Introduction to Gradient Flows in Metric Spaces (II)

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1 Introduction to Gradient Flows in Metric Spaces (II) Philippe Clément November 23, 200 Preface The aim of these Lectures Notes is to provide a gentle introduction to the theory of gradient flows in metric spaces developed in the first part of the book of Ambrosio-Gigli- Savaré [AGS]. In contrast to [LN] we do not use the notion of (local) slope of a functional defined on a metric space. As in [LN] we use Crandall-Liggett type estimates. This approach is somewhat simpler than the one used in [AGS] but it does not give the optimal rate of convergence of the approximation scheme obtained in [AGS]. The main result of these notes, Theorem 4. is a special case (α = 0; in [AGS]: λ = 0) of a result of [CD2]. Also in these notes we use recent results of [CD] which simplify the approach of [LN]. For the sake of completeness we give the corresponding estimates of Crandall-Liggett [CL7] when the operator A is m-accretive in a Banach space (X, ). The main difference between the two cases, (X, ) Banach space and the situation of Theorem 4. is that the resolvent operators J h are contractive in the accretive case but not (in general) under the assumptions of Theorem 4.. Finally it should be observed that if X is a real Hilbert space and the function φ: X (, ] is proper, lower semicontinuous and convex, then, the subdifferential φ is m-accretive, hence the first approach can be used, and the function φ satisfies the assumption (A) of Theorem 4., providing another approach to the same problem. Combining Lemma 3.2, Proposition 3. and Theorem 4., one obtains the existence and uniqueness part of the proof of Theorem of [AGS] as well as other properties of solutions to the evolution variational inequality (4.0.3) of [AGS]. Acknowledgements The author would like to thank Michael Röckner and the University of Bielefeld for their warm hospitality and for the very stimulating environment. These notes were used as a support for a series of lectures given at the University of Bielefeld during a 2 month stay in the fall of Finally I would like to thank the audience for some comments and in particular Dr. Shun-Xiang Ouyang who typed most of the manuscript.

2 An Evolution Variational Inequality on a metric space The aim of this section is to introduce an evolution variational inequality (EVI) on a metric space which will be the main subject of these notes. We motivate the definition by means of two examples. We consider in Section. a gradient flow in R N with the euclidean metric and in Section.2 a heat flow in L 2 (R n ). We show in both cases that the corresponding solutions satisfy (EVI). In this reformulation of the problem, the linear structure is not used. In Section.3 the definition of a solution to (EVI) is given and in Section.4 we prove an a priori estimate which among other things guarantees the uniqueness for the corresponding initial value problem.. Gradient flows in R N Let H := R N with the euclidean metric and let φ : R N R be differentiable. We consider the differential equation (DE) u(t) = φ(u(t)), t J, where J is an open interval of R, u(t) := d u(t) and dt φ(x) := φ x (x)., x R N. φ x N (x) A solution of (DE) is a function u : J R N which is differentiable and satisfies (DE). Moreover (.) d dt φ u(t) = φ(u(t)), u(t) = φ(u(t)) 2 = u(t) 2 0, t J, hence φ u is nonincreasing on J. In this section we require in addition that φ is Lipschitz continuous. We recall Definition.. Let (X i, d i ), i =, 2, be two metric spaces and let F : X X 2. The map F is called Lipschitz continuous if there exists M 0 such that (.2) d 2 (F (x), F (y)) Md (x, y) for every x, y X. If X = X 2, the map F is called a contraction (in X ) when M = and a strict contraction (in X ) when M <. The smallest constant for which (.2) holds will be denoted by [F ] Lip. Remark.. For any Lipschitz continuous map F we have { } d2 (F (x), F (y)) (.3) [F ] Lip = sup : x, y X, x y. d (x, y) 2

3 For any map F : X X 2 we can take the RHS (right-hand side) of (.3) as the definition of [F ] Lip. So F is Lipschitz continuous (notation: F Lip(X, X 2 )) iff [F ] Lip <. Returning to (DE), with [ φ] Lip <, we recall the following global existence and uniqueness theorem. For every x 0 R N, there exists a unique solution u of (DE) on J = R such that u(0) = x 0. We shall show that any solution of (DE) satisfies an evolution variational inequality on X = R N involving only the metric of R N and not its linear structure. We proceed in 5 steps. Step. φ is quasi-monotone. Definition.2. Let (H,, ) be a real Hilbert space and D(A) be a nonempty subset of H. A map (operator) A : D(A) H H is called monotone if (.4) Ax Ay, x y 0 for all x, y D(A), and quasi-monotone if there exists α R such that A αi is monotone, equivalently (.5) Ax Ay, x y α x y 2 for all x, y D(A). If α in (.5) can be chosen positive then A is called strongly monotone. Setting (.6) α(f ) := inf { } Ax Ay, x y : x, y D(A), x y x y 2 (possibly ) we see that A is quasi-monotone iff α(f ) >, and A is monotone iff α(f ) 0. Finally we observe that if F : H H is Lipschitz continuous, then hence F (x) F (y), x y F (x) F (y) x y [F ] Lip x y 2, x, y H; (.7) α(f ) [F ] Lip and F is quasi-monotone, since F βi is monotone for every β α(f ). Step 2. φ is quasi-convex. In view of Step there exists α R such that φ αi is monotone. Setting (.8) e(x) := 2 x 2 for x R N and noticing that e(x) = x, x R N, i.e. e = I, the identity in R N, we see that (φ αe) = φ α e = φ αi is monotone. Set (.9) ψ := φ αe, then ψ : R N R is differentiable and ψ is monotone, hence ψ is convex as a consequence of the following lemma. Lemma.. Let ψ : R N R be everywhere differentiable with ψ monotone. Then ψ is convex. 3

4 Proof. Let ψ be monotone and let x, y R N, t R. Set α(t) := ψ(( t)x + ty) ( t)ψ(x) tψ(y). Then α(0) = α() = 0 and α is differentiable, α (t) = ψ(( t)x + ty), y x + ψ(x) ψ(y). Let t < t 2. Observe that [( t 2 )x + t 2 y] [( t )x + t y] = (t 2 t )(y x). We have α (t 2 ) α (t ) = ψ(( t 2 )x + t 2 y) ψ(( t )x + t y), [( t 2 )x + t 2 y] [( t )x + t y] t 2 t 0. Hence α is nondecreasing. If α had a positive maximum ξ (0, ), then α (ξ) = 0 and in view of the mean value theorem α would be nonincreasing for t < ξ and nondecreasing for t > ξ. A contradiction. Hence α(t) 0 for t [0, ] and ψ is convex. We can rewrite (DE) as (.0) u(t) αu(t) = ψ(u(t)), t J, where ψ is convex. hence Step 3. ψ(x) is a subgradient of ψ at x. Let x, h R N. Since ψ is differentiable at x, we have ψ(x + h) ψ(x) = ψ(x), h + o( h ), [ ] lim ψ(x + th) ψ(x) = ψ(x), h. t 0 t Since ψ is convex, the function t ψ(x + th) is also convex, hence the function is nondecreasing. Therefore Equivalently, we have ψ(x), h = lim t 0 < t t t 0 ( ψ(x + th) ψ(x) ) ( ψ(x + th) ψ(x) ) (t=) ψ(x + h) ψ(x). (.) ψ(z) ψ(x) + y, z x for all z R N, where y = φ(x). Any y R N satisfying (.) is called a subgradient of ψ at x. Actually y = ψ(x) is the only subgradient of ψ at x. Indeed, from (.), choosing z = x + th, t > 0, h R N, we deduce (ψ(x + th) ψ(x)) y, h. t By taking the limit as t 0 we get ψ(x), h y, h. Replacing h by h we obtain equality. Choosing h = y ψ(x) we arrive at y = ψ(x). 4

