A GAGLIARDO NIRENBERG INEQUALITY, WITH APPLICATION TO DUALITY-BASED A POSTERIORI ESTIMATION IN THE L 1 NORM

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1 27 Kragujevac J. Math. 30 ( A GAGLIARDO NIRENBERG INEQUALITY, WITH APPLICATION TO DUALITY-BASED A POSTERIORI ESTIMATION IN THE L 1 NORM Endre Süli Oxford University Computing Laboratory, Wolfson Building, Parks Road, Oxford OX1 3QD, United Kingdom ( endre.suli@comlab.ox.ac.uk (Received October 25, 2006 Abstract. We prove the Gagliardo Nirenberg-type multiplicative interpolation inequality v L 1 (R n C v 1/2 Lip (R n v 1/2 BV(R n v Lip (R n BV(R n, where C is a positive constant, independent of v. Here Lip (R n is the norm of the dual to the Lipschitz space Lip 0 (R n := C 0,1 0 (Rn = C 0,1 (R n C 0 (R n and BV(R n signifies the norm in the space BV(R n consisting of functions of bounded variation on R n. We then use a local version of this inequality to derive an a posteriori error bound in the L 1 ( norm, with = (0, 1 n, for a finite element approximation to a boundary-value problem for a first-order linear hyperbolic equation, under the limited regularity requirement that the solution to the problem belongs to BV(. 1. INTRODUCTION The aim of this paper is to establish the following Gagliardo Nirenberg-type multiplicative interpolation inequality: there exists a constant C > 0, such that v L 1 (R n C v 1/2 Lip (R n v 1/2 BV(R n v Lip (R n BV(R n, (1

2 28 where Lip (R n is the norm of the dual to the Lipschitz space Lip 0 (R n := C 0,1 0 (R n = C 0,1 (R n C 0 (R n and BV(R n signifies the norm in the space BV(R n consisting of functions of bounded variation on R n. Here, for k N 0, C k 0(R n denotes the set of k-times continuously differentiable functions with compact support in R n and C 0 (R n = k=0 Ck 0(R n ; for the sake of notational simplicity, we write C 0 (R n instead of C 0 0(R n. We refer to text of Meyer [8], particularly Theorems 17 and 18 on p.129 in Section 2.2, for the statement and proof of the closely related improved Gagliardo Nirenberg inequality due to Cohen, Dahmen, Daubechies and DeVore, according to which there exists a constant C > 0 such that, for every function v that belongs to the intersection of the homogeneous Besov space Ḃ 1, (R n with BV(R n, v L 2 (R n C v 1/2 Ḃ 1, (R n v 1/2 BV(R n. (2 In (2, compared with the inequality (1 established here, instead of L 1 (R n the left-hand side of the inequality includes the L 2 (R n norm, while the right-hand side contains the norm Ḃ 1, (R n instead of the dual Lipschitz norm Lip (R n. We note in passing that the dual Lipschitz norm also appears in the articles by Tadmor [11] and Nessyahu & Tadmor [9], for example, in the analysis of numerical methods for scalar hyperbolic partial differential equations. We begin, in Section 2, by establishing a local version of (1. We then extend this local inequality to the whole of R n in Section 3. In the final section, Section 4, we use the local version of the inequality (1 on, where = (0, 1 n, to derive a residual-based a posteriori bound in the L 1 ( norm on the error between the analytical solution of a boundary-value problem for a first-order linear hyperbolic equation and its finite element approximation, under the limited regularity requirement that the analytical solution to the problem belongs to BV(. A posteriori error bounds are crucial building blocks of adaptive finite element algorithms, aimed at optimally distributing the computational mesh so as to accurately capture the analytical solution in a certain, prescribed, norm, or a linear or nonlinear functional of the analytical solution. The mathematics of a posteriori error estimation is an active

3 29 field of research. We shall not attempt to survey this thriving and broad subject here; instead, we refer the reader to the survey articles and monographs [1, 3, 4, 5, 6, 12] listed in the bibliography. 2. INTERIOR BOUND Suppose that R n is either a bounded open set in R n with Lipschitz continuous boundary or = R n. We begin by deriving an interior version of inequality (1. For this purpose we consider a function K C 0 (R n such that K(x 0 and K( x = K(x for all x in R n, supp(k is the unit ball B 1 = {x R n : x 1}, and R n K(ξ dξ = 1. Let us suppose that is a bounded open subset of with, let d R + { } denote the distance between and, and suppose that δ (0, d. In particular if = R n, then d = ; otherwise 0 < d <. We consider the function K δ C 0 (R n defined by K δ (x := 1 ( x δ K. n δ For any ψ L (R n, with supp(ψ, we define ψ δ = ψ K δ, where signifies convolution over R n. Observe that since supp(ψ δ supp(ψ + supp(k δ for all δ (0, d, ψ δ belongs to C 0 (R n. Since supp(ψ δ, it follows that ψ δ Lip 0 ( = C 0,1 ( C 0 (, where, for a bounded open set, C 0 ( denotes the set of all uniformly continuous functions defined on which vanish on. For = R n, Lip 0 (R n has been defined above. We begin by showing the following result.

