Autoconjugate representers for linear monotone operators Heinz H. Bauschke∗, Xianfu Wang†, and Liangjin Yao‡. February 10, 2008

Abstract Monotone operators are of central importance in modern optimization and nonlinear analysis. Their study has been revolutionized lately, due to the systematic use of the Fitzpatrick function. Pioneered by Penot and Svaiter, a topic of recent interest has been the representation of maximal monotone operators by so-called autoconjugate functions. Two explicit constructions were proposed, the first by Penot and Z˘alinescu in 2005, and another by Bauschke and Wang in 2007. The former requires a mild constraint qualification while the latter is based on the proximal average. We show that these two autoconjugate representers must coincide for continuous linear monotone operators on reflexive spaces. The continuity and the linearity assumption are both essential as examples of discontinuous linear operators and of subdifferential operators illustrate. Furthermore, we also construct an infinite family of autoconjugate representers for the identity operator on the real line.

2000 Mathematics Subject Classification: Primary 47H05; Secondary 47N10, 54A41. Keywords: Autoconjugate representer, convex function, Fenchel conjugate, Fitzpatrick function, linear monotone operator, maximal monotone operator, subdifferential operator.

1

Introduction

Throughout this paper, we assume that X is a real reflexive Banach space, with continuous dual space X ∗ , and pairing h·, ·i. ∗ Mathematics, Irving K. Barber School, UBC Okanagan, Kelowna, British Columbia V1V 1V7, Canada. E-mail: [email protected]. † Mathematics, Irving K. Barber School, UBC Okanagan, Kelowna, British Columbia V1V 1V7, Canada. E-mail: [email protected]. ‡ Mathematics, Irving K. Barber School, UBC Okanagan, Kelowna, British Columbia V1V 1V7, Canada. E-mail: [email protected].

1

The norm of X is denoted by k · k, and the norm in the dual space X ∗ by k · k∗ .  Let A : X ⇒ X ∗ be a set-valued operator, with graph graA = (x, x∗ ) ∈ X × X ∗ | x∗ ∈ Ax , with inverse operator A−1 : X ∗ ⇒ X given by gra A−1 = (x∗ , x) ∈ X ∗ × X | x∗ ∈ Ax , with  domain dom A = x ∈ X | Ax 6= ∅ , and with range ran A = A(X). Recall that A is monotone if   ∀(x, x∗ ) ∈ gra A ∀(y, y ∗ ) ∈ gra A hx − y, x∗ − y ∗ i ≥ 0. (1) A monotone operator A is maximal monotone if no proper enlargement (in the sense of graph inclusion) of A is monotone. Monotone operators are ubiquitous in Optimization and Analysis (see, e.g., [19, 31, 35, 36, 40, 45]) since they contain the key classes of subdifferential operators and of positive linear operators. In [21], Fitzpatrick introduced the following tool in the study of monotone operators. Definition 1.1 Let A : X ⇒ X ∗ . The Fitzpatrick function of A is FA : (x, x∗ ) 7→

hx, y ∗ i + hy, x∗ i − hy, y ∗ i.

sup

(2)

(y,y ∗ )∈gra A

Monotone Operator Theory has been revolutionized through the systematic use of the Fitzpatrick function; new results have been obtained and previously known result have been reproved in a simpler fashion — see, e.g., [1, 2, 3, 6, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 22, 23, 26, 27, 28, 29, 30, 33, 37, 38, 39, 41, 42, 43, 44, 46]. Before listing some of the key properties of the Fitzpatrick function, we introduce a convenient notation utilized by Penot [29]: If F : X × X ∗ → ]−∞, +∞], set F | : X ∗ × X : (x∗ , x) 7→ F (x, x∗ ), (3) and similarly for a function defined on X ∗ × X. We now define an associated operator X ⇒ X ∗ by requiring that for (x, x∗ ) ∈ X × X ∗ , x∗ ∈ G(F )x

F (x, x∗ ) = hx, x∗ i;



we also say that F is a representer for G(F ). Fact 1.2 (See [21].) Let A : X ⇒ X ∗ be maximal monotone. Then the following hold. (i) FA is proper, lower semicontinuous, and convex. (ii) FA−1 = FA| . (iii) FA ≥ h·, ·i. (iv) A = G(FA ).

2

(4)

Item (iv) of Fact 1.2 states the key property that the Fitzpatrick function FA is a representer for the maximal monotone operator A. It turns out that there are even more structured representers for A available: recall that F : X × X ∗ → ]−∞, +∞] is autoconjugate, if F ∗ = F |.

