REMARKS ON SOME MINIMIZATION PROBLEMS ASSOCIATED WITH BV NORMS

. The purpose of this paper is twofold. Firstly I present an optimal regularity result for minimizers of a 1 D convex functional involving the BV-norm, under Neumann boundary condition. This functional is a simpliﬁed version of models occuring in Image Processing. Secondly I investigate the existence of minimizers for the same functional under Dirichlet boundary condition. Surprisingly, this turns out to be a delicate issue, which is still widely open.


Introduction. Our original motivation comes from the study of minimizers of the ROF (= Rudin-Osher-Fatemi) functional
where Ω ⊂ R N is smooth and bounded, u ∈ BV (Ω) ∩ L 2 (Ω), f ∈ L 2 (Ω) is given and λ > 0 is a parameter. This functional was introduced (in a slightly different form) in [17] and it has been extensively used in Image Processing (see e.g. [5] and the references therein). After scaling we may assume that λ = 1/2 and we set It is standard that there exists a unique minimizer denoted U ∈ BV (Ω) ∩ L 2 (Ω) for the problem Inf u∈BV ∩L 2 Φ(u). ( Our first goal is to investigate the regularity of U . In variational problems one often expects a gain in regularity. For example, if we replace Ω |∇u| in (1) by 1 2 Ω |∇u| 2 , then the minimizer satisfies −∆u + u = f in Ω, ∂u ∂n = 0 on ∂Ω, and therefore u ∈ H 2 (Ω). In our situation there is a "modest" gain in regularity since U ∈ BV (Ω)∩L 2 (Ω) while f ∈ L 2 (Ω) only. Surprisingly this regularizing effect stops here as can be seen from the following simple example (see M. Bonforte and A. Figalli [2] and also T. Sznigir [20,21]). Take Ω = (0, 1) and let 0 = x 0 < x 1 < . . . < x n < x n+1 = 1. Assume that f is a step function, where 1 A denotes the characteristic function of A, and (a i ) 0≤i≤n are arbitrary constants, then for some appropriate constants (b i ) 0≤i≤n .
A regularity result due V. Caselles, A. Chambolle and M. Novaga [13] asserts that if N ≤ 7 and ∇f ∈ L ∞ loc (Ω), then ∇U ∈ L ∞ loc (Ω); moreover if Ω is convex then ∇f ∈ L ∞ (Ω) implies ∇U ∈ L ∞ (Ω). At first sight this result seems optimal. Indeed one can easily construct examples where Ω = (0, 1), f ∈ C ∞ ([0, 1]) and U / ∈ C 1 ([0, 1]), i.e., U ′ is discontinuous. (This comes from the fact that solutions of variational inequalities are not C 2 in general. And, as explained in Section 2, when N = 1, our U corresponds roughly speaking to the derivative of the solution of a variational inequality). We suspect that the result of [13] might possibly be "upgraded". Here is an improvement when N = 1.
If one insists on using the duality device in higher dimensions we are led to the functional where u ∈ X = { u ∈ L 2 (Ω; R N ); div u is a finite measure}, and f ∈ L 2 (Ω; R N ) is given. It is easy to see that Inf is achieved by a unique minimizer U ∈ X, for which we can establish the following regularity.
Problems of the type (4) were introduced in [11] but the authors did not address there the question of regularity for the minimizers.
Since we do not impose any boundary condition in (2), we expect that the minimizers U will satisfy the Neumann boundary condition Unfortunately, if N ≥ 2, we do not have sufficient information on U in order to give a meaning to (5) (the fact that ∇U ∈ L ∞ does not suffice). However when N = 1 we know that U ′ ∈ BV (by Theorem 1), and thus U ′ (0), U ′ (1) are well-defined provided f ′ ∈ BV . We will indeed prove that U ′ (0) = 0 and U ′ (1) = 0. In fact we establish a stronger conclusion: Corollary 1. Assume that Ω = (0, 1) and that f ∈ L 2 (0, 1). Then the minimizer U in (2) is constant near 0 and 1.
Open problem 2. Assume that N ≥ 2 and f ∈ C ∞ (Ω). Can one show that ∂U/∂n = 0 on ∂Ω, at least in some weak sense (to be determined)?
Next, we investigate questions similar to (2) associated with a Dirichlet boundary condition. Here is a typical example. Fix a (smooth) boundary condition g and consider the minimization problem Inf {Φ(u); u ∈ BV (Ω) ∩ L 2 (Ω), u = g on ∂Ω}. ( Recall that BV functions admit a trace in L 1 of the boundary (see e.g. [1, Theorem 3.87]) and thus the condition u = g on ∂Ω makes sense. If we take a minimizing sequence (u n ) in (6), we may assume that u n k → u in L 1 (Ω) with u ∈ BV (Ω), and it may happen that the boundary condition is "lost" in the limit, i.e., u does not satisfy u = g on ∂Ω. (Similar questions for functionals which do not involve the term Ω |u − f | 2 have been considered e.g. in [18] and [19], but the situation seems to be quite different from ours). It turns out that the existence of a minimizer in (6) is a very delicate issue: Open Problem 3. Find (necessary and sufficient) conditions on f and g such that a minimizer in (6) exists. This problem is open even for N = 1. In particular it would nice to find a statement which covers both Theorems 3 and 4 below.
