[proofplan]
Throughout, $C$ is a convex subset of a topological vector space $X$ that contains $0$ as an interior point (the standard hypothesis for the Minkowski functional to be defined and finite-valued), with
\begin{align*}
\mu_C : X &\to [0, \infty), \\
x &\mapsto \inf\{t > 0 : x \in tC\}.
\end{align*}
The argument is in four parts. Positive homogeneity is a direct rescaling of the defining infimum. Subadditivity follows from convexity of $C$ via a barycentric combination. The two-sided inclusion uses the definition of the infimum to put a small slack on the upper side and to absorb $x$ into $tC$ on the lower side. The two equality cases use openness (to upgrade $x \in C$ to $\mu_C(x) < 1$ via continuity of scalar multiplication) and closedness (to upgrade $\mu_C(x) \leq 1$ to $x \in C$ via a limiting argument).
[/proofplan]
[step:Verify positive homogeneity by rescaling the defining infimum]
Let $\lambda > 0$ and $x \in X$. The substitution $s = t/\lambda$ is a bijection from $\{t > 0 : \lambda x \in tC\}$ onto $\{s > 0 : x \in sC\}$, since $\lambda x \in tC \iff x \in (t/\lambda) C$. Hence
\begin{align*}
\mu_C(\lambda x) = \inf\{t > 0 : \lambda x \in tC\} = \inf\{\lambda s : s > 0, x \in sC\} = \lambda \mu_C(x).
\end{align*}
For $\lambda = 0$: $0 \in tC$ for every $t > 0$ (since $0 \in C$), so $\mu_C(0) = 0 = 0 \cdot \mu_C(x)$.
[/step]
[step:Establish subadditivity from convexity of $C$]
Let $x, y \in X$ and fix $\varepsilon > 0$. By definition of the infimum, there exist $s, t > 0$ with
\begin{align*}
x \in sC, \quad s < \mu_C(x) + \varepsilon, \qquad y \in tC, \quad t < \mu_C(y) + \varepsilon.
\end{align*}
Write $x = s u$ and $y = t v$ with $u, v \in C$. We express $x + y$ as a multiple of a convex combination of $u$ and $v$:
\begin{align*}
\frac{x + y}{s + t} = \frac{s}{s + t} u + \frac{t}{s + t} v \in C,
\end{align*}
where the inclusion uses convexity of $C$ with the convex coefficients $s/(s+t), t/(s+t) \in (0, 1)$. Therefore $x + y \in (s + t)C$, so
\begin{align*}
\mu_C(x + y) \leq s + t < \mu_C(x) + \mu_C(y) + 2\varepsilon.
\end{align*}
Letting $\varepsilon \downarrow 0$ gives $\mu_C(x + y) \leq \mu_C(x) + \mu_C(y)$.
[guided]
We want to show $\mu_C(x+y) \leq \mu_C(x) + \mu_C(y)$. The defining infimum makes a direct attack hard — we cannot generally find single optimal $s, t$ realising each infimum. The standard remedy is to allow an $\varepsilon$-slack: by definition of the infimum, for any $\varepsilon > 0$ there exist $s, t > 0$ with
\begin{align*}
x \in sC, \quad s < \mu_C(x) + \varepsilon, \qquad y \in tC, \quad t < \mu_C(y) + \varepsilon.
\end{align*}
Then we will prove $x + y \in (s+t)C$ and let $\varepsilon \downarrow 0$.
Why $(s+t)C$? The natural guess is to combine the two scalings additively. Writing $x = su$ and $y = tv$ with $u, v \in C$ (which is exactly what $x \in sC$ and $y \in tC$ assert), we compute
\begin{align*}
\frac{x + y}{s + t} = \frac{su + tv}{s + t} = \frac{s}{s+t} u + \frac{t}{s+t} v.
\end{align*}
This is a **convex combination** of $u$ and $v$, with weights $\alpha = s/(s+t)$ and $\beta = t/(s+t)$ that satisfy $\alpha, \beta > 0$ and $\alpha + \beta = 1$. Convexity of $C$ — invoked here as the hypothesis on $C$ — gives $\alpha u + \beta v \in C$, so $(x+y)/(s+t) \in C$, equivalently $x + y \in (s+t)C$.
