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Uniformly continuous function on a bounded set March 3, 2011

Posted by Phi. Isett in Calculus.
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Here’s a problem which is a little odd and kind of interesting; I’ll write it down before I forget about it.

Every function f : (a,b) \to {\mathbb R} which is uniformly continuous on a bounded interval is bounded.  You can prove this by bounding f(x) – f(y) by breaking the total change into a bunch of small changes (using uniform continuity) just as in one common proof of the Fundamental Theorem of Calculus (this technique also gets used in the proof of Sard’s theorem, Harnack’s inequality, and some other things).

Now replace the bounded interval (a,b) by a bounded subset S \subseteq {\mathbb R}.  Exercise: are uniformly continuous functions still bounded?

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.


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.