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

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David C. Ullrich

Posts: 3,555
Registered: 12/13/04
Re: Proof that mixed partials commute.
Posted: Nov 19, 2013 11:17 AM
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On Mon, 18 Nov 2013 18:45:53 -0500, Hetware <>

>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[
>, {Dy->0, Dx->0}]
>f_xy(x,y) = Limit[
>, {Dx->0, Dy->0}]

That's very bad notation. It's not one limit, it's the limit of
a limit. Should be


And now the big question is why

Limt(Limt(...)[x->c][y->c] = Limt(Limt(...)[y->c][x->c]

>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?

Not as far as I can see.

>I feel as though I am trying to prove the
>obvious, but it is not obvious how to prove it.

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