Minkowski functional

In mathematics, in the field of functional analysis, a Minkowski functional is a function that recovers a notion of distance on a linear space.

If K subset of a real or complex vector space X, then we define the Minkowski functional or gauge of K to be the function pK : X → [0, ∞], valued in the extended real numbers, defined by

pK(x) ≝ inf { r ∈ ℝ : r > 0 and x ∈ rK }

for every x ∈ X, where the infimum of the empty set is defined to be positive infinity ∞ (which is not a real number so that pK(x) would then not be real-valued). If the set is not empty then the infimum will necessarily be a non-negative real number. This property of being non-negative stands in contrast to other classes of functions, such as sublinear functions, that do allow negative values.

In functional analysis, K is usually assumed to have properties (e.g. such as being absorbing in X) that will guarantee that for every x ∈ X, this set { r ∈ ℝ : r > 0 and x ∈ rK } is not empty precisely because this results in pK being real-valued.

Moreover, K is also often assumed to have more properties, such as being an absorbing disk in X, since these properties guarantee that pK will be a (real-valued) seminorm on X. In fact, every seminorm p on X is equal to the Minkowski functional of any subset K of X satisfying { x ∈ X : p(x) < 1 } ⊆ K ⊆ { x ∈ X : p(x) ≤ 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.

Minkowski functionals have been applied to describe phase behavior of nanoconfined fluids.[1]

Definition
Definition and notation: 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 pK : X → [0, ∞], valued in the extended real numbers, defined by
pK(x) ≝ inf { r > 0 : x ∈ rK },

where recall that the infimum of the empty set is ∞ (i.e. inf ∅ = ∞). Here, { r > 0 : x ∈ rK } is shorthand for { r ∈ ℝ : r > 0 and x ∈ rK }.

Observe that for any x ∈ X, pK(x) ≠ ∞ if and only if { r > 0 : x ∈ rK } is not empty. Recall that we may partially extend ℝ's arithmetic operations to include ±∞, where r/±∞ ≝ 0 for all non-zero real −∞ < r < ∞. The products 0 ⋅ ∞ and 0 ⋅ −∞ remain undefined.

Some conditions making a gauge real-valued

In the field of convex analysis, the map pK taking on the value of ∞ is not necessarily an issue. However, in functional analysis we almost always want pK to be real-valued (i.e. to never take on the value of ∞), which happens if and only if the set { r > 0 : x ∈ rK } is non-empty for every x ∈ X.

In order for pK to be real-valued, it suffices for the origin of X to belong to the algebraic interior (or core) of K in X.[2] If K is absorbing in X, where recall that this implies that 0 ∈ K, then the origin belongs to the algebraic interior of K in X and thus pK is real-valued. Characterizations of when pK is real-valued are given below.

Motivating examples
Example 1

Consider a normed vector space (X, || · ||), with the norm || · || and let U be the unit ball in X. Then for every x ∈ X, we have || x || = pU(x). Thus the Minkowski functional pU is just the norm on X.

Example 2

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

K ≝ { x ∈ X : |f(x)| ≤ a  }

and let pK be the Minkowski functional of K.

Then

pK(x) = 1/a |f (x)|

for all x ∈ X.

Note that pK has the following properties:

  1. It is subadditive: pK(x + y) ≤ pK(x) + pK(y).
  2. It is homogeneous: pK(s x) = |s | pK(x) for all scalars s.
  3. It is nonnegative: pK ≥ 0.

Therefore, pK 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, pK(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 making gauges into seminorms

So that we have pK(0) = 0, we will henceforth assume that 0 ∈ K.


In order for pK to be a seminorm, it suffices for K to be a disk (i.e. convex and balanced) and absorbing in X, which are the most common assumption placed on K.

Theorem[3] — If K is an absorbing disk in a vector space X then the Minkowski functional of K, which is the map pK : X → [0, ∞) defined by

pK(x) ≝ inf { r > 0 : x ∈ rK},

is a seminorm on X. Moreover,

pK(x) = 1/sup { r > 0 : rx ∈ K } .

More generally, if K is convex and the origin to belong to the algebraic interior of K, then pK is a non-negative sublinear functional on X, which implies in particular that it is subadditive and positive homogeneous. If K is absorbing in X then one may show that p[0, 1]K is positive homogeneous i.e. that p[0, 1]K(sx) = s p[0, 1]K(x) for all real s ≥ 0, where [0, 1]K ≝ { tk : 0 ≤ t ≤ 1, k ∈ K }.[4] If q is a non-negative real-valued function on X that is positive homogeneous then the sets U ≝ { x ∈ X : q(x) < 1 } and D ≝ { x ∈ X : q(x) ≤ 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.[4]

Gauges of absorbing disks

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

Theorem — We assume that K is an absorbing subset of X. We show that:

  1. If K is convex then pK is subadditive.
  2. if K is balanced then pK is absolutely homogeneous (i.e. pK(s x) = |s| pK(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 pK(x) = pK(y) = r. For all ε > 0, we have x, y ∈ Kε ≝ (r + ε) K. Since K is convex and r + ε ≠ 0, Kε is also convex. Therefore, ½ x + ½ y ∈ Kε. By definition of the Minkowski functional pK, we have

.

But the left hand side is ½ pK(x + y) so that

pK(x + y) ≤ pK(x) + pK(y) + 2ε.

Since ε > 0 was arbitrary, it follows that pK(x + y) ≤ pK(x) + pK(y), which is the desired inequality. The general case pK(x) > pK(y) is obtained after the obvious modification.

Note: Convexity of K, together with the initial assumption that the set {r > 0: x ∈ r K} is nonempty, implies that K is absorbing.

Balancedness and absolute homogeneity

Notice that K being balanced implies that

Therefore

Algebraic properties

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

  • pK is a seminorm on X.
  • pK is a norm on X if and only if K does not contain a non-trivial vector subspace.[5]
  • For any scalar s ≠ 0, psK = 1/|s|pK.[5]
  • If J is an absorbing disk in X and J ⊆ K then pK ≤ pJ
  • If K is a set satisfying { x ∈ X : p(x) < 1 } ⊆ K ⊆ { x ∈ X : p(x) ≤ 1 } then K is absorbing in X and p = pK, where pK is the Minkowski functional associated with K (i.e. the gauge of K).[6]
    • In particular, if K is as above and q is any seminorm on X, then q = p if and only if { x ∈ X : q(x) < 1 } ⊆ K ⊆ { x ∈ X : q(x) ≤ 1 }.[6]
  • If x ∈ X satisfies pK(x) < 1 then x ∈ K.

