Maximal monotonicity of the subdifferential of a convex function: a direct proof
aa r X i v : . [ m a t h . F A ] A ug Maximal monotonicity of the subdifferentialof a convex function: a direct proof ∗ Milen Ivanov † Nadia Zlateva ‡ May, 2015
Abstract
We provide a new proof for maximal monotonicity of the subdif-ferential of a convex function.
Keywords:
Subdifferential, convex function, maximal monotonicity.
Subject Classification
Let X be a Banach space with dual X ∗ . Let f : X → R ∪ { + ∞} be a proper,convex and lower semicontinuous function.Recall that for x ∈ dom f := { x : f ( x ) = + ∞} the subdifferential of f at x in the sense of convex analysis is the set ∂f ( x ) := { p ∈ X ∗ : f ( x ) ≥ f ( x ) + p ( x − x ) , ∀ x ∈ X } , (1)while the ε -subdifferential of f at x is the set ∂ ε f ( x ) := { p ∈ X ∗ : f ( x ) ≥ f ( x ) + p ( x − x ) − ε, ∀ x ∈ X } . If x dom f , then ∂f ( x ) = ∂ ε f ( x ) = ∅ . ∗ Dedicated to Prof. R. T. Rockafellar on the occasion of his 80th anniversary. † Partially supported by Bulgarian National Scientific Fund under Grant DFNI-I02/10and by Grant 239/2016 of the Science Fund of Sofia University. ‡ Partially supported by Bulgarian National Scientific Fund under Grant DFNI-I02/10. x ∈ dom f the set ∂f ( x ) might be empty but the set ∂ ε f ( x ) isalways non-empty for any ε > ∂f is a monotone multivalued mapping from X to X ∗ , inthe sense that for all x, y ∈ X and all p ∈ ∂f ( x ), q ∈ ∂f ( y ) it holds h q − p, y − x i ≥ , which can be written also as h ∂f ( y ) − ∂f ( x ) , y − x i ≥ , ∀ x, y ∈ X. It is a well-known classical result due to Rockafellar that ∂f is in fact maximal monotone (see Theorem 2). That is, ∂f can not be extended tostrictly larger monotone mapping from X to X ∗ .This statement goes back at least as far as [8], where Minty proves it fora continuous convex function on Hilbert space (see the discussion in [11]).Moreau [5] gave a proof in Hilbert space using duality and Moreau-Yosidaapproximation.A stumbling block to generalizing Minty’s method is that for a lowersemicontinuous convex function which is not continuous, ∂f might be emptyat some points. Also controlling the norms of the subgradients appearing inthe proof is not trivial (see the discussion in [12]). On the other hand, themethod of Moreau relies heavily on the fact that Hilbert space is canonicalyisometric to its dual.The method of reducing the considerations to a line, used by Minty in [8]and generalised by Rockafellar in [11], is prevalent in the subsequent proofs,like those of Taylor [15], Borwein [1], Thibault [16], the unpublished proof ofZagrodny, using [17], and the recent one of Jules and Lasonde [6]. A textbookfollowing this line of proof is [9].The first complete proof in Banach space: that of Rockafellar [12], is amethodological break through showing that Fenchel conjugate and dualitycan be used in non-reflexive case as well. The recent proofs of Marques Alvesand Svaiter [7] and Simons [14] also use duality. It is mentioned in [18] that inmany textbooks authors prefer proving only the reflexive case (where dualitytechniques are easier due to symmetry), e.g. [18, p. 278].The famous proof of Simons [13] (see also [10]) shows how one can picka subgradient with controlled norm.Like some others, our proof starts with h ∂f ( x ) , x i ≥ , ∀ x ∈ dom ∂f. The following is proved in [6] for general lower semicontinuous functionthrough mean value inequality. We give a simple proof for the convex case.
Proposition 1.