(5)

Autoconjugate representers are readily available for two important classes of maximal monotone operators. Example 1.3 (subdifferential operator) Let f : X → ]−∞, +∞] be proper, lower semicontinuous, and convex. Then the separable sum of f and the Fenchel conjugate f ∗ , i.e., f ⊕ f ∗ : X × X ∗ → ]−∞, +∞] : (x, x∗ ) 7→ f (x) + f ∗ (x∗ ),

(6)

is an autoconjugate representer for the subdifferential operator ∂f . Example 1.4 (antisymmetric operator) Let A : X → X ∗ be continuous, linear, and antisymmetric, i.e., A∗ = −A. Then the indicator function of the graph of A, i.e., ( 0, if x∗ = Ax; ∗ ∗ ιgra A : X × X → ]−∞, +∞] : (x, x ) 7→ (7) +∞, otherwise is an autoconjugate representer for A. We now list some very pleasing and well known properties of autoconjugate functions. Fact 1.5 (Penot-Simons-Z˘ alinescu) (See [28, 29, 30, 42].) Let F : X × X ∗ → ]−∞, +∞] be autoconjugate. Then the following hold. (i) F is proper, lower semicontinuous, and convex. (ii) F ≥ h·, ·i. (iii) G(F ) is maximal monotone. (iv) If Fe : X × X ∗ → ]−∞, +∞] is autoconjugate and Fe ≤ F , then Fe = F . Unfortunately, the Fitzpatrick function FA is usually not an autoconjugate representer for A. In view of Fact 1.5(iii), it was tempting to ask whether every general maximal monotone operator possesses an autoconjugate representer. Nonconstructive existence proofs were presented by Penot [28, 29] and by Svaiter [43] in 2003 (see also Ghoussoub’s preprint [23]). The first actual construction of an autoconjugate representer for a maximal monotone operator satisfying a mild constraint qualification was provided by Penot and Z˘alinescu in 2005.

3

Fact 1.6 (Penot-Z˘ alinescu) (See [30].) Let A : X ⇒ X ∗ be maximal monotone. Suppose that the affine hull of dom A is closed. Then AA : X × X ∗ → ]−∞, +∞] (x, x∗ ) 7→ ∗inf ∗ 21 FA (x, x∗ + y ∗ ) + 12 FA∗| (x, x∗ − y ∗ ) y ∈X

(8)

is an autoconjugate representer for A. Another autoconjugate representer was very recently proposed in [9]. While this proximalaveraged based construction is more involved [4, 5, 7], it has the advantage of not imposing a constraint qualification. Fact 1.7 (See [9].) Let A : X ⇒ X ∗ be maximal monotone. Then BA : X × X ∗ → ]−∞, +∞] (x, x∗ ) 7→

inf

(y,y ∗ )∈X×X ∗

1 2 FA (x

+ y, x∗ + y ∗ ) + 12 FA∗| (x − y, x∗ − y ∗ ) + 12 kyk2 + 12 ky ∗ k2∗

(9)

is an autoconjugate representer for A. It is natural to ask “How do the autoconjugate representers AA and BA compare?” We provide two answers to this question: First, we show that if A : X → X ∗ is continuous, linear, and monotone, then AA and BA coincide; furthermore, we provide a formula for this autoconjugate representer which agrees with a third autoconjugate representer CA that is contained in the work by Ghoussoub (Theorem 3.1). Secondly, for nonlinear monotone subdifferential operators, the two autoconjugate representers may be different (Theorem 5.1). The first answer raises the question on whether autoconjugate representers for continuous linear monotone operators are unique. We answer this question in the negative by providing a family of autoconjugate representers for the identity operator Id (Theorem 4.2). However, we show that the autoconjugate representers AA and BA in this setting are characterized by a pleasing symmetry property (Theorem 4.4). We conclude by discussing discontinuous linear monotone operators. It turns out that AA may fail to be autoconjugate (Example 6.5), which underlines not only the continuity assumption in Theorem 3.1 but also the importance of the constraint qualification in Fact 1.6. The remainder of this paper is organized as follows. Section 2 contains some results on quadratic functions and another autoconjugate representer that will be used in later sections. In Section 3, we show that AA and BA coincide and provide a simple formula for it. Uniqueness of autoconjugate representations are discussed in Section 4, and a characterization in the symmetric case is also presented. In stark contrast, and as shown in Section 5, AA and BA may be different for (nonlinear) subdifferential operators. The final Section 6 reveals similar difference for discontinuous linear operators.

4

Notation utilized is standard as in Convex Analysis and Monotone Operator Theory; see, e.g., [34, 35, 45]. Thus, for a proper convex function f : X → ]−∞, +∞], we write f ∗ : x∗ 7→ supx∈Xhx, x∗ i− ∗ ∗ ∗ ∗ f (x), ∂f : X ⇒ X : x 7→ x ∈ X | (∀y ∈ X) hy − x, x i + f (x) ≤ f (y) , and dom f = x ∈ X | f (x) < +∞ , for the Fenchel conjugate, subdifferential operator, and domain of f , respectively. The strictly positive integers are N = {1, 2, . . .}.

