On the generalized mean curvature

Abstract

We study some properties of graphs whose mean curvature (in distributional sense) is a vector Radon measure. In particular, we prove that the distributional mean curvature of the graph of a Lipschitz continuous function u is a measure if and only if the distributional divergence of T u is a measure. This equivalence fails to be true if Lipschitz continuity is relaxed, as it is shown in a couple of examples. Finally, we prove a theorem of approximation in W ^sup (1,1)^ and in the sense of mean curvature of C ^sup 2^ graphs by polyhedral graphs. A number of examples illustrating different situations which can occur complete the work.[PUBLICATION ABSTRACT]

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Calc. Var. (2010) 39:491523

DOI 10.1007/s00526-010-0321-7

Calculus of Variations

On the generalized mean curvature

Elisabetta Barozzi Eduardo Gonzalez

Umberto Massari

Received: 20 May 2009 / Accepted: 22 February 2010 / Published online: 25 March 2010 Springer-Verlag 2010

Abstract We study some properties of graphs whose mean curvature (in distributional sense) is a vector Radon measure. In particular, we prove that the distributional mean curvature of the graph of a Lipschitz continuous function u is a measure if and only if the distributional divergence of T u is a measure. This equivalence fails to be true if Lipschitz continuity is relaxed, as it is shown in a couple of examples. Finally, we prove a theorem of approximation in W(1,1) and in the sense of mean curvature of C2 graphs by polyhedral graphs. A number of examples illustrating different situations which can occur complete the work.

Mathematics Subject Classication (2000) 49Q20 49Q05 53A10

1 Introduction

A function H L1(U) (U an open set of Rn+1) is said to be a variational mean curvature

of a given set E U if E locally minimizes the functional

FH (F) =

|DF| +

H(x)dx, (1)

Communicated by L. Ambrosio.

E. BarozziDipartimento di Matematica, Universit di Trento, Via Sommarive 14, 38100 Povo (TN), Italy e-mail: [email protected]

E. GonzalezDipartimento di Metodi e Modelli Matematici per le Scienze Applicate, Universit di Padova, Via Trieste 63, 35121 Padova, Italye-mail: [email protected]

U. Massari (B)

Dipartimento di Matematica, Universit di Ferrara, Via Machiavelli 35, 44100 Ferrara, Italy e-mail: [email protected]

123

492 E. Barozzi et al.

where we denote by F the characteristic function of the set F and by U|DF| the perimeter

of F in U (see for example [2,3,7,10,16]).

The motivation for this definition relies on the fact that by computing the rst variation of (1), it can be easily seen that if E is smooth in a neighborhood of a point x E U

and H is continuous at that point, then H(x) is (up to a constant factor) the classical mean curvature of E at x.

There are two basic facts about variational mean curvatures. First, the existence of a variational mean curvature HE L1(U) associated to every set E U of nite perimeter

(see [4,6]); second, the smoothness (up to a singular subset) of the boundary E of E for H L p(U), p > n + 1 (see [14,15]). This regularity result fails for p n + 1. The case

p = n +1 is particularly subtle and was studied in [12] (see also [18]). We refer to the survey

paper [11] for a more detailed account.

On the other hand, the mean curvature of non regular manifolds can be dened in a intrinsic way as follows (see, for example [1]). Let Rn be an open bounded set, let

u W(1,1)( ), dene

Mu = M = {(y, z) Rn+1 : y , z = u(y)},

and let

(y) = (1, . . . , n, n+1)(y) =

(Du(y), 1)

1 + |Du(y)|2, y

be the normal vector to Mu directed upwards.An (n + 1)-dimensional vector valued Radon measure

H = (H1, . . . , Hn, Hn+1) on M is called the generalized mean curvature of M if

divM XdHn =

X d

H =

n+1

j=1

H j X C10( , Rn+1). (2)

Here, as usual, the tangential divergence divM X of the vector eld X is dened as

divM X =

n+1

j=1

j X j, (3)

where j , j = 1, . . . , n + 1 are the tangential derivatives, that is,

j = Dj j

n+1

h=1

X jd

h Dh.

In the case when

H is absolutely continuous with respect to the Hausdorff measure Hn|M we

have

H Hn|M,

where the density

H: M Rn+1 belongs to [L1(M)]n+1. In this case (2) becomes

divM XdHn =

H XdHn X C10( , Rn+1). (4)

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On the generalized mean curvature 493

Moreover, for u C2( ), we have

H = divy

Du(y)

1 + |Du(y)|2 (y) = divyT u(y) (y), (5)

where, for short,

T u(y) =

Du(y)

1 + |Du(y)|2.

The number

H = divy

Du(y)

1 + |Du(y)|2 is called the scalar mean curvature.

It is now clear that the main motivation for definition (2) is the integration by parts given in (4) which holds in the regular case. In the case X C1( ; Rn+1), u C2( ) C1( ),

this formula becomes

divM XdHn =

H XdHn +

X NdHn1, (6)

where N is the outer normal to M on the tangent space to M.

The intimate connection between the variational mean curvature and the generalized mean curvature was shown clearly in [5]. Suppose E is locally the graph of a pseudoconvex function u Lip( ), let E be the level sets of the variational curvature HE. Then E are

locally the graphs of regular functions u and the classical mean curvature of E weakly converges to the generalized mean curvature of E (in [5] this theorem was proved with some supplementary assumptions on the function u which are removed in Sect. 3 of this paper).

The paper is organized as follows.

In Sect. 2 we give a characterization of the generalized mean curvature which holds in the case u Lip( ) (and we construct examples showing that this assumption cannot be

removed). As an application of this characterization in Sect. 3 we show the existence of generalized mean curvature for any pseudoconvex function u Lip( ) (see also [13]).

Section 4 is devoted to a collection of examples (a good collection of examples enlights about what theorems are possible).

Finally, in Sect. 5 we obtain an Approximation Theorem (in W(1,1) and mean curvature) of a C2-manifold by means of inscribed polyhedral manifolds. The rst motivation for this theorem was the somewhat surprisingly example below, which gives at once the avor of how the generalized mean curvature behaves.

Example 1

(see also Example 4 in Sect. 4).

