Minkowski functional

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

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

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

Thus every seminorm (which is a function defined by purely algebraic properties) can be associated (non-uniquely) with an absorbing disk (which is a set with certain geometric properties) and conversely, every absorbing disk can be associated with its Minkowski functional (which will necessarily be a seminorm). These relationships between seminorms, Minkowski functionals, and absorbing disks is a major reason why Minkowski functionals are studied and used in functional analysis. In particular, through these relationships, Minkowski functionals allow one to "translate" certain geometric properties of a subset of $X$ into certain algebraic properties of a function on $X.$

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

Definition

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

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

Some conditions making a gauge real-valued

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

In order for $p_{K}$ to be real-valued, it suffices for the origin of $X$ to belong to the algebraic interior or core of $K$ in $X.$ 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 subadditive: $p_{K}(x+y)\leq p_{K}(x)+p_{K}(y).$
2. It is absolutely homogeneous: $p_{K}(sx)=|s|p_{K}(x)$ for all scalars $s.$
3. It is nonnegative: $p_{K}\geq 0.$

Therefore, $p_{K}$ is a seminorm on $X,$ with an induced topology. This is characteristic of Minkowski functionals defined via "nice" sets. There is a one-to-one correspondence between seminorms and the Minkowski functional given by such sets. What is meant precisely by "nice" is discussed in the section below.

Notice that, in contrast to a stronger requirement for a norm, $p_{K}(x)=0$ need not imply $x=0.$ In the above example, one can take a nonzero $x$ from the kernel of $f.$ Consequently, the resulting topology need not be Hausdorff.

Common conditions guaranteeing gauges are seminorms

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

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

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

More generally, if $K$ is convex and the origin belongs to the algebraic interior of $K,$ then $p_{K}$ is a nonnegative sublinear functional on $X,$ which implies in particular that it is subadditive and positive homogeneous. If $K$ is absorbing in $X$ then $p_{[0,1]K}$ is positive homogeneous, meaning that $p_{[0,1]K}(sx)=sp_{[0,1]K}(x)$ for all real $s\geq 0,$ where $[0,1]K=\{tk:t\in [0,1],k\in K\}.$ 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.

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

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

Proof that the Gauge of an absorbing disk is a seminorm

Convexity and subadditivity A simple geometric argument that shows convexity of $K$ implies subadditivity is as follows. Suppose for the moment that $p_{K}(x)=p_{K}(y)=r.$ Then for all $e>0,$ $x,y\in K_{e}:=(r,e)K.$ Since $K$ is convex and $r+e\neq 0,$ $K_{e}$ is also convex. Therefore, ${\frac {1}{2}}x+{\frac {1}{2}}y\in K_{e}.$ By definition of the Minkowski functional $p_{K},$ $$p_{K}\left({\frac {1}{2}}x+{\frac {1}{2}}y\right)\leq r+e={\frac {1}{2}}p_{K}(x)+{\frac {1}{2}}p_{K}(y)+e.$$ But the left hand side is ${\frac {1}{2}}p_{K}(x+y),$ so that $$p_{K}(x+y)\leq p_{K}(x)+p_{K}(y)+2e.$$ Since $e>0$ was arbitrary, it follows that $p_{K}(x+y)\leq p_{K}(x)+p_{K}(y),$ which is the desired inequality. The general case $p_{K}(x)>p_{K}(y)$ is obtained after the obvious modification. Convexity of $K,$ together with the initial assumption that the set $\{r>0:x\in rK\}$ is nonempty, implies that $K$ is absorbing. Balancedness and absolute homogeneity Notice that $K$ being balanced implies that $$\lambda x\in rK\quad {\mbox{if and only if}}\quad x\in {\frac {r}{|\lambda |}}K.$$ Therefore $$p_{K}(\lambda x)=\inf \left\{r>0:\lambda x\in rK\right\}=\inf \left\{r>0:x\in {\frac {r}{|\lambda |}}K\right\}=\inf \left\{|\lambda |{\frac {r}{|\lambda |}}>0:x\in {\frac {r}{|\lambda |}}K\right\}=|\lambda |p_{K}(x).$$

Algebraic properties

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

• $p_{K}$ is a seminorm on $X.$
• $p_{K}$ is a norm on $X$ if and only if $K$ does not contain a non-trivial vector subspace.
• $p_{sK}={\frac {1}{|s|}}p_{K}$ 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 $K.$
  • In particular, if $K$ is as above and $q$ is any seminorm on $X,$ then $q=p$ if and only if $\{x\in X:q(x)<1\}\;\subseteq \;K\;\subseteq \;\{x\in X:q(x)\leq 1\}.$
• If $x\in X$ satisfies $p_{K}(x)<1$ then $x\in K.$

Topological properties

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

Moreover, the Minkowski functional $p_{K}$ is continuous if and only if $K$ is a neighborhood of the origin in $X.$ 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 any subset $K$ of $X.$ The more common special case where $K$ is assumed to be an absorbing disk in $X$ was discussed above.

