Lagrange multipliers on Banach spaces

Remarks

Let $h \in \ker Dg(x_0)$. By Lyusternik's theorem 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, 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.

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 $\lambda$.

Going deeper

For a student

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.