An elementary and well known theorem of differential geometry says that for any regular simple closed curve M in the plane we have

divMdH1 = 2,

where is the inner normal to M. This result does not hold for the generalized curvature if the curve M is not regular, as for example in the case of polygons. In fact, let = (1, 1)

123

494 E. Barozzi et al.

R, z = u(y) = A|y|, M = {(y, Ay), 1 < y < 0}, M+ = {(y, Ay), 0 < y < 1}, M =

M M+. For X = (X1, X2) C10((1, 1); R2) we have

divM(X1, X2)dH1 = M

divM(X1, X2)dH1 + M+

divM(X1, X2)dH1

= X(0)

(1, A)

1 + A2

+ X(0)

(1, A)

1 + A2

2A 1 + A2

X2(0) =

0, 2A

1 + A2

XdH0,

so that

H= 0,

2A 1 + A2

(0,0) = (0, 2 sin )(0,0), = arctan A.

Let

N+ =

(1, A)

1 + A2

, N =

(1, A)

1 + A2

and let = (y, z) a vector eld in C10((1, 1); R2) such that

(0, 0) =

N+ + N

N+ + N ;

we get

divMdH1 = (0, 0) (N+ + N) = N+ + N .

For a square we easily get thus

div dH1 = 42 < 2,

while for a equilateral triangle we obtain

div dH1 = 33 < 2.

Notice that 33 < 42.

2 Cartesian versus parametric

Let u W(1,1)( ); a real Radon measure dened on the Borel algebra B( ) of is the

distributional divergence of T u if

T u Ddy =

d C10( ). (7)

We have the following

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On the generalized mean curvature 495

Theorem 1 Let u W(1,1)( ); if M = Mu has generalized mean curvature

(H1, . . . , Hn, Hn+1) in R, then the distributional divergence of the vector T u is a

Radon measure in , dened by

(S) = Hn+1({(y, u(y)) : y S}) S B( ) (8) Proof For C10( ), x = (y, xn+1) R, let us dene

X(y, xn+1) = (0, 0, . . . , (y)); a straightforward computation shows that

n+1Xn+1 =

and therefore

Example 2

Now we show by an example that the reciprocal of Theorem 1 is false. Precisely, we construct a function u W(1,1)((a, b)), 0 < a < b, such that divT u is a real Radon measure on

B(a, b) while its graphic Mu R2 has not a generalized mean curvature. The curve = Mu

is the union of innitely many arcs of circle k, k N. Then, by rotation of around the

z-axis we get also a 2-dimensional example.

Let ck = (yk, zk), yk > 0 be the center of k, where the arcs k are chosen such that

(see the gure below):

(i) 2k1 and k have the same radius rk and angle k (0, 2 );

(ii) 2k is convex while 2k1 is concave

(iii) the arcs k, k N are connected in such a way that is a C1-curve.

H =

1 + |Du(y)|2

T u(y) D(y),

T u Ddy =

1 + |Du(y)|2n+1Xn+1dy = M

Xn+1dHn+1 =

n+1Xn+1dHn

dHn+1 =

123

496 E. Barozzi et al.

Let rk > 0 and k (0, 2 ) be two decreasing sequences such that

k=1

rk < +,

k = + (9)

rkk < +,

2k < + (10)

For example

rk =

1k2 , k =

1 k

is a suitable choice.Fix now

y1 > 2

rk(1 cos k)

and dene, for n 1:

y2n+1 = y1 + r1 2

k=1

rk(1 cos k) rn+1, y2n = y2n1 + 2rn cos n

z1 = 0, z2n = 2

k=1

rk sin k, z2n+1 = z2n.

Finally, dene

a = y1 + r1 2

rk(1 cos ), b = y1 + r1, =

k=1

It is easy to see that = Mu for a suitable u W(1,1)((a, b)). We compute the length

of as

H1( ) = 2

rkk < +.

Denoting by H the scalar curvature of , we get

|H(y)|dy = 2

k=1

1rk rk(1 cos k) = 2

(1 cos k) < + (11)

On the other hand we have

|H|dH1 = 2

k=1

1rk rkk = 2

k = + (12)

Now, recalling that

T u Ddy =

Hdy C10(a, b),

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On the generalized mean curvature 497

we get from (11) that the divergence of the vector eld T u is the measure Hdy with nite total variation on (a, b), while from (12) we see that Mu = does not have generalized

mean curvature. Indeed

div XdH1 =

H X dH1 X C10((a, b) R; R2)

and (12) shows the total variation of the measure H| is not nite.

Example 3

Denote by k the 2-dimensional surface obtained by rotating the curve k around the z-axis and let

k=1

Clearly = Mv, where

v(y, z) = u(), = y2 + z2, (a, b).

Consider the portion 2k1 obtained by the rotation of the arc 2k1. We know that 2k1

is a subarc of the circle

(z z2k1)2 + (y y2k1)2 = r2k.

It follows that 2k1 is the graph of the function

u() = z2k1 + r2k ( y2k1)2, (ak, bk), where for short

ak = y2k1 + rk cos k, bk = y2k1 + rk.

An easy computation shows that

u () =

( y2k1)

r2k ( y2k1)2, u () =

r2k

r2k ( y2k1)2 3

from which we get

H2( 2k1) = 2

1 + (u ())2d = 2

r2k ( y2k1)2 d

= 2

(y2k1 + rk cos t)rkdt = 2rk(y2k1k + rk sin k)

Let us notice that H2( 2k) H2( 2k1), so we get

H2( ) =

(H2( 2k1) + H2( 2k)) = 4

(y2k1krk + r2k sin k) < +.

123

498 E. Barozzi et al.

In particular this proves that v W(1,1)( ), where denotes the annular strip Bb(0, 0)

Ba(0, 0), i.e. = {(y, z) R2 : a2 < y2 + z2 < b2}. On the other hand, the scalar mean

curvature H(), a < < b, is given by

H() =

u ()

1 + (u ())2

u ()

3 + 1 + (u ())2 =

1rk

y2k1

rk , (ak, bk).

Denoting by k = Bbk (0, 0) Bak (0, 0), we easily compute

k |

H|dydz =2

1rk +

y2k1 rk

d =2(1 cos k)(x2k1+rk(1 + cos k)).

Therefore, we conclude that

|H|dydz < +,

thus showing that the distributional divergence of the vector eld T v is a Radon measure with nite total variation in . On the other hand, we have

2k1

|H|dH2 = 2

1rk +

y2k1 rk

r2k ( y2k1)2 d

= 2(y2k1k + 2rk sin k),

which yields

|H|dH2 = +,

and we conclude that = Mv does not have a generalized mean curvature.

The remain of this section is devoted to the proof of the reciprocal of Theorem 1 (see Theorem 4) when the function u W(1,1)( ) is also Lipschitz-continuous. We begin by

stating some preliminary results.