Properties

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

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

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

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

Proof

The proofs of these basic properties are straightforward exercises so only the proofs of the most important statements are given. The proof that a convex subset $A\subseteq X$ that satisfies $(0,\infty )A=X$ is necessarily absorbing in $X$ is straightforward and can be found in the article on absorbing sets. For any real $t>0,$ $$\{r>0:tx\in rK\}=\{t(r/t):x\in (r/t)K\}=t\{s>0:x\in sK\}$$ so that taking the infimum of both sides shows that $$p_{K}(tx)=\inf\{r>0:tx\in rK\}=t\inf\{s>0:x\in sK\}=tp_{K}(x).$$ This proves that Minkowski functionals are strictly positive homogeneous. For $0\cdot p_{K}(x)$ to be well-defined, it is necessary and sufficient that $p_{K}(x)\neq \infty ;$ thus $p_{K}(tx)=tp_{K}(x)$ for all $x\in X$ and all non-negative real $t\geq 0$ if and only if $p_{K}$ is real-valued. The hypothesis of statement (7) allows us to conclude that $p_{K}(sx)=p_{K}(x)$ for all $x\in X$ and all scalars $s$ satisfying $|s|=1.$ Every scalar $s$ is of the form $re^{it}$ for some real $t$ where $r:=|s|\geq 0$ and $e^{it}$ is real if and only if $s$ is real. The results in the statement about absolute homogeneity follow immediately from the aforementioned conclusion, from the strict positive homogeneity of $p_{K},$ and from the positive homogeneity of $p_{K}$ when $p_{K}$ is real-valued. $\blacksquare$

Examples

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

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

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

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

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

Positive homogeneity characterizes Minkowski functionals

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

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

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

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

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

Proof

If $f(tx)\leq tf(x)$ holds for all $x\in X$ and real $t>0$ then $tf(x)=tf\left({\tfrac {1}{t}}(tx)\right)\leq t{\tfrac {1}{t}}f(tx)=f(tx)\leq tf(x)$ so that $tf(x)=f(tx).$ Only (1) implies (3) will be proven because afterwards, the rest of the theorem follows immediately from the basic properties of Minkowski functionals described earlier; properties that will henceforth be used without comment. So assume that $f:X\to [0,\infty ]$ is a function such that $f(tx)=tf(x)$ for all $x\in X$ and all real $t>0$ and let $K:=\{y\in X:f(y)\leq 1\}.$ For all real $t>0,$ $f(0)=f(t0)=tf(0)$ so by taking $t=2$ for instance, it follows that either $f(0)=0$ or $f(0)=\infty .$ Let $x\in X.$ It remains to show that $f(x)=p_{K}(x).$ It will now be shown that if $f(x)=0$ or $f(x)=\infty$ then $f(x)=p_{K}(x),$ so that in particular, it will follow that $f(0)=p_{K}(0).$ So suppose that $f(x)=0$ or $f(x)=\infty ;$ in either case $f(tx)=tf(x)=f(x)$ for all real $t>0.$ Now if $f(x)=0$ then this implies that that $tx\in K$ for all real $t>0$ (since $f(tx)=0\leq 1$), which implies that $p_{K}(x)=0,$ as desired. Similarly, if $f(x)=\infty$ then $tx\not \in K$ for all real $t>0,$ which implies that $p_{K}(x)=\infty ,$ as desired. Thus, it will henceforth be assumed that $R:=f(x)$ a positive real number and that $x\neq 0$ (importantly, however, the possibility that $p_{K}(x)$ is $0$ or $\,\infty \,$ has not yet been ruled out). Recall that just like $f,$ the function $p_{K}$ satisfies $p_{K}(tx)=tp_{K}(x)$ for all real $t>0.$ Since $0<{\tfrac {1}{R}}<\infty ,$ $p_{K}(x)=R=f(x)$ if and only if $p_{K}\left({\tfrac {1}{R}}x\right)=1=f\left({\tfrac {1}{R}}x\right)$ so assume without loss of generality that $R=1$ and it remains to show that $p_{K}\left({\tfrac {1}{R}}x\right)=1.$ Since $f(x)=1,$ $x\in K\subseteq (0,1]K,$ which implies that $p_{K}(x)\leq 1$ (so in particular, $p_{K}(x)\neq \infty$ is guaranteed). It remains to show that $p_{K}(x)\geq 1,$ which recall happens if and only if $x\not \in (0,1)K.$ So assume for the sake of contradiction that $x\in (0,1)K$ and let $0<r<1$ and $k\in K$ be such that $x=rk,$ where note that $k\in K$ implies that $f(k)\leq 1.$ Then $1=f(x)=f(rk)=rf(k)\leq r<1.$ $\blacksquare$

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

Characterizing Minkowski functionals on star sets

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

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

Characterizing Minkowski functionals that are seminorms

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

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

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

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

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

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

Positive sublinear functions and Minkowski functionals

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

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

Correspondence between open convex sets and positive continuous sublinear functions

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

Proof

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

See also

• Asymmetric norm – Generalization of the concept of a norm
• Auxiliary normed space
• Cauchy's functional equation – Functional equation
• Finest locally convex topology – A vector space with a topology defined by convex open setsPages displaying short descriptions of redirect targets
• Finsler manifold – Generalization of Riemannian manifolds
• Hadwiger's theorem – Theorem in integral geometry
• Hugo Hadwiger – Swiss mathematician (1908–1981)
• Locally convex topological vector space – A vector space with a topology defined by convex open sets
• Morphological image processing – Theory and technique for handling geometrical structuresPages displaying short descriptions of redirect targets
• Norm (mathematics) – Length in a vector space
• Seminorm – Mathematical function
• Topological vector space – Vector space with a notion of nearness

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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). https://doi.org/10.3390/en13225976