2

Auxiliary Results

The following result in a consequence of results and proof techniques introduced by Penot, Simons, and Z˘alinescu [30, 42]. It also extends [23, Lemma 2.2]. Proposition 2.1 Let F1 and F2 be autoconjugate functions on X × X ∗ representing maximal monotone operators A1 and A2 , respectively. Suppose that [  λ PX dom F1 − PX dom F2 is a closed subspace of X, (10) λ>0

where PX : X × X ∗ → X : (x, x∗ ) 7→ x, and set F : X × X ∗ → ]−∞, +∞] : (x, x∗ ) 7→ ∗inf ∗ F1 (x, y ∗ ) + F2 (x, x∗ − y ∗ ). y ∈X

(11)

Then F is an autoconjugate representer for A1 + A2 , and the infimum in (11) is attained. Proof. Let (x, x∗ ) ∈ X × X ∗ . Using Simons and Z˘alinescu’s [42, Theorem 4.2] and the assumption that each Fi is autoconjugate, we obtain F ∗ (x∗ , x) = =

min

F1∗ (x∗1 , x) + F2∗ (x∗2 , x)

min

F1 (x, x∗1 ) + F2 (x, x∗2 )

x∗1 +x∗2 =x∗ x∗1 +x∗2 =x∗

= F (x, x∗ ).

(12)

Thus, F is autoconjugate and the infimum in (11) is attained. It remains to show that G(F ) = G(F1 ) + G(F2 ). Since autoconjugates are greater than or equal to h·, ·i (see Fact 1.5(ii)), the above implies the equivalences x∗ ∈ G(F )x ⇔ F (x, x∗ ) = hx, x∗ i ⇔ (∃ y ∗ ∈ X ∗ ) F1 (x, y ∗ ) + F2 (x, x∗ − y ∗ ) = hx, y ∗ i + hx, x∗ − y ∗ i ⇔ (∃ y ∗ ∈ X ∗ ) F1 (x, y ∗ ) = hx, y ∗ i and F2 (x, x∗ − y ∗ ) = hx, x∗ − y ∗ i ⇔ (∃ y ∗ ∈ X ∗ ) y ∗ ∈ G(F1 )(x) and x∗ − y ∗ ∈ G(F2 )(x) ⇔ (∃ y ∗ ∈ X ∗ ) y ∗ ∈ A1 x and x∗ − y ∗ ∈ A2 x ⇔ x∗ ∈ (A1 + A2 )x.

(13) 5

Therefore, G(F ) = A1 + A2 , i.e., F is a representer for A1 + A2 .



Suppose that A : X → X∗

is linear and continuous.

(14)

Then A is symmetric (resp. antisymmetric) if A∗ = A (resp. A∗ = −A). We denote the symmetric part and the antisymmetric part of A by A+ = 21 A + 21 A∗

and A◦ = 12 A − 21 A∗ ,

(15)

respectively. Throughout, we shall work with the quadratic function qA : X → R : x 7→ 21 hx, Axi,

(16)

and we will use the well known facts (see, e.g., [32]) that qA = qA+ , that ∇qA = A+ ,

(17)

and that A is maximal monotone if and only if qA is convex. The next result provides a formula ∗ that will be useful later. for qA Proposition 2.2 Let A : X → X ∗ be continuous, linear, symmetric, and monotone. Then  ∗ ∗ ∀(x, x∗ ) ∈ X × X ∗ qA (x∗ + Ax) = qA (x) + hx, x∗ i + qA (x∗ )

(18)

and ∗ qA ◦ A = qA .

(19)

Proof. Let (x, x∗ ) ∈ X × X ∗ . Then ∗ qA (x∗ + Ax) = sup hy, x∗ + Axi − qA (y) = sup hy, x∗ i − qA (y) + hy, Axi y

y



= qA (x) + sup hy, x i − qA (y) + hy, Axi − qA (x) = qA (x) + sup hy, x∗ i − qA (y − x) y

y





∗ = qA (x) + hx, x i + sup hy − x, x i − qA (y − x) = qA (x) + hx, x∗ i + qA (x∗ ),

(20)

y

which verifies (18). To see (19), set x∗ = 0 in (18).



Proposition 2.3 Let A : X → X ∗ be continuous, linear, and monotone, and let (x, x∗ ) ∈ X × X ∗ . Then ∗ ∗ FA (x, x∗ ) = 2qA ( 1 x∗ + 21 A∗ x) = 12 qA (x∗ + A∗ x). (21) + 2 + and FA∗ (x∗ , x) = ιgra A (x, x∗ ) + hx, Axi.

6

(22)

Proof. As in the proof of [3, Theorem 2.3(i)], we have FA (x, x∗ ) = sup hx, Ayi + hy, x∗ i − hy, Ayi y∈X

= 2 sup hy, 12 x∗ + 12 A∗ xi − qA+ (y) = =

y∈X ∗ 2qA ( 1 x∗ + 21 A∗ x) + 2 ∗ ∗ 1 ∗ 2 qA+ (x + A x).

