Answers 1
Step-by-step explanation:
This document contains a proof of the equality of mixed partials under a natural assumption. The
theorem is due to W. H. Young [1], but his proof is hard to follow. I hope this proof is easy to follow. It
is essentially due to Dieudonne. There is a nice exposition in Pugh’s text. This exposition is that proof,
with perhaps simpler notation.
Theorem 1. Suppose f(x, y) is defined in a neighborhood of a point (a, b). Suppose the partial derivatives
fx, fy are defined in a neighborhood of (a, b) and are differentiable at (a, b). (In particular this implies
that fx, fy are continuous at (a, b), but it is not assumed that their derivatives exist anywhere other than
at (a, b).) A short statement of the assumption is that Df = [fx, fy] is differentiable at (a, b). This is
sometimes stated as f is twice differentiable at (a, b). Then
(fx)y(a, b) = (fy)x(a, b),
sometimes stated as
fxy(a, b) = fyx(a, b).
Proof. Consider the function
∆(t) = [f(a + t, b + t) − f(a + t, b)] − [f(a, b + t) − f(a, b)].
If we let g(s) = f(a + s, b + t) − f(a + s, b)
∆(t) = g(t) − g(0).
By the mean value theorem
∆(t) = g(t) − g(0) = tg′
(ξ) (1)
= [fx(a + ξ, b + t) − fx(a + ξ, b)]t (2)
= [fx(a, b) + (fx)x(a, b)ξ + (fx)y(a, b)t + p1(t)ξ + p2(t)t]t (3)
− [fx(a, b) + (fx)x(a, b)ξ + q1(t)ξ]t (4)
= (fx)y(a, b)t
2 + [p1(t)ξ + q1(t)ξ + p2(t)t]t. (5)
Now divide by t
2 and remember that |ξ| < |t| to get
∆(t)
t
2
= (fx)y(a, b) + [p1(t)ξ + q1(t)ξ + p2(t)t]/t.
The last term goes to 0 as t → 0. Hence
(fx)y(a, b) = lim
t→0
∆(t)
t
2
.
The argument is symmetric in x and y, so
(fx)y(a, b) = lim
t→0
∆(t)
t
2
= (fy)x(a, b).
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Author:
bearqrw3
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