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On the best constant in Gaffney inequality
G. Csato, B. Dacorogna, Swarnendu Sil
To cite this version:
On the best constant in Ga¤ney inequality
G. CSATO,
B. DACOROGNA
and
S. SIL
G.C.: Facultad de Ciencias Fisicas y Matematicas
Universidad de Concepcion, Concepcion, Chile
gy.csato.ch@gmail.com
B.D.: Section de Mathématiques
EPFL, 1015 Lausanne, Switzerland
bernard.dacorogna@ep‡.ch
S.S.: Section de Mathématiques
EPFL, 1015 Lausanne, Switzerland
swarnendu.sil@ep‡.ch
August 28, 2017
Abstract
We discuss the value of the best constant in Ga¤ney inequality namely kr!k2L2 C kd!k 2 L2+ k !k 2 L2+ k!k 2 L2 when either ^ ! = 0 or y ! = 0 on @ :
1
Introduction
We start by recalling Ga¤ney inequality for vector …elds. Let Rn be a bounded open smooth
set and be the outward unit normal to @ : Then there exists a constant C = C ( ) > 0 such that for every vector …eld u 2 W1;2( ; Rn)
kruk2L2 C kcurl uk 2 L2+ kdiv uk 2 L2+ kuk 2 L2
where, on @ ; either ^ u = 0 (i.e. u is parallel to and we write then u 2 WT1;2( ; Rn))
or y u = 0 (i.e. u is orthogonal to and we write then u 2 WN1;2( ; Rn)). In the context of
di¤erential forms (identifying 1 forms with vector …elds) this generalizes to (using the notations of [14] which are summarized in the next section) the following theorem (for references see below). Theorem 1 (Ga¤ney inequality) Let 0 k n and Rn be a bounded open C2 set. Then
The aim of this article is to study the best constant in such inequality. We therefore de…ne CT( ; k) = sup !2WT1;2nf0g ( kr!k2 kd!k2+ k !k2+ k!k2 ) (1) CN( ; k) = sup !2WN1;2nf0g ( kr!k2 kd!k2+ k !k2+ k!k2 ) (2) where k k stands for the L2 norm. It is easy to see (cf. Proposition 2) that CT( ; k) ; CN( ; k) 1:
The cases k = 0 and k = n are trivial and we always have then that CT = CN = 1; so the discussion
deals with the case 1 k n 1: In our main result, we need the concept of k convexity (cf. De…nition 4), which generalizes the usual notion of convexity, which corresponds to the case of 1 convexity, while (n 1) convexity means that the mean curvature of @ is non-negative. Our main result (cf. Theorem 8) shows that the following conditions are equivalent.
(i) CT( ; k) = 1 (respectively CN( ; k) = 1).
(ii) is (n k) convex (respectively is k convex).
(iii) The sharper version of Ga¤ney inequality holds namely 8 ! 2 WT1;2 ; k (respectively
8 ! 2 WN1;2 ; k )
kr!k2 kd!k2+ k !k2: (iv) The respective supremum is not attained.
(v) CT (respectively CN) is scale invariant, namely, for every t > 0
CT(t ; k) = CT( ; k) :
The result that is (n k) convex implies that CT( ; k) = 1 was already observed by Mitrea
[27].
We also show that for general non-convex domains the constants CT( ; k) and CN( ; k) can
be arbitrarily large (see Proposition 17).
The smoothness of the domain is essential in the previous discussion. We indeed prove (see Theorem 20) that if is a polytope (convex or not), or even more generally a set whose boundary is composed only of hyperplanes, then
kr!k2= kd!k2+ k !k2; 8 ! 2 CT1 ; k :
i.e. Ga¤ney inequality is, in fact, an equality with constant CT( ; k) = 1 (the result is also valid
with T replaced by N ). This fact was already observed in [11] when n = 3 and k = 1 for polyhedra. We now comment brie‡y on the history of Ga¤ney inequality. The proof goes back to Ga¤ney [19], [20] for manifolds without boundary and for manifolds with boundary to Friedrichs [18], Morrey [29] and Morrey-Eells [31]. Further contributions are in Bolik [5] (for a version in Lp and in Hölder spaces), Csató-Dacorogna-Kneuss [14], Iwaniec-Martin [23], Iwaniec-Scott-Stro¤olini [24] (for a version in Lp), Mitrea [27], Mitrea-Mitrea [28], Morrey [30], Schwarz [33], Taylor [35] and
von Wahl [36]. Speci…cally for the inequality for vector …elds in dimension 2 and 3; we can refer to Amrouche-Bernardi-Dauge-Girault [2], Costabel [9] and Dautray-Lions [16].
A stronger version of the classical Ga¤ney inequality reads as
(and similarly with T replaced by N ) where ! 2 WTd; ;2 ; k = 8 < :! 2 L 2 ; k : 8 < : d! 2 L2 ; k+1 ! 2 L2 ; k 1 ^ ! = 0 on @ 9 = ;
and the boundary condition has to be understood in a very weak sense. Clearly WT1;2 WTd; ;2: This stronger inequality is, in fact, a regularity result and is valid for smooth or convex Lipschitz domains leading, a posteriori, to
WTd; ;2 ; k = WT1;2 ; k :
However, for non-convex Lipschitz domains one has, in general, WT1;2 6= WTd; ;2: We refer to Mitrea [27] and Mitrea-Mitrea [28]; while for vector …elds in dimension 2 and 3; see Amrouche-Bernardi-Dauge-Girault [2], Ben Belgacem-Bernardi-Costabel-Dauge [4], Ciarlet-Hazard-Lohrengel [7], Costabel [8], Costabel-Dauge [10] and Girault-Raviart [21].
Clearly Ga¤ney inequality for k = 1 is reminiscent of Korn inequality. The best constant in Korn inequality have been investigated by Bauer-Pauly [3] and Desvillettes-Villani [17]. Our results (cf. Corollary 32) allow us to recover the best constant found in [3].
We should end this introduction with a striking analogy with the classical Hardy inequality (cf., for example [26] and the bibliography therein). Indeed, classically the best constant ; when the domain is convex (and in fact (n 1) convex, see [25]), is independent of the dimension (in this case = 1=4) and the best constant is not attained; while for general non-convex domains the best constant is, in general, strictly less than 1=4: However the authors were not able to see if this connection is fortuitous or not.
2
Notations
We now …x the notations, for further details we refer to [14]. (i) A k form ! 2 k = k(Rn) is written as
! = X
1 i1< <ik n
!i1 ikdxi1
^ ^ dxik
when convenient it is identi…ed to a vector in R(nk): When necessary, we extend, in a natural way,
the de…nition of !i1 ik to any 1 i
1; ; ik n (see Notations 2.5 (iii) in [14]).
(ii) The exterior product of 2 1 and ! 2 k is ^ ! 2 k+1 and is de…ned as
^ ! = X 1 i1< <ik n 2 4 n X j=1 j!i1 ik 3 5 dxj ^ dxi1 ^ ^ dxik = X 1 i1< <ik+1 n "k+1 X =1 ( 1) 1 i !i1 i 1i +1 ik+1 # dxi1^ ^ dxik+1:
(iii) The interior product of 2 1 and ! 2 k is y ! 2 k 1 and is de…ned as
(iv) The scalar product of !; 2 k is de…ned as h!; i = X 1 i1< <ik n !i1 ik i1 ik = 1 k! X 1 i1; ;ik n !i1 ik i1 ik
the associated norm being
j!j2= X 1 i1< <ik n !i1 ik 2= 1 k! X 1 i1; ;ik n !i1 ik 2:
When k = 1 the interior and the scalar product coincide.