4 30 Lemma 1. Let ψ L (R n with supp(ψ, and define ψ δ (x = ψ K δ. Further, let B 1 denote the unit ball in R n centred at 0, and define C 1 := K(ξ dξ; B 1 then, for any δ in (0, d, ψ δ Lip( C 1 δ 1 ψ L (. Here Lip( is the norm of the space Lip 0 (, defined by w(x w(x w Lip( := sup, w Lip x, x ; x x x x 0 (. Proof. Recalling the definition of convolution, it is immediate that ψ δ (x ψ δ (x ψ L (R n K δ (x y K δ (x y dy R n for any pair of points x, x in. Noting that K δ C 0 (R n and applying the Integral Mean Value Theorem it follows that K δ (x y K δ (x y x x 1 0 K δ (θ(x y + (1 θ(x y dθ for any pair of points x, x in and any y R n. Upon integrating both sides of the last inequality with respect to y R n we deduce that K δ (x y K δ (x y dy R n ( 1 x x K δ (x y + θ(x x dθ dy, R n 0 for any pair of points x, x in. Interchanging the order of integration on the righthand side and performing the change of variables y z(y = x y + θ(x x for θ fixed in [0, 1] and x, x fixed in, recalling the translation-invariance of the Lebesgue measure on R n we find that K δ (x y K δ (x y dy x x K δ (z dz. R n R n

5 31 Now ( x K δ (x = δ 1 n ( K, δ and therefore, upon noting that supp(k = B 1, R n K δ (z dz = δ 1 C 1, with C 1 as in the statement of the Lemma. Returning to the first inequality in the proof and recalling that supp(ψ, we deduce the required result. Next we prove the following lemma. Lemma 2. Suppose that ϕ W 1,1 (; then, there exists a positive constant C 2 such that, for any δ in (0, d, ϕ ϕ δ L 1 ( C 2 δ ϕ W 1,1 (, where ϕ δ := ϕ K δ, with ϕ defined to be identically zero over the set R n \, and C 2 := B 1 max ξ B1 K(ξ, with B 1 denoting the Lebesgue measure of the unit ball B 1 in R n. Proof. Suppose, to begin with, that ϕ W 1,1 ( C (. Given x, we have the following sequence of equalities: ϕ(x ϕ δ (x = ϕ(x ϕ(yk δ (x y dy R n = ϕ(x ϕ(x yk δ (y dy R n = [ϕ(x ϕ(x y] K δ (y dy R n = δ n [ϕ(x ϕ(x y] K(y/δ dy y <δ = [ϕ(x ϕ(x δξ] K(ξ dξ. B 1 The last equality implies that ϕ(x ϕ δ (x C 3 B 1 ϕ(x ϕ(x δξ dξ, x,

6 32 where C 3 = max ξ B1 K(ξ. Upon integration over we deduce that ϕ ϕ δ L 1 ( C 3 ϕ(x ϕ(x δξ dξ dx B 1 = C 3 ϕ( ϕ( δξ L B 1 ( dξ. (3 1 In order to further bound the right-hand side in (3, observe that ϕ(x ϕ(x δξ = δξ 1 for any x in and any ξ B 1. Consequently, so that ϕ(x ϕ(x δξ δ ϕ( ϕ( δξ L 1 ( δ 0 ( ϕ(θx + (1 θ(x δξ dθ ϕ(x (1 θδξ dθ ϕ(x (1 θδξ dx dθ. (4 Upon performing the change of variables x z(x = x (1 θδξ for fixed θ in [0, 1] and ξ B 1, it follows from (4 that 1 ϕ( ϕ( δξ L 1 ( δ δ Substituting (5 into (3 gives 0 (1 θδξ ϕ(z dz dθ ϕ(z dz = δ ϕ W 1,1 (. (5 ϕ ϕ δ L 1 ( C 2 δ ϕ W 1,1 (, where C 2 = C 3 B 1. This proves the desired inequality for ϕ W 1,1 ( C (. For ϕ W 1,1 ( the inequality then follows by density of W 1,1 ( C ( in the Sobolev space W 1,1 (. Now we extend this result to functions of bounded variation. Let us suppose for this purpose that u L 1 (; we then put { } Du := sup u div v dx : v = (v 1,..., v n [C 1 0(] n, v(x 1 for x.

