A $U^{-}$-invariant section is determined on the big cell

Fix a root parametrization $u_{-\alpha }(t)={\exp }_G(tx)$ with $0\mathrm{\ne }x\in {\mathfrak{g}}_{-\alpha }$. The derived left action is $(x\cdot f)(g)=\frac{d}{dt}{\mathrm{\mid }}_{0}f(u_{-\alpha }(-t)g)$. If every $x\in {\mathfrak{n}}^{-}$ annihilates $f$, put $p_g(t)=f(u_{-\alpha }(-t)g)$. For every $t_0$, the group law gives $p_g^{\mathrm{\prime }}(t_0)=(x\cdot f)(u_{-\alpha }(-t_0)g)=0$. Thus this polynomial has zero derivative everywhere and is constant over $\mathbb{C}$. Each root subgroup fixes $f$, and their product is $U^{-}$ by [F3], so $U^{-}$ fixes $f$. Conversely, differentiating a $U^{-}$-invariant function gives ${\mathfrak{n}}^{-}f=0$.

Assume now that $f$ is invariant under $U^{-}$. For $u^{-}\in U^{-}$ and $b\in B$ the invariance gives $f(u^{-}b)=f(b)$, and the functional equation of [F1] with $g=1$ gives $f(b)=f(1\cdot b)=\lambda (b)f(1)$. Hence $f(u^{-}b)=\lambda (b)f(1)$ for all $u^{-}\in U^{-}$, $b\in B$.

Let $f,f^{\mathrm{\prime }}$ be two $U^{-}$-invariant sections of ${\mathcal{L}}_{\lambda }$ with $f(1)=f^{\mathrm{\prime }}(1)$. By step 2.1, $f$ and $f^{\mathrm{\prime }}$ agree on $\mathrm{\Omega }=U^{-}B$, which is dense open in the irreducible variety $G$ by [F2]; two regular functions on $G$ agreeing on a dense open subset agree everywhere, so $f=f^{\mathrm{\prime }}$. The linear map $f\mapsto f(1)$ is therefore injective on the space of $U^{-}$-invariant sections, which has dimension at most $1$.

Finally let $t\in T$ and put $g=t\cdot f-\lambda (t{)}^{-1}f$. By [F3] the torus normalizes every $U_{-\alpha }$, so $t\cdot f$ is again $U^{-}$-invariant: for $u^{-}\in U^{-}$ there is $u^{\mathrm{\prime }-}=t{u}^{-}{t}^{-1}\in U^{-}$ with $u^{-}\cdot (t\cdot f)=t\cdot ((t^{-1}{u}^{-}t)\cdot f)=t\cdot f$. Hence $g$ is a $U^{-}$-invariant section, and $g(1)=(t\cdot f)(1)-\lambda (t{)}^{-1}f(1)=f(t^{-1})-\lambda (t{)}^{-1}f(1)=\lambda (t^{-1})f(1)-\lambda (t{)}^{-1}f(1)=0$ by the functional equation. By step 3.1 the vanishing of $g(1)$ forces $g=0$, that is $t\cdot f=\lambda (t{)}^{-1}f$. If $f$ is nonzero, it therefore has weight $-\lambda$.

Let $h$ be any section of $T$-weight $-\lambda$. Left translation and right $B$-equivariance [F1] give $h(t{u}^{-}{t}^{-1})=h(u^{-})$ for $t\in T$. In the polynomial root coordinates on $U^{-}$ of [F3], conjugation scales each coordinate by the character $-\beta$ for a positive root $\beta$. A nonconstant monomial has character $-{\sum }_{\beta }{m}_{\beta }\beta \mathrm{\ne }0$: every positive root has nonnegative simple-root coefficients, and some $m_{\beta }>0$. Since distinct torus characters are linearly independent (restrict a finite list to a one-parameter subgroup separating their exponents), a conjugation-invariant polynomial is constant. Thus $h$ is constant on $U^{-}$ and $h(u^{-}b)=\lambda (b)h(1)$ on $\mathrm{\Omega }$. Density [F2] makes $h$ $U^{-}$-invariant on $G$. Step 3.1 now shows that the entire weight-$(-\lambda )$ space has dimension at most one; any nonzero invariant section spans it.