The following result was proved in a previous paper (see Theorem 4.1 in [5]); for the sake of completeness, we shall include here its proof.

Theorem 2 Let uh, u W(1,1)( ) such that

u in W(1,1)( ) (13)

Then, denoting for short Mh = Muh , M = Mu, we have

lim

divMh XdHn =

divM XdHn X C10( R; Rn+1) (14)

Proof Without loss of generality we can assume that

uh(y)

u(y) a.e. y

Duh(y)

Du(y) a.e. y

123

On the generalized mean curvature 499

Moreover, for almost every y , we have

h(y) =

Duh(y), 1

1 + |Duh(y)|2

(y) =

Du(y), 1

1 + |Du(y)|2

Then, for every vector eld X C10( R; Rn+1), we get

divMh XdHn =

Dj X j(y, uh(y)) jh(h DX j)(y, uh(y))

1+|Duh(y)|2dy =

Ah(y)

n+1

j=1

1 + |Duh(y)|2dy

divM XdHn =

n+1

j=1

Dj X j(y, u(y)) j( DX j)(y, u(y)) 1 + | Du(y)|2dy

A(y)

1 + |Du(y)|2dy

and therefore

divMh XdHn

divM XdHn

|Ah(y)|

1 + |Duh(y)|2 1 + |Du(y)|2dy

1 + |Du(y)|2|Ah(y) A(y)|dy

Observe that

Ah(y)

A(y) a.e. y

and that

|Ah(y)| c(X), |A(y)| c(X) y (15)

where

c(X) =

n+1

j=1

Dj X j , R + DX j , R

(16)

Then, from the inequality

1 + |p|2 1 + |q|2 |p q| p, q Rn,

123

500 E. Barozzi et al.

we get

divMh XdHn

divM XdHn

c(X)

|Duh Du|dy +

1 + |Du(y)|2|Ah(y) A(y)|dy

and (14) follows from the dominated convergence theorem and (13).

Remark Formula (14) is still true for X C1( R; Rn+1), provided that c(X) < +.

Lemma 1 Let Rn be an open bounded set with a smooth boundary , and let u

W(1,1)( ) be a function such that the divergence of the vector eld T u is a Radon measure with nite total variation in . Then the following estimate holds:

T u Ddy

|( ) + Hn1( )

, C1( ) (17)

Proof Let hk : [0, +) [0, 1] the Lipschitz-continuous function dened by

hk(t) =

1 if t 1k

kt if 0 t 1k

Dene d(y) = dist(y, ) and set

k(y) = hk(d(y)), y .

Owing to the fact that C3, there exists k0 = k0( ) N such that d C2( k) k k0,

where

k = y : d(y) <

1 k

We have

T u Ddy = lim

kT u Ddy C1( ),

and, for k k0,

kT u Ddy =

T u D(k)dy

T u Dkdy

T u (h k(d(y))Dd(y))dy.

We conclude that

kT u Ddy

||( ) + k

, k k0.

123

On the generalized mean curvature 501

On the other hand, d C2( k0) and Dd(y) = 0 and therefore, for k k0, the set k is a

regular (n 1)-manifold, so that

lim

dy = Hn1( ).

Equation 17 follows at once. Lemma 2 Let Rn be a bounded open set, let B = BR be a ball with radius R such

that B, let u Lip( ; R). Then there exists Lip(B; R), spt B,(y) =

u(y) y , such that

L L + 2

M + 2L R

d ,

where we have denoted by L and L the Lipschitz constants for u and respectively, by M

the sup{|u(y)| : y } and by d the dist( , B).

Proof The existence of a extension U Lip(Rn; R) with Lipschitz constant L and such that

U(y) = u(y) y is well known (see for example [3]). Let C(Rn) be a function

that veries 0 1 y , spt B and |D| 2d y Rn. The the function

u : B R dened by

u(y) = (y)U(y), y B has the requested properties. In fact, we have

D(y) = D(y)U(y) + (y)DU(y), from which we get

|U(y)| |U(z)| + L|y z| M + 2L R y B, z , and so

2d (M + 2L R) + L.

We prove now the following

Theorem 3 Let Rn be an open bounded set, C3, and let u Lip( ; R) such

that the divergence of the vector eld T u is a Radon measure with nite total variation in . Then there exists a sequence uh C2( ) that veries the following properties:

(i) uh u in L2( ) and a.e. in ;

(ii) Duh Du in L1( ) and a.e. in ;

(iii) Duh are uni f ormly bounded in L( );

(iv) Hh = divT uh are uni f ormly bounded in L1( )

(18)

Moreover

lim

Hhdy =

d C0( ), (19)

which says that the classical mean curvature of the graphics Mh = Muh weakly converges

to the Radon measure .

123

502 E. Barozzi et al.

Proof Let B = BR a ball with radius R that for simplicity we suppose centered at the origin

of the coordinate system and such that B and let Lip(B; R) be as in Lemma 2.

Let us dene, for 0 and v BV (B), the functional

F(v) =

1 + |Dv|2 + 2

(v )2dy +

|v |dHn1. (20)

We recall that BV (B) is the Banach space of the functions v L1(B) whose gradient Dv

in the sense of distributions is a (vector valued) Radon measure with nite total variation in B (see for example [2,3,10,16]). The functional (20) has a unique minimum u BV (B);

moreover (see for example [8,9]) u C2(B) and

divT u(y) = (u(y) (y)), y B

u(y) = (y) = 0, y B

. (21)

The inequality

F(u) F() =

1 + |D|2dy

yields

(u )2dy 2

1 + |D|2dy. (22)

In particular we get that u in L2(B) as , and therefore, recalling the semicon

tinuity of the area functional, we deduce that

lim

1 + |Du|2dy =

1 + |D|2dy. (23)

It follows that Du D as distributions and that {Du} is a bounded subset of L1( ).

As a consequence (see [3], Exercise 1.20 on p. 39) we obtain that

Du D in L1(B). (24)

With the aim to get the gradient estimate (18-iii), let us dene

+(y) = L(R |y|), y B

(y) = L(R |y|), y BOwing to the concavity of + and the convexity of and the inequality

(y) (y) +(y), y B,

from

F(u +) F(u), F(u ) F(u),

we get

(y) u(y) +(y) y B. (25)

Let us suppose rst that C1(B). Then, arguing as in the proof of the Lemma 3 in [5], we

deduce that if |Du| has a maximum at a point y0 Int(B), then

|Du(y0)| L.

