ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

24 ANALYSIS QUALS
Solution. Let E1 = {(x, t) : x ∈ X, t ∈ R, f(x) > t and g(x) < t}.
Let E2 = {(x, t) : x ∈ X, t ∈ R, f(x) < t and g(x) > t}. Then
|f − g|dµ = |f − g|dµ + |f − g|dµ
=
{f>g}
{f>g}
f − gdµ +
{ff}
g − fdµ
∞
=
{f>g}
∞
χE1(x, t)dt
−∞
= (χE1 + χE2 )(x, t)dtdµ(x),
−∞
∞
dµ(x) +
{g>f}
χE2(x, t)dt
−∞
dµ(x)
since E1 and E2 are disjoint, and χE1 vanishes off {f > g} in X, and χE2
vanishes off {g > f} in X. Then by Fubini, we have that
∞
∞
(χE1 + χE2 )(x, t)dt, dµ(x) =
−∞
(χE1 + χE2 )(x, t)dµ(x)dt.
But
χE1dµ = µ(Ft\Gt) and
−∞
χE2dµ = µ(Gt\Ft),
and as these sets are disjoint, µ((Ft\Gt)∪(Gt\Ft)) = µ(Ft\Gt)+µ(Gt\Ft).
Thus,
∞
∞
(χE1 + χE2)(x, t)dµ(x)dt = µ((Ft\Gt) ∪ (Gt\Ft))dt.
−∞
Problem 2: Let H be an infinite dimensional real Hilbert space.
a) Prove the unit sphere S of H is weakly dense in the unit ball B of H.
b) Prove there is a sequence Tn of bounded linear operators from H to
H such that Tn = 1 for all n but lim Tn(x) = 0 for all x ∈ H.
Proof. a) Let x0 ∈ B, ε > 0. Let Uε(x0) be an ε-neighborhood of x0 in
the weak topology. That is,
Uε(x0) =
−∞
x ∈ H : sup |〈x − x0, yi〉| < ε for k ∈ N, yi ∈ H
i=1,...,k
Let fi(x) = 〈x, yi〉. Then fi is a linear functional. As H is infinite
dimensional, the intersection of the kernels of a finite number of linear
functionals is a nontrivial linear subspace. Thus,
x0 +
k
ker(fi)
i=1
is contained in Uε(x0), so there is a line L through x0 contained in
Uε(x0). This line intersects S, so choose xε ∈ L ∩ S. Then
〈x0, y〉 = lim
ε→0 〈xε, y〉 for all y ∈ H.
Thus, S is weakly dense in B.
.