22.13. ANALYTIC FUNCTIONS 745

Now we apply Lemma 22.13.3 to the function

zl → Dl−1 f (x+ zlhl)(hl−1) · · ·(h1)

and concludehl → a1···1 (x,hl , · · · ,h1)

is linear. This involved taking nl = 1 to get a1···1 (x,hl , · · · ,h1). Thus from 22.13.39,

Dl−1 f (x+ zlhl)(hl−1) · · ·(h1)−Dl−1 f (x)(hl−1) · · ·(h1)

= a1···1 (x,hl , · · · ,h1)zl +o(zlhl). (22.13.40)

From this equation, it follows that

a1···1(

x,hl · · ·hi + ĥi · · ·h1

)zl−a1···1 (x,hl · · ·hi · · ·h1)zl

−a1···1(

x,hl · · · ĥi · · ·h1

)zl = o(zlhl)

because for each zl , the left side of 22.13.40 is linear in hi for each i≤ l−1. Dividing bothsides of the above by zl and then letting zl → 0, we see that anl1···1 is linear in each of thehi. Denoting zlhl by hl ,

Dl−1 f (x+hl)(hl−1) · · ·(h1)−Dl−1 f (x)(hl−1) · · ·(h1)

= a1···1 (x,hl , · · · ,h1)+o(∥hl∥)and so by Lemma 22.13.4, Dl f (x) exists and

Dl f (x)(hl) · · ·(h1) = a1···1 (x,hl , · · · ,h1).

With these lemmas, the main result can be established. This is the generalization of thewell known result for analytic functions.

Theorem 22.13.6 Let X and Y be two complex Banach spaces and let U be an open set inX. Then f : U → Y is analytic on U if and only if D f (x) exists for each x ∈U and in thiscase, f ∈C∞ (U) , and if h ∈ X , then whenever z is small enough,

f (x+ zh) = f (x)+∞

∑n=1

Dn f (x)hnzn

n!.

Proof: We know

f (x+ zh) = f (x)+∞

∑n=1

an (x,h)zn.

Differentiating, we obtain

Dk f (x+ zh)hk = k!ak (x,h)+∞

∑n=k+1

n(n−1) · · ·(n− k+1)zn−k.

Letting z = 0 this showsDk f (x)hk = k!ak (x,h)

and this proves half the theorem.Conversely, if D f (x) exists on U, it is clear that f is analytic on some ball, B(x,r) ⊆

U,z→ f (y+ zh) is analytic for y ∈ B(x,r) and small enough z. Therefore the formulainvolving the series follows.