(v) The Hodge operator associates to ! 2 k; ! 2 n k via the operation ! ^ = h !; i dx1^ ^ dxn; for every 2 n k: The interior product of 2 1 and ! 2 k can be then written as
y ! = ( 1)n(k 1)
( ^ ( !)) : We also use several times the identities
^ ( y !) + y ( ^ !) = j j2! and h ^ ; i = h ; y i : (3) (vi) The exterior derivative of ! 2 k is d! 2 k+1 and is de…ned as
d! = X 1 i1< <ik n 2 4 n X j=1 !i1 ik xj 3 5 dxj ^ dxi1 ^ ^ dxik = X 1 i1< <ik+1 n "k+1 X =1 ( 1) 1!i1 i 1i +1 ik+1 xi # dxi1^ ^ dxik+1:
When k = 1 we can identify d! with curl !:
(vii) The interior derivative (or codi¤erential) of ! 2 k is ! 2 k 1and is de…ned as
! = X 1 i1< <ik 1 n 0 @ n X j=1 !ji1 ik 1 xj 1 A dxi1 ^ ^ dxik 1:
When k = 1 we can identify ! with div !: (viii) The spaces WT1;2 ; k and W1;2
N ; k are de…ned as
WT1;2 ; k = ! 2 W1;2 ; k : ^ ! = 0 on @ WN1;2 ; k = ! 2 W1;2 ; k : y ! = 0 on @ where is the outward unit normal to @ :
(ix) The sets HT ; k and HN ; k are de…ned as
3
Some generalities
Our …rst result is the following.Proposition 2 Let Rn be a bounded open set and 0 k n: Then
CT( ; k) ; CN( ; k) 1:
Moreover
CT( ; 0) = CN( ; 0) = CT( ; n) = CN( ; n) = 1 and CT( ; k) = CN( ; n k) :
Remark 3 When k = 0 (respectively k = n), r! can be identi…ed with d! (respectively !). Therefore, for any ! 2 W1;2;
kr!k2= kd!k2 (respectively kr!k2= k !k2). Hence the statements in the proposition when k = 0 or n are trivial.
Proof Step 1. We …rst prove the main statement. Let x = (x1; ; xn) 2 and 0 < r < R be
such that
Br(x) BR(x) :
Choose a function 2 C1
0 ( ) such that = 1 in Br and 0 1 in : We extend it to Rn by
0: De…ne for every m 2 N
!m(x) = sin (mx1) (x) dx1^ ^ dxk2 C01 ; k :
Since !mvanishes on the boundary of BR; we have, by Theorem 5.7 in [14] (see also [13]),
Z BR jr!mj2= Z BR jd!mj2+ j !mj2 ) Z jr!mj2= Z jd!mj2+ j !mj2 :
To prove that CT; CN 1 it is su¢ cient to show that
lim m!1 kd!mk2+ k !mk2+ k!mk2 kr!mk2 = 1 + lim m!1 k!mk2 kr!mk2 = 1: Note that Z j!mj2 meas :
On the other hand we have that Z jr!mj2 Z Br jr!mj2= m2 Z Br cos2(mx1) dx:
Since Br is open, there exists m su¢ ciently large so that
Use now the change of variables x1= t=m to get Z jr!mj2 m meas (Br0) Z mx1+2 mx1 cos2(t) dt = m meas (Br0) : This shows that kr!mk2! 1 and thus CT; CN 1 as asserted.
Step 2. The fact that CT( ; k) = CN( ; n k) is immediate through the Hodge operator.
4
The main theorem
4.1
Statement of the theorem
We now turn to the main theorem that gives several equivalent properties of CT( ; k) = 1 (or
analogously CN( ; k) = 1). We start with a de…nition (cf. for a similar one [34]).
De…nition 4 Let Rn be an open smooth set and = @ be the associated (n 1) surface.
Let 1; ; n 1be the principal curvatures of : Let 1 k n 1: We say that is k convex if i1+ + ik 0; for every 1 i1< < ik n 1:
Remark 5 (i) When k = 1; it is easy to show that convex implies that is 1 convex. The reverse implication is also true but deeper. The result is due to Hadamard under slightly stronger conditions and as stated to Chern-Lashof [6] (see also Alexander [1]).
(ii) When 2 k n 1; the condition that is k convex is strictly weaker than saying that is convex. In particular when k = n 1 the condition means that the mean curvature of = @ is non-negative.
We will also use the following quantities.
De…nition 6 Let Rn be open and 2 C1 ; 1 : We de…ne for every 0 k n the two
maps L ; K : k(Rn) ! k(Rn) by L (!) = 0 if k = 0 and K (!) = 0 if k = n; while L (!) = X 1 i1< <ik n !i1 ikd y dxi1 ^ ^ dxik ; if k 1 K (!) = X 1 i1< <ik n !i1 ik ^ dxi1^ ^ dxik ; if k n 1:
Remark 7 (i) If is the unit normal to a surface and is extended to a neighborhood of such that j j = 1 everywhere, then (see [14] Lemma 5.5)
L ( ^ ) = ^ L ( ) and K ( y ) = y K ( ) :
The right-hand sides of these expressions do not depend on the chosen extension, see [14] Theorem 3.23. We will use this frequently henceforth.
(ii) In the remaining part of the article we will always assume that has been extended to a neighborhood of = @ so as to have j j = 1:
(iii) Note that L is linear in ! and : By de…nition it acts pointwise on !: In this way, identifying k(Rn) with vectors R(nk); the operator L can be seen as a matrix acting on !: But
We then have the main result.
Theorem 8 Let Rn be a bounded open smooth set and 1 k n 1: Then the following statements are equivalent.
(i) CT( ; k) = 1:
(ii) For every ! 2 WT1;2 ; k
e
K (!) = hK ( y !) ; y !i 0; on @ : (iii) The sharper version of Ga¤ ney inequality holds, namely
kr!k2 kd!k2+ k !k2; 8 ! 2 WT1;2 ; k : (iv) is (n k) convex.
(v) The supremum in (1) is not attained.
(vi) CT is scale invariant, namely, for every t > 0
CT(t ; k) = CT( ; k) :
Remark 9 (i) Using the Hodge operation, we obtain immediately, from the theorem, the fol-lowing equivalent relations.
- CN( ; k) = 1:
- eL (!) = hL ( ^ !) ; ^ !i 0; whenever y ! = 0 on @ : - The sharper version of Ga¤ney inequality holds, namely
kr!k2 kd!k2+ k !k2; 8 ! 2 WN1;2 ; k : - is k convex.
- The supremum in (2) is not attained. - CN is scale invariant.