5 Step 4. Some computation. Combining (.0) and (.) we obtain for every x, z R N, t J. Setting u(t) αu(t), z u(t) + ψ(u(t)) ψ(z) (.2) d 2 (x, y) := (d(x, y)) 2 = x y 2, x, y R N, and using the definition of ψ, we obtain (.3) 2 d dt d2 (u(t), z) + α u(t) 2 α u(t), z + φ(u(t)) α 2 u(t) 2 φ(z) α 2 z 2. Hence for all t J, all z R N, we have 2 d dt d2 (u(t), z) + α 2 d2 (u(t), z) + φ(u(t)) φ(z). Step 5. Integration by parts. This step consists of replacing the pointwise derivative in (.3) by a weak derivative. Multiplying (.3) by a nonnegative test function η Cc (J) (C (J) with compact support in J), and integrating by parts we arrive at (EVI) J 2 d2 (u(t), z) η(t) dt + α [ ] d 2 (u(t), z)η(t) dt φ(z) φ u(t) η(t) dt, 2 J J for every nonnegative η Cc (J). Observe that the integrals in (EVI) are well defined since u C(J; R N ), φ u C(J; R) as well as t d 2 (u(t), z), and the supports of η, η are compact in J. Notice that (EVI) makes sense if we replace (R N, d) by an arbitrary metric space (X, d), using the notation (.2), and requiring φ C(X; R) and the solution u to be continuous on J with values in X. In Section.2 we shall consider an example where we need to weaken the condition on the functional φ which requires a stronger assumption for the definition of a solution. Problem.. Let φ : R n R be everywhere differentiable. Show that if φ(x) = Ax for all x R N for some N N matrix A, then the matrix A is symmetric. Show that in this case φ(x) = 2 Ax, x + constant. Find a characterization of [ φ] Lip (.3) and α( φ) (.6) in terms of the eigenvalues of A. Problem.2. Let (H,, ) be a real Hilbert space and let A L(H) be a bounded linear n operator on H. We recall that k! Ak converges in L(H) (with respect to the operator k=0 norm) as n. We denote the limit by e A, the exponential of A. Moreover, given x 0 and f H, the function u : R H defined by (.4) u(t) = e ta x 0 + t 0 e sa f ds, t R, is continuously differentiable (C (R; H))), i.e., lim (u(t + h) u(t)) exists in H for all h t R and is denoted by u(t), and R t u(t) C(R; H). Moreover, u satisfies the linear differential equation in H: (LDE) u(t) + Au(t) = f, t R, 5

6 together with u(0) = x 0. Show that if A is symmetric, i.e., Ax, y = x, Ay for all x, y H, then the function u defined in (.4) satisfies (EVI) with J := R, φ(x) := Ax, x f, x, x H, 2 and Ax, x α := inf x H x 2 x 0 = α(a)..2 Heat flow in L 2 (R N ) as a gradient flow in L 2 (R N ) The aim of this section is to show that the heat flow in L 2 (R n ) can be viewed as a gradient flow in L 2 (R n ) with respect to the Dirichlet functional. Let (.5) p(t, x) := (2πt) N/2 e x 2 /(2t), t > 0, x R N, be the density of the gaussian probability measure on R N with mean 0 and second moment Nt, i.e., R N p(t, x) dx =, t > 0, R N x k p(t, x) dx = 0, k =,..., N, R N x 2 p(t, x) dx = Nt, t > 0. Let f 0 : R N R be a (Lebesgue) measurable function satisfying R N f 0 2 dx <. Let t > 0 and x R N. Then R N p(2t, x y) f 0 (y) dy <, hence (.6) v(t, x) := p(2t, x y)f 0 (y) dy R N is well-defined. We recall without proofs several facts about the function v : (0, ) R n R which will be used later. The function v is infinitely differentiable in (0, ) R N (notation: u C ((0, ) R N )). Clearly, if f 0 is replaced by g 0 such that g 0 = f 0 a.e. in R N then the RHS of (.6) defines the same function v so (.6) defines a map from L 2 (R N ) into C ((0, ) R N ). The function v satisfies the heat equation (.7) t v(t, x) = v(t, x), (t, x) (0, ) RN, 6

7 where v = N Set k= 2 x 2 k v. (.8) u(t)(x) := v(t, x), t > 0, x R N. For t fixed u(t) C (R N ) and R N ( u(t)(x) ) 2 dx <, hence u(t) can be viewed as an element of L 2 (R N ). Moreover, the map defined by (0, ) t u(t) L 2 (R N ) will be denoted by u. Then it appears that u C((0, ); L 2 (R N )). Even more, for every t > 0 we have (.9) lim h 0 R N ( ) u(t + h)(x) u(t)(x) h t v(t, x) 2 dx = 0. This shows that the map u : (0, ) L 2 (R N ) is (strongly) differentiable in L 2 (R N ). We denote its derivative at t by u(t). Similarly u is differentiable on (0, ) and ü(t)(x) = 2 v(t, x), (t, x) (0, ) R N. In fact u is infinitely differentiable (notation: u t 2 C ((0, ); L 2 (R N )). Moreover, for every t > 0 the function u(t) L 2 (R N ) possesses weak derivatives of all order which belong to L 2 (R N ), i.e., u(t) W k,2 (R N ) for all k N. Thus for every multi-index (α,..., α N ) x ( α α N v(t, x)) is an element of the x α x α N equivalence class of α x αn u(t) α x α N for every t > 0. As a consequence of (.7) one shows that (.20) u(t) + Au(t) = 0, t > 0, where A : W 2,2 (R n ) L 2 (R N ) L 2 (R N ) is defined by Af := where x k are weak derivatives. Moreover, we define for f W,2 (R N ) the gradient of f, where x k f := N 2 x 2 k= k x f. x N f f,, are weak derivatives. Clearly, maps W,2 (R N ) into ( L 2 (R N ) ) N. The following relation between A and is essential. We have (.2) (Af)(x)g(x) dx = R N for every f W 2,2 (R N ), g W,2 (R N ). R N N k= f(x) g(x) dx x k x k 7

8 Denoting by, the innerproduct in L 2 (R N ), we obtain by taking the innerproduct of (.20) with z u(t) where z W,2 (R N ): u(t), z u(t) = Au(t), z u(t) = u(t), z u(t) = u(t) 2 + u(t), z where denotes the norm in L 2 (R N ). Moreover, since u C ( (0, ); L 2 (R N ) ) we have u(t), z u(t) = 2 where d(f, g) := f g in L 2 (R N ). We arrive at (.22) where 2 (.23) φ(f) := 2 u(t) u(t) z 2 = 2 u(t) z 2, d dt u(t) z, u(t) z = 2 d dt u(t) z 2 = 2 d dt d2 (u(t), z) φ(z) φ u(t), t > 0 and z W,2 (R N ), R N f 2 dx for f W,2 (R N ). d dt d2 (u(t), z) Observe that t d 2 (u(t), z) is continuously differentiable as well as t Au(t), u(t) = u(t), u(t) on (0, ). Hence multiplying (.22) by nonnegative test functions η Cc (0, ) we arrive at (EVI) defined in Section. with X = L 2 (R N ) endowed with the metric induced by the innerproduct, φ : W,2 (R N ) X R defined by (.23) and α = 0. Observe that as in Section. t φ u(t) is (at least) continuous hence locally integrable on J := (0, ). The function φ is usually called a Dirichlet form [K]. It is customary to define φ on the whole of X by setting { f 2 dx for f W,2 (R N ), 2 R (.24) φ(f) := N + otherwise. D(φ) := {f L 2 (R N ) : φ(f) < } is called the effective domain of φ. From now on we shall adopt this way of writing..3 Definition of a gradient flow on a metric space Motivated by the two previous examples, we shall define a gradient flow on a metric space as follows: Definition.3. Let (X, d) be a metric space (not necessarily complete), φ : X (, + ] be proper (i.e. D(φ) := {x X : φ(x) < } ), α R. 8

