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Jeffrey Hellrung
Thursday, May 25, 2006
Math 246A, Homework 07
- Let Ω be open and connected, and let fn : Ω → R 2 be a sequence of holomorphic functions which
converges uniformly on compact sets to f. Prove that if none of the fn has a zero, then f has no zeros
or is constant equal to 0.
Solution
Suppose f 6 ≡ 0. Then the zeros of f are isolated, and for any z 0 ∈ C, there exists a circle C centered
at z 0 such that f 6 = 0 on C. In particular, |f | has a positive minimum on C, and hence 1/fn converges
uniformly to 1/f on C. Also, f
′ n converges uniformly to^ f^
′ on C, hence
lim n→∞
C
f
′ n (z)
fn(z)
dz =
C
f
′ (z)
f (z)
dz.
But each integral on the right is zero, as it counts the number of zeros of fn inside C (Rouch´e’s
Theorem). Thus the integral on the right is also zero, so by the same interpretation, f (z 0 ) 6 = 0. Since
z 0 was arbitrary, it follows that f is everywhere nonzero.
- The reciprocal 1/Γ of the Gamma function is a holomorphic function having zeros at the negative
integers. The purpose of this exercise is to introduce an alternative formula for this holomorphic
function using a Weierstrass product.
(a) Prove that
γ = lim n→∞
n ∑
k=
k
− log n
exists.
(b) Prove that the infinite product ∞ ∏
n=
z
n
e
− z n
converges uniformly on compact sets on the entire complex plane.
(c) Use the unique characterization of the Gamma function in the lecture to prove that, with γ as
above, we have
1
Γ(z)
= ze
γz
∞ ∏
n=
z
n
e
− z n (^).
Solution
(a) By considering Riemann sums, since 1/x decreases for x > 0,
n ∑
k=
k
∫ (^) n+
1
x
dx > log n,
so n ∑
k=
k
− log n > 0
for all n; i.e., the sequence is bounded from below. Further, by the Mean Value Theorem,
− log
k
xk
k
k
for some xk ∈ [1 − 1 /k, 1], so
1
k
k
and we can express
n ∑
k=
k
− log n = 1 +
n ∑
k=
k
− log k + log(k − 1) = 1 +
n ∑
k=
k
k
and so the sequence is monotonically decreasing. It follows that the sequence has a limit.
(b) We know that
∞ ∏
n=
z
n
e
− z n
converges together with ∞ ∑
n=
log
z
n
z
n
Now for a fixed R > 0, we need only consider convergence of those terms corresponding to n > R.
For these terms, we take the principal branch of log above, so that for |z| ≤ R, log(1 − z/n) may
be expanded in a Taylor series:
log
z
n
z
n
z
n
z
n
so that
log
z
n
z
n
z
n
z
n
and hence ∣ ∣ ∣log
z
n
z
n
1 2
R n
R n
which is certainly summable (1 −
R n → 1 > 0 as n → ∞), hence the original infinite product
uniformly converges on compact sets.
(c) Denote
f (z) = ze
γz
∞ ∏
n=
z
n
e
− z n (^).
We first compute
f (z + 1)
f (z)
(z + 1)e
γ(z+1)
n=
z+ n
e
− z+ n
ze γz
∞ n=
z n
e
− z n
z + 1
z
e
γ
n=
z+ n
e
− z+ n ∏ ∞ n=
z n
e
− z n
To simplify the quotient of infinite products, we note that
∞ ∏
n=
z + 1
n
e
− z+ n (^) =
∞ ∏
n=
n
z
n + 1
e
− z+ n
and ∞ ∏
n=
z
n
e
− z n (^) =
∞ ∏
n=
z
n + 1
e
− z n+
so
∏∞
n=
z+ n
e
− z+ n ∏ ∞ n=
z n
e
− z n
e
z
1 + z
∞ ∏
n=
n
e
z n+ − z+ n (^) =
e
z
1 + z
∞ ∏
n=
n
e
−z( 1 n − 1 n+1 ) − 1 n (^).
Solution
(a) We first notice that Γ(z) = Γ(z) by considering, e.g.,
Γ(z) =
e −γz
z
∞ ∏
n=
z
n
e
z/n .