(23)

This verifies (21). Furthermore, (22) follows from FA∗ (x∗ , x) = (ιgra A + h·, ·i)∗|∗ (x∗ , x) = ιgra A (x, x∗ ) + hx, Axi.  Proposition 2.4 Let F1 : X × X ∗ → ]−∞, +∞] be autoconjugate, and let A2 : X → X ∗ be continuous, linear, and antisymmetric. Then the function (x, x∗ ) 7→ F1 (x, x∗ − A2 x)

(24)

is an autoconjugate representer for G(F1 ) + A2 . Proof. Set F2 = ιgra A2 . By Example 1.4, F2 is an autoconjugate representer for A2 . Let F be as in Proposition 2.1. Then for every (x, x∗ ) ∈ X × X ∗ , we have F (x, x∗ ) = ∗inf ∗ F1 (x, x∗ − z ∗ ) + F2 (x, z ∗ ) z ∈X

= ∗inf ∗ F1 (x, x∗ − z ∗ ) + ιgra A2 (x, z ∗ ) z ∈X

= F1 (x, x∗ − A2 x). Thus, Proposition 2.1 yields that F represents G(F1 ) + A2 .

(25) 

Example 2.5 (Ghoussoub) (See also [23, Section 1].) Let f : X → ]−∞, +∞] be proper, lower semicontinuous, and convex, and let A be antisymmetric. Then the function (x, x∗ ) 7→ f (x) + f ∗ (x∗ − Ax)

(26)

is an autoconjugate representer for ∂f + A. Proof. By Example 1.3, f ⊕ f ∗ is an autoconjugate representer for ∂f . The result thus follows from Proposition 2.4.  Corollary 2.6 Let A : X → X ∗ be continuous, linear, and monotone. Then CA : X × X ∗ → ]−∞, +∞] ∗ (x, x∗ ) 7→ qA+ (x) + qA (x∗ − A◦ x) +

(27)

is an autoconjugate representer for A. In particular, if A is symmetric, then ∗ C A = qA ⊕ qA .

7

(28)

Proof. This follows from (17) and Example 2.5 (when applied to the function qA = qA+ and to the antisymmetric operator A◦ ).  We now show that the Ghoussoub representers are closed under the partial infimal convolution operation of Proposition 2.1. Proposition 2.7 Let A and B be continuous, linear, and monotone on X. Then the function F : X × X ∗ → ]−∞, +∞] : (x, x∗ ) 7→ min CA (x, x∗ − y ∗ ) + CB (x, y ∗ ) ∗ ∗ y ∈X

(29)

coincides with the autoconjugate representer CA+B for A + B. Proof. In view of Proposition 2.1, we only need to show that F = CA+B . Let (x, x∗ ) ∈ X × X ∗ . Using (29) and Corollary 2.6, we obtain ∗ ∗ F (x, x∗ ) = min qA+ (x) + qA (x∗ − y ∗ − A◦ x) + qB+ (x) + qB (y ∗ − B◦ x) + + ∗ ∗ y ∈X

∗ ∗ = qA+ (x) + qB+ (x) + (qA qB )(x∗ − A◦ x − B◦ x) + + ∗ = qA+ +B+ (x) + qA+ + qB+ (x∗ − A◦ x − B◦ x)  ∗ = q(A+B)+ (x) + q(A+B) x∗ − (A + B)◦ x +

= CA+B (x, x∗ ),

(30)

as required.

3



Coincidence

We are now ready for one of our main results. Theorem 3.1 (coincidence) Let A : X → X ∗ be continuous, linear, and monotone. Then all three autoconjugate representers AA , BA , CA for A coincide with the function ∗ (x, x∗ ) 7→ hx, x∗ i + qA (x∗ − Ax). +

Proof. The proof proceeds by proving a succession of claims. Let (x, x∗ ) ∈ X × X ∗ .

8

(31)

Claim 1: AA = CA . Using (8), (22), and (27), we obtain  AA (x, x∗ ) = ∗inf ∗ 21 FA x, x∗ + y ∗ ) + 12 FA∗ (x∗ − y ∗ , x) y ∈X  = ∗inf ∗ 21 FA x, x∗ + y ∗ ) + ιgra A (x, x∗ − y ∗ ) + qA (x) y ∈X

= 12 FA (x, 2x∗ − Ax) + qA+ (x)  ∗ = qA x∗ − 21 Ax + 21 A∗ x + qA+ (x) +  ∗ ∗ = qA x − A x + qA+ (x) ◦ +

(32)

= CA (x, x∗ ).

(33)

This verifies Claim 1. Claim 2: AA coincides with the function of (31). In view of (32) and Proposition 2.2, we see that  ∗ AA (x, x∗ ) = qA x∗ − 12 Ax + 12 A∗ x + qA+ (x) + ∗ = qA (x∗ − Ax + A+ x) + qA+ (x) + ∗ = 2qA+ (x) + hx, x∗ − Axi + qA (x∗ − Ax) + ∗ = hx, x∗ i + qA (x∗ − Ax), +

(34)

which establishes Claim 2. Claim 3: AA = BA . Using (9), (22), (21), Proposition 2.2, and Claim 2, we have BA (x, x∗ ) = =

inf

1 2 FA (x

+ y, x∗ + y ∗ ) + 21 FA∗| (x − y, x∗ − y ∗ ) + 12 kyk2 + 12 ky ∗ k2∗

inf

1 2 FA (x

+ y, x∗ + y ∗ ) + ιgra A (x − y, x∗ − y ∗ )