(ii) The condition (ii) of the theorem can be equivalently rewritten (for any ! 2 WT1;2 ; k )
as
e
K (!) = hK (!) ; !i 0; on @ since, recalling that ^ ! = 0 (since ! 2 WT1;2),
e
K (!) = hK ( y !) ; y !i = h y K (!) ; y !i = hK (!) ; ^ ( y !)i = hK (!) ; ! y ( ^ !)i = hK (!) ; !i :
Similar remarks hold for eL:
(iii) Note that if CT( ; k) = 1; then HT ; k = f0g : This follows at once from (iii) of the
4.2
Some algebraic results
Lemma 10 Let 1 k n 1 and 1; ; k2 1 with iy j = 0; if i 6= j: Then iy ( 1^ ^ i 1^ i+1^ ^ k) = 0; i = 1; ; k j 1^ ^ kj = j 1j j kj : Furthermore let Cjl= n X s2; ;sk=1 ( 1^ ^ k)js2 sk( 1^ ^ k)ls2 sk: and cj= 1^ j 1^ j+1^ ^ k; then Cjl= ((k 1)!) k X =1 c 2 j l :
Proof Step 1. We …rst establish by induction that
ky ( 1^ ^ k 1) = 0
(and similarly for all the other i). Indeed if k = 2; this is our hypothesis; so assume that the
result has been proved for k and let us prove it for k + 1: We know from Proposition 2.16 in [14] that
k+1y ( 1^ ^ k) = ( k+1y 1) ^ ( 2^ ^ k) 1^ ( k+1y ( 2^ ^ k))
applying the hypothesis of induction we have the result. Step 2. We also proceed by induction and show that
j 1^ ^ kj = j 1j j kj :
When k = 2 we have still from Proposition 2.16 in [14] that
j 1^ 2j2= j 1j2j 2j2 j 1y 2j2= j 1j2j 2j2
as wished. So let us assume that the result has been proved for k and let us establish it for k + 1: By the very same proposition as above we get
j 1j2j 1^ 2^ ^ k^ k+1j2= j 1^ 1^ 2^ ^ k+1j2+ j 1y ( 1^ ^ k+1)j2
= j 1y ( 1^ ^ k+1)j2:
Moreover since
1y ( 1^ ^ k+1) = ( 1y 1) ^ ( 2^ ^ k+1) 1^ ( 1y ( 2^ ^ k+1))
using Step 1 and the hypothesis of induction, we infer that
Combining the results we have indeed proved our claim. Step 3. Writing ( 1^ ^ k)js2 sk= det h ( sr)s=j;s2; ;sk r=1; ;k i we …nd that ( 1^ ^ k)js2 sk = k X =1 ( 1) +1 j( 1^ 1^ +1^ ^ k)s2 sk = k X =1 ( 1) +1 j c s2 sk ( 1^ ^ k)ls2 sk= k X =1 ( 1) +1 l( 1^ 1^ +1^ ^ k)s2 sk = k X =1 ( 1) +1 l c s2 sk: Observing (using Steps 1 and 2) that
n X s2; ;sk=1 c s2 sk c s2 sk = ((k 1)!)Dc; cE= 8 < : (k 1)! c 2 if = 0 if 6= we …nd that Cjl= ((k 1)!) k X =1 c 2 j l : as claimed.
We next give a way of computing the quantity K :
Lemma 11 Let 1 k n; Rn be a smooth (n 1) surface with unit normal and Rn
be a neighborhood of : Let ; 2 C1 ; k be such that, on ;
^ = ^ = 0: Then the following equation holds true, on ;
hK ( y ) ; y i + hK ( y ) ; y i = h ; y i + h ; y i hr ( y ) ; i : (4) In particular if = ^ in with
= 1^ ^ k 1
where 1; ; k 12 C1 ; 1 with, for every i; j = 1; ; k 1 and for every x 2 ;
j (x)j = 1; (x)y i(x) = 0 and i(x)y j(x) = ij;
then, in ;
Proof Step 1. Applying Lemma 5.6 in [14] we …nd
hK ( y ) ; y i = h y K ( ) ; y i = h ; y i X
I
hr I; i I
where I = (i1; ; ik) 2 Nk with 1 i1< < ik n: We thus deduce that
h y K ( ) ; y i + h y K ( ) ; y i = h ; y i + h ; y i X I hr ( I I) ; i or in other words h y K ( ) ; y i + h y K ( ) ; y i = h ; y i + h ; y i hr ( y ) ; i which is exactly (4).
Step 2. The extra statement follows from Step 1 applied to = and Lemma 10 since then y = and
hr ( y ) ; i =Dr j j2 ; E= 0: The proof is therefore complete.
4.3
Calculation of sums of principal curvatures
In the sequel Rnwill always be a bounded open smooth set with exterior unit normal : When
we say that E1; ; En 1is an orthonormal frame …eld of principal directions of @ with associated
principal curvatures 1; ; n 1; we mean that f ; E1; ; En 1g form an orthonormal basis of
Rn and, for every 1 i n 1; n X j=1 Eij lxj = iEil ) i= n X j;l=1 EilEij xlj : (5)
Lemma 12 Let E1; ; En 1 be an orthonormal frame …eld of principal directions of @ with
associated principal curvatures 1; ; n 1: Assume that E1; ; En 1 are extended locally to
a neighborhood of @ such that f ; E1; ; En 1g form an orthonormal frame …eld of Rn: Let
1 k n 1; = Ei1 ik 1 = Ei1^ ^ Eik 1 and = Ej1 jk 1 = Ej1^ ^ Ejk 1 (for k = 1; = = 1). Then h ( ^ ) ; i = ( 1+ + n 1) i1+ + ik 1 while for 6= h ( ^ ) ; i + h ( ^ ) ; i = 0: Proof The case k = 1 is immediate. Indeed the …rst equation reads as
h ( ^ ) ; i = ( ) = div ( ) = 1+ + n 1:
While nothing is to be proved for the second equation. So we discuss now the case k 2: Step 1: k 2 (…rst equation). We now prove that if = Ei1 ik 1= Ei1^ ^ Eik 1; then
(i) We …nd that h ( ^ ) ; i = (k 11)! n X s1; ;sk 1=1 2 4 n X j=1 ( ^ )js1 sk 1 xj 3 5 s1 sk 1 = A + B
where (recalling that j j = 1) A = 1 (k 1)! n X j;s1; ;sk 1=1 j s1 sk 1 xj s1 sk 1 = 1 (k 1)! n X j;s1; ;sk 1=1 j xj( s1 sk 1)2+ 1 (k 1)! n X j=1 j n X s1; ;sk 1=1 " s1 sk 1 2 2# xj = div ( ) = 1+ + n 1 and B = 1 (k 1)! n X j;s1; ;sk 1=1 "k 1 X r=1 ( 1)r sr js1 sr 1sr+1 sk 1 # xj s1 sk 1 = 1 (k 1)! n X j;s1; ;sk 1=1 "k 1 X r=1 ( 1)r sr xj js1 sr 1sr+1 sk 1 # s1 sk 1 + 1 (k 1)! n X j;s1; ;sk 1=1 "k 1 X r=1 ( 1)r sr js1 sr 1sr+1 sk 1 xj # s1 sk 1 = B1+ B2:
The result will be established once we prove that
B1= 1 (k 1)! n X j;s1; ;sk 1=1 "k 1 X r=1 ( 1)r sr xj js1 sr 1sr+1 sk 1 # s1 sk 1 = i1+ + ik 1 and B2= 1 (k 1)! n X j;s1; ;sk 1=1 "k 1 X r=1 ( 1)r sr js1 sr 1sr+1 sk 1 xj # s1 sk 1= 0:
(ii) Let us …rst prove that B2= 0 (recalling that = Ei1^ ^ Eik 1). We write
(k 1)! B2= k 1 X r=1 n X j;s1; sr 1;sr+1; ;sk 1=1 n X sr=1 ( 1)r sr s1 sk 1 ! js1 sr 1sr+1 sk 1 xj :
Observe that, for every r = 1; ; k 1;
n X sr=1 ( 1)r sr E i1^ ^ Eik 1 s1 sk 1 = y Ei1^ ^ Eik 1 s1 sr 1sr+1 sk 1 = 0;
(iii) We …nally show that B1= i1+ + ik 1 : Note that, interchanging the positions of
the indices srand sr0; we get (recalling that = Ei1^ ^ Eik 1)
B1= 1 (k 1)! n X j;s1; ;sk 1=1 "k 1 X r=1 ( 1)r sr xj js1 sr 1sr+1 sk 1 # s1 sk 1 = 1 (k 2)! n X j;s1; ;sk 1=1 h s1 xj Ei1^ ^ Eik 1 js2 sk 1 Ei1^ ^ Eik 1 s1 sk 1i :
The result follows (cf. Lemma 10) since, for every 1 j; s1 n; k 1X r=1 EijrEs1 ir = 1 (k 2)! n X s2; ;sk 1=1 Ei1^ ^ Eik 1 js2 sk 1 Ei1^ ^ Eik 1 s1 sk 1 and thus B1= k 1 X r=1 0 @ n X j;s1=1 s1 xjE j irE s1 ir 1 A = i1+ + ik 1
which is exactly what had to be proved.