9 Given an open interval J in R, we say that a function u : J X is a gradient flow (or a solution to (EVI)) on (X, d) (with respect to the pair (φ, α)), if i) u C(J; X), ii) φ u L loc (J), i.e., φ u (a,b) L (a, b) for every a, b J such that a < b, iii) u and φ u satisfy (EVI) on J. If, moreover, J = (a, b) or (a, ) with a, b R, a < b, and lim t a u(t) = x exists in (X, d), then we say that the gradient flow starts at x or has x as initial value. Remark.2. Clearly the function u defined in Section. is a gradient flow in R N (with euclidean metric) where J = R, φ C (R N ; R) and α := α( φ). Similarly the function u defined in Section.2 is a gradient flow in L 2 (R N ) where J = (0, ), φ is the Dirichlet form (.24) and α = 0. Moreover, one can show that lim u(t) = f 0 in L 2 (R N ). Hence this t 0 gradient flow has f 0 as initial value. Since u(t) W 2,2 (R N ), we could have chosen for the functional φ, its restriction to W 2,2 (R N ), namely { ( f)f dx for f W 2,2 (R N ), 2 R (.25) φ (f) := N + otherwise. It appears that φ is lower semicontinuous on L 2 (R N ) but not φ. This property will play an important role in the proof of existence of solutions to (EVI)..4 An a priori estimate The aim of this subsection is to establish an a priori estimate for solutions to (EVI). This estimate implies uniqueness for the corresponding initial value problem. Proposition.. Suppose u and v are two solutions to (EVI) (gradient flows) with respect to (φ, α) on an open interval J of R. Then the following estimate holds: (.26) d(u(t), v(t)) e α(t s) d(u(s), v(s)) for all s, t J such that s < t. In the proof of Proposition. we shall use the following lemma. Lemma.2. Let J be an open interval of R, g : J R be continuous and η C (J; R) with compact support in J. Let a, b R, a < b, and δ > 0 be such that Then we have (.27) J supp η [a, b] [a δ, b + δ] J. b [ ] g(t) η(t) dt = lim g(t + h) g(t) η(t) dt. 0<h<δ h 0 a h 9

10 Noticing that for 0 < h < δ b a we get b a g(t + h)η(t) dt = b+h a+h g(t)η(t h) dt = [ ] b g(t + h) g(t) η(t) dt = h a g(t) h b a g(t)η(t h) dt + [ η(t) η(t h) ] dt + h b+h b b+h b g(t)η(t h) dt, g(t)η(t h) dt. Since, as h 0, the integrand of the first integral of the RHS converges uniformly to g(t) η(t) and the second term tends to g(b)η(b) = 0, the lemma is proved. Lemma.3. Let J be an open interval of R, let f C(J; R) and f 2 L loc (J). Then the following assertions are equivalent. (i) J f (t) η(t) dt + J f 2(t)η(t) dt 0 for every η C c (J) nonnegative. (ii) For every t, t 2 J such that t < t 2 J f (t 2 ) f (t ) + t2 t f 2 (r) dr 0. Proof. (ii) = (i). We apply Lemma.2 with g(t) := f (t). We obtain b [ f (t) η(t) dt = lim f (t + h) f (t) ] η(t) dt 0<h<δ h h 0 (ii) lim 0<h<δ h 0 a b a ( h t+h t ) f 2 (r) dr η(t) dt = b a f 2 (t)η(t) dt observing that +h f 2 (r) dr f 2 ( ) in L (a, b) as h 0. (i) = (ii). Using a standard approximation by means of (Sobolev) mollifiers one can show that the test functions η can be chosen nonnegative absolutely continuous (in particular Lipschitz continuous) with compact support in J. Given t, t 2 J with t < t 2 we choose as test functions η n for n large enough the linear interpolation of η n (t ) = η n (t 2 ) = 0 and η n (t + ) = η n n(t 2 ) = on the interval [t n, t 2 ] and 0 outside. We obtain t2 t +/n t2 n f (t) dt n f (t) dt + f 2 (t)η n (t) dt 0. t 2 /n t t Taking the limit as n we arrive at (ii). Remark.3. Applying Lemma.3 to (EVI) we get (.28) 2 d2 (u(t 2 ), z) 2 d2 (u(t ), z) + α t2 d 2 (u(t), z) dt 2 t for every [t, t 2 ] J and z D(φ) (see Remark of [AGS]). 0 (t 2 t )φ(z) t2 t φ(u(t)) dt

11 We shall choose in (EVI) test functions η of the form η(t) := e αt η(t) where η Cc (J) nonnegative we obtain J 2 d2 (u(t), z) [ e αt( η(t) )] + α η(t) dt α + 2 d2 (u(t), z)e αt η(t) dt So the LHS reduces to J 2 d2 (u(t), z)e αt η(t) dt and by applying again Lemma.3 we obtain (.29) 2 eαt 2 d 2 (u(t 2 ), z) 2 eαt d 2 (u(t ), z) J ( t2 for every t, t 2 J with t < t 2, and any z D(φ). Now we are in a position to prove Proposition.. t J e αr( φ(z) φ(u(r)) ) dr. ) t2 e αr dr φ(z) e αr φ(u(r)) dr t Proof. Suppose u and v are two solutions of (EVI) with respect to (φ, α) (gradient flows) on an open interval J of R. Set g (t) := 2 e2αt d 2 (u(t), v(t)), t J. Clearly g C(J). We want to show that g is nonincreasing on J. In view of Lemma.3 with f := g, z := 0, and Lemma.2 with g := g, it is sufficient to prove lim 0<h<δ h 0 b a [ g (t + h) g (t) ] η(t) dt 0 for any η C c (J) nonnegative, with a, b, δ as in Lemma.2. We have g (t + h) g (t) = e 2α(t+h) 2 d2 (u(t + h), v(t + h)) e 2αt 2 d2 (u(t), v(t)) ] = e [e α(t+h) α(t+h) 2 d2 (u(t + h), v(t + h)) e αt 2 d2 (u(t), v(t + h)) ] + e [e αt α(t+h) 2 d2 (u(t), v(t + h)) e αt 2 d2 (u(t), v(t)). Using (.29) both for u and v, we obtain [ ] e α(t+h) 2 d2 (u(t + h), v(t + h)) e αt 2 d2 (u(t), v(t + h)) [( t+h ) e αr dr φ v(t + h) and [ ] e α(t+h) 2 d2 (u(t), v(t + h)) e αt 2 d2 (u(t), v(t)) [( t+h ) e αr dr φ u(t) t t t+h t t+h t ] e αr φ u(r) dr ] e αr φ v(r) dr.

12 Hence b [ g (t + h) g (t) ] η(t) dt a h b ( t+h ) e α(t+h) e αr dr φ v(t + h)η(t) dt a h t b ( t+h ) e α(t+h) e αr φ u(r) dr η(t) dt a h t b ( t+h ) + e αt e αr dr φ u(t)η(t) dt a h t b ( t+h ) e αt e αr φ v(r) dr η(t) dt. a h t Using φ v( + h) h 0 φ v in L (a, b) as well as +h w(r) dr w( ) in L (a, b) for any h w L loc (J), we obtain b [ lim g (t + h) g (t) ] η(t) dt 0<h<δ h 0 a h b a e 2αt φ v(t)η(t) dt + b 2 The Hilbert space case a b a e 2αt φ u(t)η(t) dt e 2αt φ u(t)η(t) dt b a e 2αt φ v(t)η(t) dt = 0. Let (X,, ) be a real Hilbert space with corresponding norm and metric d(, ). Let e : X R be defined by (2.) e(x) := 2 x 2, x H. Let φ : X (, + ]. We recall that φ is called proper if D(φ) := {x X : φ(x) < } is not empty. In this case we call φ convex if D(φ) is a convex subset of X and the restriction of φ to D(φ) is convex. We call φ lower semicontinuous (l.s.c.) if {x X : φ(x) c} is closed for every c R. We are now in a position to state the main existence result in the Hilbert space case when J := (0, ). Theorem 2.. Let (X,, ) be a real Hilbert space and let φ : X (, + ] be proper, l.s.c. and such that φ αe is convex for some α R. Then for every x D(φ) there exists a gradient flow u: (0, ) X on (X, d) with respect to the pair (φ, α) starting at x. Remark 2.. There are several proofs of Theorem 2., but the main ingredient in all these proofs is the m-accretivity in (H, ) of the subdifferential of the function ψ := φ αe. We first recall some definitions. Definition 2. (subdifferential). Let ϕ : (X,, ) (, + ] be proper. Let x D(ϕ) and y H. We say that y is a subgradient of ϕ at x if the following holds (2.2) y, z x + ϕ(x) ϕ(z) z D(ϕ). 2

