdorff or locally convex) and let $\displaystyle{ K }$ be an absorbing disk in $\displaystyle{ X. }$ Then $\displaystyle{ \operatorname{Int}_X K \; \subseteq \; \{x \in X : p_K(x) \lt 1\} \; \subseteq \; K \; \subseteq \; \{x \in X : p_K(x) \leq 1\} \; \subseteq \; \operatorname{Cl}_X K, }$ where $\displaystyle{ \operatorname{Int}_X K }$ is the topological interior and $\displaystyle{ \operatorname{Cl}_X K }$ is the topological closure of $\displaystyle{ K }$ in $\displaystyle{ X. }$[6] Importantly, it was not assumed that $\displaystyle{ p_K }$ was continuous nor was it assumed that $\displaystyle{ K }$ had any topological properties.

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

## Minimal requirements on the set

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

### Properties

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

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

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

1. Strict positive homogeneity: $\displaystyle{ p_K(r x) = r p_K(x) }$ for all $\displaystyle{ x \in X }$ and all positive real $\displaystyle{ r \gt 0. }$
• Positive/Nonnegative homogeneity: $\displaystyle{ p_K }$ is nonnegative homogeneous if and only if $\displaystyle{ p_K }$ is real-valued.
• A map $\displaystyle{ p }$ is called nonnegative homogeneous[7] if $\displaystyle{ p(r x) = r p(x) }$ for all $\displaystyle{ x \in X }$ and all nonnegative real $\displaystyle{ r \geq 0. }$ Since $\displaystyle{ 0 \cdot \infty }$ is undefined, a map that takes infinity as a value is not nonnegative homogeneous.
2. Real-values: $\displaystyle{ (0, \infty) K }$ is the set of all points on which $\displaystyle{ p_K }$ is real valued. So $\displaystyle{ p_K }$ is real-valued if and only if $\displaystyle{ (0, \infty) K = X, }$ in which case $\displaystyle{ 0 \in K. }$
• Value at $\displaystyle{ 0 }$: $\displaystyle{ p_K(0) \neq \infty }$ if and only if $\displaystyle{ 0 \in K }$ if and only if $\displaystyle{ p_K(0) = 0. }$
• Null space: If $\displaystyle{ x \in X }$ then $\displaystyle{ p_K(x) = 0 }$ if and only if $\displaystyle{ (0, \infty) x \subseteq (0, 1) K }$ if and only if there exists a divergent sequence of positive real numbers $\displaystyle{ t_1, t_2, t_3, \cdots \to \infty }$ such that $\displaystyle{ t_n x \in K }$ for all $\displaystyle{ n. }$ Moreover, the zero set of $\displaystyle{ p_K }$ is $\displaystyle{ \ker p_K ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left\{y \in X : p_K(y) = 0 \right\} = {\textstyle\bigcap\limits_{e \gt 0}} (0, e) K. }$
3. Comparison to a constant: If $\displaystyle{ 0 \leq r \leq \infty }$ then for any $\displaystyle{ x \in X, }$ $\displaystyle{ p_K(x) \lt r }$ if and only if $\displaystyle{ x \in (0, r) K; }$ this can be restated as: If $\displaystyle{ 0 \leq r \leq \infty }$ then $\displaystyle{ p_K^{-1}([0, r)) = (0, r) K. }$
• It follows that if $\displaystyle{ 0 \leq R \lt \infty }$ is real then $\displaystyle{ p_K^{-1}([0, R]) = {\textstyle\bigcap\limits_{e \gt 0}} (0, R + e) K, }$ where the set on the right hand side denotes $\displaystyle{ {\textstyle\bigcap\limits_{e \gt 0}} [(0, R + e) K] }$ and not its subset $\displaystyle{ \left[{\textstyle\bigcap\limits_{e \gt 0}} (0, R + e)\right] K = (0, R] K. }$ If $\displaystyle{ R \gt 0 }$ then these sets are equal if and only if $\displaystyle{ K }$ contains $\displaystyle{ \left\{y \in X : p_K(y) = 1 \right\}. }$
• In particular, if $\displaystyle{ x \in R K }$ or $\displaystyle{ x \in (0, R] K }$ then $\displaystyle{ p_K(x) \leq R, }$ but importantly, the converse is not necessarily true.
4. Gauge comparison: For any subset $\displaystyle{ L \subseteq X, }$ $\displaystyle{ p_K \leq p_L }$ if and only if $\displaystyle{ (0, 1) L \subseteq (0, 1) K; }$ thus $\displaystyle{ p_L = p_K }$ if and only if $\displaystyle{ (0, 1) L = (0, 1) K. }$
• The assignment $\displaystyle{ L \mapsto p_L }$ is order-reversing in the sense that if $\displaystyle{ K \subseteq L }$ then $\displaystyle{ p_L \leq p_K. }$[8]