By definition of $\mu_C$ as an infimum, $\mu_C(x+y) \leq s + t$. Bounding $s + t$ by the slacked sum:
\begin{align*}
\mu_C(x+y) \leq s + t < (\mu_C(x) + \varepsilon) + (\mu_C(y) + \varepsilon) = \mu_C(x) + \mu_C(y) + 2\varepsilon.
\end{align*}
Since $\varepsilon > 0$ was arbitrary, $\mu_C(x+y) \leq \mu_C(x) + \mu_C(y)$.
[/guided]
[/step]
[step:Prove the two-sided inclusion $\{\mu_C < 1\} \subset C \subset \{\mu_C \leq 1\}$]
**Upper inclusion** ($C \subset \{\mu_C \leq 1\}$): If $x \in C$, then $x \in 1 \cdot C$, so $1 \in \{t > 0 : x \in tC\}$, giving $\mu_C(x) \leq 1$.
**Lower inclusion** ($\{\mu_C < 1\} \subset C$): Suppose $\mu_C(x) < 1$. By definition of the infimum, there exists $t \in (\mu_C(x), 1)$ with $x \in tC$. Write $x = t u$ for some $u \in C$. Then
\begin{align*}
x = t u + (1 - t) \cdot 0,
\end{align*}
a convex combination of $u \in C$ and $0 \in C$ with coefficients $t \in (0, 1)$ and $1 - t \in (0, 1)$. By convexity of $C$, $x \in C$.
[/step]
[step:Show $C$ open implies $\{\mu_C < 1\} = C$]
Suppose $C$ is open. We already have $\{\mu_C < 1\} \subset C$ from the previous step, so it remains to show $C \subset \{\mu_C < 1\}$.
Let $x \in C$. The map $\lambda \mapsto \lambda x$ is continuous from $\mathbb{R}$ into $X$ (an axiom of topological vector spaces), and at $\lambda = 1$ it sends $1$ to $x \in C$. Since $C$ is open, there exists $\delta > 0$ such that $\lambda x \in C$ for all $\lambda \in (1 - \delta, 1 + \delta)$. Choose $\lambda^* = 1 + \delta/2 > 1$. Then $\lambda^* x \in C$, which rewrites as $x \in (1/\lambda^*) C$. Hence $1/\lambda^*$ is admissible in the defining infimum:
\begin{align*}
\mu_C(x) \leq \frac{1}{\lambda^*} = \frac{1}{1 + \delta/2} < 1.
\end{align*}
So $x \in \{\mu_C < 1\}$.
[guided]
Why does openness give us strict inequality? The point is that if $x \in C$ and $C$ is open, then $C$ contains not just $x$ but a small neighbourhood of $x$. In a TVS, the scalar multiplication map
\begin{align*}
m_x : \mathbb{R} &\to X, \\
\lambda &\mapsto \lambda x
\end{align*}
is continuous (this is part of the TVS axioms). Continuity of $m_x$ at $\lambda = 1$ together with the assumption that $C$ is an open neighbourhood of $m_x(1) = x$ gives $\delta > 0$ such that $m_x(\lambda) = \lambda x \in C$ for all $\lambda$ in the open interval $(1 - \delta, 1 + \delta)$.
Pick any $\lambda^* > 1$ with $\lambda^* < 1 + \delta$, e.g.\ $\lambda^* = 1 + \delta/2$. Then $\lambda^* x \in C$, which rewrites as $x \in (\lambda^*)^{-1} C$. So $t := (\lambda^*)^{-1}$ is admissible in the defining infimum:
\begin{align*}
\mu_C(x) = \inf\{t > 0 : x \in tC\} \leq (\lambda^*)^{-1} < 1.
\end{align*}
The strict inequality $(\lambda^*)^{-1} < 1$ uses $\lambda^* > 1$, which is exactly what openness allowed us to choose.
Hence $C \subset \{x : \mu_C(x) < 1\}$, completing the equality $\{\mu_C < 1\} = C$.