Topological properties

Let X be a real or complex topological vector space (TVS) (not necessarily Hausdorff or locally convex) and let K be an absorbing disk in X.

  • Int K   ⊆   { x ∈ X : pK(x) < 1 }   ⊆   K   ⊆   { x ∈ X : pK(x) ≤ 1 }   ⊆   Cl K

    where Int K is the topological interior and Cl K is the closure of K in X.[7]

    • Note that we did not assume that pK was continuous or that K had any topological properties.
  • pK is continuous if and only if K is a neighborhood of the origin in X.[7]
  • If pK is continuous then:[7]
    • Int K = { x ∈ X : pK(x) < 1}, and
    • Cl K = { x ∈ X : pK(x) ≤ 1}.

Properties (minimal requirements)

In this section, we 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 is discussed after this section. However, all results in this section may be applied to the case where K is an absorbing disk.

Notation

As before, let K be any subset of a real or complex vector space X. This article will make heavy use the following types of sets (which are defined as usual) to characterize or deduce properties of Minkowski functionals.

Notation: If A is any set of scalars then let
AK ≝ { ak : a ∈ A, k ∈ K }.
Thus for any real R and S,
(R, S)K ≝ { tk : R < t < S, k ∈ K }, and [R, S)K ≝ { tk : R ≤ t < S, k ∈ K },
and we define the sets (R, S]K, (R, ∞)K, etc. similarly. To simplify the discussion, also define
(R, ∞]K ≝ (R, ∞)K and [R, ∞]K ≝ [R, ∞)K.
Also, if I is any such interval or set of scalars then for every x ∈ X, we will let Ix denote I { x }; for instance (R, S)x ≝ { tx : R < t < S }.
Examples
  • Let X = ℝ and x ∈ X. If x > 0 then (0, 1)x is just the interval (0, x); that is, it is the open line segment lying strictly between the origin and x. Similarly, if x > 0 then (1, ∞)x = (x, ∞) is the infinite open ray starting at (but not including) x and heading directly away from the origin. If x = 0 then (0, 1)x = { 0 } = (1, ∞)x = ℝ x. If x < 0 then (0, x) = ∅ and (0, 1)x = (x, 0).
  • (0, 1) K = ∪k ∈ K (0, 1) k is the union of all open line segments lying strictly between the origin and some k ∈ K (except if k = 0 ∈ K, in which case (0, 1) k = { 0 }).
  • Let x ∈ X, C ≝ [0, 1] x, L ≝ [0, 1) x, and R ≝ (0, 1] x. If I is any of the intervals [0, 1], [0, 1), (0, 1], or (0, 1) then L = I L, C ≠ I L, R ≠ I L, and R ≠ I C. Also, L = [0, 1) C = (0, 1) C so in particular, the last two sets are not equal to C; moreover, if x ≠ 0 then also C ≠ (0, 1) C. However, we do have C = (0, 1] C = [0, 1] C and R = (0, 1] R = [0, 1] R.
Properties

If A is a set of scalars, −∞ ≤ R ≤ S ≤ ∞, and x ∈ X then we will use the following basic characerizations without comment:

  • Inclusion in a set: x ∈ (R, S)K if and only if tx ∈ (tR, tS)K for all real t > 0. Moreover, x ∈ AK if and only if sx ∈ AsK for all scalars s.
  • Exclusion from a set: x ∉ (R, S)K if and only if tx ∉ (tR, tS)K for all real t > 0; if neither R nor S is infinite then we may replace "t > 0" with "t ≥ 0" in the last condition. Moreover, x ∉ AK if and only if sx ∉ AsK for all non-zero scalars s.
    • Note that in contrast to inclusion, sx ∉ AsK for all scalars s if and only if AK = ∅.

Seminorms and Minkowski functionals

The following corollary mainly summarizes and combines the results that are established in the basic properties section below.

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

  1. Subadditive/Triangle inequality: pK is subadditive if and only if (0, 1) K is convex. If K is convex then so are (0, 1) K and (0, 1] K and pK is subadditive.
  2. Real-valued: pK is real-valued if and only if (0, ∞)K = X, in which case 0 ∈ K.
  3. Absorbing: If K is convex (or balanced) and if (0, ∞)K = X then K is absorbing in X.
    • Note that if a set A is absorbing in X and A ⊆ K then K is absorbing in X.
    • If K is convex and 0 ∈ K then [0, 1] K = K, so in particular (0, 1) K ⊆ K.
  4. Strict positive homogeneity: pK(rx) = r pK(x) for all x ∈ X and all positive real r > 0.
  5. Positive/Nonnegative homogeneity: pK is nonnegative homogeneous if and only if pK is real-valued.
  6. Absolute homogeneity: pK(u x) = pK(x) for all x ∈ X and all unit length scalars u[note 1] if and only if (0, 1) u K ⊆ (0, 1) K for all unit length scalars u, in which case (0, 1) u K = (0, 1) K for all such u and pK(s x) = |s| pK(x) for all x ∈ X and all non-zero scalars s ≠ 0. If in addition pK is also real-valued then this holds for all scalars (i.e. pK is absolutely homogeneous).
    • Recall that pK is called absolutely homogeneous if |s| pK(x) is well-defined and pK(s x) = |s| pK(x) for all x ∈ X and all scalars s (not just non-zero scalars).
    • s K ⊆ K for all unit scalars s if and only if s K = K for all unit scalars s, where the right hand side might initially appear to be a stronger condition. If this is the case then (0, 1) K = (0, 1) s K for all unit scalars s.
Proof

Most of this theorem's statements follow immediately from the results established in the basic properties section below.

