Let $F,G\in C^1(\mathbb{R}^n,\mathbb{R})$. Assume for $s\in(s_0-\varepsilon,s_0+\varepsilon)$, \begin{align} E(s) = \min F\quad\mbox{subject to}\quad G=s \end{align} is achieved at some $x(s)\in\mathbb{R}^n$. If $\nabla G(x(s))\ne 0$, then \begin{align} \nabla F(x(s)) = \lambda(s)\nabla G(x(s)) \end{align} for some $\lambda(s)\in\mathbb{R}$. $\lambda(s)$ is called the Lagrange multiplier. This is well-known and is taught in a standard Calculus course. In contrast, it's until recently that I know there is a rule says \begin{align} E'(s_0)=\lambda(s_0),\tag{1} \end{align} and I'm already a PhD student (in math). I suspect it's my problem, and asked some of my friends (also PhD students in math) whether they ever saw this rule, and the answers are totally no. So far, I still cannot find a clearly written down theorem and proof concerning this fact. Note that if $x(s)$ is itself differentiable, the proof is very simple: Since $s=G(x(s))$, $1 = \nabla G(x(s))\cdot x'(s)$. And hence \begin{align} E'(s)=\nabla F(x(s))\cdot x'(s)=\lambda(s)\nabla G(x(s))\cdot x'(s)=\lambda(s). \end{align} However, it's also not hard to give counterexamples.

I saw this rule in a paper, where the domain of $F,G$ is a Banach space and the authors justify it directly, instead of applying some general theorem, and their proof uses some convexity property of the functional they studied. Suspecting it should be true in some general setting and ought not to be justified case by case, I finally found that indeed if the minimizers $x(s)$ can be chosen so that $x(s)\to x(s_0)$ as $s\to s_0$, then (1) holds. And it seems OK too when the domain of $F,G$ is a Banach space.

The problem is: I do need this fact in my current paper, and I think it's so ``classical'' that it needs not to be proved there. Does anyone know a reference for it?


1 Answer 1


I am sure you noticed this, but here is a simple counterexample which shows that you need to assume the continuity of the minimizer $x(s)$ at $s_0$:

$$F(x_1,x_2) = x_2 \cos x_1, \qquad G(x_1,x_2) = x_2.$$

Then $E(s)=-|s|$ and, for $s\neq 0$, $\lambda(s)= - \mathrm{sgn} \; s$. For $s=0$, $\lambda(0)$ can be any number bewteen -1 and 1, depending on which minimizer you choose for the constant function $F(x_1,0) \equiv 0$.


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