A Relation between Resolvents of Subdifferentials and Metric Projections to Level Sets ()
1. Introduction
The subdifferentials of lower semi-continuous convex functionals defined on real Banach spaces play important roles in many researches of nonlinear differential equations. In fact, for example, −Laplacian or −p-Laplacian operator with a usual boundary condition is the subdifferential of lower-semicontinuous convex functional
defined in
.
Throughout this paper, let X be a real Banach space,
be the dual space, and
be the duality mapping of X. Let
be a proper lower-semicontinuous convex functional. The effective domain of
, which is denoted by
, is the following;
The level set of
for
is denoted by
, i.e.,
Since
is lower-semicontinuous and convex, the level sets are closed and convex in X.
The subgradients of
at
are the elements
satisfying
The subdifferential of
at
is the set of all subgradients of
at
, and denoted by
, i.e.,
It is known that
is a maximal monotone operator ( [1] [2] [3] ).
For every closed convex subset
, the metric projection from X onto C, which is denoted by
, is defined as below.
As is seen in Figure 1, in general,
is not unique. In this paper, we denote arbitral
by
for simplicity.
Figure 1. An example where
is not unique.
If X is reflexive, then
for
. (In fact, there is a sequence
such that
. Since X is reflexive, the bounded subset
is weakly compact. Thus, some subsequence of it converges to
. By the closed convexity of C,
.) If X is strictly convex, then
is either single or empty.
In general, if both X and
are strictly convex and reflexive, then every maximal monotone operator
satisfies the following ( [1] ).
(i) For
(equivalently, for
),
.
(ii) For
and
, there is a unique solution
to the relation
(1.1)
(iii) For
and
, let
be the unique solution of (1.1) and put
(1.2)
Then
and
satisfy the following;
is single valued and monotone,
and
are bounded, both
and
hold for
, and so on.
The subdifferential operators satisfy more properties other than (i) to (iii) above. For instance, the following (A) (B) are known.
(A) If X is a real Hilbert space H, then for
, relations
(1.3)
hold with
.
Although
in (1.3) is depending on x, while
in (1.1) is common to all
, (1.3) seems sufficiently useful to obtain solutions of
in H. (1.3) is proved without using the above properties (i) to (iii), but geometric properties of convex functionals’ graphs in Hilbert spaces ( [3] ).
(B) Let g be a given smooth functional satisfying
on X. Put
and suppose that K is convex and closed in X. Let
be the indicator functional of K, i.e.,
Then,
is a lower-semicontinuous convex functional and its subdifferential is below.
Let
be a proper lower-continuous convex functional defined on X. Then, the convex functional
is useful for conditional extremum problem on K.
For example (cf. [4] , obstacle problems), let
,
, and
, where
and
is smooth. Then,
. Since K is closed and convex,
is a lower semi-continuous convex functional, and
for
. Thus,
is useful in the obstacle problem
.
Concerning the above (A) (B), our theorem and remarks show the following.
(A)’ Same result of (A) holds under more general assumptions; (i) X is an arbitral real Banach space, (ii)
, and (iii)
exists. Here, as is mentioned above, if X is reflexive, then assumption (iii) always holds.
It seems that
in X is not solved even if X is reflexive. The author hopes that our theorem will contribute to solving this problem.
(B)’ Let
and
. Then, in general,
may fail to satisfy (1.3) in H, or, in the case of Banach space X,
(1.4)
(see Remark 2.2).
Suppose that
satisfies (1.4), and that codimension of K is finite. In general, as is seen in Figure 2 and Figure 3, if one takes arbitral
, then h may falt to satisfy (1.4). If h satisfies (1.4), then
is a kind of hyperplane tangent of
at
, because every
is the same.
Figure 2. In spaces where the unit sphere has corners,
may not be unique.
Figure 3. For
that satisfies (1.4),
is a kind of hyperplane tangent of
, and also of
.
In this paper, for simplicity,
denotes
such that (1.4) holds. Then, since
holds by
, (1.4) implies that
,
,
such that
(1.5)
On the other hand, let
be obtained by Lagrange multiplier method in the problem of minimizing
under the condition
with smooth function
. Then, Lagrange multiplier method implies that
,
such that
(1.6)
Put
and
. Suppose that K is convex. Then, by (1.6), for
,
,
(1.7)
Hence, if we put
in (1.5), then the form of (1.5) is the same as (1.7).
2. Results
As is mentioned in Section 1, let X be a real Banach space with duality mapping F, and
be a proper lower-continuous convex functional defined in X. Fix
. In the following, we denote the metric projection
by P, and Px means arbitral element of Px, for simplicity.
Let
be such that Px exists. Then,
has inner points, and any inner point of
does not included in
. Thus, Hahn-Banach theorem implies that
satisfying
(2.1)
(cf. [5] ).
Fix an arbitral
such that (2.1) holds.
Theorem 2.1. Suppose that
(2.2)
Then,
(i)
.
(ii) The inclusion
(2.3)
holds with
defined by as blow;
Remark 2.1. Assumption (2.2) holds if either
or
is dense in X.
Hence, assertion (A)’ mentioned in Section 1 follows from Theorem 2.1.
Remark 2.2. In Theorem 2.1, the assumption (2.2) is needed. In fact, if (2.2) does not hold, then there are two types of examples as below.
(i)’
(ii)’
holds, but (2.3) does not hold.
The examples of (i)’ (ii)’ are given in Section 3, and assertion (B)’ in Section 1 concerns with these examples.
3. Proofs of Results
3.1. Proof of Theorem 2.1
We verify convexity of
. Fix
and
. For
,
such that
Then, the relation
implies
Since
is arbitral,
is convex.
Now we know that both
and
are convex, and
for
. Suppose that
(3.1)
Then, by definition of subdifferential,
On the other hand, by definition of
,
Thus, the proof of Theorem 2.1 is completed if (3.1) is shown.
To verify (3.1) by contradiction, suppose that (3.1) does not hold. Then, by definition of
,
with
. By (2.2),
. Take
sufficiently small such that
Then, since
is convex,
(3.2)
On the other hand, we have verified (2.1) which contains the inclusion
. This is a contradiction to (3.2). Therefore, Theorem 2.1 is proved.
3.2. Example of (i)’ in Remark 2.2
Suppose
. Take
which satisfies
is dense in X and
are compact. (3.3)
Since
, (3.3) yields
. For example,
with bounded
,
,
.
Fix any
and
, where F is the duality mapping of X. Define the nonnegative convex functional g by
. Put
Then, since K is still an infinite dimensional linear subspace, same properties of (3.3) hold if we take
and K instead of
and X, respectively. Take
such that
Then, for
,
holds as is seen in Figure 4.
Figure 4. Since K is a hyperplane of X and
, for every
and
, one has
.
Figure 5. How
is determined differs depending on λ.
3.3. Example of (ii)’ in Remark 2.2
Let
with
. Put
By definition,
is the following.
and
Let
with
. Then, as is seen in Figure 5, the following cases hold. Case 2 satisfies (2.2).
Case 1. If
, then (1.3) does not hold. In fact, for
,
Case 2. If
, then (1.3) with
holds, since