One statement that hasn't yet been proven, and which we now prove, is that a convex subset A of X that satisfies (0, ∞)A = X is absorbing in X. Suppose that X is a vector space over the field 𝔽, where 𝔽 is ℝ or ℂ, and assume that dim X ≠ 0. Observe that a set A is absorbing in X if and only if A ∩ span { x } is absorbing in every 1-dimensional vector subspace Y ≝ span { x } = 𝔽 x, where x ≠ 0. Thus, it is necessary and sufficient to show that A ∩ Y contains an open 𝔽-ball around the origin in Y = 𝔽 x. The condition (0, ∞)A = X implies that every "open ray" in Y starting at the origin (i.e. a set of the form (0, ∞) y for some 0 ≠ y ∈ Y) contains an element of A so that in particular, A ∩ (0, ∞) y ≠ ∅ and A ∩ (0, ∞) (− y) ≠ ∅ (where (0, ∞) (− y) = (− ∞, 0) y so that A ∩ (− ∞, 0) y ≠ ∅) so that now the convexity of A ∩ Y makes it clear that for every 0 ≠ y ∈ Y, the convex set A ∩ ℝ y is a line segment (possibly open, closed, or half-closed, and possibly bounded or unbounded) containing an open sub-interval that contains the origin. If 𝔽 = ℝ then this shows that A ∩ Y is absorbing in Y = ℝ x so assume that 𝔽 = ℂ. Note that i x ∈ Y = ℂ x, where i = √−1, and that Y = ℂ x = (ℝ x) + (ℝ i x). Notice that the set (A ∩ ℝ x) ∪ (A ∩ ℝ (i x)), which is contained in A ∩ Y, is a union of two line segments intersecting at the origin (with each containing the origin in an open sub-interval) so that its convex hull, which is contained in the convex set A ∩ Y, clearly contains a quadrilateral having the origin in its interior. This shows that A ∩ Y is absorbing in Y, as desired.

Recall that the hypothesis of statement (6) allows us conclude that pK(sx) = pK(x) for all x ∈ X and all scalars s satisfying |s| = 1. Since every scalar s is of the form r ei t for some real t where r ≝ |s| ≥ 0 and ei t is real if and only if s is real, the results stated in (6) follow immediately from the aforementioned conclusion, from the strict positive homogeneity of pK, and from the positive homogeneity of pK when pK is real-valued.

Characterization of seminorms with Minkowski functionals

Note that in this next theorem, which follows immediately from the above corollary, we do not assume that K is absorbing in X but instead deduce that (0, 1) K is balanced, and that we do not assume that K is balanced (which is a property that K is often required to have) but we instead assume the weaker condition that (0, 1) s K ⊆ (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 pK is a seminorm on X if and only if the following conditions hold:

  1. (0, ∞) K = X (or equivalently, pK is real-valued);
  2. (0, 1) K is convex;
    • It suffices (but is not necessary) for K to be convex.
  3. (0, 1) u K ⊆ (0, 1) K for all unit scalars u.
    • This condition is satisfied if K is balanced or more generally if u K ⊆ K for all unit scalars u.

in which case 0 ∈ K and both (0, 1) K = { x ∈ X : pK (x) < 1 } and ∩ε > 0 (0, 1 + ε) K = { x ∈ X : pK (x) ≤ 1 } are convex, balanced, and absorbing in X.

Conversely, if f is a seminorm on X then the set V ≝ { x ∈ X : f (x) < 1 } satisfies all three of the above conditions (and thus also conclusions) and also f = pV; 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 pK is a seminorm on X.

Positive sublinear functions and Minkowski functionals

One may show that a real-valued subadditive function f : X → ℝ on a TVS X is continuous at the origin if and only if it is uniformly continuous, where if in addition f is non-negative, then f is continuous if and only if V ≝ { x ∈ X : f (x) < 1 } is an open neighborhood in X.[8] If f : X → ℝ is subadditive and satisfies f (0) = 0, then f is continuous if and only if its absolute value |f| : X → [0, ∞) is continuous.

A positive sublinear function is a real homogeneous function f : X → [0, ∞) that satisfies the triangle inequality. It follows immediately from the results below that if V ≝ { x ∈ X : f (x) < 1 } then f = pV.

Correspondence between open convex sets and positive continuous sublinear functions

Theorem[8] — Suppose that X is a TVS (not necessarily locally convex or Hausdorff) over the real or complex numbers. Then the open convex subsets of X are exactly those that are of the form z + { x ∈ X : p(x) < 1 } = { x ∈ X : p(x − z) < 1} for some z ∈ X and some positive continuous sublinear function p on X.

Proof

Let V be an open convex subset of X. If 0 ∈ V then let z ≝ 0 and otherwise let z ∈ V be arbitrary. Let p : X → [0, ∞) be the Minkowski functional of V − z where p is a continuous sublinear function on X since V − z is convex, absorbing, and open (p however is not necessarily a seminorm since it is not necessarily absolutely homogeneous). From the properties of Minkowski functionals, we have V − z = { x ∈ X : p(x) < 1  } so that V = z + { x ∈ X : p(x) < 1  }. But

z + { x ∈ X : p(x) < 1 } = { z + x : x ∈ X, p(x) < 1 } = { z + x : x ∈ X, p((z + x) − z) < 1 } = { x ∈ X : p(x − z) < 1  },

as desired. ∎

Characterizing Minkowski functionals

The next theorem shows that Minkowski functionals are exactly those functions f : X → [0, ∞] that have a certain purely algebraic property that is used widely. This theorem can be easily extended to characterize certain classes of [−∞, ∞]-valued maps (e.g. real-valued sublinear functions) in terms of Minkowski functionals. In particular, we describe later how every real homogeneous function f : X → ℝ (such as linear functionals) can be written in terms of a unique Minkowski functional having a certain property.

Theorem — Let f : X → [0, ∞] be any function. The following are equivalent:

  1. Strict positive homogeneity: f (tx) = t f (x) for all x ∈ X and all positive real t > 0.
  2. f is a Minkowski functional (i.e. there exists some subset K of X such that f = pK).
  3. f = pK where K ≝ { x ∈ X : f(x) ≤ 1 }.
  4. f = pL where L ≝ { x ∈ X : f(x) < 1 }.