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On the generalized mean curvature 503

On the other hand, from (25) we get

|Du(y)| L y B and therefore the estimate

|Du(y)| L > 0 y B, (26)

follows.

In the general case Lip(B; R) we proceed by approximating uniformly by a sequence

vh C1(B), spt vh B, whose Lipschitz constants are bounded by L. Let vh, be the

minimum of the functional

F(v) =

1 + |Dv|2 + 2

(v vh)2dy + B

|v vh|dHn1, v BV (B).

We have uh, C3(B), |Duh,(y)| L h N, > 0, y B and uh, u > 0

uniformly in B. We conclude that

|Du(y)| L > 0, y B. (27)

Finally, let us prove that (18-iv) holds. To this aim, let > 0 and dene, for y

v(y) = max{|u(y) u(y)|, }

A = {y : |u(y) u(y)| > }.

From Sards Theorem we know that, for almost all > 0 the set A is a C2 (n 1)-

manifold, and therefore

(u u)2v dy =

div T u

u u

v dy

T u D

u u

dy +

T u

u u

v dHn1,

where we have denoted by the outer normal to . Observe that

(u u)v =

1 if u(y) u(y) >

1 if u(y) u(y) <

so we get

(y) = 0 y A.

On the other hand we have v(y) = for y A and therefore

(u u)2v dy =

u u v

T u D(u u)dy +

T u

u u

v dHn1.

Now, the convexity of the function g(p) = 1 + |

p|2 implies that (T u T f ) D(u f ) 0,

123

504 E. Barozzi et al.

and recalling that

|u u|v 1,

we get at once

(u u)2v dy

T u D(u u)dy + Hn1( ).

Finally, dene

(y) =

u(y) u(y) if y A

if u(y) u(y)

;

the function is Lipschitz-continuous and thus, from (17), we get

(u u)2v dy

1 T u Ddy + Hn1( ) ||( ) + 2Hn1( ).

Letting now 0 and recalling Fatous lemma we conclude that

|divT u|dy =

(u u)2v dy ||( )

+2Hn1( ). (28)

We are now in position to prove the main theorem of this section.

Theorem 4 Let Rn be an open bounded set, C3, let u Lip( ; R). Then the

divergence of the vector eld T u is a Radon measure in if and only if M = Mu has

generalized mean curvature in R.

Proof The sufciency was already proved in Theorem 1. For the converse implication, let us begin by considering the case u C2( R). Because the tangential derivatives on M

of a vector eld X C10( ; Rn+1) depend just on X|M, we can suppose X independent of

the variable xn+1, so that Dn+1X = 0 (y, u(y)) M. Then, for j = 1, 2, . . . , n, we have

j X jdHn =

|u u|dx lim inf 0

Dj X j

1 + |Du(y)|2dy

Dj u

h=1

Dhu Dh X j

1 + |Du(y)|2 dy.

By an integration by parts in the rst integral above, we get

j X jdHn =

X j

h=1

DhuDjhu

1 + |Du(y)|2dy

Dj u

h=1

Dhu Dh X j

1 + |Du(y)|2 dy

T u D(X j Dj u)dy =

X j Dj u d,

123

On the generalized mean curvature 505

from which we get

j X jdHn = M

X j dHj,

where we have dened

Hj(S R) = S

Dj u d S B( ).

On the other hand

n+1Xn+1dHn = M

n+1( DXn+1)dHn =

Xn+1 d.

H = (H1, H2, . . . , Hn+1) veries

(2), i.e. M has generalized mean curvature. For the general case u Lip( ), let uh C2( )

be as in Theorem 3. Then, for any X C10( R; Rn+1), we have

divM XdHn = lim

Therefore, if we dene Hn+1 = , the vector measure

divMh XdHn = lim

Hh X hhdHn.

From (27), (28) we immediately obtain

Hh|dHn sup X 1 + L2(||( )

+2Hn1( )),and the proof is concluded by recalling the Riesz representation Theorem.

3 The mean curvature of a pseudoconvex function

As an application of the results stated in the previous section we shall now prove that for every pseudoconvex function u Lip( ) there exist an (n + 1)-dimensional vector valued

Radon measure

H = (H1, . . . , Hn, Hn+1) on M = Mu such that

divM XdHn =

divM XdHn

sup X lim inf

Mh |

X d

n+1

j=1

X jd

H j X C10( , Rn+1). (29)

This result, already proved in a different setting in [13], improves Theorem 4.2 in [5].

We recall here that u Lip( ) is said to be a pseudoconvex function iff

1 + |Du|2

1 + |Dv|2| v BV ( ), spt(v u) , v u, (30)

123

506 E. Barozzi et al.

or, in other words, that u is a weak supersolution of the minimal surface equation, i.e.

T u Ddy 0 C10( ), u. (31)

The following theorem holds:

Theorem 5 Let Rn be an open bounded set, C2, suppose that the mean curva

ture of is nonnegative, and let u Lip( ) be a pseudoconvex function. Then u has a generalized mean curvature

Proof For each 0, let F : BV ( ) R be the functional dened by

F(v) =

1 + |Dv|2 + 2

(v u)2dy +

|v u|dHn1, v BV ( ).

Owing to the hypothesis on , the functional F has a unique minimizer u BV ( ).

Moreover u C2( ) C( ) and

div T u = (u u) in

u = u in ss

From the minimum property of u we easily get that u u in L2( ) and that Du Du

in L1( ). As a consequence, we have that

T u Ddy = lim

div T u dy C10( ).

In order to prove the theorem it sufces to prove the existence of a constant C > 0 such that

|div T u|dy C 0.

The maximum principle and the pseudoconvexity of u implies that

u(y) < u(y) y , from which we derive

div T u(y) = (u(y) u(y)) < 0 y .

Therefore, we have

|div T u|dy =

div T udy =

T u dHn1,

so that

|div T u|dy Hn1( ),

and the theorem is established.

123

On the generalized mean curvature 507

4 Some examples

Example 4

Let = (1, 1) and let u : R be a function in C2( {0}) C( ). Moreover, let us

suppose that the following limits exists:

A = lim

t1+

u (t), B = lim

u (t), C = lim

t0+

u (t), D = lim

u (t).

Let us denote by N+(1), N(0), N+(0), N(1) the unitary vectors tangential to M = Mu

at the points 1, 0, 1 respectively, and by T+(1), T(0), T+(0), T(1) the limits of T u as

t tends to 1, 0, 1 respectively. Then, we have:1

T u dt =

Hdt + T(1)(1) + (T(0) T+(0)) (0)

T+(1)(1) C1( ) (32)

divM X dH1 =

H X dH1 + X(1) N(1) + X(0) (N(0) N+(0))

X(1) N+(1) (33)

X C1([1, 1] R; R2).