It follows that ∣ ∣Γ
1 2
2 = Γ
1 2
1 2
1 2
1 2
− iy
1 2
1 2
π sin(π( 1 2 +iy))
and
sin
π
1 2
1 2 i
e
iπ( 1 2 +iy)^ − e−iπ(^
1 2 +iy)
1 2 i (ie
−πy
πy )
= cosh(πy)
so (^) ∣ ∣ ∣ ∣
π
cosh(πy)
(b) We first show the claim for the specific case x = 0. Indeed,
|Γ(iy)| = |y|
− 1
n=
∣1 + i y n
− 1
= |y|
− 1
∏N
n=
∣1 + i y n
− 1
n=N +
∣1 + i y n
− 1
≤ |y|
− 1
∏N
n=
n |y|
∞ n=N +
= N !|y|
−N − 1
for any N ≥ 0, which is precisely the claim. We then get the claim whenever x ∈ Z−, as
Γ(x + iy) =
Γ(x + 1 + iy)
x + iy
Γ(iy)
(x + iy)(x + 1 + iy) · · · (1 + iy)
so
|Γ(x + iy)| ≤ N !|y|
−N − 1
for any N ≥ 0. This takes care of those points whose real part is a pole of Γ.
We now demonstrate the case N = 1 with x > 0:
|Γ(z)| =
∣z
− 1 e −γz
n= (1 + z/n)
− 1 e z/n
≤ |y| − 1 e −γx
n= (1 + x/n)
− 1 e x/n
= xΓ(x)|y| − 1
= Γ(x + 1)|y| − 1
For the case x < 0, let k = ⌊−x + 1⌋. Then 0 < x + k ≤ 1, and we can write
Γ(x + iy) =
Γ(x + 1 + iy)
x + iy
Γ(x + k + iy)
(x + iy)(x + 1 + iy) · · · (x + k − 1 + iy)
so
|Γ(x + iy)| ≤
Γ(x + k + 1)
|x(x + 1) · · · (x + k − 1)|
|y|
− 1 .
The general case then follows by induction, as
|Γ(x + iy)| =
Γ(x + 1 + iy)
x + iy
≤ Cx+1,N |y|
−(N +1) .
(c) We prove the claim by induction. Certainly the claim for n = 1 is true:
e
≤ 1 ≤ 2 π
e
We now suppose the claim for n, i.e.,
n
e
)n
≤ Γ(n + 1) ≤ 2 πn
n
e
)n
;
and aim to show the claim for n + 1. We first have that
n + 1
e
)n+
n
e
)n
n + 1
n
)n n + 1
e
≤ Γ(n + 1)
n + 1
n
)n n + 1
e
by assumption. Now (n + 1)Γ(n + 1) = Γ(n + 2), and by the Mean Value Theorem,
log
n
xn
n
for some xn ∈ [1, 1 + 1/n], so
log
n
n
n
)n
≤ e,
which gives us the desired inequality:
n + 1
e
)n+
≤ Γ(n + 2)
n
)n 1
e
≤ Γ(n + 2).
The second inequality is similar. We have
2 π(n + 1)
n + 1
e
)n+
= 2πn
n
e
)n
n + 1
n
)n+ n + 1
e
≤ Γ(n + 1)
n + 1
n
)n+ n + 1
e
by assumption. Using the same application of the Mean Value Theorem,
log
n
xn
n
n
n + 1
n
n + 1
n
)n+
≥ e,
which again gives us the desired inequality:
2 π(n + 1)
n + 1
e
)n+
≥ Γ(n + 2)
n
)n+ 1
e
≥ Γ(n + 2).
The claim is thus proved by induction.
(d) Let
f (n) = n
n e
−n
−∞
e
−n(e y − 1 −y) dy.
As suggested by the hint, we first show that f (n) = Γ(n). We can rewrite f as
f (n) = n
n
−∞
e
−n(e y −y) dy,
and then use a change of variables t = ne y to get
f (n) =
∞
0
t
n− 1 e
−t dt,
For y ∈ [− 1 , −x], by Taylor’s Theorem,
e
y − 1 − y ≥
y
2
y
3 e
−x ≥
y
2 −
y
2
y
2 ,
so
√ n
−x
− 1
e
−n(e y − 1 −y) dy ≤
n
−x
− 1
e
− n 3 y
2 dy,
and a change of variables to u =
n 3
y gives
n
−x
− 1
e
− n 3 y
2 dy =
−
1 3 n
α+1/ 2
−
1 3 n
1 / 2
e
−u 2 du → 0
since n 1 / 2 , n α+1/ 2 → ∞ as n → ∞, by choice of α.
