---
tag: Q7QK3PF806
kind: statement
role: theorem
name: Lagrange multipliers on Banach spaces
msc: [49K27, 46G05, 58C20]
uses: [06GZ956657, 06EH97V2A7, 07JBSZE0V4, 0WFBH1Q3WC, 3PGXH0T1K6, 03YEM7JEDN]
about: [07JBSZE0V4, 06EH97V2A7]
sources:
  - url: https://en.wikipedia.org/wiki/Lagrange_multipliers_on_Banach_spaces
same_as:
  wikidata: Q17086206
  wikipedia: Lagrange multipliers on Banach spaces
---

## Statement

Let $X$ and $Y$ be real [[Banach spaces|06GZ956657]], let $U \subseteq X$ be open, and let $f \colon U \to \R$ and $g \colon U \to Y$ be continuously [[Fréchet differentiable|06EH97V2A7]].
Let $x_0 \in U$ with $g(x_0) = 0$, and suppose that $x_0$ is a [[local extremum|0WFBH1Q3WC]] of the restriction of $f$ to $M = \set{x \in U : g(x) = 0}$.
Suppose that the derivative $Dg(x_0) \colon X \to Y$ is surjective.
Then there is a unique continuous linear functional $\lambda \in Y^*$, an element of the [[dual space|07JBSZE0V4]] of $Y$, such that
$$Df(x_0) = \lambda \circ Dg(x_0).$$

## Remarks

Let $h \in \ker Dg(x_0)$.
By [[Lyusternik's theorem|3PGXH0T1K6]] there is a curve $t \mapsto x_0 + th + r(t)$ in $M$ with $r(t) = o(t)$, and $t \mapsto f(x_0 + th + r(t))$ has a local extremum at $t = 0$, so $Df(x_0) h = 0$.
Thus $Df(x_0)$ vanishes on $\ker Dg(x_0)$, and $\lambda(Dg(x_0) h) = Df(x_0) h$ is a well defined linear map on $Y$.
It is continuous by the [[open mapping theorem|03YEM7JEDN]], and it is unique because $Dg(x_0)$ is onto.

For $X = \R^n$ and $Y = \R^m$, the functional $\lambda$ is a vector $(\lambda_1, \dots, \lambda_m)$, and the result is the [[Lagrange multiplier theorem|0VKKP8X6Q6]].

Example: an isoperimetric problem minimises $J(y) = \int_a^b L(x, y, y') \, dx$ subject to $K(y) = \int_a^b G(x, y, y') \, dx = \ell$ and fixed ends.
Here $Y = \R$, and the result gives a number $\lambda$ with $\delta J(y_0; \eta) = \lambda \, \delta K(y_0; \eta)$ for all admissible variations $\eta$.

Surjectivity is needed.
For $X = \R^2$, $Y = \R$, $f(x_1, x_2) = x_1$ and $g(x_1, x_2) = x_2^2 - x_1^3$, the point $0$ minimises $f$ on $M$, since $x_1 \ge 0$ on $M$.
But $Dg(0) = 0$, so $\lambda \circ Dg(0) = 0 \ne Df(0)$ for every $\lambda$.

## Explanations

### L1 human

At a constrained extremum, the derivative of the objective is a combination of the derivatives of the constraints.
In infinite dimensions there may be infinitely many constraints, so the coefficients form a continuous linear functional on the constraint space instead of a list of numbers.

### L2 human

This is the abstract form of the multiplier rule used for constrained problems in the calculus of variations and in optimal control.
With $Y = \R$ it gives Euler's multiplier rule for isoperimetric problems, and with $Y$ a function space it handles pointwise constraints.
The onto condition on $Dg(x_0)$ is the infinite-dimensional constraint qualification; without it the multiplier may not exist, as the example shows.