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 $$p_K : X \to [0, \infty],$$ valued in the extended real numbers, defined by $$p_K(x) := \inf \{r \in \R : r > 0 \text{ and } x \in r K\} \quad \text{ for every } x \in X,$$ where the infimum of the empty set is defined to be positive infinity $$\,\infty\,$$ (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 $$\{x \in X : p(x) < 1\} \subseteq K \subseteq \{x \in X : p(x) \leq 1\}$$ (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 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 $$\{r > 0 : x \in r K\}$$ 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 $$p_K : X \to [0, \infty],$$ valued in the extended real numbers, defined by $$p_K(x) := \inf \{r > 0 : x \in r K\},$$ where recall that the infimum of the empty set is $$\,\infty\,$$ (that is, $$\inf \varnothing = \infty$$). Here, $$\{r > 0 : x \in r K\}$$ is shorthand for $$\{r \in \R : r > 0 \text{ and } x \in r K\}.$$

For any $$x \in X,$$ $$p_K(x) \neq \infty$$ if and only if $$\{r > 0 : x \in r K\}$$ is not empty. The arithmetic operations on $$\R$$ 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 $$\{r > 0 : x \in r K\}$$ 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 $$U := \{x\in X : \|x\| \leq 1\}$$ 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:


 * 1) It is : $$p_K(x + y) \leq p_K(x) + p_K(y).$$
 * 2) It is : $$p_K(s x) = |s| p_K(x)$$ for all scalars $$s.$$
 * 3) 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 $$K.$$

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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 $$p_{[0, 1] K}(s x) = s p_{[0, 1] K}(x)$$ for all real $$s \geq 0,$$ where $$[0, 1] K = \{t k : t \in [0, 1], k \in K\}.$$ 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.

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.

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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 $$p_K(x + y) \leq p_K(x) + p_K(y) + 2 e.$$

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 $$\{r > 0 : x \in r K\}$$ is nonempty, implies that $$K$$ is absorbing.

Balancedness and absolute homogeneity

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

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

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. $$p_{s K} = \frac{1}{|s|} p_K$$ 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 $$\{x \in X : p(x) < 1\} \; \subseteq \; K \; \subseteq \; \{x \in X : p(x) \leq 1\}$$ 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 $$K.$$ If $$x \in X$$ satisfies $$p_K(x) < 1$$ then $$x \in K.$$ 
 * In particular, if $$K$$ is as above and $$q$$ is any seminorm on $$X,$$ then $$q = p$$ if and only if $$\{x \in X : q(x) < 1\} \; \subseteq \; K \; \subseteq \; \{x \in X : q(x) \leq 1\}.$$

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 $$X.$$ 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 $$\operatorname{Int}_X K = \{x \in X : p_K(x) < 1\} \quad \text{ and } \quad \operatorname{Cl}_X K = \{x \in X : p_K(x) \leq 1\}.$$

Minimal requirements on the set
This section will investigate the most general case of the gauge of 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 $$X.$$

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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 $$t > 0,$$ $$\{r > 0 : t x \in r K\} = \{t(r/t) : x \in (r/t) K\} = t \{s > 0 : x \in s K\}$$ so that taking the infimum of both sides shows that $$p_K(tx) = \inf \{r > 0 : t x \in r K\} = t \inf \{s > 0 : x \in s K\} = t p_K(x).$$ 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 $$r e^{i t}$$ for some real $$t$$ where $$r := |s| \geq 0$$ and $$e^{i t}$$ 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 $$\mathcal{L}$$ is a non-empty collection of subsets of $$X$$ then $$p_{\cup \mathcal{L}}(x) = \inf \left\{p_L(x) : L \in \mathcal{L} \right\}$$ for all $$x \in X,$$ where $$\cup \mathcal{L} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ {\textstyle\bigcup\limits_{L \in \mathcal{L}}} L.$$ </li> <li>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 $$x \in X.$$ </li> </ol>
 * Thus $$p_{K \cup L}(x) = \min \left\{p_K(x), p_L(x) \right\}$$ 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 those functions $$f : X \to [0, \infty]$$ that have a certain purely algebraic property that is commonly encountered.

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

For all real $$t > 0,$$ $$f(0) = f(t 0) = t f(0)$$ 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 $$0 < \tfrac{1}{R} < \infty,$$ $$p_K(x)= R = f(x)$$ 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 $$x = r k,$$ 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 $$[- \infty, \infty]$$-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
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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.

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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 $$V := \{x \in X : f(x) < 1\}$$ 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 $$|f| : X \to [0, \infty)$$ is continuous.

A 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

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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. $$\blacksquare$$