On constrained extrema

Assume that I and J are smooth functionals defined on a Hilbert space H. We derive sufficient conditions for I to have a local minimum at y subject to the constraint that J is constantly J(y). The first order necessary condition for I to have a constrained minimum at y is that for some constant λ, I ′ y +λJ ′ y is identically zero. Here I ′ y and J ′ y are the Fréchet derivatives of I and J at y. For the rest of the paper, we assume that y in H satisfies this necessary condition. A common misapprehension (upon which much of the stability results for capillary surfaces has been based) is to assume that if the quadratic form I ′′ y + λJ ′′ y is positive definite on the kernel of J ′ y then I has a local constrained minimum at y. This is not correct in a Hilbert space of infinite dimension; Finn [1] has supplied a counterexample in the unconstrained case, and the same difficulty will occur in the constrained case. In the unconstrained case, if (as often occurs in practice) the spectrum of I ′′ y is discrete and 0 is not a cluster point of the spectrum, then I ′′ y positive definite at a critical point y implies that I ′′ y is strongly positive, (i.e., there exists k > 0 such that I ′′ y (x) ≥ k‖x‖2 holds for all x), and this in turn does imply that y is a local minimum (see [2]). However, in the constrained case, things are not so easy. Even if I ′′ y + λJ ′′ y has a nice spectrum (in some sense), it is not clear that I ′′ y + λJ ′′ y being positive definite on the kernel of J ′ y implies that this quadratic form is strongly positive on the kernel, nor that strong positivity implies that y is a local minimum. In [3], Maddocks obtained sufficient conditions for I ′′ y + λJ ′′ y to be positive definite on the kernel of J ′ y. As Maddocks points out, this is not quite enough to say that I has a constrained minimum at y. Remarkably, essentially the same conditions as Maddocks obtained for positive definiteness do in fact imply that I has a strict local minimum at y subject to the constraint J = J(y), as we shall see. For any h ∈ H we may say J(y + h) − J(y) = J ′ y(h) + 1 2 J ′′ y (h) + 1(h)‖h‖, where 1 goes to zero as ‖h‖ goes to zero. If we consider an h for which J(y + h) = J(y), then of course 0 = J ′ y(h) + 1 2 J ′′ y (h) + 1(h)‖h‖. Now, for that h we have