Moreover, if f never takes on the value ∞ (so that the product 0 ⋅ f(x) is always well-defined) then we may add to this list:

  1. Positive homogeneity/Nonnegative homogeneity: f (tx) = t f (x) for all x ∈ X and all non-negative real t ≥ 0.
Proof

We prove only that (1) implies (3) since afterwards the rest of the theorem follows immediately from the basic properties of Minkowski functionals described later; properties that we will assume the reader is familiar with and use without mention. So assume that f : X → [0, ∞] is a function such that f (tx) = t f (x) for all x ∈ X and all real t > 0 and let K ≝ { y ∈ X : f (y) ≤ 1 }.

Note that for all real t > 0 we have f (0) = f (t0) = t f (0) so by taking t = 2 for instance, we can conclude that either f (0) = 0 or f (0) = ∞. Let x ∈ X. We must show that f (x) = pK(x).

We will now show that if f (x) = 0 or f (x) = ∞ then f (x) = pK(x), so that in particular, we will conclude that f (0) = pK(0). So suppose that f (x) = 0 or f (x) = ∞ and note that in either case we have f (tx) = t f (x) = f (x) for all real t > 0. Now if f (x) = 0 then this implies that that t x ∈ K for all real t > 0 (since f (tx) = 0 ≤ 1), which implies that pK(x) = 0, as desired. Similarly, if f (x) = ∞ then t x ∉ K for all real t > 0, which implies that pK(x) = ∞, as desired. Thus, we will henceforth assume that R ≝ f (x) a positive real number and that x ≠ 0 (note, however, that we have not yet ruled out the possibility that pK(x) is 0 or ∞).

Recall that just like f, the function pK satisfies pK(tx) = t pK(x) for all real t > 0. Since 0 < 1/R < ∞, pK(x) = R = f (x) if and only if pK(1/R x) = 1 = f (1/R x) so we may assume without loss of generality that R = 1 and it remains to show that pK(x) = 1. Since f (x) = 1, we have x ∈ K ⊆ (0, 1] K, which implies that pK(x) ≤ 1 (so in particular, we now know that pK(x) ≠ ∞). It remains to show that pK(x) ≥ 1, which recall happens if and only if x ∉ (0, 1) K. So assume for the sake of contradiction that x ∈ (0, 1) K and let 0 < r < 1 and k ∈ K be such that x = r k, where note that k ∈ K implies that f (k) ≤ 1. Then 1 = f (x) = f (rk) = r f (k) ≤ r < 1. ∎

Minkowski functionals and negative values

The definitions of strict positive homogeneity that was given for [0, ∞]-valued functions on X immediately extends, without change, to functions that are valued in other codomains.

Let f be a function on X valued in [−∞, ∞] (or even in ℂ). We say that f is strictly positively homogeneous if f (tx) = t f (x) for all x ∈ X and all positive real t > 0. If f never takes the value ±∞ then we say that f is non-negative homogeneous if f (tx) = t f (x) for all x ∈ X and all non-negative real t ≥ 0.

Note that if g, h : X → [0, ∞] are functions such that g − h is well-defined (i.e. there is no x ∈ X such that g (x) and h (x) are infinite and equal), then if both g and h are strictly positively homogeneous (which happens if and only if they are both Minkowski functionals) then so is the map f ≝ g − h : X → [−∞, ∞].

It's natural to ask if every function f : X → [−∞, ∞] that is strictly positively homogeneous is equal to the difference of two Minkowski functionals. The answer can easily be shown to be yes by using with the following definitions.

Positive and negative parts
Definition and notation: For any function f : X → [−∞, ∞], let f+ : X → [0, ∞] and f− : X → [0, ∞] be defined by
f+(x) ≝ max { 0, f (x) }
and
f−(x) ≝ max { 0, −f (x) } = − min { 0, f (x) }.
As usual, the function f+ is called the positive part of f and f− is called the negative part of f.

For every x ∈ X, at least one of f+ and f− is equal to 0 (or said differently, the function min{ f+, f− } is identically 0) and f = f+ − f−. Moreover, f+ and f− are the only [0, ∞]-valued functions with these properties. That is, if g, h : X → [0, ∞] satisfy 0 = min { g, h } and f = g − h then f+ = g and f− = h.

Characterization

Theorem — Let f : X → [−∞, ∞] be a function. Then the following are equivalent:

  1. f is strictly positively homogeneous.
  2. f+ and f− are strictly positively homogeneous (or equivalently, they are Minkowski functionals).
  3. f is equal to the (well-defined) difference of two Minkowski functionals.

Moreover, if f is real-valued then we may replace the terms "strictly positively homogeneous" with "non-negatively homogeneous" in the above characterization.

Linear functionals and Minkowski functionals

Call a map f : X → ℝ real homogeneous if f (tx) = t f (x) for all x ∈ X and all real t.

Clearly, f is real homogeneous if and only if it is non-negative homogeneous and f (− x) = − f (x) for all x ∈ X.

Call a map f : X → [−∞, ∞] additive if for all x, y ∈ X, f (x) + f (y) is well-defined and f (x + y) = f (x) + f (y).

If a map f : X → [∞, ∞] is additive then either f is constant and infinite or else it is real-valued, so it is only useful to consider real-valued additive maps.

A real linear functional on X is a map f : X → ℝ that is additive and real homogeneous.

If f : X → ℝ is additive then f (0) = 0 and for every x ∈ X, f (tx) = t f (x) for all rational real t. Thus an additive function is almost real homogeneous and it is a linear functional if and only if it is also strictly positively homogeneous. Thus, we could have equivalently defined a real linear functional as being an additive map that is equal to the difference of two real-valued Minkowski functionals.

Commuting with negation
Notation: Let η : X → X denote negation on X (i.e. η(x) ≝ − x for every x ∈ X).

We say that a function f on X commutes with negation if − f = f ∘ η. That is, if − f (x) = f (− x) for all x ∈ X.

Every ℝ-valued additive function commutes with negation, as does every real homogeneous function. The following observations shows that a function f : X → [−∞, ∞] that commutes with negation is completely determined by negation and the single [0, ∞]-valued map f+ (the same is also true of f−).

