The lowest weight space is the nilradical-invariant line

Let $v_{\mu }$ be a highest weight vector of the finite-dimensional irreducible module $V=L(\mu )$, which exists and generates $V$ by [F4]. Apply [F1] with $\lambda =\mu$: $V=U({\mathfrak{n}}^{-})v_{\mu }$, all weights of $V$ lie below $\mu$, and $V_{\mu }=\mathbb{C}{v}_{\mu }$. Applying [F3] and [F4] to $V$, its dual $V^{\ast }$ is finite-dimensional irreducible of highest weight $-w_0\mu$ with a highest weight vector $v^{\ast }$ generating $V^{\ast }$, and its weights lie below $-w_0\mu$ by [F1].

Write $V=\mathbb{C}{v}_{\mu }+{\mathfrak{n}}^{-}V$: every $y\in U({\mathfrak{n}}^{-})$ is a sum of a scalar and of nonempty products of elements of ${\mathfrak{n}}^{-}$, and a nonempty product $x_1\cdots x_k{v}_{\mu }=x_1(x_2\cdots x_k{v}_{\mu })$ lies in ${\mathfrak{n}}^{-}V$; by step 1.1 this covers all of $V$. The sum is direct: an element of ${\mathfrak{n}}^{-}V$ is a sum of vectors $xv$ with $x\in {\mathfrak{n}}^{-}$ a weight vector of weight $-\gamma$, $\gamma \mathrm{\ne }0$, by [F5], so its weight components are of weights $\nu -\gamma$ where $\nu$ is a weight of $V$; if such a component had weight $\mu$, then $\nu =\mu +\gamma$ would be a weight of $V$ strictly above $\mu$, contradicting step 1.1. Hence ${\mathfrak{n}}^{-}V\cap \mathbb{C}{v}_{\mu }=0$ and $V\mathrm{/}{\mathfrak{n}}^{-}V\cong V_{\mu }=\mathbb{C}{v}_{\mu }$ is one-dimensional.

The same computation with $V$ replaced by $V^{\ast }$, using its highest weight $-w_0\mu$ and its generating highest weight vector from step 1.1, gives $V^{\ast }=\mathbb{C}{v}^{\ast }\oplus {\mathfrak{n}}^{-}{V}^{\ast }$ and $V^{\ast }\mathrm{/}{\mathfrak{n}}^{-}{V}^{\ast }\cong (V^{\ast }{)}_{-w_0\mu }$, which is one-dimensional by [F2].

Identify $V$ with the double dual $(V^{\ast }{)}^{\ast }$ by $v\mapsto (v\mapsto f(v))$, using nondegeneracy of the evaluation pairing [F6]. For $u\in V$ and $x\in {\mathfrak{n}}^{-}$, $f\in V^{\ast }$ one has $(xf)(u)=-f(xu)$ by [F3], so the functional attached to $u$ vanishes on ${\mathfrak{n}}^{-}{V}^{\ast }$ exactly when $xu=0$ for all $x\in {\mathfrak{n}}^{-}$, that is exactly when $u\in V^{{\mathfrak{n}}^{-}}$. Therefore $V^{{\mathfrak{n}}^{-}}$ is identified with the annihilator of ${\mathfrak{n}}^{-}{V}^{\ast }$ in $(V^{\ast }{)}^{\ast }$, which is the dual space of $V^{\ast }\mathrm{/}{\mathfrak{n}}^{-}{V}^{\ast }$; this identification is $\mathfrak{h}$-equivariant, since it is given by the canonical evaluations. By step 2.2 it follows that $\dim V^{{\mathfrak{n}}^{-}}=1$ and that the single weight of this line is $-(-w_0\mu )=w_0\mu$, the negative of the weight of $V^{\ast }\mathrm{/}{\mathfrak{n}}^{-}{V}^{\ast }$.

By [F6] the pairing pairs the weight space $V_{w_0\mu }$ nondegenerately with $(V^{\ast }{)}_{-w_0\mu }$, so $\dim V_{w_0\mu }=\dim (V^{\ast }{)}_{-w_0\mu }=1$ by [F2] applied to the irreducible module $V^{\ast }$ of highest weight $-w_0\mu$. Every weight $\nu$ of $V$ satisfies $\nu \ge w_0\mu$: the pairing is nondegenerate, so $-\nu$ is a weight of $V^{\ast }$, and by [F1] applied to $V^{\ast }$ one has $-\nu$ below $-w_0\mu$, that is $\nu -w_0\mu \in Q^{+}$; hence $w_0\mu$ is the lowest weight of $V$ and $V_{w_0\mu }$ is its lowest weight space. Step 3.1 produces a one-dimensional $\mathfrak{h}$-stable line of weight $w_0\mu$ inside $V^{{\mathfrak{n}}^{-}}$, hence inside the one-dimensional space $V_{w_0\mu }$; therefore $V^{{\mathfrak{n}}^{-}}=V_{w_0\mu }$ and both are one-dimensional.