For y ∈ [−x, 0], by Taylor’s Theorem,
e
y − 1 − y ≥
y
2
y
3 ≥
y
2 −
n
3 α ,
so
√ n
−x
e
−n(e y − 1 −y) dy ≤
n
−x
e
− n 2 y 2
1 6 e x n 3 α+1 √ n
−x
e
− n 2 y 2 dy,
and, again, a change of variables to u =
n 2 y gives
e
1 6 e x n 3 α+1 √ n
−x
e
− n 2 y 2 dy = e
1 6 e x n 3 α+1 √ 2
−
1 2 nα+1/^2
e
−u 2 du →
−∞
e
−u 2 du
by the same reasoning as before.
Lastly, for y ≥ 0, e
y − 1 − y ≥
1 2 y
2 , so
n
0
e
−n(e y − 1 −y) dy ≤
n
0
e
− n 2 y 2 dy →
0
e
−u 2 du
as before. It follows that, for any ǫ > 0,
n
−∞
e
−n(e y − 1 −y) dy ≤ (1 + ǫ)
−∞
e
−u 2 du = (1 + ǫ)
2 π
for large enough n, showing this is indeed an asymptotic upper bound, and hence we can say that
n
−n e
n
nΓ(n) =
n
−∞
e
−n(e y − 1 −y) dy →
2 π,
i.e.,
Γ(n) =
2 π
n
n
e
)n
(1 + cn)
where cn → 0. Multiplication by n gives the desired formula:
Γ(n + 1) = nΓ(n) =
2 πn
n
e
)n
(1 + cn).
- Given an entire function f we shall consider the problem of finding a (natural) holomorphic function
F (z) which satisfies F (n) = f
(n) (0).
Let f denote an entire function which satisfies
|f (x + iy)| ≤ Cn(1 + |x|)
n
for all integers n, all x ∈ R, and all |y| < 2 (for example, P (z)e
−z 2 for any polynomial P is such a
function). Consider the holomorphic branch of the logarithm defined on C{t | t ≤ 0 } which satisfies
log(1) = 0 and define t
z = e
z log t
. Consider
F+(f, z) =
−∞
(x + i)
z f (x + i)dx;
F−(f, z) =
−∞
(x − i)
z f (x − i)dx;
F (z) =
Γ(z + 1)
2 πi
(F−(f, − 1 − z) − F+(f, − 1 − z)).
Prove
(a) F is a holomorphic function; and
(b) F (n) = f
(n) (0) for all positive integers n.
Solution
(a) Note that the holomorphicity of F+ and F− will imply the holomorphicity of F , so consider first
the difference quotient for F+:
h
(F+(f, z + h) − F+(f, z)) =
−∞
h
(x + i)
h − 1
(x + i)
z f (x + i)dx.
We have
(x + i)
h − 1 = e
h log(x+i) − 1 = h log(x + i)
(h log(x + i)) +
(h log(x + i))
2
so ∣ ∣(x + i)h^ − 1
∣ (^) ≤ |h|| log(x + i)|e|h||^ log(x+i)|.
Now
| log(x + i)| = |log |x + i| + i arg(x + i)| ≤ log |x + 1| + π,
so (^) ∣ ∣ (x + i)
h − 1
≤ |h| (log |x + 1| + π) |x + 1|
|h| e
π|h|
≤ |h| (log |x + 1| + π) |x + 1|e
π
for |h| < 1. Also,
|(x + i)
z | =
∣e
z log(x+i)
∣ ≤ e
|z|(log |x+1|+π) = e
π|z| |x + 1|
|z| .
It follows that
∣ ∣ ∣ ∣
h
(F+(z + h) − F+(z))
≤ e
π(|z|+1)
−∞
(log |x + 1| + π) |x + 1|
|z|+ |f (x + i)|dx,
Solution
(a) For simplicity, redefine γ to begin instead at 1 + r, traveling counterclockwise about 1. Let
γ 1 ,... , γ 4 be a partition of γ achieved by breaking γ whenever γ or 1 − γ crosses, alternately, the
negative and positive real axis. Specifically,
- γ 1 is the path from 1 + r to 1 − r, traveling counter-clockwise about 1; then the straight-line
path from 1 − r to r; and finally the path from r to −r, traveling counter-clockwise about 0.