7 33 A function u L 1 ( is said to be of bounded variation on if u BV( := Du <. The linear space of functions of bounded variation of is denoted by BV( and is equipped with the norm BV( defined by u BV( := u L 1 ( + u BV(. Thus, BV( is the set of functions u L 1 ( whose weak gradient Du is a bounded (vector-valued Radon measure with finite total variation u BV(. If u W 1,1 ( ( BV(, then u BV( = Du = u L 1 (, where u is the distributional gradient of u (cf. [10]. We recall the following approximation result of Anzellotti and Giaquinta [2]; see also Theorem 1.17 in Giusti [7] and Theorem in Ziemer [13]. Theorem 1. For each function v BV( there exists a sequence {ϕ j } j=1 of functions in W 1,1 ( C ( such that lim v ϕ j dx = 0, j Dϕ j dx = Dv. lim j Now, suppose that v BV( (extended by zero to the whole of R n and consider a sequence {ϕ j } j=1 of functions in W 1,1 ( C (, as in Theorem 1; namely, and lim v ϕ j L j 1 ( = 0 lim ϕ j W j 1,1 ( = v BV(. By Lemma 2, we also have, with ϕ j defined to be identically 0 in the set R n \, that ϕ j ϕ j K δ L 1 ( C 2 δ ϕ j W 1,1 (.

8 34 Passing to the limit as j, we deduce that v v K δ L 1 ( C 2 δ Dv = C 2 δ v BV(. Thus we have proved the following lemma. Lemma 3. Suppose that v BV(, extended to R n \ as the identically zero function; then, for each δ (0, d, v v δ L 1 ( C 2 δ v BV(, where v δ = v K δ and C 2 is the same positive constant as in Lemma 2. Now suppose that v Lip ( BV( (extended by zero onto the whole of R n. With δ (0, d as above, we then write v(xψ(x dx = v(xψ δ (x dx + v(x[ψ(x ψ δ (x] dx, where, as before, ψ L ( (extended by zero onto the whole of R n. Further, recalling that K( z = K(z for all z R n, we deduce that v(x[ψ(x ψ δ (x] dx = [v(x v δ (x]ψ(x dx. Thereby, v(xψ(x dx v Lip ( ψ δ Lip( + v v δ L 1 ( ψ L (. Recalling Lemmas 1 and 2, we find that v(xψ(x dx { } C 1 δ 1 v Lip ( + C 2 δ v BV( ψ L (. Let us, in particular, choose ψ = χ sgn(v, where χ is the characteristic function of the open set. We then deduce that v L 1 ( C 1 δ 1 v Lip ( + C 2 δ v BV( δ (0, d, v Lip ( BV(. (6 Trivially, (6 implies that v L 1 ( C 1 δ 1 v Lip ( + C 2 δ v BV( δ (0, d, v Lip ( BV(, (7

9 35 where δ = dist(,,. We begin by considering the case of d < ; the case of d =, corresponding to the choice of = R n, will be discussed in the next section. For v Lip ( BV(, v 0, we consider the mapping δ R + f(v; δ := C 1 δ 1 v Lip ( + C 2 δ v BV(. The function f(v; is strictly positive on R + and attains its minimum value at C 1 v Lip δ = δ 0 := (. C 2 v BV( We shall consider two mutually exclusive cases, depending on the size of δ 0 > 0 relative to d <. Case 1: δ 0 (0, d. On equilibrating the two terms on the right-hand side of (7 by choosing δ = δ 0, we get v L 1 ( C 4 v 1/2 Lip ( v 1/2 BV(, (8 where C 4 = 2 C 1 C 2. For future reference, we rewrite (8 in the following equivalent form: v L 1 ( C 1 C 2 v 1/2 Lip ( v 1/2 BV( + C 1 C 2 v 1/2 Lip ( v 1/2 BV(, (9 the inequality being valid for all v Lip ( BV( such that δ 0 (0, d. Since C 2 v BV( = 1 ( 1 = max C 1 v Lip ( δ 0 d, 1 ( 1 = max δ 0 d, C 2 v BV(, C 1 v Lip ( we can rewrite (9 as follows: ( 1 v L 1 ( C 1 v Lip ( max d, C 2 v BV( + C 1 C 2 v 1/2 C 1 v Lip Lip ( v 1/2 BV(. (10 ( Case 2: δ 0 [d,. In this case, the function f(v; is strictly monotonic decreasing in the interval δ (0, d, and therefore inf δ (0,d f(v, δ = f(v, d. Hence, v L 1 ( C 1 (d 1 v Lip ( + C 2 d v BV( C 1 (d 1 v Lip ( + C 2 δ 0 v BV( C 1 (d 1 v Lip ( + C 1 C 2 v 1/2 Lip ( v 1/2 BV(.