(y,y ∗ )∈X×X ∗ (y,y ∗ )∈X×X ∗

+ 12 hx − y, A(x − y)i + 21 kyk2 + 12 ky ∗ k2∗ 1 FA (x + y, 2x∗ − A(x − y)) + qA (x − y) + 12 kyk2 + 12 kx∗ − A(x − y)k2∗ y∈X 2  ∗ ∗ 1 1 ∗ inf qA x − A(x − y) + A (x + y) + qA+ (x − y) + 21 kyk2 + 21 kx∗ − A(x 2 2 + y∈X  ∗ ∗ inf qA x − Ax + A (x + y) + qA+ (x − y) + 21 kyk2 + 12 kx∗ − A(x − y)k2∗ + + y∈X ∗ inf qA (x∗ − Ax) + hx + y, x∗ − Axi + qA+ (x + y) + qA+ (x − y) + y∈X + 12 kyk2 + 21 kx∗ − A(x − y)k2∗ ∗ inf qA (x∗ − Ax) + hx + y, x∗ − Axi + 2qA+ (x) + 2qA+ (y) + y∈X + 12 k −yk2 + 21 kx∗ − A(x − y)k2∗ ∗ qA (x∗ − Ax) + hx, x∗ i + inf hy, x∗ − Axi + 2qA+ (y) + h−y, x∗ − A(x − y)i + y∈X

= inf = = =

=



9

− y)k2∗

∗ (x∗ − Ax) + hx, x∗ i + inf 2qA+ (y) + h−y, Ayi = qA + y∈X

=

∗ qA (x∗ +



− Ax) + hx, x i

= AA (x, x∗ ).

(35)

Hence BA ≥ AA . On the other hand, both AA and BA are autoconjugate (see Fact 1.6 and Fact 1.7). Altogether, Fact 1.5(iv) implies Claim 3. Finally, observe that Claims 1–3 yield the result.   Example 3.2 Suppose that X is the Euclidean plane R2 , let θ ∈ 0, π2 , and set     cos θ − sin θ 0 −1 A= and Aπ/2 = . sin θ cos θ 1 0



(36)

Then for every (x, x∗ ) ∈ R2 × R2 , AA (x, x∗ ) = BA (x, x∗ ) = CA (x, x∗ ) 1 = kx∗ − Axk2 + hx, x∗ i 2 cos θ 1 cos θ = kx∗ − (sin θ)Aπ/2 xk2 + kxk2 . 2 cos θ 2

(37)

Proof. This follows from Theorem 3.1 since A+ = (cos θ) Id, qA+ = (cos θ) 21 k · k2 , and A◦ = (sin θ)Aπ/2 . 

4

Observations on Nonuniqueness

Theorem 3.1 might nurture the conjecture that for continuous linear monotone operators, all autoconjugate representers coincide. This conjecture is false — we shall provide a whole family of distinct autoconjugate representers for the identity on R. Our constructions rests on the following result. Proposition 4.1 Let g : R → ]−∞, +∞] be such that (∀x ∈ R)

g ∗ (−x) = g(x) ≥ 0.

(38)

g(0) = 0.

(39)

Then Moreover, each of the following functions satisfies (38): ( 0, (i) the indicator function ι[0,+∞[ : x 7→ +∞,

if x ≥ 0; if x < 0; 10

(ii) the energy function 12 | · |2 ; ( (iii) for p > 1 and q > 1 such that

1 p

+

1 q

= 1, the function x 7→

1 p px , 1 q q (−x) ,

if x ≥ 0; if x < 0.

Proof. On the one hand, g(0) ≥ 0. On the other hand, g(0) = g ∗ (−0) = g ∗ (0) = supy∈R −g(y) = − inf y∈R g(y) ≤ 0. Altogether, g(0) = 0 and so (39). It is straight-forward to verify that each of the given functions satisfies (38).  Theorem 4.2 Let g : R → ]−∞, +∞] be such that for every x ∈ R, g ∗ (−x) = g(x) ≥ 0, and set q : R → R : x 7→ 21 |x|2 . Then F : R2 → ]−∞, +∞] : (x, y) 7→ q

x + y  x − y  √ +g √ . 2 2

(40)

is an autoconjugate representer for Id : R → R : x 7→ x. Proof. Let (x, y) ∈ R2 . Using the fact that q ∗ = q and the assumption on g, we see that   u−v √ √ F ∗ (y, x) = sup uy + vx − q u+v − g 2 2 (u,v)∈R2

=

sup (u,v)∈R2

= = =

sup

u+v 2 (x

+ y) −

u+v √ x+y √ 2 2



u−v 2 (x

u−v √ x−y √ 2 2

(u,v)∈R2   x−y ∗ √ √ q ∗ x+y + g − 2 2   x+y x−y √ √ q 2 +g 2

= F (x, y).

