[proofplan] The proof uses the triangular PBW factorisation to reduce arbitrary elements of $U(\mathfrak g)$ acting on the highest weight vector to elements of $U(\mathfrak n^-)$ acting on it. The positive part contributes only its scalar constant term because every positive root vector kills $v_\lambda$, and the Cartan part acts by the scalar character determined by $\lambda$. For the weight statement, PBW monomials in negative root vectors are weight vectors whose weights are obtained from $\lambda$ by subtracting nonnegative sums of positive roots; a finite linear-independence argument for distinct $\mathfrak h$-weights then rules out any other genuine weight. [/proofplan] [step:Reduce the generated module to the action of the enveloping algebra] Because $V$ is generated by $v_\lambda$ as a $\mathfrak g$-module, the induced action of the enveloping algebra gives \begin{align*} V=U(\mathfrak g)v_\lambda. \end{align*} Here $U(\mathfrak g)v_\lambda$ denotes the linear span of all vectors $uv_\lambda$ with $u\in U(\mathfrak g)$. The hypotheses of the triangular PBW factorisation [citetheorem:9359] are satisfied: $\mathfrak g$ is a finite-dimensional complex semisimple [Lie algebra](/page/Lie%20Algebra) and $\mathfrak g=\mathfrak n^-\oplus\mathfrak h\oplus\mathfrak n^+$ is the fixed triangular decomposition. Hence multiplication induces a [vector space](/page/Vector%20Space) isomorphism \begin{align*} U(\mathfrak n^-)\otimes U(\mathfrak h)\otimes U(\mathfrak n^+) \longrightarrow U(\mathfrak g). \end{align*} Therefore every element of $U(\mathfrak g)v_\lambda$ is a finite sum of vectors of the form \begin{align*} u_-u_0u_+v_\lambda, \end{align*} where $u_-\in U(\mathfrak n^-)$, $u_0\in U(\mathfrak h)$, and $u_+\in U(\mathfrak n^+)$. [/step] [step:Remove the positive and Cartan factors from each PBW product] Let $\varepsilon_+:U(\mathfrak n^+)\to\mathbb C$ be the algebra homomorphism determined by $\varepsilon_+(1)=1$ and $\varepsilon_+(x)=0$ for every $x\in\mathfrak n^+$. Equivalently, $\ker\varepsilon_+$ is the span of all positive-degree PBW monomials in elements of $\mathfrak n^+$. Since every $x\in\mathfrak n^+$ satisfies $xv_\lambda=0$, every positive-degree monomial in $U(\mathfrak n^+)$ kills $v_\lambda$. Hence \begin{align*} u_+v_\lambda=\varepsilon_+(u_+)v_\lambda \end{align*} for every $u_+\in U(\mathfrak n^+)$. Let $\lambda_U:U(\mathfrak h)\to\mathbb C$ be the algebra homomorphism extending the linear functional $\lambda:\mathfrak h\to\mathbb C$. Since $\mathfrak h$ is abelian and $hv_\lambda=\lambda(h)v_\lambda$ for every $h\in\mathfrak h$, the action of $U(\mathfrak h)$ on the line $\mathbb Cv_\lambda$ is given by \begin{align*} u_0v_\lambda=\lambda_U(u_0)v_\lambda \end{align*} for every $u_0\in U(\mathfrak h)$. Thus, for every PBW product $u_-u_0u_+$, \begin{align*} u_-u_0u_+v_\lambda = \varepsilon_+(u_+)\lambda_U(u_0)u_-v_\lambda. \end{align*} This vector lies in $U(\mathfrak n^-)v_\lambda$. Since $U(\mathfrak n^-)v_\lambda\subset U(\mathfrak g)v_\lambda=V$ is immediate from $\mathfrak n^-\subset\mathfrak g$, we obtain \begin{align*} V=U(\mathfrak n^-)v_\lambda. \end{align*} [guided] The key input is the triangular PBW factorisation. Since $\mathfrak g=\mathfrak n^-\oplus\mathfrak h\oplus\mathfrak n^+$ is the fixed triangular decomposition of the finite-dimensional complex semisimple Lie algebra $\mathfrak g$, the [Triangular PBW Factorisation] [citetheorem:9359] applies. Therefore multiplication induces a vector space isomorphism \begin{align*} U(\mathfrak n^-)\otimes U(\mathfrak h)\otimes U(\mathfrak n^+) \longrightarrow U(\mathfrak g). \end{align*} Hence every vector in $V=U(\mathfrak g)v_\lambda$ is a finite sum of vectors of the form $u_-u_0u_+v_\lambda$, where $u_-\in U(\mathfrak n^-)$, $u_0\in U(\mathfrak h)$, and $u_+\in U(\mathfrak n^+)$. Thus it is enough to understand what the positive and Cartan factors do to $v_\lambda$. Define the augmentation map \begin{align*} \varepsilon_+:U(\mathfrak n^+)\to\mathbb C \end{align*} to be the algebra homomorphism with $\varepsilon_+(1)=1$ and $\varepsilon_+(x)=0$ for every $x\in\mathfrak n^+$. The point