2 Cauchy’s Theorem and Its Applications
Theorem 5.7 Any function holomorphic in a neighborhood of a compact set K can be approximated uniformly on K by rational functions whose
singularities are in Kc.
If Kc is connected, any function holomorphic in a neighborhood of K can be approximated uniformly on K by polynomials.
We shall see how the second part of the theorem follows from the first: when Kc is connected, one can “push” the singularities to infinity thereby transforming the rational functions into polynomials.
The key to the theorem lies in an integral representation formula that is a simple consequence of the Cauchy integral formula applied to a square.
Lemma 5.8 Suppose f is holomorphic in an open set Ω, and K⊂ Ω is compact. Then, there exists finitely many segments γ1, . . . , γN in Ω− K such that
(15) f (z) =
N n=1
1 2πi
γn
f (ζ)
ζ− zdζ for all z∈ K.
Proof. Let d = c· d(K, Ωc), where c is any constant < 1/√ 2, and consider a grid formed by (solid) squares with sides parallel to the axis and of length d.
We let Q = {Q1, . . . , QM} denote the finite collection of squares in this grid that intersect K, with the boundary of each square given the positive orientation. (We denote by ∂Qm the boundary of the square Qm.) Finally, we let γ1, . . . , γN denote the sides of squares in Q that do not belong to two adjacent squares inQ. (See Figure 13.) The choice of d guarantees that for each n, γn⊂ Ω, and γndoes not intersect K; for if it did, then it would belong to two adjacent squares inQ, contradicting our choice of γn.
5These singularities are points where the function is not holomorphic, and are “poles”, as defined in the next chapter.
62 Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS
Figure 13. The union of the γn’s is in bold-face
Since for any z ∈ K that is not on the boundary of a square in Q there exists j so that z∈ Qj, Cauchy’s theorem implies
1 2πi
∂Qm
f (ζ) ζ− zdζ =
f (z) if m = j, 0 if m= j.
Thus, for all such z we have
f (z) =
M m=1
1 2πi
∂Qm
f (ζ) ζ− zdζ.
However, if Qmand Qm are adjacent, the integral over their common side is taken once in each direction, and these cancel. This establishes (15) when z is in K and not on the boundary of a square inQ. Since γn⊂ Kc, continuity guarantees that this relation continues to hold for all z∈ K, as was to be shown.
The first part of Theorem 5.7 is therefore a consequence of the next lemma.
Lemma 5.9 For any line segment γ entirely contained in Ω− K, there exists a sequence of rational functions with singularities on γ that ap-proximate the integral
γf (ζ)/(ζ− z) dζ uniformly on K.
Proof. If γ(t) : [0, 1]→ C is a parametrization for γ, then
γ
f (ζ) ζ− zdζ =
1 0
f (γ(t))
γ(t)− zγ(t) dt.
5. Further applications 63 Since γ does not intersect K, the integrand F (z, t) in this last integral is jointly continuous on K× [0, 1], and since K is compact, given > 0, there exists δ > 0 such that
sup
z∈K |F (z, t1)− F (z, t2)| < whenever|t1− t2| < δ.
Arguing as in the proof of Theorem 5.4, we see that the Riemann sums of the integral 1
0 F (z, t) dt approximate it uniformly on K. Since each of these Riemann sums is a rational function with singularities on γ, the lemma is proved.
Finally, the process of pushing the poles to infinity is accomplished by using the fact that Kc is connected. Since any rational function whose only singularity is at the point z0 is a polynomial in 1/(z− z0), it suffices to establish the next lemma to complete the proof of Theorem 5.7.
Lemma 5.10 If Kc is connected and z0∈ K, then the function/ 1/(z− z0) can be approximated uniformly on K by polynomials.
Proof. First, we choose a point z1 that is outside a large open disc D centered at the origin and which contains K. Then
1
where the series converges uniformly for z∈ K. The partial sums of this series are polynomials that provide a uniform approximation to 1/(z− z1) on K. In particular, this implies that any power 1/(z− z1)k can also be approximated uniformly on K by polynomials.