13 The collection of all subgradients of ϕ at x, which is denoted by ϕ(x), possibly empty, is called the subdifferential of ϕ at x. We may consider the map x ϕ(x) as a map from D( ϕ) X 2 X (the collection of all subsets of X), where D( ϕ) := {x D(ϕ) : ϕ }. The graph of this map is the subset G( ϕ) of X X defined by G( ϕ) := {(x, y) D(ϕ) X : y is a subgradient of ϕ at x}. It is customary to use the same notation for G( ϕ) and ϕ. Definition 2.2. A subset A of X X is called monotone if for every pair (x i, y i ) A, i =, 2, we have (2.3) y 2 y, x 2 x 0, and cyclically monotone if for every finite cyclic sequence x 0, x,..., x n = x 0 in D(A) (i.e., x H such that for every y H with (x, y) A) and every sequence y,..., y n with (x i, y i ) A, i n, we have Problem 2. ([R]). n y i, x i x i 0. i= (i) Show that if φ : (X,, ) (, + ] is proper, the subdifferential ϕ is cyclically monotone, hence monotone. (ii) Show that if A X X is not empty and cyclically monotone, there exists ϕ : X (, + ] proper, l.s.c. and convex such that A ϕ. Hint: Take (x 0, y 0 ) A and for any x X set φ(x) := sup{ y n, x x n + y n, x n x n y 0, x x 0 : (x i, y i ) A, i =,..., n, n N + }. Definition 2.3. Let (E, ) be a real Banach space. A subset A E E is called accretive ( A is called dissipative) if (2.4) x x 2 x x 2 + h(y y 2 ) for every (x i, y i ) A, i =, 2, and every h > 0. Observe that accretivity and monotonicity of A are equivalent in a real Hilbert space (H,, ). This follows from for any x, y X. x, y 0 iff x x + ty for all t > 0, Problem 2.2. Let (E, ) be a real Banach space and let A E E be accretive. Show that the two following assertions are equivalent: 3

14 (i) R(I + ha) = E for some h > 0, (ii) R(I + ha) = E for all h > 0. Here R(I + ha) denotes the range of the graph I + ha, i.e., We also write R(I + ha) := {y E : there exists x D(A) such that (x, y) I + ha}. {y E : there exists x D(A) such that y x + hax}. Definition 2.4. Let (E, ) be a real Banach space. An accretive graph A E E is called m-accretive ( A is called m-dissipative) if R(I + ha) = E for every h > 0. Example 2.. If X = R then the graph of the function x sign x is monotone but not m-accretive. However, the graph defined by is m-accretive. (x, 0) A for every x < 0 (0, y) A for every y [, ] (x, ) A for every x > 0 After these preparations we can formulate an important result of the theory of monotone operators in Hilbert spaces. Proposition 2.. Let (X,, ) be a real Hilbert space and let ϕ : X (, + ] be proper, l.s.c. and convex. Then the subdifferential ϕ is cyclically monotone and m- accretive in (X, ). Conversely, if A X X is a cyclically monotone graph which is m-accretive, then there exists a proper, l.s.c. and convex functional ϕ : X (, + ] such that A = ϕ. Moreover, if ϕ is as above and satisfies A = ϕ, then there exists some c R such that ϕ (x) = ϕ(x) + c for all x H. The first assertion in the conclusion of Proposition 2. will be proved below. Problem 2.3. Prove the second assertion of Proposition 2.. Hint: By Problem 2. there exists ϕ as in Proposition 2. such that A ϕ. Hence I + A I + ϕ and (I + A) (I + ϕ). Using the m-accretivity of A show that D ( (I + A) ) = D ( (I + ϕ) ) = X. Using the monotonicity of ϕ show that (I + ϕ) is the graph of a map from X into X. Hence (I + A) = (I + ϕ). Therefore, I + A = I + ϕ from which we conclude A = ϕ. This arguments shows that an m-accretive graph is maximal with respect to inclusion for the collection of accretive graphs in X X. In order to prove Theorem 2. we would like to solve the initial value problem for the differential inclusion u(t) αu(t) ψ(u(t)), t > 0, where ψ := φ αe; and then proceed as in Section. This can be done in the following way. Setting A := ψ + αi we observe that this operator satisfies the condition of the Crandall-Liggett theorem (See Appendix 4). 4

15 Defining { (0, ) if α 0, (2.5) I α := ( ) 0, if α < 0 α or equivalently I α = {h > 0: + αh > 0}, we have that for h I α : (I + ha) is the graph of a map from X X (everywhere defined) which we shall denote by J h : (2.6) J h x: = (I + ha) x, x H, h I α. Moreover [J h ] Lip ( + αh). It follows from Theorem I in [CL7], [Appendix 4] that (2.7) S(t)x := lim n (J t/n ) n x, x D(A) exists for t > 0, and lim t 0 S(t)x = x. We set S(0)x = x for x D(A). Moreover {S(t)} t 0 is a one-parameter semigroup of operators on D(A). It follows from Theorem II and Lemma 2.3 of [CL7] that the function u: [0, ) X defined by (2.8) u(t) := S(t)x, x D(A) satisfies: for every T > 0, u [0,T ] Lip([0, T ]; X), (hence u is differentiable a.e. since X is reflexive) and for almost all t (0, ) we have u(t) D(A) together with u(t) Au(t). In [Br73] Brezis proved that if moreover A αi is cyclically monotone, then even for x D(A), the following holds: u(t) D(A) for every [ε, T ] with ε > 0, u [0,T ] Lip([ε, T ]; X), u(t) Au(t) a.e. in (0, ). Proceeding as in Step 5 of Section. we can show that u is a gradient flow on X with respect to (φ, α), starting at x. The conclusion of Theorem. follows by observing that D(A) := D( ψ) = D(ψ) =: D(φ). If we want to use this approach in the case of a metric space, we need to be able to define J h x without using the linear structure of the Hilbert space. This can be done. Setting (2.9) Φ(h, x; y) := { 2h d2 (x, y) + φ(y), +, otherwise, y D(φ) we shall show that if h I α and x X, then (2.0) J h x is the unique global minimizer of Φ(h, x; ). More precisely, we have the following lemma. Lemma 2.. Let φ: X (, + ] be proper, let α R, h I α and ψ := φ αe. Then i) if x 0 X, x D( ψ) satisfy (2.) h (x x 0 ) αx ψ(x ), then (2.2) Φ(h, x 0 ; x ) Φ(h, x 0 ; z), for every z X. 5

16 ii) Conversly, if we suppose in addition that ψ is convex, then for any x 0 X and x D(φ) satisfying (2.2), we have x D( ψ) and (2.) holds. Proof. i) ii). Notice that x D( ψ) D(ψ). We first treat the case α = 0 (φ = ψ). Φ(h, x 0 ; z) Φ(h, x 0 ; x ) = 2h x 0 z 2 2h x 0 x 2 + ψ(z) ψ(x ) 2h x 0 z 2 2h x 0 x 2 + h x 0 x, z x = 2h x 0 z 2 = 2h 2h x 0 x 2 + h x 0 x, x 0 x + h x 0 x, z x 0 [ z x0 2 + x 0 x x 0 x, z x 0 ] = 2h (z x 0) + (x 0 x ) 2 = 2h d2 (x, z) which is nonnegative. In the general case we get an extra term in (2.) and two extra terms coming from the definition of ψ. We get Φ(h, x 0 ; z) Φ(h, x 0 ; x ) 2h d2 (x, z) α x, z x + α 2 z 2 α 2 x 2 = 2h d2 (x, z) + α2( x 2 2 x, z + z 2 ) = 2 ( h + α)d2 (x, z) 0 since h I α. ii) i). Let x 0 X, x D(φ) satisfying (2.2). Let z D(φ) and t (0, ). Then, using the convexity of ψ, we have ( t)x + tz D(ψ) and Using the definition of Φ and (2.2), we get ψ(( t)x + tz) ψ(x ) t[ψ(z) ψ(z) ψ(( t)x + tz) ψ(x ).] =Φ(h, x 0 ; ( t)x + tz) Φ(h, x 0 ; x ) 2h[ ( t)x + tz x 0 2 x x 0 2] α ( ( t)x + tz 2 x 2) 2 2h[ ( t)x + tz x 0 2 x x 0 2] α ( ( t)x + tz 2 x 2). 2 Dividing by t and letting t tend to zero we arrive at which is (2.). ψ(z) ψ(z ) h z x, x x 0 α x, z x 6