Observations — Let f : X → [−∞, ∞] be a function. Clearly, f (0) = − f (0) if and only if f (0) = 0. If x ∈ X, then − f (x) = f (− x) if and only if f+ (x) = f− (− x) and f− (x) = f+ (− x).

The following statements are equivalent:

  1. f commutes with negation (i.e. − f = f ∘ η).
  2. f− = f+ ∘ η.
  3. f+ = f− ∘ η.
  4. f = f+  −  (f+ ∘ η).
  5. f = (f− ∘ η)  −  f−.

Minkowski functionals and real homogeneous functions

The following theorem shows that there is a one-to-one correspondence, determined by the negation operation η, between (ℝ-valued) real homogeneous functions and ℝ-valued Minkowski functionals g that satisfy min{ g, g ∘ η } = 0 (that is, min{ g(x), g(− x) } = 0 for all x ∈ X).

Theorem — If f : X → ℝ is real homogeneous (such as a real linear functional) then f+ : X → [0, ∞) is a Minkowski functional satisfying 0 = min { f+,  f+ ∘ η } and f = f+ − f+ ∘ η (which implies f− = f+ ∘ η).

Conversely, every ℝ-valued Minkowski functional g : X → [0, ∞) satisfying 0 = min{ g, g ∘ η } determines a real homogeneous function f : X → ℝ defined by f ≝ g − g ∘ η, where observe that f+ = g and f− = g ∘ η.

Basic properties
Summary of properties of Minkowski functionals

We summarize some of the more important results found in the following theorem where throughout K and L are any subsets of X and x ∈ X is arbitrary.

  • pK(tx) = t pK(x) for all real t > 0 always, without any requirements at all.
    • If pK is real-valued, which happens if and only if (0, ∞) K = X, then the above equality will also hold with t = 0.
  • If 0 ≤ r ≤ ∞ then pK(x) < r if and only if x ∈ (0, r) K.
    • The proofs of the many properties listed below become straightforward exercises once this characterization is proven and then the remaining properties follow easily once it is shown that every Minkowski functional is strictly positive homogeneous.
    • Assuming that 0 < R < ∞, if x ∈ (0, R] K then pK(x) ≤ R, and the converse will hold if K contains { y ∈ X : pK(y) = 1 }.
  • pL = pK if and only if (0, 1)L = (0, 1)K.
    • It follows that pL = pK whenever L satisfies (0, 1) K ⊆ L ⊆ ∩ε > 0 (0, 1 + ε) K; in particular, L could be the first or last of these three sets or also (0, 1] K = K ∪ (0, 1) K.
  • pK is subadditive (i.e. satisfies the triangle inequality) if and only if (0, 1) K is convex. If K is convex then so are (0, 1) K and (0, 1] K = K ∪ (0, 1) K, and pK is subadditive.
  • If we let D ≝ { y ∈ X : pK(y) = 1 or pK(y) = 0 } then pD = pK and D has the particularly nice property that for any positive real R > 0, x ∈ R D if and only if pD(x) = R or pD(x) = 0.
    • Moreover, for any positive real R > 0, pD(x) ≤ R if and only if x ∈ (0, R] D.
    • If pK(0) = 0 and if there is some point at which pK takes on a positive real value, then D is necessarily not convex, balanced, or absorbing. However, the set (0, 1]D (or (0, 1)D = (0, 1)K, or others) may have some or all of these properties and since pK = pD = p(0, 1]D one may use any of these sets to try to deduce properties of this map. That is, one is always free to use whatever set L is best suited for the problem at hand (so as long L satisfies pL = pK, of course) and then change to any other such set.
  • The value(s) of pK on a set of the form Y ≝ (0, ∞) x (where x ∈ X) are completely determined by the value of pK at any one point of Y. In general, the values of pK on Y and its values on the complement X ∖ Y are completely independent.
    • This is the reason why characterizations such as: x ∉ (0, R) K if and only if K ∩ (1/R, ∞)x = ∅ are useful. The latter condition requires no change to K and involves only the open ray (0, ∞) x, which is all that the value of pK(x) depends on, while the former condition involves scaling the entire set K.
Properties of Minkowski functionals

Theorem — Let K be any subset of a real or complex vector space X and let R ≥ 0 be a real number. Then for all x ∈ X:

  1. Value at 0: pK(0) ≠ ∞ if and only if 0 ∈ K if and only if pK(0) = 0.
  2. Infinite/Real values: pK(x) = ∞ if and only if x ∉ (0, ∞) K if and only if K ∩ (0, ∞)x = ∅. Equivalently, pK(x) ≠ ∞ if and only if x ∈ (0, ∞) K. It follows that:
    • (0, ∞) K is the set of all points on which pK is real valued. In particular,
      • pK is real-valued if and only if (0, ∞) K = X, in which case 0 ∈ K.
      • pK is identically equal to ∞ if and only if K = ∅.
  3. Null space: The following are equivalent:
    1. pK(x) = 0.
    2. There exists a divergent sequence of positive real numbers (tn)∞
      n=1       → ∞       such that tn x ∈ K for all n.
      • In particular, if t x ∈ K for all real t > 0 then pK(x) = 0.
    3. (0, ∞)x ⊆ (0, 1) K.
    4. x ∈ ∩ε > 0 (0, ε)K (that is, x ∈ (0, ε)K for all ε > 0).

    So ker pK ≝ { y ∈ X : pK(y) = 0 } = ∩ε > 0 (0, ε) K and pK is identically equal to 0 if and only if (0, 1) K = X.