• Because the set $\displaystyle{ L := (0, 1) K }$ satisfies $\displaystyle{ (0, 1) L = (0, 1) K, }$ it follows that replacing $\displaystyle{ K }$ with $\displaystyle{ p_K^{-1}([0, 1)) = (0, 1) K }$ will not change the resulting Minkowski functional. The same is true of $\displaystyle{ L := (0, 1] K }$ and of $\displaystyle{ L := p_K^{-1}([0, 1]). }$
• If $\displaystyle{ D ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left\{y \in X : p_K(y) = 1 \text{ or } p_K(y) = 0 \right\} }$ then $\displaystyle{ p_D = p_K }$ and $\displaystyle{ D }$ has the particularly nice property that if $\displaystyle{ r \gt 0 }$ is real then $\displaystyle{ x \in r D }$ if and only if $\displaystyle{ p_D(x) = r }$ or $\displaystyle{ p_D(x) = 0. }$[note 1] Moreover, if $\displaystyle{ r \gt 0 }$ is real then $\displaystyle{ p_D(x) \leq r }$ if and only if $\displaystyle{ x \in (0, r] D. }$
5. Subadditive/Triangle inequality: $\displaystyle{ p_K }$ is subadditive if and only if $\displaystyle{ (0, 1) K }$ is convex. If $\displaystyle{ K }$ is convex then so are both $\displaystyle{ (0, 1) K }$ and $\displaystyle{ (0, 1] K }$ and moreover, $\displaystyle{ p_K }$ is subadditive.
6. Scaling the set: If $\displaystyle{ s \neq 0 }$ is a scalar then $\displaystyle{ p_{s K}(y) = p_K\left(\tfrac{1}{s} y\right) }$ for all $\displaystyle{ y \in X. }$ Thus if $\displaystyle{ 0 \lt r \lt \infty }$ is real then $\displaystyle{ p_{r K}(y) = p_K\left(\tfrac{1}{r} y\right) = \tfrac{1}{r} p_K(y). }$
7. Symmetric: $\displaystyle{ p_K }$ is symmetric (meaning that $\displaystyle{ p_K(- y) = p_K(y) }$ for all $\displaystyle{ y \in X }$) if and only if $\displaystyle{ (0, 1) K }$ is a symmetric set (meaning that$\displaystyle{ (0, 1) K = - (0, 1) K }$), which happens if and only if $\displaystyle{ p_K = p_{- K}. }$
8. Absolute homogeneity: $\displaystyle{ p_K(u x) = p_K(x) }$ for all $\displaystyle{ x \in X }$ and all unit length scalars $\displaystyle{ u }$[note 2] if and only if $\displaystyle{ (0, 1) u K \subseteq (0, 1) K }$ for all unit length scalars $\displaystyle{ u, }$ in which case $\displaystyle{ p_K(s x) = |s| p_K(x) }$ for all $\displaystyle{ x \in X }$ and all non-zero scalars $\displaystyle{ s \neq 0. }$ If in addition $\displaystyle{ p_K }$ is also real-valued then this holds for all scalars $\displaystyle{ s }$ (that is, $\displaystyle{ p_K }$ is absolutely homogeneous[note 3]).
• $\displaystyle{ (0, 1) u K \subseteq (0, 1) K }$ for all unit length $\displaystyle{ u }$ if and only if $\displaystyle{ (0, 1) u K = (0, 1) K }$ for all unit length $\displaystyle{ u. }$
• $\displaystyle{ s K \subseteq K }$ for all unit scalars $\displaystyle{ s }$ if and only if $\displaystyle{ s K = K }$ for all unit scalars $\displaystyle{ s; }$ if this is the case then $\displaystyle{ (0, 1) K = (0, 1) s K }$ for all unit scalars $\displaystyle{ s. }$
• The Minkowski functional of any balanced set is a balanced function.[8]
9. Absorbing: If $\displaystyle{ K }$ is convex or balanced and if $\displaystyle{ (0, \infty) K = X }$ then $\displaystyle{ K }$ is absorbing in $\displaystyle{ X. }$
• If a set $\displaystyle{ A }$ is absorbing in $\displaystyle{ X }$ and $\displaystyle{ A \subseteq K }$ then $\displaystyle{ K }$ is absorbing in $\displaystyle{ X. }$
• If $\displaystyle{ K }$ is convex and $\displaystyle{ 0 \in K }$ then $\displaystyle{ [0, 1] K = K, }$ in which case $\displaystyle{ (0, 1) K \subseteq K. }$
10. Restriction to a vector subspace: If $\displaystyle{ S }$ is a vector subspace of $\displaystyle{ X }$ and if $\displaystyle{ p_{K \cap S} : S \to [0, \infty] }$ denotes the Minkowski functional of $\displaystyle{ K \cap S }$ on $\displaystyle{ S, }$ then $\displaystyle{ p_K\big\vert_S = p_{K \cap S}, }$ where $\displaystyle{ p_K\big\vert_S }$ denotes the restriction of $\displaystyle{ p_K }$ to $\displaystyle{ S. }$
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 $\displaystyle{ A \subseteq X }$ that satisfies $\displaystyle{ (0, \infty) A = X }$ is necessarily absorbing in $\displaystyle{ X }$ is straightforward and can be found in the article on absorbing sets.