We do not know any result concerning this problem when N ≥ 2. If N = 1 we have an answer in two special cases: Theorem 3. Fix a 0 , a 1 ∈ R, and consider the minimization problem A minimizer exists if and only if a 0 = a 1 = a with |a| ≤ 2, (8) and in this case the unique minimizer is the constant function a.
The above result shows that a minimizer in (6) exists only under very restrictive assumptions. A partial result in the same spirit was established by T. Sznigir [20,Theorem 3.16]. An interesting application of Theorem 3 to the study of some "regularized interpolation" problems is presented in [6]. Here is another case where the Inf in (6) is attained.
2. Proofs of Theorem 1, 2 and Corollary 1 via duality. We will first transform problem (2) with N = 1, and problem (4) with general N ≥ 1, into a variational inequality following a duality technique introduced in [3, Section I.1.3] (see also [9]); a similar idea was rediscovered in [11], but our approach is simpler. Related change of unknown in 1D are used in [22] and [2]. For the convenience of the reader we present this device in Lemma 1 below (see also the proof of Theorem 2).
We now turn to the proof of (20). As usual we denote by Sign the monotone graph which is the inverse of γ defined in (16), i.e., From (15) we see that But (14) and (19).
We now turn to the proof of (12). We have, by (25)-(26), and thus, by (19), (Recall that V n ∈ H 2 ∩ H 1 0 and U n ∈ H 1 ). Returning to (20) we obtain as above, for every u ∈ BV , lim sup Choosing in particular u = U in (32) we conclude that lim sup we infer that lim Passing to the limit in (31)(using (18)) yields the desired conclusion (12).
Proof of Theorem 1. We know that solutions of variational inequalities such as (10) enjoy the property that V ′′ ∈ BV loc provided f is sufficiently smooth; see Brezis-Kinderlehrer [BK, Theorem 3]. Keeping (11) in mind we will deduce that U ′ ∈ BV loc . In our situation we can even extend this result up to the boundary so that V ′′ ∈ BV and therefore U ′ ∈ BV , which is the desired conclusion. For this purpose we adapt the proof from [BK]. (A similar technique, applied to evolution equations, has been rediscovered in [16]). We first assume that f is smooth and we consider, as in [3], an approximation of the graph γ (defined in (16)) by smooth monotone functions γ ε such that γ ε (0) = 0. Let V ε be the (smooth) solution of the Let (θ δ ) be a smooth approximation of the graph Sign defined in (24) (e.g. θ δ (t) = t |t 2 +δ 2 | 1/2 ). Differentiating (33) and multiplying by Integrating (34) on (0, 1), using the monotonicity of θ δ , and (33), we find As δ → 0 (with fixed ε) we see that Returning to (33) we deduce from (36) that Clearly V ε → V in H 1 0 (0, 1) as ε → 0, where V is defined in Lemma 1. We conclude that (V ′ + f ) ′ ∈ BV . Keeping in mind (11) we find that the minimizer U of (2) satisfies U ′ ∈ BV and Thus estimate (37) has been established for smooth f . By a density argument (as in (23)-see also Lemma 2 below) we reach the same conclusion assuming only f ′ ∈ BV .
Proof of Theorem 2. We follow the same strategy as in the proof of Theorem 1 with some substantial modifications. We start as above with the solution V of the variational inequality From classical results on variational inequalities (see e.g. [3], [7]) we know that V satisfies the following properties: where and γ is defined in (16), and in particular V is smooth in N, ∆V ∈ BV (Ω).
We claim that U is the unique minimizer of (4) and this will complete the proof of Theorem 2. Fix u ∈ L 2 (Ω; R N ) such that div u is a finite measure .
We need to check that From (40) we have V ∈ Sign Z= Sign (div U ) by (47). Hence and Therefore On the other hand Next we claim that This needs to be justified carefully since we only know that div u is a finite measure and Ω V div u is to be understood in the sense of duality of measures and continuous functions. It suffices to invoke the following: This fact can be established as in [4, proof of Theorem 9.17].

Minimizing under Dirichlet condition. Proofs of Theorems 3 and 4.
We now turn to the minimization problem (6). When N = 1 it takes the form where f ∈ L 2 (0, 1), a 0 and a 1 are given.
Since problem (54) is delicate and need not have a solution we replace it by a relaxed problem which always admits a solution.