− y) − q

−q

u+v √ 2



u+v √ 2

−g



−g

u−v √ 2

u−v √ 2





(41)

 Hence F is autoconjugate. In view of (39), we have (x, y) ∈ gra G(F ) ⇔ y ∈ G(F )x ⇔ F (x, y) = √  √  √  xy ⇔ q (x + y)/ 2 + g (x − y)/ 2 = xy ⇔ 14 (x + y)2 + g (x − y)/ 2 = xy ⇔ 41 (x − y)2 + √  g (x − y)/ 2 = 0 ⇔ x − y = 0 ⇔ (x, y) ∈ gra(Id).  ∗ = C Remark 4.3 Consider Theorem 4.2. If we set g = q = 12 | · |2 , then F = q ⊕ q = qId ⊕ qId Id by Corollary 2.6. Thus, this pleasingly symmetric choice of g gives rise to AId = BId = CId . Proposition 4.1 provides other choices of g that lead to different autoconjugate representers for Id.

Having settled the nonuniqueness of autoconjugate representers, it is natural to ask “What makes the autoconjugate representers of Theorem 3.1 special?” The next result provides a complete answer for a large class of linear operators.

11

Theorem 4.4 Let A : X → X ∗ be continuous, linear, monotone, and symmetric. Let F : X×X ∗ → ]−∞, +∞] be such that ran A is closed. Then   F is autoconjugate, F = CA ⇔ (42) F (0, 0) = 0,    ∀(x, y) ∈ X × X F (x, Ay) = F (y, Ax). ∗ )(0, 0) = 0. Let x and y Proof. “⇒”: By Corollary 2.6, F is autoconjugate and F (0, 0) = (qA ⊕ qA ∗ )(x, Ay) = q (x) + q ∗ (Ay) = q (x) + q (y) = be in X. Using (19), we have F (x, Ay) = (qA ⊕ qA A A A A ∗ ∗ qA (y) + qA (Ax) = (qA ⊕ qA )(y, Ax) = F (y, Ax).

“⇐”: Let (x, x∗ ) ∈ X × X ∗ . We proceed by verifying the next two claims. Claim 1: x∗ ∈ / ran A ⇒ F (x, x∗ ) = +∞. ∗ Assume that x ∈ / ran A. The Separation Theorem yields z ∈ X such that hz, x∗ i > 0

(43)

and maxhz, ran Ai = 0. Since A is symmetric, we deduce that Az = 0. This implies (∀ρ ∈ R) F (ρz, 0) = F (ρz, A0) = F 0, A(ρz) = F (0, 0) = 0. Thus (∀ρ ∈ R) F (x, x∗ ) = F (x, x∗ ) + F (ρz, 0) = F (x, x∗ ) + F ∗ (0, ρz) ≥ hx, 0i + hρz, x∗ i = ρhz, x∗ i.

(44)

In view of (43), we see that Claim 1 follows by letting ρ → +∞ in (44). Claim 2: x∗ ∈ ran A ⇒ F (x, x∗ ) ≥ CA (x, x∗ ). Assume that x∗ ∈ ran A, say x∗ = Ay. Then 2F (x, x∗ ) = 2F (x, Ay) = F (x, Ay) + F (y, Ax) = F (x, Ay) + F ∗ (Ax, y) ≥ hx, Axi + hy, Ayi and hence, using (19), ∗ ∗ F (x, x∗ ) ≥ qA (x) + qA (y) = qA (x) + qA (Ay) = (qA ⊕ qA )(x, x∗ ).

(45)

This and (28) yield Claim 2. Note that Claim 1 and Claim 2 yield F ≥ CA . Therefore, Fact 1.5(iv) implies that F = CA .

5



Autoconjugate Representers for ∂(− ln)

Theorem 3.1 showed that three ostensibly different autoconjugate representers are in fact identical for continuous linear monotone operators. It is tempting to consider a subdifferential operator ∂f , and to compare A∂f , B∂f , and f ⊕ f ∗ . It turns out that these autoconjugate representers for ∂f may all be different. To aid in the construction of this example, it will be convenient to work in this section with the negative natural logarithm function ( − ln(x), if x > 0; f : R → ]−∞, +∞] : x 7→ (46) +∞, if x ≤ 0, 12

and with the set  C = (x, x∗ ) ∈ R × R | x∗ ≤ − x1 < 0 .

(47)

It is well known that f ∗ (x) = −1 + f (−x)

(∀x ∈ R)

(48)

and straight-forward to verify that √1 C 2 1 2C

 1 = (x, x∗ ) ∈ R × R | x∗ ≤ − 2x <0 ,  1 = (x, x∗ ) ∈ R × R | x∗ ≤ − 4x <0 ,

(49)

√1 C 2

(51)

(50)

and ( 12 C ( ]0, +∞[ × ]−∞, 0[ .