  4. pK(x) < R if and only if x ∈ (0, R) K if and only if K ∩ (1/R, ∞)x ≠ ∅. Equivalently, pK(x) ≥ R if and only if x ∉ (0, R) K if and only if K ∩ (1/R, ∞)x = ∅. It follows that:
    • { y ∈ X : pK(y) < R } = (0, R) K.
    • This set equality and the two equivalences remain true if R is replaced by ∞.[note 2]
  5. pK(x) ≤ R if and only if x ∈ (0, R + ε) K for every ε > 0. It follows that:
    • { y ∈ X : pK(y) ≤ R } = ∩ε > 0 (0, R + ε) K
      • Note that ∩ε > 0 (0, R + ε) K means ∩ε > 0 ((0, R + ε) K) and not (∩ε > 0 (0, R + ε)) K = (0, R] K.
    • (0, R] K ⊆ ∩ε > 0 (0, R + ε) K and if R > 0 then these sets are equal if and only if K contains { y ∈ X : pK(y) = 1 }.
      • Thus, if x ∈ R K or x ∈ (0, R] K then pK(x) ≤ R (but the converse is not necessarily true).
  6. pK(x) = R if and only if x ∉ (0, R) K and x ∈ (0, R + ε)K for every ε > 0. It follows that:
    • If x ∈ R K and x ∉ (0, R) K then pK(x) = R (but the converse is not necessarily true).
    • pK(x) = 1 if and only if x ≠ 0, K ∩ (1, ∞)x = ∅, and K ∩ (ε, 1]x ≠ ∅ for all ε < 1.
      • So if R > 0 then pK(x) = R if and only if x ≠ 0, K ∩ (1/R, ∞)x = ∅, and K ∩ (ε, 1/R]x ≠ ∅ for all ε < 1/R.[note 3]
  7. Strict positive homogeneity: pK(tx) = t pK(x) for all real t > 0. It follows that:
    • 0 ⋅ pK(x) is well-defined if and only if pK(x) ≠ ∞ (or equivalently, x ∈ (0, ∞) K), in which case pK(tx) = t pK(x) for t = 0 if and only if 0 ∈ K.
    • Positive/Nonnegative homogeneity: pK is nonnegative homogeneous[note 4] if and only if pK is real-valued.
    • If 0 ∈ K then (0, ∞) K is the unique maximal set of points in X on which pK is nonnegative homogeneous. If 0 ∉ K then this set is empty.
  8. Gauge equality: Let L be a subset of X. The following are equivalent:
    1. pL = pK.
    2. (0, 1) L = (0, 1) K.
    3. (0, r) L = (0, r) K for some (or equivalently, for all) real r > 0.
    4. ∩ε > 0 (0, 1 + ε) L = ∩ε > 0 (0, 1 + ε) K.
    5. ∩ε > 0 (0, r + ε) L = ∩ε > 0 (0, r + ε) K for some (or equivalently, for all) real r > 0.

    It follows that:

    • If (0, 1) K ⊆ L ⊆ ∩ε > 0 (0, 1 + ε) K then pL = pK. However, the converse is in general false.
      • In particular, if (0, 1) K ⊆ L ⊆ (0, 1] K then pL = pK.
    • If L is not a subset of ∩ε > 0 (0, 1 + ε) K then pL ≠ pK.
  9. Gauge comparison: Let L be a subset of X. Then pK ≤ pL if and only if (0, 1) L ⊆ (0, 1) K.
  10. Kernel ∪ Unit Sphere can be used in place of K: Let D ≝ { y ∈ X : pK(y) = 1 or pK(y) = 0 } where this set can be written without reference to pK as: D = ∩r > 0 (0, r)K  ∪  [( ∩ε > 0 (0, 1 + ε) K) ∖ (0, 1) K ]. Then:
    • pD = pK.
      • If pK is valued 0 at any point x of X (a very mild condition), then this equality would not hold if D had instead been defined to be only the unit sphere S ≝ { y ∈ X : pK(y) = 1 } (without including the kernel) because we would then have pS(x) = ∞ rather than pK(x) = 0.
    • It is in general false that x ∈ R D if and only if pD(x) = R (e.g. consider when pK is a norm or a seminorm). The following is the correct statement: If R > 0 then x ∈ R D if and only if pD(x) = R or pD(x) = 0.
    • K ⊆ (0, 1] K ⊆ (0, 1] D and if R > 0 then (0, R] D = { y ∈ X : pD(y) ≤ R }.
    • (0, 1) K ⊆ D if and only if the range of pK is contained in { 0, ∞ }.
  11. Restriction: Let S be a real or complex vector subspace of X and let pK ∩ S : S → [0, ∞] denote the Minkowski functional of K ∩ S on S. Then pK|S = pK ∩ S, where pK|S denotes the restriction of pK to S.
  12. Subadditive/Triangle inequality: The following are equivalent:
    1. pK satisfies the triangle inequality on X.[note 5]
    2. (0, 1) K is convex.
    3. ∩ε > 0 (0, 1 + ε) K is convex.
    • If K is convex then pK is subadditive and (0, 1] K = K ∪ (0, 1) K is also convex.
    • Assumptions that can be made when checking for subadditivity: If pK is real-valued, then pK is subadditive if and only if for all non-zero y, z ∈ X such that ℝ y ≠ ℝ z and pK(x + y) = 1, we have 1 ≤ pK(x) + pK(y).
    • If x, y ∈ X then pK(x + y) ≤ pK(x) + pK(y) holds whenever any of the following conditions are satisfied: y ∈ (∞, −1)x ∪ (−1, ∞)x, or x = 0, or y = 0, or pK(x) = ∞, or pK(y) = ∞, or pK(x + y) = 0. If pK(0) = 0 then it also holds when y = − x.
      • In particular, pK is always subadditive on any set of the form [0, ∞) x (i.e. on any "ray").
    • If pK(0) = 0 (or if the real dimension of X is at least 2) then pK is subadditive on X if and only if its restriction to every 2-dimensional real vector subspace of the form S = ℝy + ℝz is subadditive (where y, z ∈ X).
  13. Invariance under scalar multiplication: If s ≠ 0 is a scalar then ps K(y) = pK(1/s y) for all y ∈ X. If r is real and 0 < r < ∞ then pr K(x) = pK(1/r x) = 1/r pK(x). It follows that:
    • If s ≠ 0 is a scalar then the following are equivalent:
    1. pK(s y) = pK(y) for all y ∈ X.
    2. pK(1/s y) = pK(y) for all y ∈ X.
    3. pK(sn y) = pK(y) for all y ∈ X and all integers n.
    4. ps K = pK.
    5. p1/s K = pK.
    6. psn K = pK for all integers n.
    7. (0, 1) s K = (0, 1) K.
    • If s ≠ 1 is a positive real number then (0, 1) s K = (0, 1) K if and only if the range of pK is contained in { 0, ∞ }. Consequently, in general, this characterization will typically only be useful when s has unit length.
    • pK is symmetric (i.e. pK(− y) = pK(y) for all y ∈ X) if and only if pK = p − K, which happens if and only if (0, 1) K = − (0, 1) K.
      • Note that for any set S, − S ⊆ S if and only if − S = S, in which case (0, 1) S = − (0, 1) S.
    • pK(u y) = pK(y) for all y ∈ X and all unit length scalars u if and only if and only if pK = pu K for all unit length scalars u, which happens if and only if (0, 1) u K = (0, 1) K for all unit length scalars u.
      • Note that for any set S, u S ⊆ S for all unit length scalars u if and only if u S = S for all such u, in which case (0, 1) u S = (0, 1) S for all such u.
    • If s is a scalar such that sK = K, then pK(s x) = pK(x) if and only if s ≠ 0, K ≠ { 0 }, or x = 0 (see footnote for proof).[note 6]
Proof

