742 CHAPTER 22. THE DERIVATIVE

Proof: Let C be small enough that the circles tmC for all m = 1, · · · , l and C have radiusless than δ

l . First assume tm ̸= 0 for all m. Then

ak1···kl (x, tlhl , · · · , t1h1)

=

(1

2πi

)l ∫C· · ·∫

C

f(x+∑

lm=1 wmtmhm

)∏

lm=1 (wmtm)

km+1 ·

l

∏m=1

tkm+1m dw1 · · ·dwl

Here we just multiplied and divided by ∏lm=1 tkm+1

m .

=

(1

2πi

)l ∫tlC· · ·∫

t1C

f(x+∑

lm=1 umhm

)∏

lm=1 (um)

km+1 du1 · · ·dul

l

∏m=1

tkmm

= ak1···kl (x,hl , · · · ,h1)l

∏m=1

tkmm .

Formally, wi ∈C and so tiwi ≡ ui ∈ tiC. Then tidwi = dui and so dwi = (1/ti)dui. This iswhy ∏

lm=1 tkm+1

m gets changed to ∏lm=1 tkm

m .If tm = 0 for any m, the result of both sides in the above equals zero due to the fact that∫

C

1

wkm+1m

dwm = 0

whenever km ≥ 1.To verify 22.13.38, use 22.13.37 to conclude∥∥ak1···kl (x,hl · · ·h1)

∥∥≤∣∣∣∣∣∣∣∣ak1···kl

(x,

hl

∥hl∥· · · h1

∥h1∥

)∣∣∣∣∣∣∣∣ l

∏m=1||hm||km

and∣∣∣∣∣∣ak1···kl

(x, hl∥hl∥· · · h1∥h1∥

)∣∣∣∣∣∣ is bounded by

M

(2π)l

∫C· · ·∫

C

1

∏lm=1 |wm|km+1 d |w1| · · ·d |wl | ≡C.

Lemma 22.13.3 Suppose

g(x+ zh) = g(x)+∞

∑m=1

bm (x,h)zm

for all z small enough. Then

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