17 Having shown that we can define J h x in terms of the metric d, the functional φ and α R, it is important to investigate whether the condition φ αe convex can also be expressed in terms of these data. The condition φ αe means that for every y 0, y D(φ) and t (0, ) we have Observe that φ(( t)y 0 + ty ) αe(( t)y 0 + ty ) ( t)φ(y 0 ) + tφ(y ) α( t)e(y 0 ) αte(y ). (2.3) e(( t)y 0 + ty ) = ( t)e(y 0 ) + te(y ) t( t)e(y 0 y ) holds for every t R and y 0, y X. Indeed e(( t)y 0 + ty ) + t( t)e(y 0 y ) = 2 ( t)2 y t2 y 2 + ( t)t y 0, y =( t)e(y 0 ) + te(y ). + 2 ( t)2 y t2 y 2 ( t)t y 0, y It follows that φ αe is convex iff φ satisfies for t (0, ), y 0, y D(φ), (2.4) φ(( t)y 0 + ty ) ( t)φ(y 0 ) + tφ(y ) α 2 d2 (y 0, y ). In view of (2.3) again the function 2 d2 (x, ) satisfies for x, y 0, y X and t R: 2 d2 (x, ( t)y 0 + ty ) = e(( t)(y 0 x) + t(y x)) = ( t)e(y x) + t(y x) t( t)e(y y 0 ). Therefore, we have (2.5) Φ(h, x; ( t)y 0 + ty ) = 2h d2 (x, ( t)y 0 + ty ) + φ(( t)y 0 + ty ) [ ] [ ] ( t) 2h d2 (x, y 0 ) + φ(y 0 ) + t 2h d2 (x, y ) + φ(y ) =( t)φ(h, x; y 0 ) + tφ(h, x; y ) 2 ( h + α)t( t)d2 (y 0, y ) 2 ( h + α)t( t)d2 (y 0, y ) for every h > 0, x X, y 0, y D(φ), t (0, ) iff φ αe is convex. Observe that if h I α, then the function Φ(h, x; ) is strictly convex. We conclude this section by showing that in the Hilbert space case, the functional Φ(h, x; ) possesses a global minimizer if the following conditions are satisfied: φ is proper and l.s.c. α R such that φ αe is convex x H and h I α. 7

18 In a first step we shall assume φ bounded below. Then clearly Φ(h, x; ) is also bounded below. Set γ := inf{φ(h, x; y): y D(φ)}. Let {y n } D(φ) be a minimizing sequence i.e. lim n Φ(h, x; y n ) = γ. We claim that {y n } is a Cauchy-sequence in H. Indeed from (2.5) with t = 2 and h + α > 0 we obtain for m, n, noticing that 2 (y m + y n ) D(φ), 2 ( h + α)d2 (y m, y n ) 2 Φ(h, x; y m) + 2 Φ(h, x; y n) 2 Φ(h, x; y m + y n ) 2 2 [Φ(h, x; y m) γ] + 2 [Φ(h, x; y n) γ], which tends to zero as m, n. In view of the completeness of (X, d) there exists ȳ X such that d(y n, ȳ) 0 as n. Since φ is l.s.c, Φ(h, x; ) is also l.s.c., hence Φ(h, x; ȳ) lim Φ(h, x; y n ) = γ <. n Therefore ȳ D(φ) and γ Φ(h, x; ȳ) γ which completes the proof in this case. Now we remove the additional assumption φ bounded below by using the fact that a proper, l.s.c convex function in a Hilbert space is bounded below by a continuous affine function of the form y a + b, y where a R, b H. [See Appendix 5]. Then it is easy to verify that Φ(h, x; ) is bounded below. This implies that J h (see (2.9)) is well-defined and by Lemma (2.), J h x D( ψ) and satisfies (2.6) h (x J hx) αj h x ψ(x), h I α. In particular, when α = 0, (2.6) implies that R(I + h ψ) = X for every h > 0, hence ψ is m-accretive in (X, ). (first assertion of Proposition 2.). Moreover, when α is not necessarily equal to 0, we may define for h I α : (2.7) φ h (x) := Φ(h, x; J h x) = 2h x J hx 2 + φ(j h x). The function φ h is called the Yosida-Moreau approximation of φ. (see Section 3 of [LN]). (Notice that φ n (x) φ(j n x) in general (!)). Problem 2.4. Let (X,, ) and (E,, E ) be real Hilbert spaces. Let T : D(T ) X E be a closed densely defined operator. Let { 2 φ(x) := T x 2 E, x D(T ), +, otherwise. Show that the functional φ is quadratic (i.e. φ(x + y) + φ(x y) = 2(φ(x) + φ(y)), x, y D(φ), and φ(λx) = λ 2 φ(x), λ R, x D(φ)), convex and l.s.c. Let T T : D(T T ) X E be defined by D(T T ) := {x D(T ): T x D(T )} and T T x := T (T x), x D(T T ). Show that φ = T T. 3 Semi-discrete approximation In this section, we shall consider the problem of minimization of the functional Φ(h, x; ) in the case of a metric space in order to define J h x. Observe that in the Hilbert space 8

19 case, J h x has the following interpretation. Let A be the operator defined by A := ψ +αi and let u be a solution to u(t) + Au(t) 0, t > 0 with initial value x. Then J h x can be interpreted as the approximation of u at h obtained by using the Euler implicit scheme (u(h) u(0)) + Au(h) 0. Iterating we obtain a so-called semi-discrete approximation n of u: x, J h x, (J h ) 2 x,..., (J h ) n x. Given t > 0 and setting h := t, the next step consists of n proving that (J t/n ) n x tends to u(t). In our situation, we shall prove, as in Crandall-Ligett thereom, that lim n (J t/n ) n exists and that this limit is a gradient flow starting at x. This will be done in the next sections. Let (X, d) be a metric space and φ: X (, + ] be proper. Motivated by the results of the preceding section, we can formulate the condition introduced in [AGS] which plays the role of the convexity of φ αe in the Hilbert case. (H ) There exists α R such that for every x, y 0, y D(φ), there exists a map γ : [0, ] D(φ) satisfying γ(0) = y 0, γ() = y for which the following inequality holds: [ ] 2h d2 (x, γ(t)) + φ(γ(t)) ( t) 2h d2 (x, y 0 ) + φ(y 0 ) (3.) [ ] ( ) + t 2h d2 (x, y ) + φ(y ) h + α 2 t( t)d2 (y 0, y ), for every t belonging to (0, ) and every h > 0 with + αh > 0. Remark 3.. In (2.6), γ(t) = ( t)y 0 + ty, t [0, ], which is independent of x. From now on, we shall always assume that the functional φ is proper and satisfies (H ). So when we use α R, it will always refer to the α of assumption (H ). In particular, when we say h I α it means α of (H ). Given x X and h I α we shall define Φ(h, x; y) as in (2.9) so we have Φ: I α X X (, ]. Lemma 3.. Let φ: X (, ] be proper and satisfy (H ). Let h I α and x D(φ). Then (i) If γ := inf{φ(h, x; y): y X} >, then, for every y, z D(φ), (3.2) 8 ( h + α)d2 (y, z) 2 (Φ(h, x; y) γ) + (Φ(h, x; z) γ). 2 (ii) If x D(φ) is a global minimizer of Φ(h, x; ), then for every z D(φ), (3.3) Φ(h, x; z) Φ(h, x; x) 2 ( h + α)d2 ( x, z). Proof. Notice that γ < since Φ(h, x; ) is proper. (i) Use (H ) with y 0 := y, y := z, t = 2 and observe γ Φ(h, x; γ( 2 )). (ii) Use (H ) with y 0 := z, y := x. We obtain 2h d2 (x, x) + φ( x) = Φ(h, x; x) Φ(h, x; γ(t)) ( t)[ 2h d2 (x, z) + φ(z)] + t[ 2h d2 (x, x) + φ( x)] 2 ( h + α)t( t)d2 ( x, z), 9