For any real $\displaystyle{ t \gt 0, }$ $\displaystyle{ \{r \gt 0 : t x \in r K\} = \{t(r/t) : x \in (r/t) K\} = t \{s \gt 0 : x \in s K\} }$ so that taking the infimum of both sides shows that $\displaystyle{ p_K(tx) = \inf \{r \gt 0 : t x \in r K\} = t \inf \{s \gt 0 : x \in s K\} = t p_K(x). }$ This proves that Minkowski functionals are strictly positive homogeneous. For $\displaystyle{ 0 \cdot p_K(x) }$ to be well-defined, it is necessary and sufficient that $\displaystyle{ p_K(x) \neq \infty; }$ thus $\displaystyle{ p_K(tx) = t p_K(x) }$ for all $\displaystyle{ x \in X }$ and all non-negative real $\displaystyle{ t \geq 0 }$ if and only if $\displaystyle{ p_K }$ is real-valued.

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

### Examples

1. If $\displaystyle{ \mathcal{L} }$ is a non-empty collection of subsets of $\displaystyle{ X }$ then $\displaystyle{ p_{\cup \mathcal{L}}(x) = \inf \left\{p_L(x) : L \in \mathcal{L} \right\} }$ for all $\displaystyle{ x \in X, }$ where $\displaystyle{ \cup \mathcal{L} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ {\textstyle\bigcup\limits_{L \in \mathcal{L}}} L. }$
• Thus $\displaystyle{ p_{K \cup L}(x) = \min \left\{p_K(x), p_L(x) \right\} }$ for all $\displaystyle{ x \in X. }$
2. If $\displaystyle{ \mathcal{L} }$ is a non-empty collection of subsets of $\displaystyle{ X }$ and $\displaystyle{ I \subseteq X }$ satisfies $\displaystyle{ \left\{x \in X : p_L(x) \lt 1 \text{ for all } L \in \mathcal{L}\right\} \quad \subseteq \quad I \quad \subseteq \quad \left\{x \in X : p_L(x) \leq 1 \text{ for all } L \in \mathcal{L}\right\} }$ then $\displaystyle{ p_I(x) = \sup \left\{p_L(x) : L \in \mathcal{L}\right\} }$ for all $\displaystyle{ x \in X. }$