- γ 2 is the path −r to r, traveling counter-clockwise about 0; then the straight-line path from
r to 1 − r; and finally the path from 1 − r to 1 + r, traveling clockwise about 1.
- γ 3 is the path from 1 + r to 1 − r, traveling clockwise about 1; then the straight-line path
from 1 − r to r; and finally the path from r to −r, traveling clockwise about 0.
- γ 4 is the path from −r to r, traveling clockwise about 0; then the straight-line path from r
to 1 − r; and finally the path from 1 − r to 1 + r, traveling counter-clockwise about 1.
(b) Let ak, bk be the starting and ending points, respectively, of γk. By assumption, then, fk(bk) =
fk+1(ak+1), so 4 ∏
k=
fk(bk)
fk(ak)
f 4 (b 4 )
f 1 (a 1 )
hence we aim to show this is equal to 1. If we let log k denote the branch of log inducing fk, then
we have
fk(s) = e
(x−1) logk s e
(y−1) logk (1−s) ,
so 4 ∏
k=
fk(bk)
fk(ak)
= e
(x−1)
P 4 k= logk
bk ak e
(y−1)
P 4 k= logk
1 −bk 1 −ak .
By considering each γk individually, we have
log 1
b 1
a 1
= log
r
1 + r
1 − b 1
1 − a 1
= log
1 + r
r
log 2
b 2
a 2
= log
1 + r
r
1 − b 2
1 − a 2
= log
r
1 + r
log 3
b 3
a 3
= log
r
1 + r
− iπ, log 3
1 − b 3
1 − a 3
= log
1 + r
r
− iπ;
log 4
b 4
a 4
= log
1 + r
r
− iπ, log 4
1 − b 4
1 − a 4
= log
r
1 + r
− iπ.
Thus each sum of logarithms cancels to 0, and we indeed have that f 4 (b 4 ) = f 1 (a 1 ).
(c) Restricting our attention to s ∈ [r, 1 − r], we can state the following equalities:
log 2 s = log 1 s + 2πi, log 2 (1 − s) = log 1 (1 − s);
log 3 s = log 1 s + 2πi, log 3 (1 − s) = log 1 (1 − s) − 2 πi;
log 4 s = log 1 s, log 4 (1 − s) = log 1 (1 − s) − 2 πi.
The differences in 2πi’s arise whenever s or 1 − s has circled the origin between the portion of
γk on the real axis and the portion of γk+1 on the real axis. Using the above equalities, we can
write, again for s ∈ [r, 1 − r],
f 1 (s) = e (x−1) log 1 s e (y−1) log 1 (1−s) ;
f 2 (s) = e
(x−1) log 1 s e
(y−1) log 1 (1−s) e
2 πi(x−1) = f 1 (s)e
2 πix ;
f 3 (s) = e
(x−1) log 1 s e
(y−1) log 1 (1−s) e
2 πi(x−1) e
− 2 πi(y−1) = f 1 (s)e
2 πi(x−y) ;
f 4 (s) = e
(x−1) log 1 s e
(y−1) log 1 (1−s) e
− 2 πi(y−1) = f 1 (s)e
− 2 πiy ;
and so, as r ↓ 0,
I →
0 (−f 1 (s) + f 2 (s) − f 3 (s) + f 4 (s)) ds
−1 + e
2 πix − e
2 πi(x−y)
− 2 πiy
1 0 f 1 (s)ds
1 − e 2 πix
1 − e − 2 πiy
1 0
s x− 1 (1 − s) y− 1 ds
= 4 e
πi(x−y) sin(πx) sin(πy)
0 s
x− 1 (1 − s)
y− 1 ds
so
B(x, y) =
0
s
x− 1 (1 − s)
y− 1 ds = lim r↓ 0
Ie
πi(y−x)
4 sin(πx) sin(πy)
- Prove that for any of the following complex manifolds all holomorphic bijections to itself (meromorphic
in case of S) are given by linear fractional transformations. (Determining the exact transformations
doing this in each case is another exercise.)
(a) The Riemann sphere.
(b) The open unit disc. (Hint: Given an automorphism f , prove that |f (z)| tends to 1 as |z| tends
to 1. Apply a reflection principle to the (locally) harmonic function log |f (z)| to extend to an
automorphism of the full Riemann sphere.)
(c) The punctured complex plane C{ 0 }.
(d) The upper half-plane {ℑ(z) > 0 }.
(e) The complex plane.
Solution
(a)
(b)
(c)
(d)
(e)