It now suffices to prove that 1/(z− z0) can be approximated uniformly on K by polynomials in 1/(z− z1). To do so, we use the fact that Kcis
64 Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS
and since the sum converges uniformly for z∈ K, the approximation by partial sums proves our claim.
This result allows us to travel from z0to z1through the finite sequence {wj} to find that 1/(z − z0) can be approximated uniformly on K by polynomials in 1/(z− z1). This concludes the proof of the lemma, and also that of the theorem.
6 Exercises
1. Prove that
∞
0
sin(x2) dx =
∞
0
cos(x2) dx =
√2π 4 . These are the Fresnel integrals. Here,∞
0 is interpreted as limR→∞R
0 . [Hint: Integrate the function e−z2 over the path in Figure 14. Recall that
∞
−∞e−x2dx =√ π.]
R Reiπ4
0
Figure 14. The contour in Exercise 1
2. Show that
∞
0
sin x x dx = π
2. [Hint: The integral equals 2i1 ∞
−∞eix−1
x dx. Use the indented semicircle.]
3. Evaluate the integrals
∞
0
e−axcos bx dx and
∞
0
e−axsin bx dx , a > 0
by integrating e−Az, A =√
a2+ b2, over an appropriate sector with angle ω, with cos ω = a/A.
6. Exercises 65 4. Prove that for all ξ∈ C we have e−πξ2=
∞
−∞
e−πx2e2πixξdx.
5. Suppose f is continuously complex differentiable on Ω, and T ⊂ Ω is a triangle whose interior is also contained in Ω. Apply Green’s theorem to show that
T
f (z) dz = 0.
This provides a proof of Goursat’s theorem under the additional assumption that f is continuous.
[Hint: Green’s theorem says that if (F, G) is a continuously differentiable vector field, then
T
F dx + G dy =
Interior of T
∂G
∂x −∂F
∂y
dxdy.
For appropriate F and G, one can then use the Cauchy-Riemann equations.]
6. Let Ω be an open subset ofC and let T ⊂ Ω be a triangle whose interior is also contained in Ω. Suppose that f is a function holomorphic in Ω except possibly at a point w inside T . Prove that if f is bounded near w, then
T
f (z) dz = 0.
7. Suppose f :D → C is holomorphic. Show that the diameter d = supz, w∈D|f(z) − f(w)| of the image of f satisfies
2|f(0)| ≤ d.
Moreover, it can be shown that equality holds precisely when f is linear, f (z) = a0+ a1z.
Note. In connection with this result, see the relationship between the diameter of a curve and Fourier series described in Problem 1, Chapter 4, Book I.
[Hint: 2f(0) =2πi1
|ζ|=r f (ζ)−f (−ζ)
ζ2 dζ whenever 0 < r < 1.]
8. If f is a holomorphic function on the strip−1 < y < 1, x ∈ R with
|f(z)| ≤ A(1 + |z|)η, η a fixed real number
for all z in that strip, show that for each integer n≥ 0 there exists An≥ 0 so that
|f(n)(x)| ≤ An(1 +|x|)η, for all x∈ R.
[Hint: Use the Cauchy inequalities.]
66 Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS 9. Let Ω be a bounded open subset ofC, and ϕ : Ω → Ω a holomorphic function.
Prove that if there exists a point z0∈ Ω such that ϕ(z0) = z0 and ϕ(z0) = 1 then ϕ is linear.
[Hint: Why can one assume that z0= 0? Write ϕ(z) = z + anzn+ O(zn+1) near 0, and prove that if ϕk= ϕ◦ · · · ◦ ϕ (where ϕ appears k times), then ϕk(z) = z + kanzn+ O(zn+1). Apply the Cauchy inequalities and let k→ ∞ to conclude the proof. Here we use the standard O notation, where f (z) = O(g(z)) as z→ 0 means that|f(z)| ≤ C|g(z)| for some constant C as |z| → 0.]
10. Weierstrass’s theorem states that a continuous function on [0, 1] can be uni-formly approximated by polynomials. Can every continuous function on the closed unit disc be approximated uniformly by polynomials in the variable z?
