\section{Finite theta functions and rational positive polytopes}\label{sec:convexity}
Define the rational alternating isomorphism $\Lambda:\LL_\QQ\to\LL_\QQ^*$ by
\begin{equation}\label{eq:lambda}
 \Lambda=\begin{pmatrix}0&\Delta^{-1}\\-\Delta^{-1}&\Delta^{-1}B\end{pmatrix},
 \qquad\Lambda^{-1}=\begin{pmatrix}B\Delta&-\Delta\\\Delta&0\end{pmatrix}.
\end{equation}
It satisfies $\Lambda P=R$ and $P^{\mathsf T}\Lambda=-R^{\mathsf T}$. The matrices are inverse because their block products have diagonal blocks $I$ and off-diagonal blocks zero; the lower-right block of $\Lambda$ is $-\Omega$, which is skew-symmetric. Thus $\Lambda$ is alternating. For a nonzero finite Laurent polynomial $A=\sum_a c_a z^a$, its tropicalization at $w\in\LL_\RR^*$ is $A^{\mathrm{trop}}(w)=\min_{c_a\ne0}\pair a w$.

\begin{lemma}\label{lem:finiteconvex}
If $\tht_{m,+}$ is finite in the initial positive chamber, then
\[
 h(p)=\tht_{m,+}^{\mathrm{trop}}(\Lambda p)
\]
is concave along every broken line, with its affine time parameter.
\end{lemma}
\begin{proof}
Retain auxiliary charge variables and write the finite lift as
\[
 A=\sum_{n\in E}c_n z^{m+Pn}t^n,
 \qquad E\subset\NN^J\text{ finite},\quad c_n>0.
\]
Set $K_0=\max_{n\in E}|n|_1$. For a series, its weight at $w$ is the infimum of $\pair a w$ over its nonzero monomials $z^at^n$. If every correction of a wall has nonnegative $w$-weight, its automorphism and inverse do not decrease this infimum. Applying both directions gives equality. When correction weights are strictly positive, each input monomial attaining the minimum retains its coefficient: no higher-weight term can contribute to it. These statements hold for path products degree by degree.

For general $p$, join a general positive-chamber point to $p$ by a straight segment avoiding joints. Every crossed wall of charge $v$ has $\pair{Rv}{p}<0$, and hence
\[
 \pair{Pv}{\Lambda p}=-\pair{Rv}{p}>0.
\]
Transportation therefore gives weight $h(p)$ for $\tht_{m,p}$, and a minimum attained by a monomial of $A$ of charge at most $K_0$. For any direction approaching a nongeneric point, a fixed monomial of the finite set $E$ is minimal sufficiently near that point. Degreewise coefficient stabilization and continuity of $h$ preserve the equality in the one-sided limit.

On a straight segment, $h$ is a minimum of finitely many affine functions and is concave. Consider a bend at $q$ with incoming and outgoing exponents $a_-,a_+$. All walls at $q$ have parallel normals. Since $R$ is injective, their charges are parallel as well. Write their primitive charge as $v$ and their primitive integral normal as $\nu=cRv$, $c>0$. Normalize other normals by raising their functions to positive integer powers. The combined positive wall function is a series $f$ in $Z=z^{Pv}t^v$. Put $A_0=\pair{a_-}{Rv}\ne0$ and $\sigma=\operatorname{sgn}A_0$. The bend has the form
\[
 a_+=a_-+ck\sigma A_0Pv,\qquad k\ge0.
\]
Choose an actual degree $J_0\ge k$ in $f$, together with a finite factor witness for it. This is possible because $k$ is an average of degrees selected from powers of $f$.

Choose a charge cutoff $K_1\ge K_0+J_0|v|_1\max_{n\in E}|\nu(m+Pn)|$ containing the witness. In a sufficiently small ball about $q$, this finite diagram has only the common wall hyperplane. The theta truncations on its two open half-balls are fixed positive finite polynomials $A_-,A_+$, related by
\begin{equation}\label{eq:blocks}
 A_+\equiv E_{\sigma\nu,f}(A_-)\pmod{\text{charge}>K_1}.
\end{equation}
Here $E_{\sigma\nu,f}(z^a)=z^af^{\sigma\nu(a)}$. For either half-ball, the monomials common to $A$ and its theta truncation realize $h$ at every general point, by the preceding transportation argument. Every other monomial in that truncation has weight at least $h$. Both statements are comparisons of finitely many continuous linear functions, so extend to the closed half-ball.

