# Reference for Minkowski functional when 0 is not in the interior

The Minkowski functional on a normed linear space $$E$$ is usually defined for convex (or sometimes even non convex) subsets $$C$$ of $$E$$ such that $$0 \in \operatorname{int}(C)$$. Is there any standard reference for the definition on convex sets such that $$0 \in C$$ but not necessarily $$0 \in \operatorname{int}(C)$$? For Minkowski functionals of convex sets convexity and positive homogeneity still hold on a convex cone after all.

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I will try to give a concise answer here (omitting the proofs) and leave some references in the end. There will be three parts in this answer, the first two rather introductory. Throughout this answer, $$\mathbb{K} = \mathbb{R}$$ or $$\mathbb{C}$$ and all vector spaces are considered over $$\mathbb{K}$$ by default.

## Part I: General context for Minkowski functionals

You do not even need a topology on your vector space to define Minkowski functionals.

Let $$X$$ be a vector space. We call a subset $$S \subseteq X$$

• circled if for any $$\lambda \in \mathbb{K}$$ with $$|\lambda| \leqslant 1$$ one has $$\lambda S \subseteq S$$;
• absolutely convex if it is convex and circled;
• absorbing if for any $$x \in X$$ there is some $$C > 0$$ such that $$x \in \lambda S$$ for any $$\lambda \in \mathbb{K}, \, |\lambda| \geqslant C$$.

Remark. Observe that if $$S$$ is circled or absorbing, it automatically contains $$0$$. However, we still can not talk about interiors because there is no topology on $$X$$.

Definition. Let $$S \subseteq X$$ be any absorbing subset. The Minkowski functional of $$\boldsymbol{S}$$ is the function

$$\boldsymbol{p_S} \colon X \to [0, +\infty), \; x \mapsto \inf\{\lambda \geqslant 0 \,\colon x \in \lambda S\}.$$

The absorbing condition guarantees it is well-defined.

Proposition 1. Let $$S \subseteq X$$ be an absorbing subset.

1. $$p_S(\lambda x) = \lambda p_S(x)$$ for all $$x \in X, \lambda \geqslant 0$$;
2. If $$S$$ is circled, then $$p_S(\lambda x) = |\lambda| p_S(x)$$ for all $$x \in X, \lambda \in \mathbb{K}$$;
3. If $$S$$ is convex, then $$p_S(x+y) \leqslant p_S(x) + p_S(y)$$ for all $$x,y \in X$$;
4. If $$S$$ is absolutely convex, then $$p_S$$ is a seminorm.

The main application of Minkowski functionals is to produce seminorms on a vector space out of its intrinsic geometry. Thus, you need to consider not arbitrary subsets, but absolutely convex absorbing ones.

## Part II: Locally convex vector spaces

This question of yours (as well as many other questions in functional analysis) should be studied in a context far more general than that of normed spaces.

Definition. A topological vector space is a vector space $$X$$ endowed with a topology such that both addition $$X \times X \to X$$ and multiplication by scalars $$\mathbb{K} \times X \to X$$ are continuous maps (where $$\mathbb{K}$$ carries the standard topology).

Here is a little exercise to get accustomed to this definition (we will use it below).

Exercise. Show that any neighborhood of $$0$$ in a topological vector space is automatically absorbing.

Now, general topological vector spaces are not so nice from the topological and analytical point of view. We should impose a little more restrictions on the topology. This leads to the notion of a locally convex vector space. There are two equivalent definitions. In fact, Minkowski functionals play a significant role in the proof of this equivalence. Before giving these definitions and disscussing their equivalence, we need to understand how to build a topology out of a family of seminorms.

Definition (temporary). A polynormed space is a vector space $$X$$ endowed with a family of seminorms $$P$$. We write $$(X, P)$$ or just $$X$$.

