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 or of $K$ is defined to be the function p_K : X \to , \infty valued in the extended real numbers, defined by
$p_K(x) := \inf \ \quad \text x \in X,$
where the infimum of the empty set is defined to be positive infinity $\,\infty\,$ (which is a real number so that $p_K(x)$ would then 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

$\ \subseteq K \subseteq \$

(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 defined by purely algebraic properties) can be associated (non-uniquely) with an absorbing disk (which is a 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 properties of a subset of $X$ into certain 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 $\$ 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 of $K$ or the associated with or induced by $K$ as being the function p_K : X \to , \infty valued in the extended real numbers, defined by

$p_K(x) := \inf \,$

(recall that the infimum of the empty set is $\,\infty$, that is, $\inf \varnothing = \infty$). Here, $\$ is shorthand for $\.$

For any $x \in X,$ $p_K(x) \neq \infty$ if and only if $\$ is not empty.
The arithmetic operations on $\R$ can be extended to operate on $\pm \infty,$ where $\frac := 0$ 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 $\$ 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 or of $K$ in $X.$
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 := \$ 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 $\mathbb.$
Let $f : X \to \mathbb$ be any linear functional on $X$ (not necessarily continuous).
Fix $a > 0.$
Let $K$ be the set
$K := \$
and let $p_K$ be the Minkowski functional of $K.$
Then
$p_K(x) = \frac , f(x), \quad \text x \in X.$
The function $p_K$ has the following properties:

#It is : $p_K(x + y) \leq p_K(x) + p_K(y).$
#It is : $p_K(s x) = , s, p_K(x)$ for all scalars $s.$
#It is : $p_K \geq 0.$

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 $K.$

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 $p_$ is positive homogeneous, meaning that $p_(s x) = s p_(x)$ for all real $s \geq 0,$ where , 1K = \.
If $q$ is a nonnegative real-valued function on $X$ that is positive homogeneous, then the sets $U := \$ and $D := \$ satisfy , 1U = U and , 1D = D;
if in addition $q$ is absolutely homogeneous then both $U$ and $D$ are balanced.

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.

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, $\frac x + \frac y \in K_e.$
By definition of the Minkowski functional $p_K,$
$p_K\left(\frac x + \frac y\right) \leq r + e = \frac p_K(x) + \frac p_K(y) + e.$

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

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 $\$ is nonempty, implies that $K$ is absorbing.

Balancedness and absolute homogeneity

Notice that $K$ being balanced implies that
$\lambda x \in r K \quad \mbox \quad x \in \frac K.$

Therefore
$p_K (\lambda x) = \inf \left\ = \inf \left\ = \inf \left\ = , \lambda, p_K(x).$

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.
*$p_ = \frac p_K$ for any scalar $s \neq 0.$
*If $J$ is an absorbing disk in $X$ and $J \subseteq K$ then $p_K \leq p_J.$
*If $K$ is a set satisfying $\ \; \subseteq \; K \; \subseteq \; \$ then $K$ is absorbing in $X$ and $p = p_K,$ where $p_K$ is the Minkowski functional associated with $K;$ that is, it is the gauge of $K.$
* In particular, if $K$ is as above and $q$ is any seminorm on $X,$ then $q = p$ if and only if $\ \; \subseteq \; K \; \subseteq \; \.$
*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

$\operatorname_X K \; \subseteq \; \ \; \subseteq \; K \; \subseteq \; \ \; \subseteq \; \operatorname_X K,$

where $\operatorname_X K$ is the topological interior and $\operatorname_X K$ is the topological closure of $K$ in $X.$
Importantly, it was 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 $X.$
If $p_K$ is continuous then
$\operatorname_X K = \ \quad \text \quad \operatorname_X K = \.$

Minimal requirements on the set

This section will investigate the most general case of the gauge of 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 $X.$

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 $t > 0,$

$\ = \ = t \$

so that taking the infimum of both sides shows that

$p_K(tx) = \inf \ = t \inf \ = t p_K(x).$

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 $p_K(x) \neq \infty;$ thus $p_K(tx) = t p_K(x)$ for all $x \in X$ and all real $t \geq 0$ if and only if $p_K$ is real-valued.

