Let
H be a Hilbert space. It is well-known that if
Z is a convex and closed subset of
H, then for every
x∈H, there is a unique element
z∈Z such that
‖x−z‖=d(x,Z). Let's denote this unique element
z by
PZx.
Problem. Let Z1⊇Z2⊇Z3⊇⋯ be a chain of closed convex subsets of a Hilbert space H, show that:
(a) If ⋂∞n=1Zn≠∅, then ‖x−PZnx‖→‖x−PZx‖.
(b) If ⋂∞n=1Zn=∅, then ‖x−PZnx‖→∞.
Since
Zn's are not subspace of
H,
PZn itself is not a linear map, and therefore standard tools from functional analysis cannot be used. Nevertheless, by assuming
Zn's are further a closed subspace, it is instructive to see a weak-convergence argument to conclude the result in (a) --- a slightly complicated solution.
Some also asked me can
⋂∞n=1Zn=∅ happen? Below is an example to show there is a chain of closed convex subsets with empty intersection:
Example. Let
Zn={(1,1√2,…,1√n,a1,a2,…):(a1,a2,…)∈ℓ2}, then
Zn is convex, a closed subset of
ℓ2, and descending in
n. However, if there is
(x1,x2,…)∈⋂∞n=1Zn, then necessarily
xn=1/√n for every
n, and this element is not in
ℓ2, a contradiction. So
⋂∞n=1Zn=∅.
Next what is
d(x,Zn) in this particular example? By definition for
z∈Zn we have
‖x−z‖2=n∑k=1|xk−1√k|2+∑k>n|xk−zk|2, hence for fixed
x∈ℓ2 the smallest possible
‖x−z‖ is
d(x,Zn)=√n∑k=1|xk−1√k|2, of course this diverges to infinity as
n→∞ for any fixed
x∈ℓ2. What's more,
PZnx=(1,1√2,…,1√n,xn+1,xn+2,…).◼