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 longwinded proofs of this theorem. These proofs involved geometric arguments, auxiliary functions, the meanvalue 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 ==> xc < delta_F exists
G(y)G(c) < epsilon_G ==> yc < delta_G exists
x(t)c < epsilon_x ==> tc < delta_x exists
y(t)c < epsilon_y ==> tc < delta_y exists
Now epsilon_F ==> delta_F can be used as delta_F = epsilon_x which implies delta_x > tc 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.