20 t (0, ). Hence ( t)[ 2h d2 (x, x) + φ( x)] ( t)[ 2h d2 (x, z) + φ(x)] 2 ( h + α)td2 ( x, z)]. Divding by t and letting t tend to we arrive at (3.3). Remark 3.2. i) It is obvious that if x D(φ) satisfies (3.3) it is a global minimizer of Φ(h, x; ). Moreover, it follows form (3.3) that Φ(h, x; ) possesses at most one global minimizer. ii) The RHS of (3.3) (which would simply be equal to zero without (H )) will play an essential role in the proof of existence of a gradient flow). iii) Noticing that h + α > 0 in (3.2) we see that any minimizing sequence {x n} D(φ) is a Cauchy-sequence in (X, d). For the existence theorem, we will assume (X, d) to be complete which insures the existence of a limit, denoted by x X. Assuming moreover φ, hence also Φ(h, x; x) to be l.s.c., we obtain Φ(h, x; x) lim Φ(h, x; x n ) = γ <. n Hence x D(Φ(h, x; )) = D(φ). Therefore < γ Φ(h, x; x) γ < which implies that x is a (the) global minimizer of Φ(h, x; ). In view of the preceding remarks, it follows that under these additional assumptions the boundedness from below of Φ(h, x; ) implies the existence of a minimizer. Therefore we introduce as in [AGS, (2.4.0)], the next and last assumption on φ, namely: (H 2 ) There exist x D(φ), r > 0 and m R such that φ(y) m for every y X satisfying d(x, y) r. Lemma 3.2. ([AGS, Lemma 2.4.8, p. 52]) Let φ : X (, + ] be proper and satisfy (H ) and (H 2 ). Let α be as in (H ) and x, r, m be as in (H 2 ). Then for every y X (3.4) { φ(y) m if d(x, y) r, φ(y) c bd(x, y) + 2 αd2 (x, y) if d(x, y) > r, where c := φ(x ) and b := r (φ(x ) m ) 2 α +r with α + := max(α, 0). Proof. The first part of (3.4) is simply (H 2 ). We prove the second part. Assume y D(φ) with d(x, y) > r. From (H ) with x := x, y 0 := x, y := y and t := r d(x,y) (0, ) we find y := γ(t) D(φ) independent of h I α such that (3.5) [ ] 2h d2 (x, y ) + φ(y ) ( t) 2h d2 (x, x ) + φ(x ) [ ] ( ) + t 2h d2 (x, y) + φ(y) h + α 2 t( t)d2 (x, y) for every h I α. Multiplying by h (> 0) and letting h tend to zero in (3.5) we get 2 d2 (x, y ) t2 2 d2 (x, y) = 2 r2, 20

21 hence by (H 2 ) (3.6) φ(y ) m. Using (3.6), the nonnegativity of the first term in (3.5) and d(x, x ) = 0 we obtain φ(y) φ(x ) t (φ(x ) m ) ( h + α ) t 2 d2 (x, y) + α 2 d2 (x, y). In case α 0 we let h tend to + and in case α < 0 we let h tend to α. Using the definition of t we obtain (3.4). As a simple consequence of Lemma 3.2 we obtain Corollary 3.. Let φ : X (, + ] be as in Lemma 3.2, and α R be as in (H ). Then for every h > 0 satisfying + α > 0, for every x X, M > 0 there exist β > 0 and h δ R such that (3.7) Φ(h, x; y) βd 2 (x, y) + δ Proof (sketch). Use for every x X such that d(x, x) M and for every y X. d 2 (x, y) ( ε 2 )d 2 (x, y) M 2 (/ε 2 ) and for 0 < ε, η <. d 2 (x, y) ( + η 2 )d 2 (x, y) + ( + /η 2 )d 2 (x, x) Under the assumptions of Corollary 3. the function y Φ(h, x; y) is bounded from below. We define φ h (x) as its infimum on X. Definition 3.. Let φ be as in Lemma 3.2, h + α > 0 with h > 0 and α as in (H ). (3.8) φ h (x) := inf Φ(h, x; y). y X Remark 3.3. ) φ h is a map from X into R. 2) The notation φ h is consistent with the notation of Section 2. Indeed, in Section 2 φ h (x) := Φ(h, x; J h x) where J h x is the unique minimizer of y Φ(h, x; y). In this section the existence and uniqueness of such a minimizer will be obtained only for x D(φ). Lemma 3.3. Let φ : X (, + ] be proper, l.s.c. and satisfy (H ), (H 2 ). Then for every h I α the function φ h : X R is continuous and for every x D(φ) the function X y Φ(h, x; y) possesses a unique global minimizer element of D(φ) which we denote by J h x. Moreover the map D(φ) x J h x D(φ) is continuous. Proof.. Continuity of φ h. Let x n, x X, n, be such that Φ(h, x n ; y), n, hence lim d(x n, x) = 0. Let y D(φ), then φ h (x n ) n lim φ h(x n ) lim Φ(h, x n ; y) = Φ(h, x; y). n n 2

22 Taking the infimum over y D(φ) we get Let y n D(φ), n, be such that lim φ(x n) φ h (x) <. n Φ(h, x n ; y n ) φ h (x n ) + n, n. In view of Corollary 3. there exists C > 0 such that d(x, y n ) C, n. We have φ h (x) Φ(h, x; y n ), n, hence { φ h (x) lim Φ(h, x; y n ) = lim n n 2h d2 (x, y n ) } h d(x n, x)d(x, y n ) + φ(y n ) (since d(x, y n ) is bounded) { ( = lim d(x, yn ) d(x, x n ) ) 2 + φ(yn )} 2h n lim n { } 2h d2 (x n, y n ) + φ(y n ) lim φ h (x n ). n 2. Global minimizer. Let x D(φ) and let {y n } n D(φ) be a minimizing sequence, i.e. lim Φ(h, x; y n ) = n φ h (x). As in the proof in Section 2, in view of the lower semicontinuity of Φ(h, x; ) and the completeness of (X, d), it is sufficient to prove that (y n ) n is a Cauchy sequence. If y denotes the limit, note that Φ(h, x; y) < hence y D(φ). In order to show that (y n ) is a Cauchy sequence we use assumption (H ) with x := x n, y 0 := y n, y := y m, t =, where 2 (x n ) n D(φ) such that lim d(x n, x) = 0. Let C > 0 be such that d(x n, x) C, n n. From (H ) we obtain the existence of y n,m D(φ) satisfying Φ(h, x n ; y n,m ) 2 Φ(h, x n; y n ) + 2 Φ(h, x n; y m ) 8 Since Φ(h, x n ; y n,m ) φ h (x n ), we get ( h + α ) d 2 (y n, y m ). (3.9) d 2 (y n, y m ) 4( h + α ) [( Φ(h, xn ; y n ) φ h (x n ) ) + Φ(h, x n ; y m ) φ h (x n ) ], for m, n. Next we show that the right-hand side of (3.9) tends to zero as m, n. By Corollary 3. we see that any minimizing sequence is bounded, in particular there exists C 2 > 0 such that d(x, y n ) C 2, n. It follows that Φ(h, x n ; y n ) Φ(h, x; y n ) = 2h d2 (x n, y n ) d 2 (x, y n ) 2h d(x n, x) ( d(x n, y n ) + d(x, y n ) ) 2h (C + 2C 2 )d(x n, x) 0 as n. In view of the continuity of φ h, we get Φ(h, x n ; y n ) φ h (x n ) Φ(h, x n ; y n ) Φ(h, x; y n ) + Φ(h, x; y n ) φ h (x) + φ h (x) φ h (x n ) 0. 22