Step 2: k 2 (second equation). We …nally establish that if
= Ei1 ik 1 = Ei1^ ^ Eik 1 and = Ej1 jk 1 = Ej1^ ^ Ejk 1:
and 6= ; then
h ( ^ ) ; i + h ( ^ ) ; i = 0: This amounts to showing that X = 0 where
X = n X s1; ;sk 1=1 2 4 n X j=1 ( ^ )js1 sk 1 xj 3 5 s1 sk 1+ n X s1; ;sk 1=1 2 4 n X j=1 ( ^ )js1 sk 1 xj 3 5 s1 sk 1: (6) We write X = A + B where A = n X j;s1; ;sk 1=1 j s1 sk 1 xj s1 sk 1+ n X j;s1; ;sk 1=1 j s1 sk 1 xj s1 sk 1 = 2 n X j j xj n X s1; ;sk 1=1 s1 sk 1 s1 sk 1+ n X j=1 j n X s1; ;sk 1=1 [ s1 sk 1 s1 sk 1] xj = 0
which leads to B = B1+ B2 where B1= n X j;s1; ;sk 1=1 "k 1 X r=1 ( 1)r sr xj js1 sr 1sr+1 sk 1 # s1 sk 1 + n X j;s1; ;sk 1=1 "k 1 X r=1 ( 1)r sr xj js1 sr 1sr+1 sk 1 # s1 sk 1 B2= n X j;s1; ;sk 1=1 "k 1 X r=1 ( 1)r sr js1 sr 1sr+1 sk 1 xj # s1 sk 1 + n X j;s1; ;sk 1=1 "k 1 X r=1 ( 1)r sr js1 sr 1sr+1 sk 1 xj # s1 sk 1:
It remains to prove, in order to show that X = 0 where X is as in (6), that B1= B2= 0:
(i) We start with the fact that B2= 0: We rewrite the de…nition as
B2= k 1X r=1 n X j;s1; sr 1;sr+1; ;sk 1=1 " n X sr=1 ( 1)r sr s1 sk 1 # js1 sr 1sr+1 sk 1 xj + k 1 X r=1 n X j;s1; sr 1;sr+1; ;sk 1=1 " n X sr=1 ( 1)r sr s1 sk 1 # js1 sr 1sr+1 sk 1 xj :
Since, for every r = 1; ; k 1;
n X sr=1 ( 1)r sr s1 sk 1 = ( y )s1 sr 1sr+1 sk 1 = 0 n X sr=1 ( 1)r sr s1 sk 1 = ( y )s1 sr 1sr+1 sk 1 = 0; we …nd that indeed B2= 0:
(ii) We …nally prove that B1= 0: Note that, interchanging the positions of the indices sr and
sr0; we obtain B1= (k 1) n X j;s1; ;sk 1=1 s1 xj js2 sk 1 s1 sk 1+ js2 sk 1 s1 sk 1 :
The result follows if we can show that, if 6= where
Since both identities are established similarly, we prove only the …rst one, namely n X j;s1; ;sk 1=1 s1 xj h ( 1^ ^ k 1)js2 sk 1( 1^ ^ k 1)s1s2 sk 1 i = 0: (7)
Since 6= we assume, up to reordering, that 16= 1: We claim that
Cjs1 = n X s2; ;sk 1=1 h ( 1^ ^ k 1)js2 sk 1( 1^ ^ k 1)s1s2 sk 1 i = ((k 2)!) j1 s1 1 h 2^ ^ k 1; 2^ ^ k 1i : (8) which leads to n X j;s1; ;sk 1=1 s1 xj h ( 1^ ^ k 1)js2 sk 1( 1^ ^ k 1)s1s2 sk 1 i = ((k 2)!) h 2^ ^ k 1; 2^ ^ k 1i n X j;s1=1 s1 xj j 1 s11 = ((k 2)!) h 2^ ^ k 1; 2^ ^ k 1i 1 n X s1=1 s1 1 s11 = 0
(where 1 is the principal curvature corresponding to 1) which is exactly (7). It remains to show
(8). We have ( 1^ ^ k 1)js2 sk 1= k 1 X r=1 ( 1)r+1 jr cr s2 sk 1 ( 1^ ^ k 1)s1s2 sk 1= k 1 X t=1 ( 1)t+1 s1 t (bt)s2 sk 1: where cr= 1^ r 1^ r+1^ ^ k 1 and bt= 1^ t 1^ t+1^ ^ k 1:
We therefore have that Cjs1 = k 1 X r;t=1 ( 1)r+t jr s1 t n X s2; ;sk 1=1 cr s2 sk 1 (bt)s2 sk 1 = ((k 2)!) k 1 X r;t=1 ( 1)r+t jr s1 t D cr; bt E :
Invoking Lemma 10 and the fact that 16= 1; we obtain that, unless r = t = 1;
D cr; bt
4.4
Formulas for
L and K in terms of principal curvatures
We …rst prove the symmetry of L and K ; which essentially follows from the symmetry of the second fundamental form of a hypersurface. We use Remark 7 (i)–(iii) in the following lemma and its proof. In particular, recall that we have extended in a neighborhood of such that for any
;
[L ( ^ ) = ^ L ( ) and K ( y ) = y K ( )] on :
Lemma 13 Let Rn be a smooth n 1 dimensional hypersurface with unit normal and let
1 k n 1: Then at every point x0of and for every ; 2 k(Rn) the following two identities
hold
hL ( ^ ) ; ^ i = hL ( ^ ) ; ^ i hK ( y ) ; y i = hK ( y ) ; y i :
Proof We only prove the …rst one, the second one follows by duality ([14] Lemma 5.3).
Step 1. Note that since 2 k(Rn) ; i.e. has constant coe¢ cients, L ( ) = d ( y ) : Let us
prove that we can assume
(x0) = (1; 0; ; 0) = e1:
Choose A 2 O(n) such that A ( (x0)) = e1 and set = A ( ): It has the property that
A 1x
0 = e1:
Set
~ = A ( ) and ~ = A ( ) :
It now follows from Theorem 3.10 (note that A = A] if A 2 O (n)) and Proposition 2.19 in [14] that
h ^ L ( ) ; ^ i = A ( ) ^ d A]( )y A ( ) ; A ( ) ^ A ( ) =D ^ d ( y ~) ; ^ ~E: Since also ~ 2 k(Rn) ; i.e. has constant coe¢ cients, d ( y ~) = L (~) : This proves the claim of
Step 1.
