744 CHAPTER 22. THE DERIVATIVE

Proof: If l = 1, the conclusion is obvious and is nothing more than the definition ofthe derivative.

f (x+h)− f (x)−a(x,h) = o(∥h∥)

and so from the definition of the derivative, a(x,h) = D f (x)h.Next let n = 2. By assumption,

D f (x+h)(h1)−D f (x)(h1)−a(x,h,h1) = o(||h||).

Let L(x) be defined byL(x)(h)(h1)≡ a(x,h,h1).

Then L(x) ∈L (U,L (X ,Y )) because

||L(x)|| ≡ sup||h||≤1

||L(x)(h)|| ≡ sup||h||≤1

sup||h1||≤1

||L(x)(h)(h1)|| ≤C.

Also||D f (x+h)−D f (x)−L(x)h||

≡ sup||h1||≤1

||D f (x+h)(h1)−D f (x)(h1)−L(x)(h)(h1)||

= sup||h1||≤1

||D f (x+h)(h1)−D f (x)(h1)−a(x,h,h1)||= o(||h||)

and so L(x) = D2 f (x). Continuing in this way, we verify the conclusion of the lemma.

Lemma 22.13.5 If f is analytic on U, then f ∈C∞ (U). Also

Proof: By Lemma 22.13.3 applied to g = f and Lemma 22.13.2, D f (x) exists and

D f (x)(h) = a1 (x,h).

These lemmas implied that h→ a1 (x,h) was linear. Suppose Dl−1 f (x) exists for l ≥ 2.

f

(x+

l

∑m=1

zmhm

)=

∞

∑nl=0· · ·

∞

∑n1=0

an1···nl (x,hl , · · · ,h1)zn11 · · ·z

nll .

Differentiate with respect to z1, · · · ,zl−1 to obtain

Dl−1 f

(x+

l−1

∑m=1

zmhm + zlhl

)(hl−1) · · ·(h1) =

∞

∑nl=0

∞

∑nl−1=1

· · ·∞

∑n1=1

anlnl−1···n1 (x,h1 · · ·hl)

(l−1

∏m=1

nm

)zn1−1

1 · · ·znl−1−1l−1 znl

l .

Take zi = 0 for i = 1, · · · , l−1. Then

Dl−1 f (x+ zlhl)(hl−1) · · ·(h1) =∞

∑nl=0

anl1···1 (x,hl , · · · ,h1)znll . (22.13.39)