170 CHAPTER 6. SPECTRAL THEORY

where α, β ∈ (0, 1).If the terms f (x+ t, y) and f (x, y + s) are interchanged in 6.17, ∆ (s, t) is unchanged

and the above argument shows there exist γ, δ ∈ (0, 1) such that

∆ (s, t) = fyx (x+ γt, y + δs) .

Letting (s, t) → (0, 0) and using the continuity of fxy and fyx at (x, y) ,

lim(s,t)→(0,0)

∆(s, t) = fxy (x, y) = fyx (x, y) . ■

The following is obtained from the above by simply fixing all the variables except for thetwo of interest.

Corollary 6.7.2 Suppose U is an open subset of Fn and f : U → R has the propertythat for two indices, k, l, fxk

, fxl, fxlxk

, and fxkxlexist on U and fxkxl

and fxlxkare both

continuous at x ∈ U. Then fxkxl(x) = fxlxk

(x) .

Thus the theorem asserts that the mixed partial derivatives are equal at x if they aredefined near x and continuous at x.

Now recall the Taylor formula with the Lagrange form of the remainder. What followsis a proof of this important result based on the mean value theorem or Rolle’s theorem.

Theorem 6.7.3 Suppose f has n + 1 derivatives on an interval, (a, b) and let c ∈ (a, b) .Then if x ∈ (a, b) , there exists ξ between c and x such that

f (x) = f (c) +

n∑k=1

f (k) (c)

k!(x− c)

k+f (n+1) (ξ)

(n+ 1)!(x− c)

n+1.

(In this formula, the symbol∑0

k=1 ak will denote the number 0.)

Proof: It can be assumed x ̸= c because if x = c there is nothing to show. Then thereexists K such that

f (x)−

(f (c) +

n∑k=1

f (k) (c)

k!(x− c)

k+K (x− c)

n+1

)= 0 (6.18)

In fact,

K =−f (x) +

(f (c) +

∑nk=1

f(k)(c)k! (x− c)

k)

(x− c)n+1 .

Now define F (t) for t in the closed interval determined by x and c by

F (t) ≡ f (x)−

(f (t) +

n∑k=1

f (k) (t)

k!(x− t)

k+K (x− t)

n+1

).

The c in 6.18 got replaced by t.