The rst formula follows immediately by a straightforward integration by parts. For the second one, let X C1([1, 1] R; R2), X(t, z) = X(t) (t, z) [1, 1] R. We

compute

1X1dH1 =

X1 u

1 + u 2

u 1 + u 2

1 + u 2dt

1 1 + u 2

dt.

Integrating by parts this yields

1X1dH1 =

1 X1 u u

1 (1 + u 2)3

+X1(1) 1

1 + D2

X1(0)

11 + C2

1 1 + B2

X1(1)

1 1 + A2

X1H1dH1 + X1(1)

11 + D2

X1(0)

11 + C2

1 1 + B2

X1(1)

1 1 + A2

123

508 E. Barozzi et al.

Similarly,

2 X2dH1 =

1 1 + u 2

u 1 + u 2

1 + u 2dt =

u 1 + u 2

X2 dt,

and, integrating by parts again, we get

2 X2dH1 =

1 X2 u

1 (1 + u 2)3

+X2(1) D

1 + D2

X2(0)

C1 + C2

B 1 + B2

X2(1)

A1 + A2 =

X2H2dH1 + X2(1)

D 1 + D2

X2(0)

C1 + C2

B 1 + B2

X2(1)

A 1 + A2

and (33) follows at once. This shows that u has a generalized mean curvature

H in (a, b)R

and consequently that the divergence of T u is a Radon measure in (1, 1), precisely

H = H|M(0,0) (0,0) (N(0) + N+(0)) = Hdt 0 (T(0) T+(0))

In particular, for u(y) = A|y| we have again the Example 1 in Sect. 1.

Example 5

Let P1, P2, P3 be three independent points in the plane and let T = [P1, P2, P3] be the closed

triangle with vertices at P1, P2, P3, i.e. the set of all points P R2 which can be expressed

P =

k=1

k = 1, k 1.

Then, let P0 be an interior point to T and consider the triangles T1 = [P0, P2, P3], T2 = [P0, P1, P3], T3 = [P0, P1, P2]. For h = 0 consider the piecewise linear and continuous

function u : T R dened byu(P0) = h, u|T 0, u|Tk linear k = 1, 2, 3,

and let M = Mu be the graphic of u. Let Si j, i = j be the edge which is in common to the

triangles u(Ti), u(Tj), let Ni j be the outer normal (with respect to u(Ti)) to Si j on the tangent space to u(Ti). For X C10(T ; R3), support X T , the integration by parts formula (6)

gives immediately

divT XdH2 =

k Pk, where

k=1

3 divTk XdH2 =

i = j

Si j

X N

i j

+ N ji

dH1.

123

On the generalized mean curvature 509

This show that in this case the generalized mean curvature

H of M = Mu is given by

i = j

Ni j + N ji dH1|Si j .

Therefore in this case

H is a singular measure supported on the edges Si j. We remark also that there is not curvature concentrated at the vertex (0, 0, h). An analogous result is easily obtained for a continuous u : T R, u C1(Ti) i = 1, 2, 3; in this case we get

H =

Ni j + N ji dH1|Si j ,

where Hk is the classical mean curvature of Mk = u(Tk) and k is the normal vector to Mk

directed upwards, k = 1, 2, 3.

Example 6

It is easy to see the existence of Lipschitz functions u for which the distributional divergence of T u is not a Radon measure, as the following example shows. Let us begin by considering the function u : [0, +) R dened by

u1(t) =

3 Hkk dH2|Tk +

i = j

0 for t [0, 1/2) (1, +)

3t 3/2 for t [1/2, 3/4)

3 3t for t [3/4, 1]

For h 1 let

uh(t) =

2h1

2h1t

, t [0, +),

and dene

uh(t), t [0, 1].

Then u Lip([0, 1]) with Lipschitz constant equal to 3, but the distribution divT u is not a

Radon measure. In fact, it is not difcult to compute that in this case we have

sup

u(t) =

1 T u D dt; C10(0, 1), 1

= +.

Example 7

Let u be the function dened in the example above and let

g(y, z) = u(), = y2 + z2 1.

Let us show now that the distributional divergence of T g is a Radon measure in B =

$(y, z) R2 ; y2 + z2 < 1%. For = 2h+1, = 3 2h1, h N, the function g is

differentiable and we have

gy = u ()

y , gz = u ()

z ,

123

510 E. Barozzi et al.

and therefore

T g(y, z) =

u ()

y ,

, div T g(y, z) = u ()

1 .

Let C10(B); we have

T g D dydz =

h=1

B32h1

T g D dydz

B2h

T g D dydz+

B2h+1 B32h+1

and therefore, integrating by parts we obtain

T g D dydz =

divT g dydz

h=1

3 10

B32h1

dH1

B2h+1

dH1

B2h

dH1

From this last equality we get the estimate

T g D dydz

|div T g| dydz +

h=1

610 6 2h1 + 2h + 2h+1

42 10

Example 8

Let

Cn [0, 1]

be the Cantor set, where as usual Cn is the union of 2n closed intervals with length 3n. The Hausdorff dimension dimH (X) of a set X R is dened by dimH (X) = inf{s : Hs(X) =

0}. For the Cantor set we have

dimH (C) = =

ln 2 ln 3.

C =

For n N let gn : [0, 1] R be the absolutely continuous function

gn(t) =

3n 2n

Cn(s)ds.

The functions gn are non decreasing and, owing to the fact that

|gn+1(t) gn(t)|

13 2n+1

t [0, 1],

123

On the generalized mean curvature 511

the sequence gn is uniformly convergent in [0, 1] to a non decreasing continuous function

g : [0, 1] R, which is known as the Cantor-Vitali function. We dene a convex function

f C1([0, 1]) by letting

f (t) =

g(s) ds, t [0, 1].

From Sect. 3 we know that the distributional divergence of T f is a Radon measure. We give here a direct proof of this fact. We have

T f (t) =

f (t)

1 + ( f (t))2 =

g(t)

1 + g2(t) t [0, 1],

so that, for C10([0, 1]), we obtain1

T f D dt =

1 g(t)

1 gn(t)

0 1 + g(t)2 (t)dt = lim n+

0 1 + gn(t)2 (t)dt.