10 36 Now, since d δ 0, we have that ( 1 1 = max d d, 1 = max δ 0 ( 1 d, C 2 v BV( C 1 v Lip ( so, once again, ( 1 v L 1 ( C 1 v Lip ( max d, C 2 v BV( + C 1 C 2 v 1/2 C 1 v Lip Lip ( v 1/2 BV(. (11 ( Combining Cases 1 and 2, as expressed by inequalities (10 and (11, we see that, irrespective of the relative magnitudes of δ 0 and d, we have ( 1 v L 1 ( C 1 v Lip ( max d, C 2 v BV( + C 1 C 2 v 1/2 C 1 v Lip Lip ( v 1/2 BV(, (12 ( for all v Lip ( BV(\{0}, where and where d := dist(, (0,., 3. GLOBAL BOUND We shall now consider the case when = R n, and extend the result stated in (12 from to the whole of R n. On taking d = in (7 and equilibrating the two terms on the right-hand side of (7 by choosing δ = δ 0 (0, d = (0,, we have that v L 1 ( C 4 v 1/2 Lip (R n v 1/2 BV(R n v Lip (R n BV(R n. Now, let us assume that { j } j=1 is a nested sequence of bounded open sets, 1 2 R n such that R n = j=1 j. On taking = j in the last inequality, we see that v L 1 ( j C 4 v 1/2 Lip (R n v 1/2 BV(R n v Lip (R n BV(R n, j = 1, 2,..., where C 4 = 2 C 1 C 2 is independent of j. Defining v j (x := χ j (xv(x, j = 1, 2,..., where χ j is the characteristic function of j, we can restate this inequality as follows: v j L 1 (R n C 4 v 1/2 Lip (R n v 1/2 BV(R n v Lip (R n BV(R n, j = 1, 2,....

11 37 As v(x = lim j v j (x for a.e. x R n, it follows by Fatou s Lemma that v L 1 (R n lim inf j v j L 1 (R n. Consequently, passing to the limit over j gives v L 1 (R n C 4 v 1/2 Lip (R n v 1/2 BV(R n v Lip (R n BV(R n, (13 which is the desired multiplicative interpolation inequality. 4. APPLICATION IN A POSTERIORI ERROR ANALYSIS Suppose that = (0, 1 n, and let Γ signify the union of all (n 1-dimensional open faces of. Let us denote by ν the unit outward normal vector to Γ. Suppose that b [C 0,1 ( ] n, c C 0,1 (, and f BV(. We shall suppose that there exists a positive contact c 0 such that c(x + 1 b(x c 2 0 for all x, and that the components b i, i = 1,..., n, of the vector field b are strictly positive on. We then consider the first-order linear hyperbolic partial differential equation Lu := (bu + cu = f in, (14 supplemented by the boundary condition u Γ = g, (15 where Γ = {x Γ : b(x ν(x < 0} and g L 1 (Γ ; analogously, we define Γ + = {x Γ : b(x ν(x > 0}. The weak formulation of the boundary-value problem amounts to finding u BV( such that u b v dx + (b ν u v ds + Γ + c u v dx = f v dx (b ν g v ds (16 Γ for all v C 0,1 (. Now, let {T h } h>0 denote a shape-regular family of partitions of into disjoint open simplices κ whose union is ; for κ T h we define h κ := diam(κ and let h =

12 38 max κ Th h κ. Further, let V hp denote the set of all continuous piecewise polynomials of degree p 1 defined on T h. that The finite element approximation of (16 is defined as follows: find u h V hp such u h b v h dx + (b ν u h v h ds + c u h v h dx Γ + = f v h dx (b νg v h ds v h V hp. (17 Γ On denoting the expression on the left-hand side of (17 by B(u h, v h, it is easily seen that (w h, v h V hp V hp B(w h, v h R is a bilinear functional, and B(v h, v h c 0 v h 2 L 2 ( v h V hp. Since V hp is a finite-dimensional linear space it then follows that problem (17 has a unique solution u h V hp. To derive an a posteriori bound on the error u u h in the L 1 ( norm where, we begin by establishing an a posteriori error bound in the Lip ( norm using a duality argument which involves the formal adjoint L : z b z + cz of the differential operator L. Let ψ Lip 0 (, and let z C 0,1 ( denote the corresponding (classical solution to the hyperbolic boundary-value problem L z = ψ in subject to z Γ+ = 0. Consequently, z C 0,1 ( also satisfies the following identity: w b z dx + (b νw z ds + c w z dx = w ψ dx w BV(. (18 Γ +