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

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

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

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

### Positive homogeneity characterizes Minkowski functionals

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

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

1. Strict positive homogeneity: $\displaystyle{ \; f(t x) = t f(x) }$ for all $\displaystyle{ x \in X }$ and all positive real $\displaystyle{ t \gt 0. }$
• This statement is equivalent to: $\displaystyle{ f(t x) \leq t f(x) }$ for all $\displaystyle{ x \in X }$ and all positive real $\displaystyle{ t \gt 0. }$
2. $\displaystyle{ f }$ is a Minkowski functional: meaning that there exists a subset $\displaystyle{ S \subseteq X }$ such that $\displaystyle{ f = p_S. }$
3. $\displaystyle{ f = p_K }$ where $\displaystyle{ K := \{x \in X : f(x) \leq 1\}. }$
4. $\displaystyle{ f = p_V \, }$ where $\displaystyle{ V \,:= \{x \in X : f(x) \lt 1\}. }$

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

1. Positive/Nonnegative homogeneity: $\displaystyle{ f(t x) = t f(x) }$ for all $\displaystyle{ x \in X }$ and all nonnegative real $\displaystyle{ t \geq 0. }$
Proof

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

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

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

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

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

This theorem can be extended to characterize certain classes of $\displaystyle{ [- \infty, \infty] }$-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 $\displaystyle{ f : X \to \R }$ (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[10] — Let $\displaystyle{ f : X \to [0, \infty] }$ be any function and $\displaystyle{ K \subseteq X }$ be any subset. The following statements are equivalent:

1. $\displaystyle{ f }$ is (strictly) positive homogeneous, $\displaystyle{ f(0) = 0, }$ and $\displaystyle{ \{x \in X : f(x) \lt 1\} \; \subseteq \; K \; \subseteq \; \{x \in X : f(x) \leq 1\}. }$
2. $\displaystyle{ f }$ is the Minkowski functional of $\displaystyle{ K }$ (that is, $\displaystyle{ f = p_K }$), $\displaystyle{ K }$ contains the origin, and $\displaystyle{ K }$ is star-shaped at the origin.
• The set $\displaystyle{ K }$ is star-shaped at the origin if and only if $\displaystyle{ t k \in K }$ whenever $\displaystyle{ k \in K }$ and $\displaystyle{ 0 \leq t \leq 1. }$ A set that is star-shaped at the origin is sometimes called a star set.[9]

### Characterizing Minkowski functionals that are seminorms

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

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

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

in which case $\displaystyle{ 0 \in K }$ and both $\displaystyle{ (0, 1) K = \{x \in X : p(x) \lt 1\} }$ and $\displaystyle{ \bigcap_{e \gt 0} (0, 1 + e) K = \left\{x \in X : p_K(x) \leq 1\right\} }$ will be convex, balanced, and absorbing subsets of $\displaystyle{ X. }$

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

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

### Positive sublinear functions and Minkowski functionals

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

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

Correspondence between open convex sets and positive continuous sublinear functions

Theorem[11] — Suppose that $\displaystyle{ X }$ 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 $\displaystyle{ X }$ are exactly those sets that are of the form $\displaystyle{ z + \{x \in X : p(x) \lt 1\} = \{x \in X : p(x - z) \lt 1\} }$ for some $\displaystyle{ z \in X }$ and some positive continuous sublinear function $\displaystyle{ p }$ on $\displaystyle{ X. }$

Proof

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

## Notes

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

## References

1. Narici & Beckenstein 2011, p. 109.
2. Narici & Beckenstein 2011, p. 119.
3. Jarchow 1981, pp. 104-108.
4. Narici & Beckenstein 2011, pp. 115-154.
5. Schaefer 1999, p. 40.
6. Narici & Beckenstein 2011, p. 119-120.
7. Kubrusly 2011, p. 200.
8. Schechter 1996, p. 316.
9. Schechter 1996, p. 303.
10. Schechter 1996, pp. 313-317.
11. Narici & Beckenstein 2011, pp. 192-193.