假设这个问题的解是 \left. \begin{aligned} &f(x)=f(x(t))\\ &f(x^{*})=f(x(0))\\ &\nabla f(x^{*})=\nabla f(x^{*}) \cdot \nabla x(0) = 0 \end{aligned} \right. ​ f ( x ) = f ( x ( t ) ) f ( x ∗ ) = f ( x ( 0 ) ) ∇ f ( x ∗ ) = ∇ f ( x ∗ ) ⋅ ∇ x ( 0 ) = 0 ​

最后一个等式之所以是0, 是因为对于一元可导函数 M=\{y|\sum_{i=1}^{m}y_i\nabla h_{i}(x^{*})=0^{(m)}\}, y\in R^{m}, \nabla h_{i}(x^{*}) \in R^m, 0^{(m)}\in R^m M = { y ∣ i = 1 ∑ m ​ y i ​ ∇ h i ​ ( x ∗ ) = 0 ( m ) } , y ∈ R m , ∇ h i ​ ( x ∗ ) ∈ R m , 0 ( m ) ∈ R m

接下来, 我们需要证明 \dot{\mathbf{x}}(0)=\mathbf{y}+\nabla \mathbf{h}\left(\mathbf{x}^{*}\right)^{T} \dot{\mathbf{x}}(0)=\mathbf{y} x ˙ ( 0 ) = y + ∇ h ( x ∗ ) T x ˙ ( 0 ) = y

最后得证,
UTF8gbsn多约束的拉格朗日乘子问题.f(x)h1(x)=0⋮hm(x)=0\left. \begin{aligned} \quad & f(x)\\ \quad& h_1(x)=0\\ & \quad \quad \vdots\\ & h_m(x)=0 \end{aligned} \right.​f(x)h1​(x)=0⋮hm​(x)=0​假设这个问题的解是x∗x^{*}x∗.