11. Let f be a holomorphic function on the disc DR0 centered at the origin and of radius R0.
(a) Prove that whenever 0 < R < R0 and|z| < R, then
f (z) = 1 2π
2π
0
f (Reiϕ)Re
Reiϕ+ z Reiϕ− z
dϕ.
(b) Show that
Re
Reiγ+ r Reiγ− r
= R2− r2 R2− 2Rr cos γ + r2.
[Hint: For the first part, note that if w = R2/z, then the integral of f (ζ)/(ζ− w) around the circle of radius R centered at the origin is zero. Use this, together with the usual Cauchy integral formula, to deduce the desired identity.]
12. Let u be a real-valued function defined on the unit discD. Suppose that u is twice continuously differentiable and harmonic, that is,
u(x, y) = 0 for all (x, y)∈ D.
(a) Prove that there exists a holomorphic function f on the unit disc such that Re(f ) = u.
Also show that the imaginary part of f is uniquely defined up to an additive (real) constant. [Hint: From the previous chapter we would have f(z) = 2∂u/∂z. Therefore, let g(z) = 2∂u/∂z and prove that g is holomorphic.
Why can one find F with F= g? Prove that Re(F ) differs from u by a real constant.]
7. Problems 67 (b) Deduce from this result, and from Exercise 11, the Poisson integral repre-sentation formula from the Cauchy integral formula: If u is harmonic in the unit disc and continuous on its closure, then if z = reiθ one has
u(z) = 1 2π
2π
0
Pr(θ− ϕ)u(ϕ) dϕ
where Pr(γ) is the Poisson kernel for the unit disc given by
Pr(γ) = 1− r2 1− 2r cos γ + r2.
13. Suppose f is an analytic function defined everywhere inC and such that for each z0∈ C at least one coefficient in the expansion
f (z) =
∞ n=0
cn(z− z0)n
is equal to 0. Prove that f is a polynomial.
[Hint: Use the fact that cnn! = f(n)(z0) and use a countability argument.]
14. Suppose that f is holomorphic in an open set containing the closed unit disc, except for a pole at z0 on the unit circle. Show that if
∞ n=0
anzn
denotes the power series expansion of f in the open unit disc, then
n→∞lim an an+1 = z0.
15. Suppose f is a non-vanishing continuous function onD that is holomorphic in D. Prove that if
|f(z)| = 1 whenever|z| = 1, then f is constant.
[Hint: Extend f to all ofC by f(z) = 1/f(1/z) whenever |z| > 1, and argue as in the Schwarz reflection principle.]
7 Problems
1. Here are some examples of analytic functions on the unit disc that cannot be extended analytically past the unit circle. The following definition is needed. Let
68 Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS f be a function defined in the unit discD, with boundary circle C. A point w on C is said to be regular for f if there is an open neighborhood U of w and an analytic function g on U , so that f = g on D ∩ U. A function f defined on D cannot be continued analytically past the unit circle if no point of C is regular for f .
(a) Let
f (z) =
∞ n=0
z2n for|z| < 1.
Notice that the radius of convergence of the above series is 1. Show that f cannot be continued analytically past the unit disc. [Hint: Suppose θ = 2πp/2k, where p and k are positive integers. Let z = reiθ; then
|f(reiθ)| → ∞ as r → 1.]
(b)∗Fix 0 < α <∞. Show that the analytic function f defined by
f (z) =
∞ n=0
2−nαz2n for|z| < 1
extends continuously to the unit circle, but cannot be analytically continued past the unit circle. [Hint: There is a nowhere differentiable function lurking in the background. See Chapter 4 in Book I.]
2.∗Let
F (z) =
∞ n=1
d(n)zn for|z| < 1
where d(n) denotes the number of divisors of n. Observe that the radius of con-vergence of this series is 1. Verify the identity
∞ n=1
d(n)zn=
∞ n=1
zn 1− zn. Using this identity, show that if z = r with 0 < r < 1, then
|F (r)| ≥ c 1
1− rlog(1/(1− r))
as r→ 1. Similarly, if θ = 2πp/q where p and q are positive integers and z = reiθ, then
|F (reiθ)| ≥ cp/q
1
1− rlog(1/(1− r))
as r→ 1. Conclude that F cannot be continued analytically past the unit disc.
