WA 1: Solutions to some problems
1. Problem 1.1
Solution. Let s=x+iy. Since
n−s=e−sln n=e−xln ne−iy ln n=n−xe−iy ln n
and |e−iy ln n|= 1, we have (n−s)∞
n=1 ∈`2if and only if
∞
X
n=1
|n−s|2=
∞
X
n=1
(n−x)2=
∞
X
n=1
1
n2x<∞.
Applying the integral test, we see that R∞
1
dt
t2x<∞if and only if 2x > 1, thus, the
series P∞
n=1
1
n2xconverges if and only if Res >1
2.
2. Problem 1.2
Solution. The only difficult part was to check the property hf, fi>0 for f6= 0.
We have
hf, f i=Zb
a
(|f(t)|2+|f0(t)|2)dt ≥Zb
a
|f(t)|2dt.
Since f6= 0, there is a point t0∈[a, b] such that f(t0)6= 0. Since fis continuous
on [a, b] (and thus, continuous at t0), there is a closed interval I⊆[a, b] containing
t0such that f(t)6= 0 for every point t∈I. Let m= mint∈I|f(t)|2. Since |f|2is
continuous, it attains its minimum on the closed interval I, and therefore, m > 0.
Therefore,
hf, f i ≥ Zb
a
|f(t)|2dt ≥ZI
|f(t)|2dt ≥m`(I)>0,
where `(I) denotes the length of I.
7. (an extra credit problem) Problem 1.13
Solution. We have
kgk2=1
2πi Z∂D
1
(z−α)(z−β)(z−α)(z−β)
dz
z1/2
=1
2πi Z∂D
z dz
(z−α)(z−β)(1 −αz)(1 −βz )1/2
,
where we used the identity zz = 1 for z∈∂D. Since the function
f(z) = z
(1 −αz)(1 −βz )
is analytic on clos D, the fraction
z
(z−α)(z−β)(1 −αz)(1 −βz )=f(z)
(z−α)(z−β)
1
