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.  
 A: 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.


*

*$p_S(\lambda x) = \lambda p_S(x)$ for all $x \in X, \lambda \geqslant 0$;

*If $S$ is circled, then $p_S(\lambda x) = |\lambda| p_S(x)$ for all $x \in X, \lambda \in \mathbb{K}$;

*If $S$ is convex, then $p_S(x+y) \leqslant p_S(x) + p_S(y)$ for all $x,y \in X$;

*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 allows 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.


*

*For any $x \in X$ the family $\{U_{p; \varepsilon}(x) \, \colon \; p \in P, \, \varepsilon > 0 \}$ is a subbase of neighborhoods of $x$;

*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 $x$;

*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 $\tau(P)$;

*$(X, \tau(P))$ 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:


*

*[algebraic-analytic definition] The topology of $X$ is generated by some family of seminorms.

*[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):
[

References


*

*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.

*Well, if you accidentally know Russian, all that I've written here is a brief summary of these wonderful lecture notes by A. Pirkovskii.

*Finally, all of this can be found on wiki.