3. Morera’s theorem states that if f is continuous inC, and
Tf (z) dz = 0 for all triangles T , then f is holomorphic inC. Naturally, we may ask if the conclusion still holds if we replace triangles by other sets.
7. Problems 69 (a) Suppose that f is continuous onC, and
(16)
C
f (z) dz = 0
for every circle C. Prove that f is holomorphic.
(b) More generally, let Γ be any toy contour, andF the collection of all trans-lates and ditrans-lates of Γ. Show that if f is continuous onC, and
γ
f (z) dz = 0 for all γ∈ F
then f is holomorphic. In particular, Morera’s theorem holds under the weaker assumption that
Tf (z) dz = 0 for all equilateral triangles.
[Hint: As a first step, assume that f is twice real differentiable, and write f (z) = f (z0) + a(z− z0) + b(z− z0) + O(|z − z0|2) for z near z0. Integrating this expan-sion over small circles around z0yields ∂f /∂z = b = 0 at z0. Alternatively, suppose only that f is differentiable and apply Green’s theorem to conclude that the real and imaginary parts of f satisfy the Cauchy-Riemann equations.
In general, let ϕ(w) = ϕ(x, y) (when w = x + iy) denote a smooth function with 0≤ ϕ(w) ≤ 1, and
R2ϕ(w) dV (w) = 1, where dV (w) = dxdy, and
denotes the usual integral of a function of two variables in R2. For each > 0, let ϕ(z) = −2ϕ( −1z), as well as
f(z) =
R2
f (z− w)ϕ(w) dV (w),
where the integral denotes the usual integral of functions of two variables, with dV (w) the area element of R2. Then f is smooth, satisfies condition (16), and f→ f uniformly on any compact subset of C.]
4. Prove the converse to Runge’s theorem: if K is a compact set whose complement if not connected, then there exists a function f holomorphic in a neighborhood of K which cannot be approximated uniformly by polynomial on K.
[Hint: Pick a point z0 in a bounded component of Kc, and let f (z) = 1/(z− z0).
If f can be approximated uniformly by polynomials on K, show that there exists a polynomial p such that|(z − z0)p(z)− 1| < 1. Use the maximum modulus principle (Chapter 3) to show that this inequality continues to hold for all z in the component of Kc that contains z0.]
5.∗There exists an entire function F with the following “universal” property: given any entire function h, there is an increasing sequence{Nk}∞k=1of positive integers, so that
n→∞lim F (z + Nk) = h(z) uniformly on every compact subset ofC.
70 Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS (a) Let p1, p2, . . . denote an enumeration of the collection of polynomials whose coefficients have rational real and imaginary parts. Show that it suffices to find an entire function F and an increasing sequence{Mn} of positive integers, such that
(17) |F (z) − pn(z− Mn)| < 1
n whenever z∈ Dn,
where Dn denotes the disc centered at Mn and of radius n. [Hint: Given h entire, there exists a sequence{nk} such that limk→∞pnk(z) = h(z) uni-formly on every compact subset ofC.]
(b) Construct F satisfying (17) as an infinite series
F (z) =
∞ n=1
un(z)
where un(z) = pn(z− Mn)e−cn(z−Mn)2, and the quantities cn> 0 and Mn>
0 are chosen appropriately with cn→ 0 and Mn→ ∞. [Hint: The function e−z2 vanishes rapidly as |z| → ∞ in the sectors {| arg z| < π/4 − δ} and {|π − arg z| < π/4 − δ}.]
In the same spirit, there exists an alternate “universal” entire function G with the following property: given any entire function h, there is an increasing sequence {Nk}∞k=1 of positive integers, so that
k→∞lim DNkG(z) = h(z)
uniformly on every compact subset of C. Here DjG denotes the jth (complex) derivative of G.