# Lagrange multipliers on Banach spaces

In the field of calculus of variations in mathematics, the method of Lagrange multipliers on Banach spaces can be used to solve certain infinite-dimensional constrained optimization problems. The method is a generalization of the classical method of Lagrange multipliers as used to find extrema of a function of finitely many variables.

## The Lagrange multiplier theorem for Banach spaces

Let X and Y be real Banach spaces. Let U be an open subset of X and let f : UR be a continuously differentiable function. Let g : UY be another continuously differentiable function, the constraint: the objective is to find the extremal points (maxima or minima) of f subject to the constraint that g is zero.

Suppose that u0 is a constrained extremum of f, i.e. an extremum of f on

$g^{-1}(0)=\{x\in U\mid g(x)=0\in Y\}\subseteq U.$ Suppose also that the Fréchet derivative Dg(u0) : XY of g at u0 is a surjective linear map. Then there exists a Lagrange multiplier λ : YR in Y, the dual space to Y, such that

$\mathrm {D} f(u_{0})=\lambda \circ \mathrm {D} g(u_{0}).\quad {\mbox{(L)}}$ Since Df(u0) is an element of the dual space X, equation (L) can also be written as

$\mathrm {D} f(u_{0})=\left(\mathrm {D} g(u_{0})\right)^{*}(\lambda ),$ where (Dg(u0))(λ) is the pullback of λ by Dg(u0), i.e. the action of the adjoint map (Dg(u0)) on λ, as defined by

$\left(\mathrm {D} g(u_{0})\right)^{*}(\lambda )=\lambda \circ \mathrm {D} g(u_{0}).$ ## Connection to the finite-dimensional case

In the case that X and Y are both finite-dimensional (i.e. linearly isomorphic to Rm and Rn for some natural numbers m and n) then writing out equation (L) in matrix form shows that λ is the usual Lagrange multiplier vector; in the case n = 1, λ is the usual Lagrange multiplier, a real number.

## Application

In many optimization problems, one seeks to minimize a functional defined on an infinite-dimensional space such as a Banach space.

Consider, for example, the Sobolev space ${\textstyle X=H_{0}^{1}([-1,+1];\mathbb {R} )}$ and the functional ${\textstyle f:X\rightarrow \mathbb {R} }$ given by

$f(u)=\int _{-1}^{+1}u'(x)^{2}\,\mathrm {d} x.$ Without any constraint, the minimum value of f would be 0, attained by u0(x) = 0 for all x between −1 and +1. One could also consider the constrained optimization problem, to minimize f among all those uX such that the mean value of u is +1. In terms of the above theorem, the constraint g would be given by

$g(u)={\frac {1}{2}}\int _{-1}^{+1}u(x)\,\mathrm {d} x-1.$ However this problem can be solved as in the finite dimensional case since the Lagrange multiplier $\lambda$ is only a scalar.