A family of seminorms on a vector space allow us to turn it into a topological vector space as follows. Let $$(X, P)$$ be a polynormed space. For each $$x \in X, \, \varepsilon > 0, \,$$ and $$\, p_1, \ldots, p_n \in P$$ ($$n$$ also varies) define

$$U_{p_1, \ldots, p_n; \varepsilon} (x) = \{y \in X \, \colon \; p_i(y-x) < \varepsilon \;\; \forall \, i = 1, \ldots, n\}.$$

Clearly, $$U_{p_1, \ldots, p_n; \varepsilon} (x) = \bigcap_{i=1}^n U_{p_i; \varepsilon} (x)$$. This is a generalization of an $$\varepsilon$$-ball in a normed space.

Define the topology $$\boldsymbol{\tau(P)}$$ on $$X$$ generated by $$P$$ to be the topology with a subbase

$$\{U_{p; \varepsilon}(x) \, \colon \; x \in X, \, p \in P, \, \varepsilon > 0 \}.$$

Proposition 2. Let $$(X, P)$$ be a polynormed space.

1. For any $$x \in X$$ the family $$\{U_{p; \varepsilon}(x) \, \colon \; p \in P, \, \varepsilon > 0 \}$$ is a subbase of neighborhoods of $$x$$;
2. For any $$x \in X$$ the family $$\{U_{p_1, \ldots, p_n; \varepsilon}(x) \, \colon \; n \in \mathbb{N}, \, p_i \in P, \, \varepsilon > 0 \}$$ is a base of neighborhoods of $$x$$;
3. The family $$\{U_{p_1, \ldots, p_n; \varepsilon}(x) \, \colon \; x\in X, \, n \in \mathbb{N}, \, p_i \in P, \, \varepsilon > 0 \}$$ is a base for $$\tau(P)$$;
4. $$(X, \tau(P))$$ is a topological vector space.

Remark. If your family $$P$$ consists of one norm, this is nothing but a normed space with its usual normed topology.

The thing is, the same topology can be generated by different families of seminorms, and we care not as much about $$P$$ as about $$\tau(P)$$.

Now we are ready to give the two definitions announced above.

Theorem 3. Let $$X$$ be a topological vector space. TFAE:

1. [algebraic-analytic definition] The topology of $$X$$ is generated by some family of seminorms.
2. [geometric definition] There is a base of neighborhoods of $$0$$ consisting of convex sets.

Such $$X$$ is called a locally convex space.

Remark. The second (geometric) definition reveals the mystery behind this name.

Proof (sketch): $$(1) \Rightarrow (2):$$ It is easy to check that each set $$U_{p_1, \ldots, p_n; \varepsilon}(0)$$ is convex (even absolutely convex). $$(2) \Rightarrow (1):$$ Here we first need to prove the following technical

Lemma 4: If there is a base of neighborhoods of $$0$$ consisting of convex sets, then there is a base at $$0$$ consisting of absolutely convex sets.

After that, one takes a base $$\mathcal{U}$$ of absolutely convex neighborhoods of $$0$$ and takes the desired family $$P$$ of seminorms to be the family of corresponding Minkowski functionals: $$P = \{\, p_V \, \colon \; V \in \mathcal{U}\}$$. Now it remains to show that $$\tau(P)$$ is the initial topology on $$X$$. $$\quad \square$$

Let $$X$$ be a locally convex space (e.g., a normed space). There are many absolutely convex absorbing sets in $$X$$ (as we already know, they are objects of linear nature and do not depend on topology). Each of them gives you a good Minkowski functional (good $$=$$ a seminorm). But for a general such set, its Minkowski seminorm may have nothing to do with the (already existing and fixed) topology on $$X$$. At the same time, there are some very special absolutely convex absorbing sets in $$X$$, those that are open (hence neighborhoods of $$0$$). And they are exactly those, whose Minkowski seminorms respect the topology on $$X$$. By this we simply mean that they are continuous as maps $$X \to [0, +\infty)$$. This can be reformulated as follows. If $$p$$ is a seminorm on $$X$$, it is continuous of and only if for every family of seminorms $$P$$ generating the topology of $$X$$ the family $$P \cup \{p\}$$ generates the same topology. In other words, we can add it to any generating family and the topology will not get finer. We can summarize it in a nice diagram (let me just draw it by hand):