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 $X$. $\quad \square$


Part III: Your question

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):

[1]


References

  1. The main reference is V. Bogachev, O. Smolyanov, Topological Vector Spaces and Their Applications, Springer (2017). You will only need 1.1 - 1.3 there.
  2. Well, if you accidentally know Russian, all that I've written here is a brief summary of these wonderful lecture notes by A. Pirkovskii.
  3. Finally, all of this can be found on wiki.
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