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Uniformly continuous function on a bounded set
*March 3, 2011*

*Posted by Phi. Isett in Calculus.*

Tags: Advanced Calculus

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Tags: Advanced 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 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 by a bounded subset . Exercise: are uniformly continuous functions still bounded?

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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.*

Tags: constrained optimization, economics, Karush-Kuhn-Tucker, Kuhn-Tucker, Lagrange multiplier theorem, Lagrange multipliers, proof of Kuhn-Tucker

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Tags: constrained optimization, economics, Karush-Kuhn-Tucker, Kuhn-Tucker, Lagrange multiplier theorem, Lagrange multipliers, proof of Kuhn-Tucker

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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 be a real-valued, continuously differentiable function (called the “objective function” — the thing we want to maximize or minimize), and let be continuously differentiable (called the “constraint function” for reasons which will soon be clear).

If is an extremizer of the restriction of to the (codimension ) “constraint manifold” given by , then there is a unique linear functional satisfying the equality of linear maps .

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

**Proof:**

We know that must satisfy some first order condition: namely (by further restricting to trajectories on ) that for any which can be realized as a velocity at by a path within the constraint manifold. An exercise in the implicit function theorem shows that these velocities include all of (these are all such velocities — they are the directions in which one can move without changing , and together they are called the “tangent space” of at under generic circumstances).

Finally, now that , it is a trivial exercise in linear algebra to show that there is a unique linear functional defined by the identity . Basically, we have shown that the zero set of the linear map is contained in the zero set of , 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 is contained in a level set of .

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 with the system of inequalities . We simply replace the role of trajectories within the constraint manifold by trajectories going *into* the constraint set.