23 Finally Φ(h, x m ; y m ) Φ(h, x n ; y m ) = 2h d2 (x m, y m ) d 2 (x n, y m ) 2h d(x m, x n ) 2(C + C 2 ) 0 as m, n. Since φ h (x n ) φ h (x m ) 0, it follows that the right-hand side of (3.9) tends to zero. Next we prove the uniqueness of the minimizer. Since every minimizing sequence is a Cauchy sequence it is easy to see that the minimizer is unique (construct a new minimizing sequence from two minimizing sequences (u n ) and (v n ) converging respectively to u and v. Then the new minimizing sequence converges to w = u = v). Next we prove the continuity of x J h x. Let x n, x D(φ), n be such that lim n d(x n, x) = 0. We have to show that lim n J h x n = J h x. In view of part 2. of the proof of Lemma 3.3 it is sufficient to prove that {J h x n } is a minimizing sequence for Φ(h, x; ). Indeed, if it is the case, {J h x n } is a Cauchy sequence whose limit is the global minimizer of Φ(h, x; ) which is denoted by J h x. First we show that {J h x n } is bounded i.e. there exists y X (hence for every y X) d(y, J h x n ) is bounded in R. Using (3.3) with x := x n and some z D(φ) ( ) we get ( ) (3.0) 2 h + α d 2 (J h x n, z) 2h d2 (x n, z) + φ( z) φ h (x n ), n. Since x n x and φ h is continuous, the RHS of (3.0) is bounded, hence {J h x n } is bounded. Let {y n } be a minimizing sequence for Φ(h, x; ). We know that {y n } is a Cauchy sequence, hence also a bounded sequence. Now we are ready to show that {J h x n } is also a minimizing sequence for Φ(h, x; ). Set Φ(h, x; J h x n ) = I (n) + I (n) 2 where I (n) := 2h d2 (x n, J h x n ) + φ(j h x n ), I (n) 2 := [ d 2 (x, J h x n ) d 2 (x n, J h x n ) ]. 2h Let {y n } D(φ) be a minimizing sequence for Φ(h, x; ) i.e. lim n Φ(h, x; y n ) = γ := inf y X Φ(h, x; y) >. Then by definition of J h x n we have I (n) 2h d2 (x n, y n ) + φ(y n ) = I (n) 3 + I (n) 4 where I (n) 3 := 2h d2 (x, y n )+φ(y n ) and I (n) 4 := 2h( d 2 (x n, y n ) d 2 (x, y n ) ). Hence Φ(h, x; J h x n ) I (n) 3 + ( I (n) 4 + I (n) ) ( (n) 2. We claim that limn I 4 + I (n) ) 2 0. Indeed I (n) 4 = 2h (d(x n, y n ) d(x, y n ))(d(x n, y n ) + d(x, y n )) 2h d(x n, x)[d(x n, y n ) + d(x, y n )]. Since {y n } is a minimizing sequence for Φ(h, x; ), it is bounded (see part 2. of the proof above). Moreover, {x n } is also bounded since d(x n, x) 0, hence [d(x n, y n ) + d(x, y n )] is bounded and lim n I (n) 4 0. Similarly I (n) 2 d(x, x 2h n)[d(x, J h x n ) + d(x n, J h x n )] and in view of the boundedness of {J h x n } we obtain lim n I (n) 2 0. Consequently lim n (I(n) 4 + I (n) 2 ) lim I (n) 4 + lim I (n) 2 0. n n Since lim n I (n) 3 = γ, we obtain lim n Φ(h, x; J h x n ) γ inf n Φ(h, x; J h x n ). Therefore lim n Φ(h, x; J h x n ) = γ which completes the proof of the continuity of y J h y. 23

24 We summarize the results of this section in Proposition 3.. Assume (X, d) complete metric space, φ: X (, + ] proper, l.s.c., (H ) and (H2), x D(φ) and h > 0, + αh > 0. Then the functional Φ(h, x; ) defined in (2.9) possesses a unique global minimizer in X denoted by J h x. Moreover J h x D(φ) and satisfies the variational inequality. (VI) ( ) 2 h + α d 2 (J h x, z) 2h d2 (x, z) + φ h (x) φ(z) for every z D(φ), where φ h is defined in (3.8). Proof. The first assertion follows from Lemma 3.3 and (VI) is a reformulation of (3.3) when x D(φ). When x D(φ) we approximate x by a sequence x n D(φ), replace x by x n in (VI) and pass to the limit using the continuity of both J h and φ h. 4 Existence of solutions to EVI The aim of this section is to establish the existence of solutions to (EVI) via the semidiscrete approximation in the case α = 0. For the general case we refer the reader to [AGS], [LN] (where Crandall-Liggett type estimates are used involving the local slope of φ denoted by φ (x)) and [CD2] (where Crandall-Liggett type estimates are used not involving φ (x)). In this section we shall make the following assumptions: (X, d) is a complete metric space, φ: X (, + ] is proper and l.s.c.. (A) There exists h 0 > 0 such that for every h (0, h 0 ] and every x D(φ), the following variational inequality: find y D(φ) satisfying (4.) for all z D(φ), possesses a solution. 2h [d2 (y, z) d 2 (x, z)] + 2h d2 (y, x) + φ(y) φ(z) Remark 4.. i) (4.) is (VI) with α = 0. It follows from Proposition 3. that if φ satisfies (H ) with α = 0 and (H 2 ), then φ satisfies (A). ii) Clearly if y D(φ) satisfies (4.), then y is a global minimizer of Φ(h, x; ). Since 2h d2 (y, z) + Φ(h, x; y) Φ(h, x; z) for every z D(φ), this global minimizer is unique (choose z D(φ) global minimizer). 24

25 Definition 4.. Let h (0, h 0 ] and x D(φ). We denote by J h x the unique solution to the variational inequality (4.) and by J h the map from D(φ) into D(φ) defined by x J h x. We set J 0 x := x and Jh 0x := x, J h nx := J h(j n h x) for every x D(φ) and n. In particular Jh = J h. As a consequence of (A) we have: there exist x D(φ), C, C 2 R such that (4.2) φ(z) C + C 2 d(x, z) for every z D(φ). Indeed setting x = x D(φ), h = h 0 in (4.) we get φ(z) φ(j h0 x ) + 2h 0 d 2 (x, J h0 x ) 2h 0 [d(j h0 x, z) + d(x, z)] d(x, J h0 x ). Then (4.2) follows from the triangle inequality. Crandall-Liggett estimates Our first goal is to show that the sequence {J m t/m x}, t > 0, x D(φ), t/m < h 0 is a Cauchy-sequence in (X, d). We proceed as in [CL7]. This is possible thanks to the following Lemma 4.. Let x, y D(φ), γ, δ > 0, then (4.3) d 2 (J γ x, J δ y) γ γ + δ d2 (J γ x, y) + δ γ + δ d2 (x, J δ y). Proof. Neglecting the third term in (4.), which is negative, we get [ d 2 (J γ x, z) d 2 (x, z) ] + φ(j γ x) φ(z) 2γ [ d 2 (J δ y, ẑ) d 2 (y, ẑ) ] + φ(j δ y) φ(ẑ) 2δ for every z, ẑ D(φ). Setting z := J δ y, ẑ := J γ x and adding the two inequalities we obtain ( γ + ) δ 2 d2 (J γ x, J δ y) 2δ d2 (J γ x, y) + 2γ d2 (x, J δ y). Multiplying by 2 γδ γ+δ we arrive at (4.3). Remark 4.2. In case (X, ) is a real Banach space and A X X is m-accretive [See Appendix 4], and J h := (I + ha), h > 0 we have J γ x J δ y γ γ + δ J γx y + δ γ + δ x J δy for every x, y X and γ, δ > 0. Indeed by accretivity of A we get J γ x J δ y J γx J δ y + γδ [ γ + δ γ (x J γx) δy)] δ (y J γ γ + δ J γx y + δ γ + δ x J δy, 25