Step 2. We now assume that (x0) = e1: By linearity it is su¢ cient to show the claim for
= dxi1
^ ^ dxik and = dxj1
^ ^ dxjk:
We distinguish two cases.
Case 1: i1 = 1 or j1 = 1: We consider only the case i1 = 1; the other one being handled
similarly. Then ^ = 0 (which implies L ( ^ ) = 0) and therefore hL ( ^ ) ; ^ i = 0 = hL ( ^ ) ; ^ i and the symmetry is proved.
Case 2: i1; j1> 1: We have to show that
h ^ d ( y ) ; ^ i = h ^ d ( y ) ; ^ i :
Since j1> 1; we get at x0 that = e1y (e1^ ) and the same for : So we have to show that
Clearly we can assume that (i1; ; ik) 6= (j1; ; jk) : By a direct calculation one obtains d ( y ) = n X s=1 k X =1 ( 1) 1 xisdxs^ dxi1 ^ ^ ddxi ^ ^ dxik;
whereba means that a has been omitted. Two possibilities may then happen.
Case 2.1. fi1; ; ikg and fj1; ; jkg di¤er in more than one index (considered as sets). Then
for any s = 1; ; n D
dxs^ dxi1^ ^ ddxi ^ ^ dxik; dxj1^ ^ dxjk
E = 0: It follows that hd ( y ) ; i = 0 and by symmetry (9) follows.
Case 2.2. There exist ; 2 f1; ; kg such that i1; ; bi ; ; ik = j1; ; bj ; ; jk :
Without loss of generality and hence
(j1; ; jk) = i1; ; i 1; j ; i ; ; bi ; ; ik :
We immediately …nd
hd ( y ) ; i = ( 1) + xij :
In the same way, using that
(i1; ; ik) = j1; ; bj ; ; j ; i ; j +1; ; jk ;
one obtains
hd ( y ) ; i = ( 1) + xji :
So to prove the symmetry we have to show that
i xj =
j
xi or equivalently r ej ; ei = r ei ; ej :
This last equality follows from the symmetry of the second fundamental form of the hypersurface at the point x0; because ei and ej are tangent vectors.
We now improve Lemma 13 (for a di¤erent proof of the next result see Lemma 37).
Lemma 14 Let Rn be a smooth n 1 dimensional hypersurface with unit normal and let
1 k n 1: Let E1; ; En 1be an orthonormal set of principal directions of with associated
principal curvatures 1; ; n 1: Then, at every point x02 and for every ; 2 k(Rn) ; the
following two identities hold
Remark 15 When k = n 1; the …rst formula reads as hL ( ^ ) ; ^ i = h ; E1^ ^ En 1i h ; E1^ ^ En 1i n 1 X j=1 j
while, when k = 1; the second one reads as
hK ( y ) ; y i = h ; i h ; i
n 1
X
j=1 j:
Proof We only prove the second statement. The …rst one can be deduced from the other one by duality, using for instance [14] Lemma 5.3. Since both sides of the equation are bilinear in ( ; ) it is su¢ cient to show the identity for basis vectors of k(Rn) : We choose basis vectors of the type
(a) = Ei1^ ^ Eik; = Ej1^ ^ Ejk
or of the type
(b) = ^ Ei1^ ^ Eik 1; = ^ Ej1^ ^ Ejk 1:
If either one of or is of the type (a), then one immediately obtains that both sides of (11) are zero and the equation is trivially satis…ed (see Lemma 10 and (3)). So we only need to consider the case that and are both of type (b). We distinguish two cases. We also let x02 :
Case 1: (i1; ; ik 1) = (j1; ; jk 1) : In that case the right hand side of (11) is equal to
X 1 l1< <lk 1 n 1 ; ^ El1 lk 1 ; ^ El1 lk 1 X j =2fl1; ;lk 1g j= X j =2fi1; ;ik 1g j:
We now use Lemma 11 with = Ei1 ^ ^ Eik 1: We can assume that f ; E1; : : : ; En 1g are
extended to an orthonormal basis in a neighborhood of x0: Note that from (3) and Lemma 10 we
get that y ( ^ ) = ^ ( y ) = : So Lemma 11 gives
hK ( y ) ; y i = hK ( ) ; i = h ( ^ ) ; i : (12) We get the result appealing to Lemma 12 namely
h ( ^ ) ; i = ( 1+ + n 1) i1+ + ik 1 =
X
j =2fi1; ;ik 1g
j:
Case 2: (i1; ; ik 1) 6= (j1; ; jk 1): The right hand side of (11) is now 0: So we have to
show that
hK ( y ) ; y i = 0
Let = Ei1^ ^ Eik 1 and = Ej1^ ^ Ejk 1: It follows from Lemma 11 (using (3) as in
Case 1) that
hK ( y ) ; y i + hK ( y ) ; y i = h ( ^ ) ; i + h ( ^ ) ; i hr ( y ) ; i : Recall that f ; E1; : : : ; En 1g are extended to an orthonormal basis in a neighborhood of x0 and
therefore r ( y ) = 0: Thus it follows from Lemmas 12 and 13 that
4.5
Proof of the main theorem
For the equivalence (i) , (iv), we give below a proof which is elementary and self-contained. A second proof can be obtained from Theorem 28 and the remark following it. Still another proof, more in the language of di¤erential geometry, can be given using Theorem 35. These two other proofs are independent of the one given below and of the previous analysis..
Proof (Theorem 8). We know from Theorem 5.7 in [14] (see also [13] or Theorem 33 for a slightly di¤erent way of expressing the identity) that, for every ! 2 WT1;2 ; k [ W
1;2 N ; k ; Z jd!j2+ j !j2 jr!j2 = Z @ (hL ( ^ !) ; ^ !i + hK ( y !) ; y !i) : (13) Step 1: (i) ) (ii). Assume that CT( ; k) = 1: This means that, for every ! 2 WT1;2 ; k ;
0 kd!k2+ k !k2 kr!k2+ k!k2= k!k2+ K (!) where K (!) = Z @ e K (!) = Z @ hK ( y !) ; y !i :
Next let ' 2 WT1;2 ; k be such that ' = ! on @ and ' 0 in outside an neighborhood
of @ : Note that, since ' = ! on @ ; then
K ( y ') = K ( y !) on @ : We thus have, by (13) and since ' = ! on @ ;
0 k'k2+ K (') = k'k2+ K (!) : Since k'k2is as small as we want, we deduce that
K (!) = Z @ e K (!) = Z @ hK ( y !) ; y !i 0; 8 ! 2 WT1;2 ; k : (14) We now prove (ii) from the above inequality. Choose ! 2 WT1;2 ; k ; 2 C1 and =
! 2 WT1;2 ; k : Invoking (14) we …nd that 0 K ( ) = Z @ e K ( ) = Z @ 2K (!) :e
Since is arbitrary, we have the claim, i.e. eK (!) 0: Step 2: (ii) ) (iii). From (13) we have
kd!k2+ k !k2 kr!k2= K (!) = Z
@
e K (!) and thus the result, since K (!) 0 (because eK (!) 0).