Furthermore, [8, Example 3.4] yields ∗

∀(x, x ) ∈ R × R



( √ 1 − 2 −xx∗ , F∂f (x, x ) = +∞, ∗

if x ≥ 0 and x∗ ≤ 0; otherwise,

(52)

and ∗| = −1 + ιC . F∂f

(53)

Theorem 5.1 The functions A∂f , B∂f , and f ⊕f ∗ have domains √12 C, 12 C, and ]0, +∞[×]−∞, 0[, respectively. Consequently, they are three different autoconjugate representers for ∂f . Proof. Using (8), (52), (53) and (49), we see that  ∗| dom A∂f = (x, 12 x∗1 + 21 x∗2 ) ∈ R × R | (x, x∗1 ) ∈ dom F∂f and (x, x∗2 ) ∈ dom F∂f  = (x, 12 x∗1 + 21 x∗2 ) ∈ R × R | x ≥ 0, x∗1 ≤ 0, and (x, x∗2 ) ∈ C  1 <0 = (x, x∗ ) ∈ R × R | x∗ ≤ − 2x =

√1 C, 2

(54)

as claimed. Similarly, by (9), (52), and (53), dom B∂f = = =

∗| 1 1 2 dom F∂f + 2 dom F∂f  1 1 2 [0, +∞[ × ]−∞, 0] + 2 C 1 2 C.

(55)

Furthermore, by (46) and (48), dom(f ⊕ f ∗ ) = (dom f ) × (dom f ∗ ) = ]0, +∞[ × ]−∞, 0[ .

(56)

We thus have verified the statements concerning the domains. Fact 1.6, Fact 1.7, and Example 2.5 imply that all three functions are autoconjugate representers for ∂f . In view of (51), these functions are all different since their domains are also all different.  13

Remark 5.2 Using (52) and (53), one may verify that ( √  − −1 − 2xx∗ , if (x, x∗ ) ∈ ∀(x, x∗ ) ∈ R × R A∂f (x, x∗ ) = +∞, otherwise.

√1 C; 2

(57)

However, we do not have an explicit formula for B∂f .

6

Discontinuous Symmetric Operators

In this final section, we investigate discontinuous symmetric operators. Specifically, we assume throughout this section that A : X ⇒ X ∗ is maximal monotone, at most single-valued, dom A is a linear subspace, and A|dom A is linear and symmetric. Put differently, we assume that A : dom A → X ∗

is linear, symmetric, and maximal montone.

It is convenient to extend the definition of qA in (16) to this more general setting via ( 1 hx, Axi, if x ∈ dom A; qA : X → R : x 7→ 2 +∞, otherwise.

(58)

(59)

A key tool is the function f : X → ]−∞, +∞] : x 7→

sup hx, Ayi − 21 hy, Ayi,

y∈dom A

which was introduced by Phelps and Simons. Fact 6.1 (Phelps-Simons) (See [32].) The following hold. (i) f is proper, lower semicontinuous, and convex. (ii) A = ∂f . (iii) dom A ⊆ dom f ⊆ dom A and (∀x ∈ dom A) f (x) = 12 hx, Axi. (iv) A is continuous ⇔ dom A = X ⇔ dom f = X. Corollary 6.2 The following hold. (i) f + ιdom A = qA . (ii) f = qA ⇔ dom f = dom A. ∗∗ = f . (iii) qA ∗ (iv) If A is one-to-one, then f = qA −1 .

14

(60)

Proof. (i): Clear from Fact 6.1(iii). (ii): Since dom qA = dom A, this item is a consequence of (i). (iii): Using Fact 6.1(i)&(ii) and a result by J. Borwein (see [11, Theorem 1] or [45, Theo∗∗ . (iv): If A is one-to-one, rem 3.1.4(i)]), we see that f = f ∗∗ = (f + ιdom ∂f )∗∗ = (f + ιdom A )∗∗ = qA then (∀x ∈ X) f (x) =

sup y ∗ ∈dom A−1

hx, y ∗ i− 12 hA−1 y ∗ , y ∗ i =

sup y ∗ ∈dom qA−1

∗ hx, y ∗ i−qA−1 (y ∗ ) = qA −1 (x). (61)

This completes the proof.



Proposition 6.3 We have: dom A = dom f ⇔ every sequence (xn )n∈N in dom A such that (xn )n∈N and (hxn , Axn i)n∈N are convergent must satisfy lim xn ∈ dom A. Proof. “⇒”: Assume that (xn )n∈N is a sequence in dom A such that (xn )n∈N converges to x ∈ X and (hxn , Axn i)n∈N is also convergent. Using Fact 6.1(i) and Corollary 6.2(i), we have x ∈ dom f and thus x ∈ dom A. “⇐”: In view of Fact 6.1(iii), it suffices to show that dom f ⊆ dom A. To this end, let x ∈ dom f . In view of Corollary 6.2(iii), there exists a sequence (xn )n∈N in dom qA = dom A such xn → x and 1  2 hxn , Axn i = qA (xn ) → f (x). By assumption, x ∈ dom A, as required. Theorem 6.4 Let B : X ∗ → X be continuous, linear, symmetric, monotone, and one-to-one. Suppose that A = B −1 . Then ∗ AA = qA ⊕ qB = (qB + ιdom A ) ⊕ qB .

