Date: Nov 18, 2013 6:45 PM
Author: Hetware
Subject: Proof that mixed partials commute.
In _Calculus and Analytic Geometry_ 2nd ed.(1953), Thomas provides:

Theorem. If the function w=f(x,y) together with the partial derivatives

f_x, f_y, f_xy and f_yx are continuous, then f_xy = f_yx.

Both Thomas and Anton (1980) provide rather long-winded proofs of this

theorem. These proofs involved geometric arguments, auxiliary

functions, the mean-value theorem, epsilon error variables, a

proliferation of symbols, and a generous helping of obscurity.

Starting from the definition of partial differentiation, and using the

rules of limits, along with a modest amount of basic algebra, I came up

with this:

f_x(x,y) = Limit[[f(x+Dx,y)-f(x,y)]/Dx, Dx->0]

f_yx(x,y) = Limit[[f_x(x,y+Dy)-f_x(x,y)]/Dy, Dy->0]

= Limit[

[[f(x+Dx,y+Dy)-f(x,y+Dy)]-[f(x+Dx,y)-f(x,y)]]/DyDx

, {Dy->0, Dx->0}]

f_xy(x,y) = Limit[

[[f(x+Dx,y+Dy)-f(x+Dx,y)]-[f(x,y+Dy)-f(x,y)]]/DxDy

, {Dx->0, Dy->0}]

The only caveat is that the rules for limits, such as /the product of

limits is equal to the limit of the products/, are stated in terms of a

single variable. For example:

Limit[F(t) G(t), t->c] = Limit[F(t), t->c] Limit[G(t), t->c]

Whereas I am assuming

Limit[F(x) G(y), {x->c, y->c}] = Limit[F(x), x->c] Limit[F(y), y->c].

I argue as follows. The statement that x->c as y->c can be formalized by

treating x and y as functions of t such that

Limit[x(t), t->c] = Limit[y(t), t->c] = c

|F(x)-F(c)| < epsilon_F ==> |x-c| < delta_F exists

|G(y)-G(c)| < epsilon_G ==> |y-c| < delta_G exists

|x(t)-c| < epsilon_x ==> |t-c| < delta_x exists

|y(t)-c| < epsilon_y ==> |t-c| < delta_y exists

Now epsilon_F ==> delta_F can be used as delta_F = epsilon_x which

implies delta_x > |t-c| exists. So

Limit[F(x(t)), t->c] = Limit[F(x), x->c], etc.

It follows that

Limit[F(x(t)) G(y(t)), t->c]

= Limit[F(x(t)), t->c] Limit[G(y(t)), t->c]

= Limit[F(x), x->c] Limit[G(y), y->c]

Am I making sense here? I feel as though I am trying to prove the

obvious, but it is not obvious how to prove it.