Step 3: (iii) ) (i). This is trivial, once coupled with Proposition 2. Step 4: (ii) ) (iv). We choose in (ii), for 1 i1< < ik 1 n 1;
From the assumption and the second conclusion in Lemma 14, we then obtain 0 hK ( y !) ; y !i = X
j =2fi1; ;ik 1g
j:
Step 5: (iv) ) (ii). This follows from the second conclusion of Lemma 14.
Step 6: (iii) ) (v). The fact that the supremum is not attained follows from (iii), since, for every ! 2 WT1;2 ; k ;
kr!k2 kd!k2+ k !k2 kd!k2+ k !k2+ k!k2 hence the result.
Step 7: (v) ) (i). In order to prove the statement, we show that if CT > 1; then there exists
a maximizer. We divide the proof into three substeps.
Step 7.1. Let !s2 WT1;2 ; k nf0g be a maximizing sequence (and hence !sis not a constant
form), i.e. lim s!1 kr!sk2 kd!sk2+ k !sk2+ k!sk2 = CT:
Without loss of generality, up to replacing !s by !s= kr!sk ; we can assume that kr!sk = 1 and
hence lim s!1 h kd!sk2+ k !sk2+ k!sk2 i = 1 CT < 1: (15)
In particular k!sk is bounded and thus, up to a subsequence that we do not relabel, there exists
! 2 WT1;2 ; k such that
!s* ! in W1;2:
We prove in the next substeps that ! is a maximizer.
Step 7.2. We …rst show that ! 6= 0: Suppose, for the sake of contradiction, that ! = 0; then from (15) we get lim s!1 h kd!sk2+ k !sk2 i = 1 CT < 1: (16)
From (13), we infer that there exists c1= c1( ) such that
kd!sk2+ k !sk2= 1 + Z @ hK ( y ! s) ; y !si 1 c1 Z @ j! sj2:
Since (cf. Proposition 5.15 in [14]) there exists c2= c2( ) such that for every > 0
Z @ j! sj2 kr!sk2+ c2 k!sk2= + c2 k!sk2 we deduce that kd!sk2+ k !sk2 1 c1 c1c2 k!sk2: Letting s ! 1 we …nd lim s!1 h kd!sk2+ k !sk2 i 1 c1
and, since is arbitrary, we …nd a contradiction with (16).
(i) We have, recalling that ! 2 WT1;2 ; k ; 1 = kr!sk2= kr (!s !)k2+ kr!k2+ 2 Z hr (!s !) ; r!i kr (!s !)k2+ CT kd!k2L2+ k !k 2 L2+ k!k 2 L2 + 2 Z hr (!s !) ; r!i : Since !s* ! in W1;2; we …nd that 1 = lim s!1kr (!s !)k 2 + kr!k2 lim s!1kr (!s !)k 2 + CT kd!k2L2+ k !k 2 L2+ k!k 2 L2 :
(ii) Since !s ! 2 WT1;2 ; k ; we have
kr (!s !)k2 CT kd (!s !)k2L2+ k (!s !)k2L2+ k(!s !)k2L2 = CT kd!sk2+ k !sk2+ k!sk2+ kd!k2L2+ k !k 2 L2+ k!k 2 L2 CT Z 2 [hd!s; d!i + h !s; !i + h!s; !i] :
Passing to the limit, recalling (15) and that !s* ! in W1;2; we get
lim s!1kr (!s !)k 2 1 CT kd!k2L2+ k !k 2 L2+ k!k 2 L2 :
(iii) Combining (i) and (ii) we obtain 1 = lim s!1kr (!s !)k 2 + kr!k2 lim s!1kr (!s !)k 2 + CT kd!k2L2+ k !k 2 L2+ k!k 2 L2 1
which implies that
kr!k2= CT kd!k2L2+ k !k 2 L2+ k!k 2 L2 as wished.
Step 8: (i) ) (vi). Since (i) (and thus (iii)) holds, we …nd, for ! 2 WT1;2 t ; k and setting
! (x) = u (x=t) ; kr!k2L2(t )= Z t jr! (y)j 2 dy = tn 2 Z jru (x)j2dx = tn 2kruk2L2( ) tn 2kduk2L2( )+ tn 2k uk 2 L2( )= kd!k 2 L2(t )+ k !k 2 L2(t )
which shows that CT(t ; k) = CT( ; k) = 1:
Step 9: (vi) ) (i). Without loss of generality we can assume that t < 1: We reason by con-tradiction and assume that CT( ; k) > 1: Invoking (v) we have that there exists u 2 WT1;2 ; k
such that CT( ; k) = kruk2L2( ) kduk2L2( )+ k uk 2 L2( )+ kuk 2 L2( ) : Setting ! (x) = u (x=t) ; we obtain that ! 2 WT1;2 t ; k and
Since t < 1; we get CT( ; k) < kr!k2L2(t ) kd!k2L2(t )+ k !k 2 L2(t )+ k!k 2 L2(t ) CT(t ; k)
which is our claim.
Theorem 8 (combined with Remark 9) has as an immediate corollary the following. Corollary 16 Let Rn be a bounded open smooth set and k = 1: Then
(i) CT( ; 1) = 1 if and only if the mean curvature of @ is non-negative;
(ii) CN( ; 1) = 1 if and only if is convex.
5
Some examples
We now deal with some special cases where we can make CT; CN arbitrarily large.
Proposition 17 Let 1 k n 1: Then there exists a set k B (a …xed ball of Rn) such that
CT( k; k) ; CN( k; n k) are arbitrarily large.
Remark 18 Except for the case k = 1; the sets k that we construct are not smooth. However it
is easy to modify slightly these sets so as to make them smooth, while preserving the proposition. Proof Since CT( ; k) = CN( ; n k) ; it is su¢ cient to prove the result for CT( ; k) : For the
sake of clarity we deal with the case k = 1 separately.
Step 1 ( k = 1). We let, for x 2 Rn; jxj denote the usual Euclidean norm. Let 0 < r < 1 and 1= fx 2 Rn : r < jxj < 1g : We then choose 2 C1([r; 1]) arbitrary and
! (x) = (jxj) n X i=1 xidxi2 WT1;2 1; 1 : Clearly !xji = ij+ 0xixj jxj ; d! = 0 and ! = div (x ) = n + jxj 0 leading to j !j2= (n + jxj 0)2 and jr!j2= n 2+ 2 jxj 0+ jxj2( 0)2:
Choose (s) = s n (with this choice we have ! = 0). We therefore have (denoting by n the
measure of the unit sphere of Rn) that
Z 1 jr!j2= n Z 1 r n2 n s n 1ds = n n 2 n s n n 1 r = n(n 1) r n 1 while Z 1 jd!j2+ j !j2+ j!j2 = n Z 1 r s n+1ds = ( n n 2 r n+2 1 if n > 2 nlog r if n = 2:
Thus, for r su¢ ciently small, we deduce that CT( 1; 1) is arbitrarily large as wished.
Step 2 ( 2 k n 1). We divide the proof into two parts. Step 2.1. Let us introduce some notations.
1) We write for x = (x1; ; xn) 2 Rn
jxjk =
q x2
1+ + x2n k+1:
2) Let 0 < r < 1: The set k Rn is then chosen as
k= fx 2 Rn: r < jxjk< 1 and 0 < xn k+2; ; xn< 1g :
3) We …nally let 2 C1([r; 1]) to be chosen below,
'k(x) = n k+1X
i=1
xidxi and !k(x) = (jxjk) 'k(x) ^ dxn k+2^ ^ dxn 2 k:
Step 2.2. Observe the following facts.