Let X be a Banach space and let f : X → R ∪ { + ∞} be aproper, convex and lower semicontinuous function.If f : X → R ∪ { + ∞} satisfies h ∂f ( x ) , x i ≥ , ∀ x ∈ dom ∂f (2) then ∈ ∂f (0) .Proof. For a > g a ( x ) := a k x k .If p ∈ ∂g a ( x ) then by definition p (0 − x ) ≤ g a (0) − g a ( x ), that is, p ( x ) ≥ a k x k . (3)Let f a ( x ) := f ( x ) + g a ( x ) . (4)Since g a is continuous, the Sum Theorem, see for example [18], implies thatdom ∂f a = dom ∂f and ∂f a ( x ) = ∂f ( x ) + ∂g a ( x ) , ∀ x ∈ dom ∂f. (5)From (2), (3) and (5) it follows that h ∂f a ( x ) , x i ≥ a k x k , ∀ x ∈ dom ∂f. Consequently, ∀ p ∈ ∂f a ( x ) ⇒ k p k ≥ a k x k . (6)On the other hand, f a is bounded below for each a >
0. Indeed, take x ∈ dom f . Since f is lower semicontinuous there is δ > f ( x + δB X ) > −∞ , where B X is the closed unit ball. Take c ∈ R such that3 < inf f ( x + δB X ). By Hahn-Banach Theorem, see for example [4], we canseparate the epigraph of f , that is the set { ( x, t ) : f ( x ) ≤ t } , from the set( x + δB X ) × ( −∞ , c ]. So, there is ( p, r ) ∈ X ∗ × R \ (0 ,
0) such that p ( x ) + rt ≥ p ( y ) + rs, ∀ x ∈ dom f, ∀ t ≥ f ( x ) , ∀ y ∈ x + δB X , ∀ s ≤ c. Setting x = x we see that sup { rs : s ≤ c } < ∞ which is only possibleif r ≥
0. But if r = 0 then setting x = x and y = x + h we see that p ( x ) ≥ p ( x ) + p ( h ) for all h ∈ δB X which implies p = 0, contradiction.Therefore, r > r as well use t = f ( x ) and s = c to obtain for q = p/r that q ( x ) + f ( x ) ≥ q ( y ) + c, ∀ x ∈ dom f, ∀ y ∈ x + δB X , Set y = x and rearrange to get for r = q ( x ) + cf ( x ) ≥ − q ( x ) + r ≥ −k q kk x k + r, ∀ x ∈ X. Therefore, f a ( x ) ≥ r + k x k ( a k x k − k q k ) and f a is bounded below.Let x n be a minimising sequence for f a . So, f a ( x n ) < f a ( x n ) + ε n for some ε n →
0. Equivalently, 0 ∈ ∂ ε n f a ( x n ).From Brøndsted-Rockafellar Theorem, see [3], it follows that there are y n ∈ x n + √ ε n B X and p n ∈ ∂f a ( y n ) such that k p n k ≤ √ ε n . From this and(6) it follows that k y n k ≤ √ ε n a . Therefore, x n →
0. Since f a is lower semicontinuous, 0 is the global minimumof f a .In other words, f a ( x ) ≥ f a (0) ⇐⇒ f ( x ) ≥ f (0) − a k x k , ∀ x ∈ X. Since a > f , or, equivalently,0 ∈ ∂f (0).The Rockafellar’s Theorem follows by an easy and well known argument(see for example [9], p. 59): Theorem 2. (Rockafellar [12]) Let X be a Banach space and let f : X → R ∪ { + ∞} be a proper, convex and lower semicontinuous function. Then ∂f is a maximal monotone mapping from X to X ∗ . roof. Let ( y, q ) ∈ X × X ∗ be in monotone relation to the graph of ∂f , thatis h ∂f ( x ) − q, x − y i ≥ , ∀ x ∈ dom ∂f. (7)Consider the function ¯ f ( x ) := f ( x + y ) − q ( x ) . It is immediate to check that (7) implies (2) for ¯ f . By Proposition 1 we get0 ∈ ∂ ¯ f (0) which easily translates to q ∈ ∂f ( y ). Therefore, ∂f cannot beproperly extended in a monotone way. References [1] J. M. Borwein, A note on ǫǫ