(62)

∗ BA = A∗∗ A = qB ⊕ qB

(63)

and ∗ = dom A. are both representers for A. Furthermore, AA = BA ⇔ dom qB ∗ +ι Proof. Since qA = qB dom A by Corollary 6.2(i)&(iv), it suffices to verify the left equality in (62). ∗ Let (x, x ) ∈ X × X ∗ . Using (8) and Fact 1.2(ii), we see that 1 AA (x, x∗ ) = inf F (x, x∗ + y ∗ ) + 21 FA∗ (x∗ − y ∗ , x) ∗ 2 A y

1 F (x∗ + y ∗ , x) + 21 FB∗ (x, x∗ − y ∗ ) = inf ∗ 2 B y

1 = inf F (x∗ + y ∗ , x) + ∗ 2 B y

1 2

 ιgra B (x∗ − y ∗ , x) + hx∗ − y ∗ , B(x∗ − y ∗ )i .

(64)

If x ∈ / ran B = dom A, then (64) shows that AA (x, x∗ ) = +∞, as required. So assume that x ∈ ran B = dom A. In view of (64), (21), and (19), we deduce that AA (x, x∗ ) = 21 FB (2x∗ − Ax, x) + 12 hx, Axi  ∗ 1 ∗ 1 = qB 2 x + 2 B(2x − Ax) + qA (x) ∗ = qB (Bx∗ ) + qA (x)

= qB (x∗ ) + qA (x). 15

(65)

Hence (62) holds. Using (62), Corollary 6.2(iii)&(iv), (28), and Theorem 3.1, we see that | ∗∗ ∗∗ ∗∗ ∗ ∗ ∗ ∗ A∗∗ = qA ⊕ qB = qB ⊕ qB = (qB ⊕ qB ) = BB = BB = BB −1 = BA , A = (qA ⊕ qB )

(66)

∗∗ ⊕ q ⇔ q = q ∗∗ ⇔ dom q ∗∗ = dom A so that (63) holds. Furthermore, AA = BA ⇔ qA ⊕ qB = qA B A A A ∗ ⇔ dom qB = dom A by Corollary 6.2. 

Example 6.5 Suppose that X is the Hilbert space `2 (N) of square-summable sequences; thus, X ∗ = X. Set (67) B : X → X : (ξk )k∈N 7→ ( k1 ξk )k∈N and suppose that A = B −1 . Then ran B = dom A is dense in X, but it is not closed (since, e.g., ( k1 )k∈N ∈ X r (ran B)). Now set   1 1 1 1 , 24/3 , . . . , n4/3 , 0, 0, . . . . (68) x = k4/3 and (∀n ∈ N) xn = 14/3 k∈N On the one P hand, (xn )n∈N lies A and xn → x ∈ X r (dom A). On the other hand, Pnin dom 1 1 k hxn , Axn i = nk=1 k4/3 = k=1 k5/3 → ζ(5/3) ∈ R. Altogether, Proposition 6.3 implies that k4/3 ∗ dom A ( dom qB . Therefore, by Theorem 6.4, AA is neither lower semicontinuous nor equal to BA . While AA is still a representer for A, it cannot be autoconjugate. Remark 6.6 Several comments are in order. (i) Without the constraint qualification, Fact 1.6 fails (see Example 6.5, where dom A is a subspace that is not closed). (ii) It is conceivable that A∗∗ A is always an autoconjugate representer for A — this would sharpen Fact 1.6 and it would be consistent with Theorem 6.4. (iii) Suppose that B is as in Theorem 6.4, that A = B −1 , and that dom A = ran B is a dense subspace of X with dom A 6= X. We do not know whether (dom f ) r (dom A) 6= ∅ must hold (as it does in Example 6.5), i.e. (see Proposition 6.3), whether there exists a sequence (xn )n∈N in dom A such that (xn )n∈N converges to some point x ∈ X r (dom A), yet (hxn , Axn i)n∈N converges to a real number. In contrast, there does exist a point x ∈ X r (dom A) such that every sequence (xn )n∈N in dom A converging to x must have hxn , Axn i → +∞. (Indeed, since dom A 6= X, it follows from Fact 6.1(iv) that dom f 6= X. Take x ∈ X r (dom f ) and assume that (xn )n∈N lies in dom A and converges to x. Then +∞ = f (x) ≤ lim f (xn ) = lim 12 hxn , Axn i by Fact 6.1(i)&(iii)). Thus for every sequence (x∗n )n∈N in X ∗ such that Bx∗n → x ∈ / ran B, it ∗ ∗ ∗ ∗ ∗ ∗ follows that kBxn k · kxn k∗ ≥ hBxn , xn i = hBxn , A(Bxn )i → +∞. Since 0 ∈ dom f and so x 6= 0, we deduce kx∗n k∗ → +∞, which is a well known result from Functional Analysis (see [25, Corollary 17.G]).

16

Acknowledgment Heinz Bauschke was partially supported by the Canada Research Chair Program and by the Natural Sciences and Engineering Research Council of Canada. Xianfu Wang was partially supported by the Natural Sciences and Engineering Research Council of Canada.

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