13 39 Thus, for any z h V hp, (u u h ψ dx = (u u h b z dx + (b ν(u u h z ds + c (u u h z dx Γ + = (u u h b (z z h dx + (b ν(u u h (z z h ds Γ + + c (u u h (z z h dx = f (z z h dx (b ν g (z z h dx Γ [ ] u h b (z z h dx + (b ν u h (z z h ds + c u h (z z h dx Γ + = (f (bu h cu h (z z h dx (b ν (g u h (z z h dx. Γ On defining the internal residual R = f (bu h cu h on and the boundary residual R Γ = b ν (g u h on Γ, we then deduce the error representation formula (u u h ψ dx = R (z z h dx + R Γ (z z h dx z h V hp. Γ Hence, with h signifying the positive piecewise constant function defined on T h such that h(x = h κ for all x κ and all κ T h, (u u h ψ dx hr L 1 ( h 1 (z z h L ( + hr Γ L 1 (Γ h 1 (z z h L (Γ Thus, ( hr L 1 ( + hr Γ L 1 (Γ h 1 (z z h L ( z h V hp. (u u h ψ dx ( hr L 1 ( + hr Γ L 1 (Γ inf h 1 (z z h L z h V (. hp Using a standard approximation property of the finite element space V hp in the L ( norm, we deduce that ( (u u h ψ dx K approx hr L 1 ( + hr Γ L 1 (Γ z Lip(,

14 40 where K approx is a positive constant, dependent only on the shape-regularity of the family {T h } h>0. For the sake of simplicity, we shall assume henceforth that b is a constant vector with positive entries. Since ψ Lip 0 ( vanishes on Γ, it follows by hyperbolic regularity theory that c 0 z L ( ψ L ( and c 0 z L ( ψ L ( + c L ( z L ( ψ L ( + c 1 0 c L ( ψ L (. Furthermore, because ψ Γ = 0, we have that ψ L ( diam( ψ L (, and therefore, where z Lip( K stab ψ Lip(, ( K stab = c c 1 0 diam( c L (. Thus, (u u h ψ dx K ( stabk approx hr L 1 ( + hr Γ L 1 (Γ ψ Lip( for all ψ Lip 0 (, whereby, ( u u h Lip ( K stab K approx hr L 1 ( + hr Γ L 1 (Γ =: ApostLip. Furthermore, u u h BV( u BV( + u h BV( K stab f BV( + u h BV( =: Apost BV, where we have used that u BV( K stab f BV(. We note that, trivially, BV( L 1 ( and, when R n is bounded, as is the case in this section, L 1 ( Lip (; therefore also BV( Lip ( and thereby Lip ( BV( = BV(.

15 41 It now follows from (12 that u u h L 1 ( ( C 1 u u h 1/2 max C1 u u Lip ( d h 1/2, C Lip ( 2 u u h 1/2 BV( + C 1 C 2 u u h 1/2 u u Lip ( h 1/2 C 1 Apost 1/2 Lip max + C 1 C 2 Apost 1/2 Lip Apost 1/2 BV, in any domain, with d = dist(,. BV( ( C1 Apost 1/2, C d Lip 2 Apost 1/2 BV We note, in particular, that if C1 d Apost 1/2 Lip C 2 Apost 1/2 BV, that is if Apost Lip C 2(d 2 then the following conditional a posteriori error bound holds: the condition being inequality (19. C 1 Apost BV, (19 u u h L 1 ( 2 C 1 C 2 Apost 1/2 Lip Apost 1/2 BV, Since the quantity Apost BV featuring in the right-hand side of (19 is expected to be, at best, of size O(1 as h 0, while the left-hand side of (19 is anticipated to decay as O(h 2s with some s (0, 1/2], we expect to be able to set d := dist(, = O(h s as h 0. This is the desired a posteriori bound on the error between the analytical solution u BV( and its finite element approximation u h V hp, in terms of the computable domain and boundary residuals, the numerical solution u h and the data. Acknowledgements: I wish to express my sincere gratitude to Ricardo Nochetto (University of Maryland, Christoph Ortner (University of Oxford and Maria Schonbek (University of California at Santa Cruz for helpful comments and stimulating discussions on the subject of this paper.

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