For n N the function

un(t) =

gn(t)

1 + gn(t)2, t [0, 1],

is continuous in [0, 1] and differentiable in [0, 1]Cn. Therefore, we can integrate by parts,

getting thus

1 gn(t)

0 1 + gn(t)2 (t)dt =

un(t) (t)dt =

u n(t)(t)dt.

On the other hand, we have

u n(t) =

g n(t)

1 + gn(t)2 3

3n 2n

1 + gn(t)2 3

Cn (t) t [0, 1] Cn,

from which we obtain the estimate

un(t) (t)dt

3n 2n

1 + gn(t)2 3

cn(t)dt

3n 2n

3n = .

We conclude that

T f D dt

and this immediately implies the existence of a Radon measure on [0, 1] such that

T f D dt =

123

512 E. Barozzi et al.

Notice that1

d = lim

3n 2n

Cn(t)(t)dt,

1 + gn(t)2 3

from which we derive that

support [0, 1] C

d = 0,

i.e., the measure = divT u is concentrated on the Cantor set C.

5 Approximation theorem

The approximation of a smooth surface with a triangulated mesh appears in many applications, see for example [17]. In this section we shall prove that when u is smooth, the surface M = Mu can be uniformly approximated with polyhedral surfaces with convergence of the

area and the generalized mean curvature. Although the proof in the cases n = 1 and n > 1

is essentially the same, for the sake of the reader we expose the two cases separately. The case n = 1

Let u C2([a, b]) be a smooth function, let be the unitary normal vector eld to

M = Mu directed upwards, i.e.

(y) = (y, u(y)) =

(u (y), 1)

1 + u (y)2, y [a, b].

Let us x a point y [a, b] and for h [a y, b y] dene

(y, h) =

u(y+h)u(y)

h for h = 0

u (y) for h = 0

The function (y, ) belongs to C1([a y, b y]) and has the following properties:

(y, h) (y, 0) =

1 h

'u (y + t) u (y)(dt (34)

(y, h) =

u (y + h)h [u(y + h) u(y)]

h2 for h [a y, b y], h = 0

(y, 0) = lim

(35)

(y, h) =

u (y)

2 for h = 0

(y, h) (y, 0) =

1 h2

t[u (y + t) u (y)]dt (36)

For h [a y, b y], h = 0, we denote by W(y, h) the unitary normal vector eld to the

segment with endpoints (y, u(y)) and (y + h, u(y + h)) directed upwards, i.e.

W(y, h) =

( (y, h), 1)

1 + ( (y, h))2.

123

On the generalized mean curvature 513

By dening

W(y, 0) = lim

W(y, h) = (y),the function W(y, ) is dened and continuous all over the interval [a y, b y].

We need the following

Lemma 3 Under the notations and assumptions above, we have

W(y, h) = (y) +

with

uniformly on y.

Proof For a xed point y [a, b] the function W(y, ) belongs to C1([a y, b y]) and

therefore

W(y, h) = W(y, 0) +

+ 'W (y, t) W (y, 0)( dt.

For the sake of brevity we introduce the function G : R R2 dened as

G(s) =

(s, 1)

1 + s2

W (y, h) = G (u (y))

Therefore, in order to prove (38), we compute

W (y, t) W (y, 0) = G ( (y, t)) (y, t) G ( (y, 0)) (y, 0)

= G ( (y, t)) (y, t) G ( (y, t)) (y, 0)

+G ( (y, t)) (y, 0) G ( (y, 0)) (y, 0)

= G ( (y, t)) ' (y,

t) (y, 0)( + (y, 0)

'G ( (y, t)) G ( (y, 0))(.

(y)

2 h + (y, h), (37)

lim

(y, h)h = 0 (38)

W (y, t)dt = W(y, 0) + W (y, 0)h

getting thus

W(y, h) = G( (y, h)) and W (y, h) = G ( (y, h)) (y, h). We obtain then

W(y, 0) = G(u (y)) = (y) and

W (y, 0) = lim

u (y)

2 =

(y)

2 .

This shows that

(y, h) =

'W (y, t) W (y, 0)( dt.

123

514 E. Barozzi et al.

Now, letting

M = sup $|u (s)|,

s [a, b]%

, M = sup $|u (s)|,

s [a, b]%

A = sup $|G (s)|, |

s| M %

, B = sup $|G (s)|, |

s| M %

we get

G ( (y, t))

' (y, t) (y, 0)( + (y, 0)

'G ( (y, t)) G ( (y, 0))(

1 h2

0 t[u (y + t) u (y)]dt

2 B

1 h

0 [u (y + t) u (y)]dt

These estimates together with the uniform continuity of u , u in [a, b] gives immediately

the desired statement (38).

Consider now a sequence of partitions Pk, k N, of the interval [a, b],

= {a = y0, y1, . . . , yNk = b}, Nk N (in order to avoid a heavy notation we abbreviate yi = yki). Let

Pk = max {hi = yi yi1, i = 1, . . . , Nk} be the norm of the partition Pk. We assume that

) (i) Pk

(ii) c > 0 : Nk Pk c k N

(39)

The above conditions (i), (ii) are satised if there exists a positive constant c such that

Pk c inf{h j , j = 1, . . . , Nk} k N.

This is the case if for example the points yj are equidistant, i.e., if

h j =

b a

Nk j = 1, . . . , Nk.

Let pk : [a, b] R be the piecewise linear function such that pk(yi) = u(yi) yi Pk,

let k = pk([a, b]) be the polygonal with vertices at the points Pi = (yi, u(yi)), i =

0, 1, . . . , Nk, and let Wi = W(yi, hi+1), i = 0, 1, . . . , Nk 1. Notice that Wi is simply the

unitary normal vector eld to the segment [Pi, Pi+1]. Finally, let k C1([a, b]R; R2) be

an unitary vector eld only depending on the rst variable, i.e., k(y, z) = k(y) (y, z)

[a, b] R, and such that

(40)

With the notations and assumptions introduced above, we have the following approximation theorem for curves:

k(P0) = k(a) = W0

k(PNk ) = k(b) = WNk1 k(Pi) = k(yi) =

Wi1+Wi Wi1+Wi , i = 1, . . . , Nk 1.

123

On the generalized mean curvature 515

Theorem 6 Let

H = (H1, H2) be the classical curvature of the smooth curve M = Mu, let

Hk = (Hk1, Hk2) be the generalized curvature of k and let H be the scalar curvature of M.