(i) If is the outward unit normal to k; then
^ !k= 0 on @ :
Indeed one sees that this is the case by distinguishing between the lateral boundaries jxjk = r; 1
where
= 1 jxjk
(x1; ; xn k+1; 0; ; 0) ) ^ 'k = 0
and the horizontal boundaries xs= 0; 1 (n k + 2 s n) where
= es ) ^ dxn k+2^ ^ dxn= 0:
(ii) We have, for 1 i1< < ik n; that
!i1 ik k (x) = 8 > < > : q x2 1+ + x2n k+1 xi1 if 1 i1 n k + 1 (i2; ; ik) = ((n k + 2) ; ; n) 0 otherwise. and thus, if 1 i; j n k + 1; @!i (n k+2)k n @xj = ij+ 0xixj jxjk
and all the other partial derivatives are 0: (iii) This leads to d!k = 0: Indeed if we set
0(s) = s (s) and (x) = (jxj k)
we see that
(jxjk) 'k(x) = d (x) ) d!k = 0:
(iv) We now prove that
j !kj2= ((n k + 1) + jxjk 0) 2
Indeed, since ( !k)i1 ik 1 = k X =1 ( 1) 1 X i 1<j<i @!i1 i 1ji ik 1 k @xj ; we have ( !k)i1 ik 1= 0 unless (i1; ; ik 1) = ((n k + 2) ; ; n) ; while
( !k)(n k+2) n= n k+1X j=1 @!jn k+2k n @xj = n k+1X j=1 @ ( (jxjk) xj) @xj = (n k + 1) + jxjk 0: (v) We next observe that
jr!kj2= (n k + 1) 2+ 2 jxjk 0+ jxj 2 k( 0)
2
: (vi) Finally choose (s) = s (n k+1) (with this choice we have !
k = 0). We therefore have
( n k+1 denoting the measure of the unit sphere of Rn k+1) that
Z k jr!kj2= n k+1 Z 1 r (n k + 1) (n k) s n+k 2ds = n k+1(k n) s n+k 1 1r = n k+1(n k) r n+k 1 1 while Z k jd!kj2+ j !kj2+ j!kj2 = n k+1 Z 1 r s n+kds = ( n k+1 n k 1 r n+k+1 1 if n > k + 1 n k+1log r if n = k + 1:
Therefore, when r ! 0; we …nd (writing for the asymptotic behavior) kr!kk2 kd!kk2+ k !kk2+ k!kk2 ( (n k 1)(n k) r2 if n > k + 1 1 r2log r if n = k + 1:
Thus, for r su¢ ciently small, we deduce that CT( k; k) is arbitrarily large as wished.
6
The case of polytopes
De…nition 19 Rn is said to be a generalized polytope, if there exist 0; 1; ; M bounded
open polytopes such that, for every i; j = 1; ; M with i 6= j;
i 0; i\ j = ; and = 0n M
S
i=1 i :
In this case i; i = 1; ; M; are called the holes.
Theorem 20 Let Rn be a generalized polytope. Then the following identity holds
kr!k2= kd!k2+ k !k2; 8 ! 2 CT1 ; k [ CN1 ; k :
Remark 21 (i) Note that we do not make any assumption on the topology of the domain and that holes are allowed. The identity shows that there are no non-trivial harmonic …elds with vanishing tangential (or normal) component which are of class C1: However, in presence of holes,
there are non-trivial harmonic …elds with weaker regularity (this is, of course, a problem only on the boundary, since harmonic …elds are C1 in the interior).
Before proceeding with the proof, we need to introduce a few notations that would help us keep track of the signs in the proof.
Notation 22 (i) For 16 k 6 n; we write
Tk = f(i1; ; ik) 2 Nk: 16 i1< < ik 6 ng:
For I = (i1; ; ik) 2 Tk; we write dxI to denote dxi1^ ^ dxik:
(ii) For i 2 I; we write Ibi = (i1; ;bi; ; ik); where bi denotes the absence of the named
index i: Note that, Iib
p 2 T
k 1; for all 1 6 p 6 k: Similarly, for i; j 2 I; i < j; we write
Iijb = (i1; ;bi; ; bj; ; ik):
(iii) Given I 2 Tk and i; j =2 I; i 6= j; we write [iI] to denote the increasing multiindex formed
by the index i and the indices in I: In other words [iI] is the permutation of the indices such that [iI] 2 Tk+1: Furthermore, we de…ne the sign of [i; I] ; denoted by sgn [i; I] ; as
dx[iI]= sgn [i; I] dxi^ dxI:
Similarly, [ijI] is the permutation of the indices such that [ijI] 2 Tk+2 and sgn [i; j; I] is given by dx[ijI]= sgn [i; j; I] dxi^ dxj^ dxI:
We need a few lemmas for the theorem.
Lemma 23 Let n 2 and 1 k n 1 be integers. Then for any I 2 Tk+1 and every i; j 2 I
with i 6= j; sgn h i; Iijb i sgn h j; Iijb i = sgn i; Ibi sgn h j; Ibj i and for any I 2 Tk 1 and every i; j =2 I with i 6= j;
sgn [i; [jI]] sgn [j; [iI]] = sgn [i; I] sgn [j; I] : Remark 24 When k = 1 both equations read as
sgn [i; j] sgn [j; i] = 1:
Indeed elements I 2 Tk 1 are as if they were absent, i.e. [jI] = j and sgn [i; I] = 1: Proof Since Ibi=hjIijbiand Ibj =hiIijbi; we have the identity
sgn i; Ibi sgn h j; Iijb i = sgn h i; j; Iijb i = sgn h j; i; Iijb i = sgn h j; Ibj i sgn h i; Iijb i :
This proves the …rst identity. The second one is just the …rst one, where I 2 Tk 1 plays the role
of Iijb: This …nishes the proof.
The next lemma gives a pointwise identity.
Remark 26 When k = 1 the lemma reads as jd!j2+ j !j2 jr!j2= 2X i<j @!i @xi @!j @xj @!j @xi @!i @xj while for k = 2 jd!j2+ j !j2 jr!j2 = 2 X i<j<k @!ij @xj @!ik @xk @!ij @xk @!ik @xj +@! ij @xk @!jk @xi @!ij @xi @!jk @xk +@! ik @xi @!jk @xj @!ik @xj @!jk @xi : Proof We calculate d! = X I2Tk+1 X i2I sgn i; Ibi @! Ibi @xi ! dxI and ! = X I2Tk 1 X i =2I sgn [i; I]@! [iI] @xi ! dxI: We start by evaluating jd!j2; we …nd jd!j2= X I2Tk+1 X i2I @!Ibi @xi 2 + 2 X I2Tk+1 X i;j2I i<j sgn i; Ibi sgn h j; Ibji @! Ibi @xi @!Ibj @xj :
Rewriting the terms in two di¤erent ways, we obtain, jd!j2= X I2Tk X i =2I @!I @xi 2 + 2 X I2Tk+1 X i;j2I i<j sgn i; Ibi sgnhj; Ibji @! Ibi @xi @!Ibj @xj (19) jd!j2= X I2Tk X i =2I @!I @xi 2 + 2 X I2Tk 1 X i;j =2I i<j
sgn [i; [jI]] sgn [j; [iI]]@!
