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

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Roland Franzius

Posts: 419
Registered: 12/7/04
Re: Proof that mixed partials commute.
Posted: Nov 21, 2013 2:25 AM
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Am 19.11.2013 00:45, schrieb 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[
> [[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.
>



Since you mention the existence and continuity for all partial
derivatives up to second order you have to use the equicontinuity as a
condition that limits of continuous functions are continuous.

Equicontinuity means exactly what you have written implicitely, namely
that your epsilon/delta-_functions_ of the other variable are constants,
independent of the free other variables.

Generally

lim 1/h( f(x+h,y)+f(x,y))

is a continuously indexed limit of functions in the free variable y.

A general limit of continous/differentiable functions may develop
singularities as lim_n->oo (x^n) because for every n you need another
epsolon/delta-bound.

The commutativity of independent differential operators is at the heart
of functional analysis with the rules

d^2 =0 for exterior differential forms or

f_xy - f_yx=0 for the algebraic setting

or rot grad f=0

for the vector analysis setting. So its a much better learning strategy
to look for smoothness definitions in the contexts of function spaces in
contrast to painfully filtering a general sufficient condition for the
interchageability of pointwise limits for real functions in the supremum
norm.

--

Roland Franzius



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