of this map is to separate the scalar part of $u_+$ from all products involving at least one element of $\mathfrak n^+$. Every such positive-degree product kills $v_\lambda$, because the rightmost factor from $\mathfrak n^+$ already kills $v_\lambda$. Therefore \begin{align*} u_+v_\lambda=\varepsilon_+(u_+)v_\lambda. \end{align*} Next define \begin{align*} \lambda_U:U(\mathfrak h)\to\mathbb C \end{align*} as the algebra homomorphism extending $\lambda:\mathfrak h\to\mathbb C$. Since $\mathfrak h$ is abelian, $U(\mathfrak h)$ is generated by commuting elements of $\mathfrak h$, and each $h\in\mathfrak h$ acts on $v_\lambda$ by the scalar $\lambda(h)$. Hence every polynomial expression in elements of $\mathfrak h$ acts by the corresponding scalar: \begin{align*} u_0v_\lambda=\lambda_U(u_0)v_\lambda. \end{align*} Combining the two reductions, every PBW product satisfies \begin{align*} u_-u_0u_+v_\lambda=\varepsilon_+(u_+)\lambda_U(u_0)u_-v_\lambda. \end{align*} So the positive and Cartan factors leave only a scalar, and the remaining vector is produced entirely by $U(\mathfrak n^-)$. Since $U(\mathfrak n^-)\subset U(\mathfrak g)$, this proves \begin{align*} V=U(\mathfrak n^-)v_\lambda. \end{align*} [/guided] [/step] [step:Compute the weights of PBW monomials in negative root vectors] Choose a PBW basis of $U(\mathfrak n^-)$ consisting of ordered monomials in root vectors from the negative root spaces $\mathfrak g_{-\alpha}$ with $\alpha\in\Phi^+$. For $x\in\mathfrak g_{-\alpha}$, the root-shifting property [citetheorem:9362] gives \begin{align*} xV_\nu\subset V_{\nu-\alpha} \end{align*} for every $\nu\in\mathfrak h^*$. Since $v_\lambda\in V_\lambda$, induction on the length of a PBW monomial shows that any monomial \begin{align*} x_1x_2\cdots x_s \end{align*} with $x_j\in\mathfrak g_{-\beta_j}$ and $\beta_j\in\Phi^+$ sends $v_\lambda$ either to $0$ or to a vector of weight \begin{align*} \lambda-\sum_{j=1}^s\beta_j. \end{align*} The empty monomial sends $v_\lambda$ to a vector of weight $\lambda$. By the standard root-system fact that each positive root is a nonnegative integer combination of the simple roots $\alpha_1,\dots,\alpha_r$, every $\beta_j\in\Phi^+$ has an expansion \begin{align*} \beta_j=\sum_{i=1}^r c_{ij}\alpha_i \end{align*} with $c_{ij}\in\mathbb Z_{\ge 0}$. Therefore every nonzero PBW monomial vector has weight \begin{align*} \lambda-\sum_{i=1}^r m_i\alpha_i \end{align*} for some $m_1,\dots,m_r\in\mathbb Z_{\ge 0}$. [/step] [step:Exclude weights not appearing among the PBW monomial weights] Let $\mu\in\mathfrak h^*$ be a weight of $V$, and choose a nonzero vector $w\in V_\mu$. Since $V=U(\mathfrak n^-)v_\lambda$, the vector $w$ is a finite linear combination of PBW monomial vectors. Let $S\subset\mathfrak h^*$ be the finite set of weights of the nonzero PBW monomial vectors appearing in one such expression for $w$. We use the elementary fact that eigenspaces for distinct $\mathfrak h$-weights are linearly independent. Indeed, if $\eta_1,\dots,\eta_N\in\mathfrak h^*$ are distinct and $z_j\in V_{\eta_j}$ satisfy \begin{align*} \sum_{j=1}^N z_j=0, \end{align*} then all $z_j=0$. To prove this, choose $h_0\in\mathfrak h$ such that $\eta_1(h_0)\neq \eta_j(h_0)$ for every $j>1$; such an element exists because the complement of the finite union of hyperplanes $\{h\in\mathfrak h: \eta_1(h)=\eta_j(h)\}$ is nonempty. Applying $h_0-\eta_1(h_0)$ to the relation gives \begin{align*} \sum_{j=2}^N \bigl(\eta_j(h_0)-\eta_1(h_0)\bigr)z_j=0, \end{align*} and each coefficient $\eta_j(h_0)-\eta_1(h_0)$ is nonzero. Repeating the same argument inductively on the remaining distinct weight groups forces $z_2=\cdots=z_N=0$, and then $z_1=0$ as well. Since $w\ne 0$ lies in $V_\mu$ and also in the sum of the weight spaces indexed by $S$, [linear independence](/page/Linear%20Independence) of distinct weight spaces implies $\mu\in S$. By the preceding step, every element of $S$ has the form \begin{align*} \lambda-\sum_{i=1}^r m_i\alpha_i \end{align*} with $m_i\in\mathbb Z_{\ge 0}$. Hence $\mu$ has this form. This proves the asserted restriction on the weights of $V$. [/step]