Set $p_-=q+\epsilon a_-$ and $p_+=q-\epsilon a_+$ for small $\epsilon>0$. Choose a monomial $z^bt^{n_b}$ common to $A$ and $A_-$ realizing $h(p_-)$ for all sufficiently small $\epsilon$. Then
\[
 \pair b{\Lambda q}=h(q),\quad |n_b|_1\le K_0,\quad
 h(p_-)=h(q)+\epsilon\pair b{\Lambda a_-}.
\]
Write $d=\nu(b)=c\pair b{Rv}$, and let $A_\pm^{(d)}$ be the sum of terms of $A_\pm$ whose exponents $a$ satisfy $\nu(a)=d$. Wall transformations preserve this pairing, so the corresponding block of \eqref{eq:blocks} is multiplication by $f^{\sigma d}$. Its positive-power direction has no cancellation.

We construct an actual exponent $b_+$ of $A_+$ with
\begin{equation}\label{eq:crossineq}
 b_+-b\in\RR Pv,\qquad
 \pair{b_+}{\Lambda a_+}\ge\pair b{\Lambda a_-}.
\end{equation}
If $A_0\pair b{Rv}>0$, then $\sigma d=|d|>0$. For $\sigma=1$, choose the constant term of $f^{|d|}$, giving $b_+=b$. For $\sigma=-1$, choose degree $|d|J_0$, giving $b_+=b+|d|J_0Pv$. Both charges are at most $K_1$.

If $A_0\pair b{Rv}<0$, use the reverse positive block $A_-^{(d)}=A_+^{(d)}f^{|d|}$. A nonzero source of $b$ has $b_+=b-rPv$ and charge $n_b-rv\in\NN^J$, for some integer $r\ge0$. If $\sigma=-1$, then $\pair{Pv}{\Lambda p_-}>0$; its constant-term contribution to $A_-$ would have smaller weight than $b$ unless $r=0$. If $\sigma=1$, the same pairing is negative. The degree $|d|J_0$ contribution is within the cutoff, and minimality of $b$ forces $r\ge |d|J_0$.

In all four cases write $b_+=b+ck'\sigma \pair b{Rv}Pv$. The choices give $k'=0$ or $J_0$ in the first pair of cases, and $k'=r/|d|$ in the second pair. The identities $\pair{Pv}{\Lambda a_+}=-A_0$ and $\pair b{\Lambda Pv}=\pair b{Rv}$ follow from $\Lambda P=R$ and $\pair{Rv}{Pv}=0$. They give
\[
 \pair{b_+}{\Lambda a_+}-\pair b{\Lambda a_-}
 =c(k-k')\sigma A_0\pair b{Rv}\ge0.
\]
If $d=0$, the block is unchanged and $b_+=b$ gives equality. This proves \eqref{eq:crossineq}. Since $\pair{Rv}{q}=0$, also $\pair{b_+}{\Lambda q}=h(q)$. The finite half-ball inequality gives
\[
 h(p_+)\le h(q)-\epsilon\pair{b_+}{\Lambda a_+}
 \le h(q)-\epsilon\pair b{\Lambda a_-}.
\]
Thus the derivative in actual time cannot increase at the bend. This proves concavity on the whole line. Continuity transfers it to fixed-record nongeneric limits.
\end{proof}

\begin{proposition}\label{prop:rational}
If every theta function is Laurent finite, there is a bounded rational positive polytope with the origin in its interior.
\end{proposition}
\begin{proof}
Choose an integral basis $a_1,\ldots,a_d$ of $\LL$ and put
\[
 f(p)=\min_{j,\,\sigma=\pm1}\tht_{\sigma a_j,+}^{\mathrm{trop}}(\Lambda p).
\]
By Lemma~\ref{lem:finiteconvex}, this is concave on broken lines. The straight-line terms imply
\[
 f(p)\le-\max_j|\pair{a_j}{\Lambda p}|.
\]
Since $\Lambda$ is invertible, $\{f\ge-1\}$ is bounded. It is an intersection of finitely many rational half-spaces, all strict at the origin. Concavity along broken lines implies the decreasing inequality for structure constants and hence positivity by \cite[Lemmas~8.4 and 8.7]{GHKK}. Explicitly, apply the final-slope inequality to the two lines in a near-output pair, using a linear piece active at the output; adding gives $f(r)\ge f(p)+f(q)$.
\end{proof}

This proves the implication (1)$\Rightarrow$(2) in Theorem~\ref{thm:main}. For the converse, Proposition~\ref{prop:finite} followed by Proposition~\ref{prop:rational} allows us to replace the original polytope by a rational one. Henceforth $\Xi$ is this rational polytope and $g=-f$ its gauge. Lemma~\ref{lem:gauge} and coercivity continue to hold.