22.13. ANALYTIC FUNCTIONS 743

Proof: Recall that

f

(x+

l

∑m=1

zmhm

)=

∞

∑kl=0· · ·

∞

∑k1=0

ak1···kl (x,hl , · · · ,h1)zk11 · · ·z

kkl

and so one can write the following where gnm is defined in the following expression.

g(x+ z1h1 + z2h2) =∞

∑m=0

∞

∑n=0

gmn (x,h1,h2)zm1 zn

2.

Thus,

g(x+ z1h1) =∞

∑m=0

gm0 (x,h1,h2)zm1 = g(x)+

∞

∑m=1

bm (x,h)zm1 ,

g(x+ z2h2) =∞

∑n=0

g0n (x,h1,h2)zn2 = g(x)+

∞

∑n=1

bn (x,h)zn2

which implies

gm0 (x,h1,h2) = bm (x,h1) , g0n (x,h1,h2) = bn (x,h2) .

Now let z1 = z2 = z. Then

g(x+ z(h1 +h2)) = g(x)+∞

∑n=0

bn (x,h1 +h2)zn

= g(x)+ z(g10 (x,h1,h2)+g01 (x,h1,h2))+higher order terms in z.

Therefore,

b1 (x,h1 +h2) = g10 (x,h1,h2)+g01 (x,h1h2)

= b1 (x,h1)+b1 (x,h2)

Lemma 22.13.4 Suppose a(x,hl , · · · ,h1) is multilinear, (hi→ a(x,hl , · · · ,h1) is linear),

||a(x,hl , · · · ,h1)|| ≤Cl

∏m=1||hm||,

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

−a(x,hl , · · · ,h1) = o(||hl ||).

Then Dl f (x) exists and

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