Then, we have

lim

k d

Hk= M

H dH1 = M

H dH1 (41)

Proof From the integration by parts formula (6) and Theorem 2 we get

H dH1 =

divM dH1 = lim

div k dH1

div k dH1 = (b)

PNk PNk1

(a)

P1 P0

Nk1

i=1

(yi)

Pi Pi1

Pi+1 Pi

div k k dH1 =

Nk1

i=1

i(yi)

Pi Pi1

Pi+1 Pi

Noticing that

Pi Pi1

Pi+1 Pi

Pi+1 Pi =

ik(yi), |i| 2,

we obtain

div k dH1

div k k dH1

Nk1

k=1

|((yi) k(yi)) k(yi)| + (b)

PNk PNk1

+ (a)

P1 P0

Now, definition of W(y, h) and (37) yields

Wi + Wi+1 = W(yi, hi+1) + W(yi, hi) = 2(yi) +

(yi)

2 (hi+1 hi) + i,

where for brevity i = (yi, hi+1) + (yi, hi). Now, recalling that = 0, we compute

Wi + Wi1 2 = 4 + (yi) 2

(hi+1 hi)2

4 + 2i + 4(yi) i + (yi)(hi+1 hi)i

= 4 + i,

123

516 E. Barozzi et al.

where the meaning of i is obvious. From this, we derive

((yi) i(yi)) i(yi) = (yi) i(yi)1=(yi)

2(yi)+

(yi)

2 (hi+1hi)+ i 4 + i =

2 + i(yi)

= 4 + i

1 =

i(yi) 4 + i +

2 4 + i

4 + i =

i(yi) 4 + i

4 + i(2 + 4 + i) =

Then Lemma 3 shows that for every > 0 there exists k N such that k > k we have

< i = 1, . . . , Nk,

and therefore

div k dH1 k

div k k dH1

2 Nk Pk + (b)

PNk PNk1

+ (a)

P1 P0

The theorem follows at once by noticing that

lim

(b)

PNk PNk1

PNk PNk1 =

lim

(a)

P1 P0

P1 P0 =

The general case n N

Our aim now is to extend Theorem 6 to an arbitrary dimension n. Let us begin by establishing the following lemma:

Lemma 4 Let be an open bounded subset of Rn, let P , let V Rn be an unitary

vector, i.e., V = 1, and let d = dist(P, ). Dene a function : (d, d) R by

(t) = )

u(P + tV ) u(P)

t if t = 0 Du(P) V if t = 0

Then C1(d, d) and therefore

(t) = (0) + (0)t + (t) (42)

where

(t)t = 0,

uniformly on P as P varies on compact subsets of .

Proof From the definition we have that is a continuous function on (d, d). Moreover,

we have

(t) =

[Du(P + tV ) V ] t [u(P + tV ) u(P)]

t2 for t = 0,

lim

123

On the generalized mean curvature 517

while, for t = 0, we compute

(0) = lim

(t) (0)

t = lim

u(P + tV ) u(P) t Du(P) V

= lim

[Du(P + tV ) Du(P)] V

2t =

2 lim

h=1

V h

k=1

Dhku(P + tV )V k

12Q(P, V ), where we have denoted by Q(P, V ) the quadratic form

Q(P, V ) =

h,k=1

Dhku(P)V hV k.

Then we have

2Q(P, V ) = (0),

and therefore C1(d, d). For , 0 < d < dist( , ), P , t (d, d)

we can then write

(t) (0) =

lim

(t) =

' (s) (0)( ds.

Now, the mean value theorem ensures the existence of a point (0, s) such that

(s) (0) =

1 s2

(s)ds = (0)t +

[Du(P +sV ) V ] s[Du(P +sV )Du(P)]12Q(P, V )s2

s {Q(P + V, V ) Q(P, V )} .

Finally, owing to the uniform continuity of the second derivatives of u in , we get that for every > 0 there exists > 0 such that

| (s) (0)| < s : |s| < P , and thus, from

(t) = (P, t) =

' (s) (0)( ds,

we obtain the estimate

| (t)| |t| t (d, d), P .

Let V1, V2, . . . , Vn be n unitary and linearly independent vectors in Rn, let P , let

t Rn such that t < d = dist(P, ), t j = 0 j = 1, 2, . . . , n. In the following we

will denote by (P, t) the linear afne manifold in Rn+1 which passes by the points

(P, u(P)) ,

P + t1V1, u(P + t1V1),

P + t2V2, u(P + t2V2) , . . . ,

P + tnVn, u(P + tnVn),

123

518 E. Barozzi et al.

and by W(P, t) the unitary normal vector to (P, t) directed upwards. We have the following

Lemma 5 Let u C2( ), let Bd = Bd(0) be the ball centered at 0 with radius d =

dist(P, ); then the function W(P, ) dened above can be extended to a function belong

ing to C1(Bd) such that

W(P, t) = (P) +

j=1

j (P)t j + (P, t), (43)

where

j =

t j (P, 0), j = 1, 2, . . . , n,

and

lim

(P, t)

t =

uniformly with respect to P as P varies on compact subsets of and with respect to V1, V2, . . . , Vn varying in Sn.

Proof From the equation

xn+1 = u(P) +

h=1

Ah(P, t)

xh Ph

of the hyperplane (P, t) we get

W(P, t) =

(A(P, t), 1)

1 + |A(P, t)|2,

where for short A denotes the vector

A(P, t) = A1(P,

t), A2(P, t), . . . , An(P, t) .

Let Vj =

V 1j, V 2j, . . . , V nj

; we have

u(P + t j Vj) u(P) =

h=1

Ah(P, t)t j V hj j = 1, 2, . . . , n,

and therefore

h=1

u(P + t j Vj) u(P)t j = j (t j) (44)

Let V be the matrix Vjh = V hj and let V1 be its inverse, that is, Vjk = V 1kj; multiplying

equation (44) by V 1kj we obtain

h=1

Ah(P, t)VjhV 1kj = j (t j)V 1kj,

from which we get

j=1

Ah(P, t)V hj =

j (t j)V 1kj =

h=1

Ah(P, t)

j=1

VjhV 1kj = Ak(P, t).

123

On the generalized mean curvature 519

From Lemma 4 we derive that, for every k = 1, 2, . . . , n, the function Ak(P, ) con be

extended to a function belonging to C1(Bd) and, moreover, that

lim

Ak(P, t) = Ak(P, 0) =

j=1

j (0)V 1kj =

j=1

Du(P) Vj

V 1kj

Dhu(P)VjhV 1kj = Dku(P).