The hypothesis of statement (7) allows us to conclude that $p_K(s x) = p_K(x)$ for all $x \in X$ and all scalars $s$ satisfying $, s, = 1.$
Every scalar $s$ is of the form $r e^$ for some real $t$ where $r := , s, \geq 0$ and $e^$ 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 $\mathcal$ is a non-empty collection of subsets of $X$ then $p_(x) = \inf \left\$ for all $x \in X,$ where $\cup \mathcal ~\stackrel~ L.$
#* Thus $p_(x) = \min \left\$ for all $x \in X.$
#If $\mathcal$ is a non-empty collection of subsets of $X$ and $I \subseteq X$ satisfies
$\left\ \quad \subseteq \quad I \quad \subseteq \quad \left\$
then $p_I(x) = \sup \left\$ for all $x \in X.$

The following examples show that the containment $(0, R] K \; \subseteq \; (0, R + e) K$ could be proper.

Example: If $R = 0$ and $K = X$ then $(0, R] K = (0, 0] X = \varnothing X = \varnothing$ but $(0, e) K = X = X,$ which shows that its possible for $(0, R] K$ to be a proper subset of $(0, R + e) K$ 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 , 1K \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 , 1K = 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 $x \in (0, 1 + e) K.$
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 those functions f : X \to , \infty/math> that have a certain purely algebraic property that is commonly encountered.

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

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 f : X \to , \infty/math> is a function such that $f(t x) = t f(x)$ for all $x \in X$ and all real $t > 0$ and let $K := \.$

For all real $t > 0,$ $f(0) = f(t 0) = t f(0)$ 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 $f(x) = \infty;$ in either case $f(t x) = t f(x) = f(x)$ for all real $t > 0.$
Now if $f(x) = 0$ then this implies that that $t x \in K$ for all real $t > 0$ (since $f(t x) = 0 \leq 1$), which implies that $p_K(x) = 0,$ as desired.
Similarly, if $f(x) = \infty$ then $t x \not\in K$ 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 $p_K(t x) = t p_K(x)$ for all real $t > 0.$
Since $0 < \tfrac < \infty,$ $p_K(x)= R = f(x)$ if and only if $p_K\left(\tfrac x\right) = 1 = f\left(\tfrac x\right)$ so assume without loss of generality that $R = 1$ and it remains to show that $p_K\left(\tfrac x\right) = 1.$
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 $x = r k,$ where note that $k \in K$ implies that $f(k) \leq 1.$
Then $1 = f(x) = f(r k) = r f(k) \leq r < 1.$ $\blacksquare$

This theorem can be extended to characterize certain classes of \infty, \infty/math>-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 $f : X \to \R$ (such as linear functionals) can be written in terms of a unique Minkowski functional having a certain property.

Characterizing Minkowski functionals on star sets

Characterizing Minkowski functionals that are seminorms

In this next theorem, which follows immediately from the statements above, $K$ is 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 $(0, 1) s K \subseteq (0, 1) K$ 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.

Positive sublinear functions and Minkowski functionals

It may be shown that a real-valued subadditive function $f : X \to \R$ 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 := \$ is an open neighborhood in $X.$
If $f : X \to \R$ is subadditive and satisfies $f(0) = 0,$ then $f$ is continuous if and only if its absolute value , f, : X \to 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 := \$ 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

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 $p_K^([0, 1)) = (0, 1) K,$ from which it follows that $V - z = \$ and so
$V = z + \.$
Since $z + \ = \,$ this completes the proof. $\blacksquare$

See also

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Notes

References

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Further reading

* F. Simeski, A. M. P. Boelens, and M. Ihme. "Modeling Adsorption in Silica Pores via Minkowski Functionals and Molecular Electrostatic Moments". ''Energies'' 13 (22) 5976 (2020). .

{{Convex analysis and variational analysis

Convex analysis
Functional analysis
Hermann Minkowski