Equality of Mixed Partials
∂2 f
∂x∂y
Theorem. If the partial derivatives
and
2
∂2 f
∂y∂x
∂ f
∂x∂y (x0 , y0 )
=
exist and are continuous at (x0 , y0 ), then
2
∂ f
∂y∂x (x0 , y0 )
Proof: Here is an outline of the proof. The details are given as footnotes at the end of the
outline. Fix x0 and y0 and define(1)
1
F (h, k) = hk
f (x0 + h, y0 + k) − f (x0 , y0 + k) − f (x0 + h, y0 ) + f (x0 , y0 )
Then, by the mean value theorem,
h
2
F (h, k) = h1 ∂f
∂y (x0 + h, y0 + θ1 k) −
3 ∂ ∂f
∂x ∂y (x0
=
4 1
k
h
+ θ3 h, y0 + k) −
5 ∂ ∂f
(x0
∂y ∂x
+ θ3 h, y0 + θ4 k)
=
i
+ θ3 h, y0 )
i
+ θ2 h, y0 + θ1 k)
∂f
(x0
∂x
F (h, k) =
+ θ1 k)
∂f
∂y (x0 , y0
∂f
(x0
∂x
for some 0 < θ1 , θ2 , θ3 , θ4 < 1. All of θ1 , θ2 , θ3 , θ4 depend on x0 , y0 , h, k. Hence
∂ ∂f
(x0
∂x ∂y
+ θ2 h, y0 + θ1 k) =
∂ ∂f
(x0
∂y ∂x
+ θ3 h, y0 + θ4 k)
for all h and k. Taking the limit (h, k) → (0, 0) and using the assumed continuity of both partial
derivatives at (x0 , y0 ) gives
∂ ∂f
∂ ∂f
(x0 , y0 ) = ∂y
(x0 , y0 )
∂x ∂y
∂x
The Details
2
2
∂ f
∂ f
(x0 , y0 ) and ∂y∂x
(x0 , y0 )
(1) We define F (h, k) in this way because both partial derivatives ∂x∂y
are defined as limits of F (h, k) as h, k → 0. For example,
h
i
∂f
1 ∂f
∂ ∂f
(x
,
y
)
=
lim
(x
,
y
+
k)
−
(x
,
y
)
0
0
0
0
0
0
∂y ∂x
∂x
k→0 k ∂x
h
i
f (x0 +h,y0 )−f (x0 ,y0 )
(x0 ,y0 +k)
= lim k1 lim f (x0 +h,y0 +k)−f
−
lim
h
h
k→0
=
h→0
h→0
f (x0 +h,y0 +k)−f (x0 ,y0 +k)−f (x0 +h,y0 )+f (x0 ,y0 )
lim lim
hk
k→0 h→0
= lim lim F (h, k)
k→0 h→0
Similarly,
∂ ∂f
(x0 , y0 )
∂x ∂y
1
h→0 h
= lim
∂f
(x0
∂y
h
lim
+ h, y0 ) −
∂f
(x0 , y0 )
∂y
i
f (x0 +h,y0 +k)−f (x0 +h,y0 )
(x0 ,y0 )
− lim f (x0 ,y0 +k)−f
k
k
k→0
k→0
0 )−f (x0 ,y0 +k)+f (x0 ,y0 )
lim lim f (x0 +h,y0 +k)−f (x0 +h,y
hk
h→0 k→0
1
h→0 h
= lim
=
h
i
= lim lim F (h, k)
h→0 k→0
c Joel Feldman.
2004. All rights reserved.
1
(2) The mean value theorem says that, for any differentiable function ϕ(x), the slope of the line
joining the points x0 , ϕ(x0 ) and x0 + k, ϕ(x0 + k) on the graph of ϕ is the same as the
slope of the tangent to the graph at some point between x0 and x0 + k. This is, there is some
0 < θ1 < 1 such that
y
y = ϕ(x)
ϕ(x0 +k)−ϕ(x0 )
k
dϕ
dx (x0
=
+ θ1 k)
x0
x
x0 +θ1k x0 +k
Applying this with x replaced by y and ϕ replaced by G(y) = f (x0 + h, y) − f (x0 , y) gives
G(y0 +k)−G(y0 )
k
=
=
dG
(y0
dy
∂f
∂y (x0
+ θ1 k)
for some 0 < θ1 < 1
+ h, y0 + θ1 k) −
∂f
∂y (x0 , y0
Hence, for some 0 < θ1 < 1,
h
i
h
1 ∂f
0)
=
(x0 + h, y0 + θ1 k) −
F (h, k) = h1 G(y0 +k)−G(y
k
h ∂y
(3) Define H(x) =
∂f
(x, y0
∂y
+ θ1 k)
∂f
(x0 , y0
∂y
+ θ1 k)
i
+ θ1 k). By the mean value theorem
1
h
F (h, k) =
=
=
h
i
H(x0 + h) − H(x0 )
dH
(x0 + θ2 h)
dx
∂ ∂f
∂x ∂y (x0 + θ2 h, y0
for some 0 < θ2 < 1
+ θ1 k)
(4) Define A(x) = f (x, y0 + k) − f (x, y0 ). By the mean value theorem
F (h, k) =
=
=
(5) Define B(y) =
∂f
(x0
∂x
1
k
A(x0 +h)−A(x0 )
h
1 dA
k dx (x0
1
k
h
∂f
(x0
∂x
+ θ3 h)
i
for some 0 < θ3 < 1
+ θ3 h, y0 + k) −
∂f
(x0
∂x
+ θ3 h, y0 )
i
+ θ3 h, y). By the mean value theorem
F (h, k) =
=
=
c Joel Feldman.
h
2004. All rights reserved.
1
k
h
i
B(y0 + k) − B(y0 )
dB
dy (y0 + θ4 k)
∂ ∂f
(x0 + θ3 h, y0
∂y ∂x
for some 0 < θ4 < 1
+ θ4 k)
2