It follows that W(P, 0) = (P). The uniform convergence to zero of

(P, t)

h, j=1

, which is a

consequence of the uniform continuity of the second derivatives of u on compact subsets of , is easily proved proceeding along the lines of Lemma 3.

Now we pass to the approximation of a smooth manifold M = Mu by polyhedral manifolds

k. Let us begin by recalling some geometric elementary concepts which will be useful in the sequel. We say that n + 1 points of Rn are independent if they span an afne n-dimensional

subspace of Rn. Let P0, P1, . . . , Pn be n + 1 independent points. The n-simplex spanned

by P0, P1, . . . , Pn is the convex hull of {P0, P1, . . . , Pn}, i.e., the set T Rn of all points

P Rn which can be written as

P =

j=0

j = 1, j 0.

The points Pj, j = 0, 1, . . . , n are called the vertices of T . We denote T by [P0, P1, . . . , Pn].

A simplex spanned by a subset of {P0, P1, . . . , Pn} is a face of T . An n-simplex T has exactly

n + 1 (n 1)-faces and exactly (n+1)n2 (n 2) faces.

Finally, we say that a family P = $T

j Pj, where

j=0

1, T2, . . . , Tq

of n-simplices of Rn is a simplicial

decomposition of a set E iff

(a) E = T1 T2 Tn;

(b) for i = j the intersection of any two simplices of P is either empty or a face of each of

them.

For k N let Pk = *

T k1, T k2, . . . , T kNk +

be a simplicial decomposition of k =

j=1

T kj such

that

(i) k k N and lim

k+ Hn

( k) = 0;(ii) there exists a constant c (independent of k) such that

Nk Pk n c k N, (45) where Nk is the number of simplices in Pk and Pk is the norm of the mesh, i.e.

Pk = max{diam(T ), T Pk}.

Remark In order to ensure the existence of simplicial families Pk with the properties

requested above, observe that for a cube Q Rn we can produce a simplicial decomposition

given by c(n) = 2n1n! congruent n-simplices by the following recursive procedure:(1) if n = 2 we divide Q in c(2) = 2 2 = 4 congruent 2 simplices.

123

520 E. Barozzi et al.

(2) if n = 3 we rst divide everyone of the 6 = 3! faces of Q in 4 triangles and taking the

barycenter of Q as a 4th vertex we get c(3) = 22 3! 3-simplices.

(3) in the general case, by dividing the 2n faces of Q into c(n 1) (n 1)-simplices and

taking the barycenter of Q as a (n + 1)th vertex, we getc(n) = 2n c(n 1) = 2n 2n2(n 1)! = 2n1 n!

n-simplices.

Now, for xed k N, let us divide Rn in a mesh of disjoints cubes with sides of length 1k,

and divide everyone of these cubes in 2n1n! n-simplices. Let Fk be the family of all these

simplices. Then the family Pk = {T Fk, T } and the set k = ,{T,

T Pk}

: T kj R be the linear function dened

pk,j (Pk,js) = u(Pk,js) s = 0, 1, . . . , n + 1, and let pk : k R be the piecewise linear function such that pk(y) = pk,j (y) y T kj, j = 1, 2, . . . , Nk.

Finally, let k = pk( k) and jk = pk(T kj).

Let T ki, T kj be two adjacent n-simplices in Pk, and, accordingly with the notations intro

ducing Lemma 5, let Wk,i = W(Pi, ti) and Wk,j = W(Pj, tj ) be the unitary vectors normal

to ik, jk respectively. Then dene a unitary vector eld

k :

if P ik jk

Denoting by Sk the union of all the d-faces, d < n 1, belonging to the n-simplices in Pk

we have Hn1(Sk) = 0, and therefore there is an open set Ak Rn such that Sk Ak and

Hn1(Ak) <

1 k .

Let k : k Rn+1 be a C1( k) vector eld, k 1, such that

k(y) = k(y) y

j=1

u in W1,1( ) and

verify the requested properties (i) and (ii) above. For T kj = [Pk,j0, Pk,j1, . . . , Pk,jn+1] Pk, let pk,

j=1

jk Rn+1

k(P) =

Wk,j if P jk k

Wk,i + Wk,j

jk Ak.

We are now able to prove the following

Theorem 7 Let u C2( ) C1( ) and suppose that the scalar mean curvature H of M = Mu belongs to L1( ). Then pk

HdHn =

divM dHn = lim

div k k dHn (46)

123

On the generalized mean curvature 521

Proof The rst assertion follows at once from the true definition of the piecewise linear functions pk. With regard to (46), observe that, owing to the fact that

HdHn = lim

div k dHn,

it sufces to prove that

lim

div k ( k) dHn = 0. (47)

Let N jk be the outer normal to jk which lies on the plane containing jk; from the integration by parts formula we get

div k ( k) dHn = jk

( k) N jk dHn1.

Let us denote, for brevity, 1 jk = jk Ak, 2 jk = jk Ak. We have

div k ( k) dHn = 1 jk

( k) N jk dHn1 + 2 jk

( k) N jk dHn1.

We have

2 jk

( k) N jk dHn1

2Hn1(2 jk) = 2Hn1( jk Ak).

Letting k = {(P, u(P)), P k}, we have

1 jk

( k) N jk dHn1 =

1 jk k

( k) N jk dHn1

m=1,m = j

1 jk mk

( k) N jk dHn1

1 jk k

( Wk,j ) N jk dHn1

m=1,m = j

1 jk mk

Wk,j + Wk,m

N jk dHn1

123

522 E. Barozzi et al.

Thus we get

j=1

( k) N jk dHn1

4k +

j=1

1 jk k

( Wk,j ) N jk dHn1

j=1

m=1,m = j

1 kj mk

Wk,j + Wk,m

(N jk + Nmk) dHn1

The rst integral in the right-hand side of this inequality goes to 0 as k . For the second

integral, recalling Lemma 5 and reasoning as in the case n = 1 we derive that > 0 there

exists k N such that

Wk,j + Wk,m

(N jk + Nmk)

|tj | in jk j = 1, 2, . . . , Nk k > k.

Letting L = Du , , we conclude that

j=1

m=1,m = j

1 kj mk

Wk,j + Wk,m

(N jk + Nmk) dHn1

|tj |L Hn1(T jk) L Nk Pk n.

Recalling the hypotheses on the simplicial decompositions Pk the theorem follows

at once.

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On the generalized mean curvature 523

14. Massari, U.: Esistenza e regolarit delle ipersuperci di curvatura media assegnata in Rn. Arch. Ration. Mech. Anal. 55, 357382 (1974)

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123

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