[/guided]
[/step]
[step:Show $C$ closed implies $\{\mu_C \leq 1\} = C$]
Suppose $C$ is closed. We already have $C \subset \{\mu_C \leq 1\}$, so it remains to show $\{\mu_C \leq 1\} \subset C$.
Let $x \in X$ with $\mu_C(x) \leq 1$.
*Case 1: $\mu_C(x) < 1$.* By the lower inclusion proved above, $x \in C$.
*Case 2: $\mu_C(x) = 1$.* For each $n \in \mathbb{N}$, let $t_n = 1 + 1/n > 1 = \mu_C(x)$. Since $t_n > \mu_C(x)$, by definition of the infimum there exists some $s_n \in [\mu_C(x), t_n)$ with $x \in s_n C$. Equivalently, $u_n := x/s_n \in C$. As $n \to \infty$, $s_n \to 1$. By continuity of scalar multiplication in the TVS $X$, the sequence
\begin{align*}
u_n = \frac{1}{s_n} x \to \frac{1}{1} x = x \quad \text{in } X.
\end{align*}
Each $u_n \in C$ and $C$ is closed, so the limit $x \in C$.
*Case 3: $\mu_C(x) = 0$.* By positive homogeneity, $\mu_C(2x) = 2 \mu_C(x) = 0 < 1$, so $2x \in \{\mu_C < 1\} \subset C$ by the lower inclusion. Then
\begin{align*}
x = \frac{1}{2} (2x) + \frac{1}{2} \cdot 0
\end{align*}
is a convex combination of $2x \in C$ and $0 \in C$, so $x \in C$ by convexity.
In all three cases $x \in C$, so $\{\mu_C \leq 1\} \subset C$.
[guided]
The goal is: if $C$ is closed, $\mu_C(x) \leq 1$ forces $x \in C$. The strategy is **approximation from inside**: produce a sequence in $C$ converging to $x$ and invoke closedness.
We split on $\mu_C(x) < 1$, $\mu_C(x) = 1$, and $\mu_C(x) = 0$ (the last is a degenerate case where the construction in Case 2 needs care).
**Case $\mu_C(x) < 1$.** We already proved $\{\mu_C < 1\} \subset C$, so $x \in C$. No new work.
**Case $\mu_C(x) = 1$.** We cannot conclude $x \in C$ directly from $\{\mu_C < 1\} \subset C$ — the lower inclusion is **strict** on the level set $\{\mu_C = 1\}$. Instead, approximate from outside: pick $s_n \downarrow 1$ such that $x \in s_n C$, i.e.\ $u_n := x/s_n \in C$. Why can we pick such $s_n$? By definition of the infimum: for every $t > \mu_C(x) = 1$, there exists $s \in [\mu_C(x), t)$ with $x \in sC$ (take $s = t - \eta$ small enough). Picking $t_n = 1 + 1/n$ and a corresponding $s_n \in [1, 1 + 1/n)$, we get $s_n \to 1$.
Now $u_n = (1/s_n) x \to (1/1) x = x$ by continuity of the scalar multiplication map $\lambda \mapsto \lambda x$ in the TVS $X$ (a TVS axiom), and the fact that $1/s_n \to 1$ in $\mathbb{R}$. Each $u_n \in C$, and $C$ is closed, so the limit $x \in C$.
**Case $\mu_C(x) = 0$.** The construction above breaks down: $\mu_C(x) = 0$ means we would pick $s_n \in [0, t_n)$, and as $t_n \downarrow 0$ we get $s_n \to 0$, but then $u_n = x/s_n$ blows up rather than converging to $x$. We use a different argument: by positive homogeneity, $\mu_C(2x) = 2 \mu_C(x) = 0 < 1$, so $2x \in \{\mu_C < 1\} \subset C$. Then $0 \in C$ (the standing hypothesis) and convexity give $x = \frac{1}{2}(2x) + \frac{1}{2} \cdot 0 \in C$.
Either way, $x \in C$, completing the equality $\{\mu_C \leq 1\} = C$.
[/guided]
[/step]
