23.3 Penalizaton And Projection Operators
In this section, X will be a reflexive Banach space such that X, X′ has a strictly
convex norm. Let K be a closed convex set in X. Then the following lemma is
Lemma 23.3.1 Let K be closed and convex nonempty subset of X a reflexive Banach
space which has strictly convex norm. Then there exists a projection map P such that
Px ∈ K and for all y ∈ K,
be a minimizing sequence for
y ∈ K
is bounded. Hence there is a subsequence, still denoted by
that yn → w ∈ K.
How many closest points to x are there? Suppose w1 is another one. Then
contradicting the assumption that both w,w1 are closest points to x. Therefore, Px
consists of a single point. ■
Denote by F the duality map such that
. This is described earlier but
there is also a very nice treatment which is somewhat different in [?]
. Everything can be
generalized and is in [?]
but here I will only consider this case. First here is a useful
Proposition 23.3.2 Let F be the duality map just described. Let ϕ
Proof: This follows from
Next is a really nice result about the characterization of Px in terms of
Proposition 23.3.3 Let K be a nonempty closed convex set in X a reflexive Banach
space in which both X,X′ have strictly convex norms. Then w ∈ K is equal to Px if and
for every y ∈ K.
Proof: First suppose the condition. Then for y ∈ K, it follows from the above
proposition about the subgradient,
and so since this holds for all y it follows that
for all y which says that w = Px.
Next, using the subgradient idea again, for θ ∈
then for y ∈ K
Now divide by θ and let θ ↓ 0 and use the hemicontinuity of F given above.
Definition 23.3.4 An operator of penalization is an operator f : X → X′ such
that f = 0 on K, f is monotone and nonzero off K as well as demicontinuous.
(Strong convergence goes to weak convergence.) Actually, in applications, it is
usually easy to give an ad hoc description of an appropriate penalization operator.
Proposition 23.3.5 Let K be a closed convex nonempty subset of X a reflexive Banach
space such that X,X′ have strictly convex norms. Then
is an operator of penalization. Here P is the projection onto K. This operator of
penalization is demicontinuous.
Proof: First, observe that f
is 0 on
and nonzero off K
. Why is it
The first term is ≥
0 because F
is monotone. As to the second, it equals
and both of these are ≥ 0 because of Proposition 23.3.3 which characterizes the
Now why is this hemicontinuous? Let xn → x.Then Pxn is clearly bounded.
Taking a subsequence, it can be assumed that Pxn → ξ weakly. Is ξ = Px?
It follows that
. However, from convexity and strong lower
semicontinuity implying weak lower semicontinuity,
and so ξ = Px because there is only one value in Px. This has shown that, thanks to
uniqueness of Px, xn → x implies Pxn → Px weakly.
Next we show that f is demicontinuous. Suppose xn → x. Then from what
was just shown, Pxn → Px weakly. Thus xn − Pxn → x − Px weakly. Then
from Proposition 23.3.3
which characterizes the projection map. It follows that, since F
monotone hemicontinuous and bounded, it is also pseudomonotone and so for all
is bounded. If it converges to
It follows that
is arbitrary, it follows that ξ
weakly. Thus this is demicontinuous.