= Solution
Write $u=\sum_\lambda a_\lambda e(\lambda)$ with finite support. The <root reflection> is an involution, and $s_\alpha u=-u$ implies $a_{s_\alpha\lambda}=-a_\lambda$. Any fixed <weight> has coefficient zero. For each nonfixed orbit choose its representative $\lambda$ with
$$
m=\lambda(H_\alpha)>0,\qquad
s_\alpha\lambda=\lambda-m\alpha.
$$
Integrality of $m$ follows because $\lambda$ belongs to the <weight lattice>; positivity is ensured by our choice of orbit representative. Pairing the orbit coefficients gives
$$
u=\sum_{\text{chosen }\lambda}a_\lambda[e(\lambda)-e(\lambda-m\alpha)].
$$
For each pair, elementary finite telescoping shows
$$
e(\lambda)-e(\lambda-m\alpha)
=(1-e(\alpha))\left[-\sum_{r=1}^me(\lambda-r\alpha)\right].
$$
There are finitely many chosen orbits, and each inner sum is finite. This proves the <root-binomial divisibility of reflection anti-invariants>:
$$
\boxed{\frac{u}{1-e(\alpha)}
=-\sum_{\text{chosen }\lambda}a_\lambda\sum_{r=1}^{\lambda(H_\alpha)}
e(\lambda-r\alpha)\in\mathbb Z[\Lambda_W].}
$$
For precision, the geometric-series expression is interpreted coefficientwise. Its coefficient at a <weight> $\nu$ is $\sum_{n\geq0}a_{\nu-n\alpha}$, which is a finite sum because $u$ has finite support. On each coset of $\mathbb Z\alpha$, reflection pairs the coefficients with opposite signs, so their total is zero. These cumulative sums vanish both before the smallest occupied <weight> and after the largest occupied <weight>. Only finitely many cosets are occupied, hence the product of the formal series with $u$ has finite support and equals the boxed quotient.
The individual terms $e(n\alpha)u$ need not vanish for large $n$. For example, $u=e(\alpha)-e(-\alpha)$ is anti-invariant, and its quotient is $-1-e(-\alpha)$. Thus \b[“finite sum” means a finite result after coefficientwise cancellation], not an actually truncated geometric series. No analytic convergence assumption is involved.
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