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Hello World, the Lagrange Multiplier Theorem, and Kuhn-Tucker conditions July 26, 2009

Posted by Phi. Isett in Calculus, economics, Uncategorized.
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This is a blog which was started on a whim in order to host a very extensive reply to a blog entry of Terence Tao.  This entry is not the reply — the next entry is.

I may continue this blog to contain pieces of mathematics that I find pretty, important and/or not so commonly known.  In any case, I’d feel silly devoting an entire blog to a single reply.  So here.. I’ll take this opportunity to include my favorite proof of the Lagrange multiplier theorem, and later on maybe I’ll post a nice proof of Stirling’s formula, a group algebraic proof of the Riemann-Lebesgue lemma, or something…

The Lagrange Multiplier Theorem helps us solve constrained maximization / minimization problems, making it (among other things) extremely important in economics.  A (weak form of) the Lagrange Multiplier theorem can be stated as follows:

Let f : \mathbb{R}^n \to \mathbb{R} be a real-valued, continuously differentiable function (called the “objective function” — the thing we want to maximize or minimize), and let g : \mathbb{R}^n \to \mathbb{R}^m be continuously differentiable (called the “constraint function” for reasons which will soon be clear).

If x_0 is an extremizer of the restriction of f to the (codimension m) “constraint manifold” given by M \equiv \{ x : g(x) = 0 \}, then there is a unique linear functional \lambda : \mathbb{R}^m \to \mathbb{R} satisfying the equality of linear maps Df(x_0) = \lambda \circ Dg(x_0).

We assume m \leq n — things tend to be more complicated when the number of constraints exceeds the number of degrees of freedom.

Proof:

We know that f must satisfy some first order condition: namely (by further restricting f to trajectories on M) that Df(x_0)v = 0 for any v \in \mathbb{R}^n which can be realized as a velocity at x_0 by a path within the constraint manifold. An exercise in the implicit function theorem shows that these velocities include all of \mbox{ Ker } Dg(x_0) (these are all such velocities — they are the directions in which one can move without changing g, and together they are called the “tangent space” of M at x_0 under generic circumstances).

Finally, now that \mbox{ Ker } Dg(x_0) \subseteq \mbox{ Ker } Df(x_0), it is a trivial exercise in linear algebra to show that there is a unique linear functional \lambda : \mathbb{R}^m \to \mathbb{R} defined by the identity Df(x_0) = \lambda \circ Dg(x_0).  Basically, we have shown that the zero set of the linear map Dg(x_0) is contained in the zero set of Df(x_0), but because the two maps are linear, their level sets are all translations of the zero level set, and therefore we conclude that every level set of Dg(x_0) is contained in a level set of Df(x_0).

Does anyone know how to turn off the annoying preview thing for hyperlinks..? Never mind, got it.

Edit (29 Aug, 2009):

The same ideas can be used to give a proof of the Kuhn-Tucker conditions for a constrained maximization problem where  we replace the system of equalities \{ g(x) = 0 \} with the system of inequalities \{ g_i(x) \leq 0 | i =1, \ldots, m \}.  We simply replace the role of trajectories within the constraint manifold by trajectories going into the constraint set.

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