ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

10 ANALYSIS QUALS
(a) Prove that the linear functional L : f ↦→ f(1) is bounded.
(b) Find the element g ∈ H representing L.
(c) Show that f ↦→ ReL(f) achieves its maximal value on the set
B := {f ∈ H : f ≤ 1 and f(0) = 0} ,
that this maximum occurs at a unique point, and determine this maximal
value.
Solution. (a) Using Cauchy-Schwarz,
∞
|L(f)| = ˆf(k)
k=0
≤
∞
k=0
ˆ
f(k)
∞
1 + |k| 2
=
1 + |k| 2
k=0
ˆ
∞
1
f(k) ≤
1 + |k|
k=0
2
1/2
∞
(1 + |k|
k=0
2
) ˆ
f(k)
∞
1
=
1 + |k| 2
1/2 f .
k=0
2
1/2 (b) Let ˆg(k) = 1
1+|k| 2 , g(z) = ∞ k=0 ˆg(k)zk . Then g(z) converges for all
z ∈ D, and
∞
k=0
so g ∈ H. Then
(1 + |k| 2 )(1 + |k| 2 ) −2 < ∞,
〈f, g〉 =
=
∞
(1 + |k| 2 ) ˆ f(k)ˆg(k)
k=0
∞
k=0
1 + |k| 2
1 + |k| 2 ˆ f(k)
= f(1) = L(f)
(c) B is a compact, convex set in a Hilbert space, so a bounded linear
functional on it must achieve a maximum at a unique point. We claim
the maximizer is f(z) = z √ 2 . This is in B, and the handwavey argument
of why this is the maximizer is that we want all possible weight on the
smallest possible k, since the computation of the norm gives increasing
penalty to larger k. Since f(0) = 0, ˆ f(0) = 0, so we cannot put any
weight on k = 0, and thus we put all possible weight on k = 1.
Problem 7: Suppose that f : C → C is continuous on C and holomorphic on
C − R. Prove that f is entire.
Solution. Let Γ be a triangle in C. If Γ does not cross the real axis, then as f
is holomorphic off R, Cauchy integral theorem guarantees that
f(z)dz =
Γ
0.
Now suppose Γ crosses the real axis. For k > 0, we cut a 1/k neighborhood
out about the real axis and let Γ (1)
k and Γ(2)
k be the resulting polygons above
and below the real axis, respectively, as shown in the figure.