Most numbered statements in the theorem above have an unnumbered sub-list. We prove only the numbered statements and some of the less obvious statements found in the unnumbered sub-lists. All other statements are either direct consequences of the numbered statement under which it is found, or else will become straightforward exercises once all the numbered statements are proven.

Proof of (1) and (2): Straightforward exercises.

Proof of (3): The equivalence of (a) and (d) will follow immediately from (4) (or from (5)), and the proofs of the remaining equivalences in (3) are straightforward exercises. ∎

Proof of (4): Let LHS (i.e. Left Hand Side) be the statement "pK(x) < R" and let RHS be the statement "x ∈ (0, R) K". We must show that LHS is true if and only if RHS is true. If K = ∅ then LHS and RHS are both false so henceforth assume that K ≠ ∅. If R = 0 then LHS is clearly false and RHS is also false since (0, R) K = ∅ so henceforth assume that R > 0. If pK(x) = ∞ then LHS is false (since R < ∞) and (2) gives x ∉ (0, ∞) K so that in particular, RHS is also false. So henceforth assume that pK(x) ≠ ∞. Observe that LHS is true if and only if pK(x) ≝ inf { r ∈ ℝ : r > 0 and x ∈ rK } < R, which happens if and only if there exists some real r such that 0 < r < R and x ∈ rK, where this can be restated more succinctly as x ∈ (0, R) K, which is exactly the statement RHS. Thus LHS is true if and only if RHS is true, which completes the proof of (4). ∎

Proof of (5): Note that pK(x) ≤ R if and only if for every pK(x) < R + ε for every ε > 0, which by (4) happens if and only if x ∈ (0, R + ε)K for every ε > 0, as desired. The proof of the claim that if R > 0 then (0, R] K = ∩ε > 0 (0, R + ε) K if and only if this is true for R = 1 will be a straightforward exercise once (7) is proven. Then the claim that this equality holds for if and only if K contains { y ∈ X : pK(y) = 1 } will be easily seen with the proof of (10). ∎

Proof of (6): The main characterization in (6) is just a combination of (4) and (5). The characterization of pK(x) = 1 is now readily verified, and its generalization to pK(x) = R for positive R will follow immediately once (7) is proven. ∎

Proof of (7): Fix 0 < t < ∞. If x = 0 then (1) implies that pK(tx) = t pK(x) so assume x ≠ 0. Note that pK(x) = ∞ if and only if x ∉ (0, ∞) K if and only if tx ∉ (t0, t∞) K = (0, ∞) K if and only if pK(tx) = ∞, in which case pK(tx) = ∞ = t pK(x) so we're done. So assume S ≝ pK(x) < ∞, which (as was just shown) implies that pK(tx) < ∞ and K ≠ ∅. Since pK(x) = S, by (6) we have x ∉ (0, S) K and x ∈ (0, S + ε)K for every ε > 0 so that multiplying by t > 0 gives tx ∉ (0, tS) K and tx ∈ (0, tS + tε)K for every ε > 0. Since ε > 0 was arbitrary, it follows that tx ∉ (0, tS) K and tx ∈ (0, tS + ε)K for every ε > 0, which is equivalent to pK(tx) = tS, as desired. ∎

Proof of (8): The characterizations of when pL = pK are now deduced easily from the following observation: if f and g are any [0, ∞]-valued functions on X, then f = g if and only if { y ∈ X : f(y) < r } = { y ∈ X : g(y) < r } for all real 0 < r < ∞ (importantly, note that it is not necessary to check that these set equalities hold for r = 0 or r = ∞). Note that this characterization of f = g remains true if all instances of "< r" are replaced by "≤ r". Using (4) or (7), it is easy to see that pL = pK holds if and only if { y ∈ X : pL(y) < r } = { y ∈ X : pK(y) < r } for some real r > 0 (in which case this will be true for all positive reals). Similarly, (7) shows that this characterization of pL = pK remains true if "< r" is replaced by "≤ r". For the statement involving the set D, it is readily verified that (0, 1) D = (0, 1) K so that pL = pK, and the rest now follows with the help of (5). ∎

Proof of (9): Immediate.

Proof of (10): Recall that (0, 1)K = { y ∈ X : pK(y) < 1 } so by using (7), we have

(0, 1)D = { ty ∈ X : 0 < t < 1, pK(y) = 1 or pK(y) = 0 } = { ty ∈ X : 0 < t < 1, pK(ty) = t or pK(ty) = 0 } = (0, 1)K,

so (8) implies that pD = pK. The rest of (10) follows easily. ∎

Proof of (11): Immediate.

Proof of (12): It is straight forward to show that if pK is subadditive then (0, 1) K = { y ∈ X : pK(y) < 1} and { y ∈ X : pK(y) ≤ 1 } = ∩ε > 0 (0, R + ε) K are convex.