26 using the fact that (J γ x, γ (x J γx)) A and (J δ x, δ (y J δy)) A. Next we set for x D(φ), γ, δ > 0: (4.4) a i,j := 2 d2 (J i γx, J j δ x), i, j N 0. Using (4.3) and a double induction in i and j we obtain (4.5) a i,j γ γ + δ a i,j + δ γ + δ a i,j, i, j N >0 Remark 4.3. Similarly if (4.6) b i,j := J i γx J j δ x in Remark 4.2 we arrive at (4.7) b i,j γ γ + δ b i,j + δ γ + δ b i,j, i, j. In view of Lemma A3. of Appendix 3 we can estimate b i,j in terms of b i,0 and b 0,j. We have b i,0 = Jγx i x = (Jγx i Jγ i x) + (Jγ i x Jγ i 2 x) + + J γ x x i Jγ k x Jγ k x. k= Using the fact that [J γ ] Lip we obtain J k γ x J k γ x J γ x x for every k, hence b i,0 i J γ x x. Next we make the stronger assumption x D(A). Noticing again that (J γ x, γ (x J γx)) A, we obtain by accretivity of A: x J γ x (x J γ x) + γ[y γ (x J γx)] = γ y for any y Ax. It follows that { } (4.8) Ax := sup γ x J γx : γ > 0 inf{ y : y Ax} <, (4.9) x J γ x γ Ax, (4.0) b i,0 (iγ) Ax, i. (4.) b 0,j (jδ) Ax, j. Notice that instead of x D(A) we could choose x X such that Ax <. Applying Lemma A3. with K := Ax, r = we obtain for m, n, (4.2) J m γ x Jδ n x Ax (mγ nδ) 2 + γδ(m + n). In particular choosing γ := t, δ := t, t > 0 we see that {J m m n t/mx} is a Cauchy-sequence. We denote its limit by u(t). Moreover u(t) x u(t) J m t/m x + J m t/m x x u J m t/m x + t Ax 26

27 for every m. Hence u(t) x t Ax. In particular lim t 0 u(t) = x. Finally we obtain some regularity in t by setting γ := t, δ := s, 0 < s < t. We obtain m m and by what preceeds also u(t) u(s) Ax t s, s, t > 0 (4.3) u(t) u(s) Ax t s, s, t 0. Hence u Lip([0, ); X) and (4.4) [u] Lip Ax. In order to find estimates for a m,n we proceed as in Remark 4.3, so we need to estimate a i,0 and a 0,j. We have d 2 (J i γx, x) ( i k= d(j k γ x, J k γ x)) 2 i i k= d 2 (J k γ x, J k γ x) by the Cauchy-Schwarz inequality. We cannot use as in Remark 4.3 the inequality [J h ] Lip. However we have Lemma 4.2. Let x D(φ). Then i) (4.5) φ(j h x) h d2 (J h x, x) + φ(j h x) φ(x) for every h (0, h 0 ). ii) Given T > 0 there exists K := K(x, T ) > 0 such that (4.6) for every h (0, h 0 ) and l satisfying 2 d2 (J l hx, x) K(hl) (4.7) lh T. Proof. i) (4.5) follows from (4.) where z := x, y := J h x. ii) From (4.5) we obtain where hl T, l. Adding we get d 2 (J h x, x) h (φ(x) φ(j h x)), d 2 (J 2 hx, J h x) h ( φ(j h x) φ(j 2 hx) ),. d 2 (Jhx, l J l h x) h ( φ(j l h x) φ(jhx) ) l, (4.8) l k= d 2 (J k hx, J k h x) h ( φ(x) φ(j l hx) ). Using (4.2) we have (4.9) φ(j l hx) C + C 2 d(x, x) + C 2 d(x, J l hx). 27

28 We estimate d 2 (J l hx, x) ( l k= d(j k hx, J k h x)) 2 l l k= d 2 (J k hx, J k h x) by the Cauchy-Schwarz inequality. Using (4.7), (4.8) and (4.9): l since l k= d 2 (Jhx, k J k h x) (lh) [φ(x) + C + C 2 d(x, x)] + 2 C2 2T (lh) + 2 d2 (x, Jhx), l [ (lh) C 2 d(x, Jhx) l (lh) 2lh d2 (x, Jhx) l + lh ] 2 C 2 2 and (lh) 2 (lh)t, l. Hence 2 d2 (Jh l x, x) K(x, T )(lh), where (4.20) K(x, T ) := φ(x) + C + C 2 d(x, x) + 2 C2 2T. Combining (4.5), (4.6), (4.7) and Lemma A3., we obtain Lemma 4.3. Let x D(φ), T > 0. Then, for every t [0, T ], J t/m x is well-defined when m > T h 0, and (4.2) lim m J m t/mx exists. Notice that for t = 0, Jt/m m x = x. Moreover, if u(t) is defined as the limit in (4.2), then the function u is /2-Hölder continuous on [0, T ] and (4.22) φ(u(t)) φ(x)(< ) for every t (0, T ]. Proof. i) existence of the limit in (4.2). Let t (0, T ]. Let m, n > T h 0, γ := t, δ := t, m n then γ, δ (0, h 0 ), Jγx, i J j δ x, i m, j n are well-defined, γm = δn = t T. Hence {a i,j } defined in (4.4) satisfy Lemma A3. with K = K(x, T ) in (4.20) and r =, in view of (4.5) and Lemma 4.. Therefore, 2 d2 (Jt/mx, m Jt/nx) n Kt m + n, which implies the limit in (4.2) exists in (4.2) thanks to the completeness of X. Choosing γ := t, δ := s with 0 < s < t T, m > T m m h 0, we obtain by Lemma A3. 2 d2 (Jt/mx, m Js/mx) m K (t s) ts m, hence d 2 (u(t), u(s)) 2K t s. Finally from Lemma 4., (4.6), (4.7) we obtain 2 d2 (Jt/m m x, x) Kt which implies d 2 (u(t), x) 2Kt. Therefore d(u(t), u(s)) 2K t s /2 for every 0 s < t T. Finally we prove (4.22). In view of (4.5) φ(j t/m x) φ(x), hence φ(jt/m 2 x) φ(j t/mx) φ(x), m > T h 0. Since φ is l.s.c., we get φ(u(t)) lim m φ(jt/m m x) φ(x). 28

29 Remark 4.4. It follows from the proof of Lemma 4.3 that 2 d2 (u(t), J n t/n x) Kt n. This rate of convergence is not optimal. Indeed in [AGS, Theorem 4.0.4] under assumptions (H ) with α = 0 and (H 2 ) it is shown that d 2 (u(t), J n t/n x) t n (φ(x) φ t/n(x)). For related results, we refer the reader to the interesting paper [NS]. In the next lemma we establish an estimate which is useful for weakening the condition x D(φ). Lemma 4.4. Let x D(φ), y D(φ) and T > 0. Then J t/m x and J t/m y are well-defined for every t (0, T ] and m > T h 0 and we have (4.23) lim m d(j m t/mx, J m t/my) d(x, y). Proof. As in the proof of of Lemma 4.3 we see that J t/m x, J t/m y are well-defined for t [0, T ] and m > T h 0. For t = 0 (4.23) is trivial. Let t (0, T ] and m > T h 0. Set γ := t m. Deleting the third term in (4.) we obtain 2γ (d2 (Jγ k y, z) d 2 (Jγ k y, z)) φ(z) φ(jγ k y), 2γ (d2 (Jγ k x, ẑ) d 2 (Jγ k x, ẑ)) φ(ẑ) φ(jγ k x), k =,..., m. Setting z := Jγ k x and ẑ := Jγ k y, adding and multiplying by γ, we obtain d 2 (J k γ x, J k γ y) d 2 (J k γ Summing over k from to m we get x, J k y) 2γ[φ(J k y) φ(jγ k y)]. (4.24) d 2 (J m γ x, J m γ y) d 2 (x, y) + 2γ[φ(y) φ(j m γ y)]. γ Next we replace γ by t/m and observe that lim m Jt/m m y exists by Lemma 4.3. In view of the lower semicontinuity of φ, {φ(jt/m m y)} is bounded, therefore taking the lim m in (4.24) we arrive at (4.23). In the next lemma we assume x D(φ). Lemma 4.5. Let x D(φ) and T > 0. Then, for m > T h 0 and t [0, T ], J t/m x is well-defined and (4.25) lim m J m t/mx exists. Moreover, if u(t) is defined as the limit in (4.25), then (4.26) u C([0, T ]; X), γ (4.27) u(t) D(φ) for t (0, T ], and u satisfies (4.28) 2 d2 (u(t 2 ), z) 2 d2 (u(t ), z) (t 2 t )[φ(z) φ u(t 2 )] for all 0 < t < t 2 and all z D(φ). 29