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Topic: Proof that mixed partials commute.
Replies: 20   Last Post: Nov 22, 2013 11:57 PM

 Messages: [ Previous | Next ]
 Hetware Posts: 148 Registered: 4/13/13
Proof that mixed partials commute.
Posted: Nov 18, 2013 6:45 PM

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.

Date Subject Author
11/18/13 Hetware
11/18/13 William Elliot
11/18/13 Hetware
11/19/13 Robin Chapman
11/19/13 David C. Ullrich
11/19/13 Brian Q. Hutchings
11/19/13 Hetware
11/20/13 Robin Chapman
11/20/13 Hetware
11/20/13 David C. Ullrich
11/20/13 Hetware
11/20/13 Ki Song
11/21/13 Robin Chapman
11/21/13 David C. Ullrich
11/20/13 Robin Chapman
11/20/13 Hetware
11/21/13 Robin Chapman
11/21/13 Robin Chapman
11/21/13 David C. Ullrich
11/22/13 ross.finlayson@gmail.com
11/21/13 Roland Franzius