Assume that K is convex, which recall is true if and only if sK + tK = (s + t)K for all non-negative real s and t. We will show that pK is subadditive. Let x, y ∈ X and note that if R ≝ pK(x) is infinite or S ≝ pK(y) is infinite then we're done, so assume that both are finite. We want to show that pK(x + y) ≤ R + S, which is true if and only if x + y ∈ (0, R + S + ε)K for every ε > 0. Fix ε > 0. Since pK(x) ≤ R, we have x ∈ (0, R + ε/2)K and similarly we have y ∈ (0, S + ε/2)K so that x + y ∈ (0, R + ε/2)K + (0, S + ε/2)K = (0, R + S + ε)K, where the last equality is due to K being convex. ∎

If K is convex then pK is subadditive, which implies that (0, 1) K is convex, where this fact can now be used to easily prove that (0, 1] K = K ∪ (0, 1) K is also convex.

If y, z ∈ X are such that pK(y + z) > pK(xy) + pK(z), then pK is not subadditive on ℝy + ℝz, so it suffices to check subadditivity on sets of the form ℝy + ℝz.

If y ∈ (0, ∞)x then pK(x + y) ≤ pK(x) + pK(y) is immediate from pK being strictly positively homogeneous, while if y = − tx = t(− x) with 1 ≠ t > 0 then this inequality follows from the appropriate choice of one of the equalities: x + y = (1 − t)x = (t − 1)(− x)). That pK is subadditive on any set of the form [0, ∞) x is now immediate. If Q ≝ pK(x + y) > 0 is real (we assume that Q > 0 since if Q = 0 then we're done) then pK(x + y) = Q ≤ pK(x) + pK(y) if and only if pK(1/Qx + 1/Qy) = 1 ≤ pK(1/Qx) + pK(1/Qy) so in this case, it suffices to prove the triangle inequality under the assumption that pK(x + y) = 1. ∎

Proof of (13): The proofs are straightforward exercises that follow easily from the above characterizations. ∎

Examples
  1. If ℒ is a non-empty collection of subsets of X then p∪ℒ (x) = inf { pL(x) : L ∈ ℒ } for all x ∈ X, where ∪ℒ ≝ ∪L ∈ ℒ L.
    • Thus pK∪L(x) = min { pK(x), pL(x) } for all x ∈ X.
  2. If ℒ is a non-empty collection of subsets of X and I ⊆ X satisfies
    { x ∈ X : pL(x) < 1 for all L ∈ ℒ } ⊆ I ⊆ { x ∈ X : pL(x) ≤ 1 for all L ∈ ℒ }

    then pI (x) = sup { pL(x) : L ∈ ℒ } for all x ∈ X.

We now give examples where the containment (0, R] K ⊆ ∩ε > 0 (0, R + ε) K is proper.

Example: Note that if R = 0 and K = X then (0, R] K = (0, 0] X = ∅ but ∩ε > 0 (0, ε) K = ∩ε > 0 X = X, which shows that its possible for (0, R] K to be a proper subset of ∩ε > 0 (0, R + ε) K when R = 0. ∎

We now show that the containment can be proper when R = 1 with an example that may be generalized to any real R > 0. Assuming that [0, 1] K ⊆ K, the following example is representative of how it happens that x ∈ X satisfies pK(x) = 1 but x ∉ (0 ,1] K.

Example: Let x ∈ X be non-zero and let K ≝ [0, 1) x where note that [0, 1] K = K and x ∉ K. Since x ∉ (0, 1) K = K we have pK(x) ≥ 1. That pK(x) ≤ 1 follows from observing that for every ε > 0 we have (0, 1 + ε) K = (0, 1 + ε)([0, 1) x) = [0, 1 + ε) x, which contains x. Thus pK(x) = 1 and x ∈ ∩ε > 0 (0, 1 + ε) K. However, (0, 1] K = (0, 1]([0, 1) x) = [0, 1) x = K so that x ∉ (0, 1] K, as desired. ∎

See also

Notes
  1. u is having unit length means that |u| = 1.
  2. This is the content of statement (2). In the equality {{math|1=K ∩ (1/R, ∞)x = ∅}, we take 1/∞ ≝ 0 and 1/0 ≝ ∞ so that (1/0, ∞) ≝ { t ∈ ℝ : ∞ < t < ∞ } = ∅.
  3. One may restrict ε to range over any non-degenerate interval of the form (a, 1/R) or [a, 1/R).
  4. Definition: pK being nonnegative homogeneous on a set S means that r pK(y) is well-defined and pK(ry) = r pK(y) for all y ∈ S and all non-negative real r ≥ 0. If S is not mentioned then it is understood that S ≝ X.
  5. Definition: Recall that pK is subadditive and satisfies the triangle inequality on a subset S of X if pK(y + z) ≤ pK(y) + pK(z) for all y, z ∈ S. If mention of S is omitted then it is assumed that S ≝ X.
  6. Proof: Let LHS be the statement "pK(sx) = pK(x)" and let RHS be the statement "s ≠ 0, K ≠ { 0 }, or x = 0". We want to show that LHS is true if and only if RHS is true. (⇐) Assume that RHS is true. Clearly, LHS is true if K = ∅ or if x = 0 so assume henceforth that K ≠ ∅ and x ≠ 0. Observe that if s = 0 then sK = K implies that K ⊆ { 0 }, where K ≠ ∅ now implies that K = { 0 }, but this contradicts the statement RHS, which we assumed was true. Thus we must have s ≠ 0. Note that pK(x) = ∞ ⇔ x ∉ (0, ∞) K ⇔ sx ∉ (0, ∞) sK = (0, ∞) K (since s ≠ 0) ⇔ pK(sx) = ∞ so we may now assume that both pK(x) and pK(sx) are finite. Let R ≝ pK(x). Then x ∉ (0, R) K and x ∈ (0, R + ε)K for every ε > 0 so that (since sx ≠ 0) sx ∉ (0, R) sK and sx ∈ (0, R + ε)sK for every ε > 0, where now using sK = K allows us to conclude that pK(sx) = R, as desired. (⇒) Assume that RHS is false so that s = 0, K = { 0 }, and x ≠ 0. Because s = 0 and 0 ∈ K, we have 0 = pK(sx). Since K = { 0 } and x ≠ 0, we have pK(x) = ∞. Thus 0 = pK(sx) ≠ pK(x) = ∞, which shows that LHS is false, as desired. ∎

References
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  6. Schaefer 1999, p. 40.
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