Under certain conditions themixed partial derivatives will always be equal. This
astonishing fact was first observed by Euler around 1734. It is also called Clairaut’s
theorem.
Theorem 6.7.1Suppose f : U ⊆ F^{2}→ ℝ where U is an open set on which f_{x},f_{y}, f_{xy}and f_{yx}exist. Then if f_{xy}and f_{yx}are continuous at the point
(x,y)
∈ U, it follows
fxy (x,y) = fyx (x,y).
Proof:Since U is open, there exists r > 0 such that B
is unchanged and the
above argument shows there exist γ,δ ∈
(0,1)
such that
Δ (s,t) = fyx (x + γt,y + δs) .
Letting
(s,t)
→
(0,0)
and using the continuity of f_{xy} and f_{yx} at
(x,y)
,
(s,tli)→m(0,0)Δ (s,t) = fxy (x,y) = fyx (x,y).■
The following is obtained from the above by simply fixing all the variables except for the two
of interest.
Corollary 6.7.2Suppose U is an open subset of F^{n}and f : U → ℝ has the property thatfor two indices, k,l, f_{xk}, f_{xl},f_{xlxk}, and f_{xkxl}exist on U and f_{xkxl}and f_{xlxk}are bothcontinuous at x ∈ U. Then f_{xkxl}
(x )
= f_{xlxk}
(x)
.
Thus the theorem asserts that the mixed partial derivatives are equal at x if they are defined
near x and continuous at x.
Now recall the Taylor formula with the Lagrange form of the remainder. What
follows is a proof of this important result based on the mean value theorem or Rolle’s
theorem.
Theorem 6.7.3Suppose f has n + 1 derivativeson an interval,
(a,b)
and let c ∈
(a,b)
. Then ifx ∈
(a,b)
, there exists ξ between c and x such that
∑n f(k)(c) k f-(n+1)(ξ) n+1
f (x) = f (c) + k! (x− c) + (n +1)! (x− c) .
k=1
(In this formula, the symbol∑_{k=1}^{0}a_{k}will denote the number 0.)
Proof:It can be assumed x≠c because if x = c there is nothing to show. Then there exists K
such that
( ∑n (k) )
f (x)− f (c)+ f--(c)(x− c)k + K (x− c)n+1 = 0 (6.18)
k=1 k!
From 6.20 and the continuity of H, if v is small enough,
f (x + v) ≥ f (x) + 1δ2|v |2 − 1δ2|v |2 = f (x)+ δ2|v|2.
2 4 4
This shows the first claim of the theorem. The second claim follows from similar reasoning.
Suppose H
(x )
has a positive eigenvalue λ^{2}. Then let v be an eigenvector for this eigenvalue. From
6.20,
1 1 ( )
f (x+tv) = f (x)+-t2vT H (x )v+ -t2 vT (H (x+tv)− H (x))v
2 2
which implies
f (x+tv) = f (x )+1 t2λ2|v|2+1 t2 (vT (H (x+tv)− H (x ))v )
2 2
≥ f (x )+1 t2λ2|v|2
4
whenever t is small enough. Thus in the direction v the function has a local minimum at x. The
assertion about the local maximum in some direction follows similarly. ■
This theorem is an analogue of the second derivative test for higher dimensions. As in one
dimension, when there is a zero eigenvalue, it may be impossible to determine from the
Hessian matrix what the local qualitative behavior of the function is. For example,
consider
4 2 4 2
f1(x,y) = x + y ,f2(x,y) = − x + y .
Then Df_{i}
(0,0)
= 0 and for both functions, the Hessian matrix evaluated at
(0,0)
equals
( )
0 0
0 2
but the behavior of the two functions is very different near the origin. The second has a saddle
point while the first has a minimum there.