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Mathematics-Online lexicon: Annotation to

Lagrange Condition

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z overview

If the scalar function $ f$ has a local extremum at a point $ x_\star$ subject to the constraints $ g_i(x)=0$, then there exist Lagrange multipliers $ \lambda_i$, such that

$\displaystyle f^\prime(x_\star) =
\lambda^{\operatorname t}g^\prime(x_\star)

This characterization requires that $ f$ and $ g$ are continuously differentiable in a neighborhood of $ x_\star$ and that the Jacobi matrix $ g^\prime(x_\star)$ has maximal rank.

For a single constraint, the Lagrange condition has the simple form

$\displaystyle \operatorname{grad} f(x_\star)
\operatorname{grad} g(x_\star) \,,

if $ \operatorname{grad} g(x_\star)\neq 0$, i.e., the level sets of $ f$ and $ g$ touch at the extreme point.

The Lagrange condition is not sufficient to decide if $ x_*$ is an extremum, or to determine its type. This requires additional information.

The global extrema of a function can be obtained by comparing the function values at the points which satisfy the Lagrange condition, the points on the boundary of the admissible set, and points where the rank of $ g^\prime$ is not maximal.

Let $ n$ denote the number of variables and $ m$ the number of constraints.

For $ m\ge n$ there is nothing to show since an arbitrary $ n$-vector can always be represented as linear combination of $ n$ linear independent rows of $ g^\prime$.

For $ m<n$ denote by $ x=(u,v)\in\mathbb{R}^m\times\mathbb{R}^{n-m}$ a partition of the variables, where, after permutation, we can assume that $ g_u(u_\star,v_\star)$ is invertible. Then, by the implicit function theorem, the constraints can locally be solved for $ u$ in terms of $ v$:

$\displaystyle g(u,v)=0 \Leftrightarrow u=\varphi(v),\quad
v\approx v_\star

As a consequence, the gradient of the function $ v \mapsto f(\varphi(v),v)$ vanishes at an extremum:

$\displaystyle f_u(u_\star,v_\star)\varphi^\prime(v_\star) + f_v(u_\star,v_\star) = 0

Moreover, differentiating the contraints $ g(\varphi(v),v)=0$, it follows that

$\displaystyle \varphi^\prime(v) = -g_u(u,v)^{-1} g_v(u,v)\,.$    

Substituting $ \varphi^\prime$ into the expression for the gradient, we see that, with

$\displaystyle \lambda = f_u(u_\star,v_\star) g_u(u_\star,v_\star)^{-1}

the equations

$\displaystyle f_u = \lambda g_u,\quad
f_v = \lambda g_v \,,

are valid at the point $ (u_\star,v_\star)$. These identities correspond to the $ u$- and $ v$-components of the Lagrange condition $ f^\prime=\lambda g^\prime$.


  automatisch erstellt am 26.  1. 2017