Kac Moody Kostant multiplicity formula

Put $\lambda =\mathrm{\Lambda }+\rho$ and $\beta =\mathrm{\Lambda }-\mu$. A contributing $w$ must satisfy $\gamma =w\lambda -(\mu +\rho )\in Q^{+}$ by F2. F1's height bound gives ${\delta }_w=\lambda -w\lambda \in Q^{+}$ and $\mathrm{ht}({\delta }_w)\ge \mathrm{\ell }(w)$. Since $\beta ={\delta }_w+\gamma$, no $w$ contributes unless $\beta \in Q^{+}$. If it does, $\mathrm{\ell }(w)\le \mathrm{ht}(\beta )$, so only finitely many words in the finite simple-reflection alphabet, and hence finitely many elements, can contribute.

Expand F1's inverse product using F2. The term indexed by $(w,\gamma )$ has exponent $w(\mathrm{\Lambda }+\rho )-\rho -\gamma$ and coefficient $\det (w)K(\gamma )$. Equating this exponent to $\mu$ forces exactly the argument of $K$ in the statement. The coefficient extraction is finite by step 1.1, so gives the displayed identity. For $\mu =\mathrm{\Lambda }$, step 1.1 forces $\mathrm{\ell }(w)=0$, and the value is $K(0)=1$. Outside the cone every summand and the corresponding weight space are zero. There is no analytic summation or choice of an infinite family.