CAUCHY'S THEOREM FOR RECTANGLES

Chapter 9
CAUCHY’S THEOREM
FOR RECTANGLES
The fundamental Cauchy theorem for holomorphic functions will be proven in
steps. The original proof by Cauchy required not only f ′ to exist, but also to be
continuous. The condition was relaxed by Goursat (1900), who proved Cauchy’s
theorem for triangular paths. A simpler proof was then found for rectangles,
and is given here.
Theorem 9.0.8. If f is holomorphic on a domain, and R is a rectangle in the
domain, with boundary ∂R:
∫
∂R
dζf(ζ) = 0 (9.1)
Proof. Fix ǫ > 0 arbitrarily small; we’ll show that the integral I(R) is such that
|I(R)| < ǫ.
Let L be the length of the diagonal of R, and choose an orientation of the
boundary. Cut the sides of R in halves, and obtain four rectangles R 1k (the index
1 stands for generation); let I(R 1k ) be the values of the integral on the oriented
boundary of the rectangles. It is I(R) = ∑ k I(R 1k) because of cancellations of
integrals on shared sides (having opposite orientations).
Denote I 1 the integral with largest modulus among I(R 11 )...I(R 14 ), and R 1
the corresponding rectangle. Then |I(R)| ≤ ∑ k |I(R 1k)| ≤ 4|I 1 |.
The partition is repeated again on R 1 and produces four rectangles R 2k , and
I 1 = ∑ k I(R 2k). Select the rectangle R 2 with largest value |I(R 2k )|, and let I 2
be the value of the integral on its boundary. It is |I| ≤ 4 2 |I 2 |.
A sequence of rectangles R 1 ⊂ . . . ⊂ R n ⊂ R n+1 ⊂ . . . is generated, and at
generation n:
|I| ≤ 4 n |I n |
Let a be a point in the interesection of all rectangles of the sequence: a ∈ ⋂ k R k.
Since f(z) is holomorphic it is true that given the ǫ we started with, there ∃ δ
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1 2
3 4
Figure 9.1: The integral on a rectangular path can be obtained as the sum of
four integrals on the dashed paths. The contributions from overlapping but
oppositely oriented sides cancel.
such that
f(z) − f(a)
z − a
= f ′ (a) + r(z, a)
with remainder |r(z, a)| < ǫ for all z such that |z − a| < δ. In R n two points
have separation at most L/2 n , therefore a rephrasing is: there ∃n (generation)
such that f(z) = f(a) + f ′ (a)(z − a) + r(z, a)(z − a) where |r(z, a)| < ǫ for all
z ∈ R n . This gives the estimate:
∫
∣ ∣∣∣
|I n | =
∣ dz[f(a) + f ′ (a)(z − a) + r(z, a)(z − a)]
∣ = dz r(z, a)(z − a)
∣
∂R n
∫∂R n
∫
≤
|dz||r(z, a)(z − a)| ≤ ǫ 4L
∂R n
2 n sup |z − a| < 4ǫ L2
z∈∂R n
4 n
(Note that ∫ ∂R
dz(cz + d) = 0 on any rectangle, by use of the primitive). Then
|I| ≤ 4ǫ L 2 for arbitrary ǫ, so it equals zero.
The hypothesis of the theorem can be weakened to allow for functions that
are holomorphic up to a point in the domain (or a finite collection of points).
This will be useful to study Cauchy-type integrals.
Proposition 9.0.9. Let f(z) be a function that is continuous on a domain D
and holomorphic on D/{a}. Then ∫ dζf(ζ) = 0 for any rectangle R in the
∂R
domain.
Proof. If a /∈ R nothing changes in the proof of 9.0.8. If a ∈ R the key ingredients
for a proof are: |f(z)| < M (because f is continuous on a compact
set R) and Theorem 9.0.8. The trick is to decompose R = R a ∪ r 1 ∪ ... ∪ r k
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where R a is a square with side ǫ that contains a (as interior or boundary point
if a ∈ ∂R), and r i are rectangles that fit the partition. The contour integral is
I(R) = I(R a ) + ∑ i I(r i). Darboux’s inequality gives I(R a ) ≤ 4Mǫ, and the
terms I(r i ) = 0, by Theorem 9.0.8. Then I(R) = 0.
Example. [Fourier transform of the Gaussian function]
An important application of the theorem for rectangles is the Fourier transform
of the Gaussian function:
∫ ∞
−∞
dx
√ e −x2 +ikx = √ 1 e −k2 /4 , k ∈ R (9.2)
2π 2
Consider the integral ∫ ∂R dζe−ζ2 on the rectangle with corners ±R, ±R+i(k/2).
The integral is zero because the function is entire, and is the sum of four integrals
evaluated on the sides. For R → ∞ only integrals on [−R, R] and [−R +
i(k/2), R + i(k/2)] survive and give the final result.
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