We start with a basic inequality familiar to the experts (see e.g. [8, Appendix 18.8]) Lemma 2. Let (u n ) be a bounded sequence in BV such that u n → u in L 1 , u n (0) → α 0 and u n (1) → α 1 . Then u ∈ BV and lim inf Clearly v n , v ∈ BV (R) and Since Combining this with (56) yields (55).
Given u ∈ BV (0, 1) set Our next lemma asserts that F is lower semi-continuous on BV ; more precisely: Lemma 3. Let (u n ) be a bounded sequence in BV (0, 1) such that u n → u in L 2 (0, 1), then u ∈ BV ((0, 1)) and Proof. Passing to a subsequence we may assume that u n (0) → α 0 and u n (1) → α 1 . From Lemma 2 we have From Lemma 3 we deduce that Denote by U the unique minimizer in (59).  On the other hand, set for δ > 0 small Clearly U δ ∈ BV and, for a.e. δ, Problem (54) usually admits no minimizer, while Problem (59) always admits a minimizer denoted U . If (by chance!) Problem (54) admits a minimizer U , then U = U . Indeed we have (by assumption) U ∈ BV, U (0) = a 0 , U (1) = a 1 and Thus F ( U ) = A = B (by Lemma 4) and U is a minimizer for (59); from uniqueness we deduce that U = U .
On the other hand if we happen to know that the minimizer U of (59) satisfies U (0) = a 0 and U (1) = a 1 , then U is a minimizer of (54). Indeed we have Conclusion. The existence of a minimizer for Problem (54) boils down to the question whether U satisfies U (0) = a 0 and U (1) = a 1 .
Proof of Theorem 3. In view of the above discussion we introduce the relaxed problem Inf where (recall that f = 0 in Theorem 3). We denote again by U the minimizer in (63). We distinguish two cases. Case 1. a 0 a 1 ≤ 0.
In Case 1 we have U = 0. Indeed for every u ∈ BV we have On the other hand G(0) = |a 0 | + |a 1 | and therefore 0 is the unique minimizer in (63). We conclude in Case 1 that our original problem (7) admits a solution only if a 0 = a 1 = 0 and in this case U = 0 is the minimizer of (7).
We now turn to Case 2. Without loss of generality we may assume that Our next result gives a complete description of U in this case.
Let ξ ∈ (0, 1) be any point of continuity of U . Recall that U ∈ BV and thus U is continuous except at a countable numbers of points. Using the constant function u ≡ U (ξ) as a testing function in (63) yields and thus we conclude that |U (ξ)| 2 = 1 0 |U | 2 for a.e. ξ. In view of (65) we deduce that U is a constant. Therefore we are led to the minimization of the function An easy inspection show that Min Proof of Theorem 4. We use a totally different strategy. We rely instead on regularity estimates as in the proof of Theorem 1. Set so that j ε (t) → |t| as ε → 0, j ε (t) ≤ |t| + εt 2 ∀t, and j ′′ ε (t) ≥ 2ε ∀t. By standard variational theory the minimization problem admits a unique minimizer U ε . Moreover U ε satisfies the equation where β ε = j ′ ε , and U ′′ ε has the same regularity as f , e.g. U ′′ ε ∈ L 2 if f ∈ L 2 . We now proceed as in the proof of Theorem 1, namely we set so that, by (68), we have and Since β ε is strictly monotone and bijective we may introduce its inverse function Combining (72) and (71) yields This is the dual problem to (67). Note that, by (70) so that V ε satisfies a Neumann condition (by contrast, in Theorem 1, U satisfied a Neumann condition and V satisfied a Dirichlet condition).
We now divide the proof in 5 steps.
Let (f n ) be a sequence of smooth functions such that f n → f in W 1,1 (0, 1) as n → ∞, f n (0) = f (0), and f n (1) = f (1) for all n. (The existence of such a sequence is a consequence of the density of C ∞ c (0, 1) in W 1,1 0 (0, 1)). Denote by U n ∈ W 1,∞ (0, 1) the minimizer given by Step 1, corresponding to f n in place of f . From (77) in Step 2 we have Therefore (U n ) is a Cauchy sequence in W 1,1 (0, 1). We may thus assume that U n → U in W 1,1 (0, 1) as n → ∞, for some U ∈ W 1,1 (0, 1) satisfying U (0) = a 0 and U (1) = a 1 . By construction we know that Passing to the limit as n → ∞ completes the proof of Step 3.
Let (f n ) be a sequence of smooth functions such that f n → f in f ∈ C([0, 1]) as n → ∞, f n (0) = f (0), f n (1) = f (1) for all n. By Steps 1 and 2 we see that the corresponding sequence (U n ) in W 1,∞ (0, 1) satisfies U n (0) = a 0 , U n (1) = a 1 for all n, and U n − U m L ∞ ≤ f n − f m L ∞ ∀m, n.
Step 5. We now turn to the last assertion in Theorem 4, i.e., U ′ ∈ BV if f ′ ∈ BV .