Proof:

Since g is continuously differentiable and $\nabla g(x^*)$has full rank, there exists a neighborhood N of x^* on which $\nabla g(x)$has full rank and therefore 132#6exists. I can therefore define $f:N\rightarrow{\bf {\rm R}}^n$ by

133#7

and, since g is continuously differentiable, f is continuous. It follows that the IVP

134#8

has a solution x that exists on some interval [-a,a]. Since $z\in{\cal N}(\nabla g(x^*)^T)$, it follows that f(x^*)=z and hence $\dot{x}(0)=z$. It remains only to show that x is feasible, that is, that

135#9

I will use the mean value theorem to show that this hold for each component of g, that is, that for each 136#10,

137#11

By the MVT, for each i and 138#12, there exists t^* between 0 and t such that

139#13

By construction, x(0)=x^*, so g(x(0))=0. Moreover,

140#14

However, by construction, 141#15 and hence

142#16

Therefore, g(x(t))=0. QED

Regularity is thus a sufficient condition for the existence of Lagrange multipliers. It is also a necessary and sufficient condition for the uniqueness of the multipliers. This is because, given a local optimum x^*, the Lagrange multiplier $\lambda^*$ is determined by the linear system

143#17

and the solution to this system (if it exists) is unique provided the columns of $\nabla g(x^*)$ form a linearly independent set. (In Example 3.2, the two columns of $\nabla g(x^*)$ were identical, hence linearly dependent. Therefore, x^* was not a regular point, and no Lagrange multiplier existed.)