[jI]
@xi
@![iI]
@xj
: (20)
We next evaluate j !j2; we get j !j2= X I2Tk 1 X i =2I @![iI] @xi 2 + 2 X I2Tk 1 X i;j =2I i<j sgn [i; I] sgn [j; I]@! [iI] @xi @![jI] @xj :
Rewriting the terms in two di¤erent ways, we …nd j !j2= X I2Tk X i2I @!I @xi 2 + 2 X I2Tk 1 X i;j =2I i<j sgn [i; I] sgn [j; I]@! [iI] @xi @![jI] @xj (21) j !j2= X I2Tk X i2I @!I @xi 2 + 2 X I2Tk+1 X i;j2I i<j sgnhi; Iijbisgnhj; Iijbi @! Ibj @xi @!Ibi @xj : (22)
Invoking (20) and (21), we …nd jd!j2+ j !j2 jr!j2 = X I2Tk 1 X i;j =2I i<j sgn [i; I] sgn [j; I]@! [iI] @xi @![jI] @xj
+ sgn [i; [jI]] sgn [j; [iI]]@!
[jI]
@xi
@![iI] @xj
:
Using Lemma 23, the last two identities establish (17) and (18), respectively. Lemma 27 Let Rn be a bounded open Lipschitz set and ! 2 C1 ; k : Then
kd!k2+ k !k2 kr!k2= Z @ h ^ !; d!i Z @ X I2Tk+1 X i;j2I sgn i; Ibi sgn h j; Ibj i !Ibj i@! Ibi @xj (23) kd!k2+ k !k2 kr!k2= Z @ h y !; !i Z @ X I2Tk 1 X i;j =2I sgn [i; I] sgn [j; I] ![iI] j@! [jI] @xi : (24) Proof We divide the proof in three steps.
Step 1. Integrate the equations of Lemma 25 to get kd!k2+ k !k2 kr!k2= Z X I2Tk+1 X i;j2I i6=j sgn i; Ibi sgnhj; Ibji @! Ibi @xi @!Ibj @xj @!Ibj @xi @!Ibi @xj ! (25) kd!k2+ k !k2 kr!k2= Z X I2Tk 1 X i;j =2I i<j sgn [i; I] sgn [j; I] @! [iI] @xi @![jI] @xj @![jI] @xi @![iI] @xj : (26)
Step 2. Noting that (the following argument uses the fact that ! 2 C2 but by density the identity (27) is valid for ! 2 C1)
@!Ibi @xi @!Ibj @xj @!Ibj @xi @!Ibi @xj = @ @xj !Ibj @! Ibi @xi @ @xi !Ibj @! Ibi @xj
we integrate by parts (25), bearing in mind that is Lipschitz, to obtain kd!k2+ k !k2 kr!k2= Z @ X I2Tk+1 X i;j2I i6=j sgn i; Ibi sgn h j; Ibj i !Ibj j @! Ibi @xi !Ibj i @! Ibi @xj : (27)
Step 3. Since is Lipschitz, is de…ned for a.e. x 2 @ : Thus, for a.e. x 2 @ ; we …nd
^ ! = X I2Tk+1 0 @X j2I sgn h j; Ibj i j!Ibj 1 A dxI:
But, since ! 2 C1 ; k ; for every x 2 ; we have
We therefore deduce the pointwise identity, for a.e. x 2 @ ; h ^ !; d!i = X I2Tk+1 X i;j2I sgn i; Ibi sgnhj; Ibji!Ibj j @! Ibi @xi = X I2Tk+1 X i2I !Ibi i@! Ibi @xi + X I2Tk+1 X i;j2I i6=j sgn i; Ibi sgnhj; Ibji!Ibj j @! Ibi @xi : We then have X I2Tk+1 X i;j2I i6=j sgn i; Ibi sgnhj; Ibji!Ibj j @! Ibi @xi = h ^ !; d!i X I2Tk+1 X i2I !Ibi i@! Ibi @xi : (28)
Substituting (28) in (27), we obtain (23). Analogous calculations yield (24), starting from integra-tion by parts of (26).
We now prove a theorem for piecewise C2 Lipschitz domains, which can be viewed as a gener-alization of Theorem 3.1.1.2 in [22] valid for k = 1:
Theorem 28 Let Rn be a bounded open Lipschitz set with piecewise C2 boundary, i.e. @ =
[N
s=1 s; where the s are C2 and relatively open subset of @ and @ n [Ns=1 s has zero surface
measure. Let E1; ; En 1 be an orthonormal frame …eld of principal directions of [Ns=1 s with
associated principal curvatures 1; ; n 1: Then
kd!k2+ k !k2 kr!k2= N X s=1 Z s n 1X l=1 ljEl^ !j2 ! for every ! 2 CT1 ; k (29) kd!k2+ k !k2 kr!k2= N X s=1 Z s n 1 X l=1 ljEly !j2 ! for every ! 2 CN1 ; k : (30)
Remark 29 (i) By standard regularization, the theorem is valid for ! 2 WT1;2 ; k (respectively
! 2 WN1;2 ; k ) if @ is (fully) C2:
(ii) Note that the two identities above are the same as those appearing at the end of Theorem 33.
Proof Step 1. We …rst show (29). Since @ n [Ns=1 s has zero surface measure, we obtain from
(23) kd!k2+ k !k2 kr!k2 = Z @ h ^ !; d!i N X s=1 Z s X I2Tk+1 X i;j2I sgn i; Ibi sgnhj; Ibji!Ibj i @!Ibi @xj : (31)
We argue on each of the s: Since E1; ; En 1 denote a frame of tangent vectors at each point
of s; we can write ej=Pn 1l=1 EljEl+ j : Thus, for any j = 1; ; n;
Step 1.1. We set A = X I2Tk+1 X i;j2I sgn i; Ibi sgnhj; Ibji!Ibj i @! Ibi @xj
and note, in view of (32), that A = B + C where
B = X I2Tk+1 X i2I sgn i; Ibi i@! Ibi @ ! 0 @X j2I sgnhj; Ibji j!Ibj 1 A C = X I2Tk+1 X i;j2I sgn i; Ibi sgn h j; Ibj i !Ibj i n 1 X l=1 Elj @! Ibi @El ! : (i) We …rst observe that
B = X I2Tk+1 ^ @!@ I ( ^ !)I = ^ !; ^ @! @ : (ii) We next prove that
C = n 1X l=1 El^ !; @ @El( ^ !) n 1X l=1 ljEl^ !j2:
Indeed note that, for any I 2 Tk+1 and any l = 1; ; n 1; @ @El( ^ !) I = @ @El X i2I sgn i; Ibi i!Ibi ! =X i2I sgn i; Ibi i@! Ibi @El +X i2I sgn i; Ibi !Ibi @ i @El :
For any j 2 I; multiplying by sgn h
j; Ibj i
!IbjEj
l and summing over l = 1; ; n 1 and j 2 I; we
deduce (recalling that lEli= @ i=@El) that
which is our claim, since C =PI2Tk+1CI:
Combining (i) and (ii) we have obtained that
A = ^ !; ^@! @ + n 1X l=1 El^ !; @ @El( ^ !) n 1 X l=1 ljEl^ !j2:
Step 1.2. Combining (31) and Step 1.1, we just proved that kd!k2+ k !k2 kr!k2 = N X s=1 Z s " h ^ !; d!i ^ !; ^ @! @ n 1 X l=1 El^ !; @ @El( ^ !) + n 1X l=1 ljEl^ !j2 # :
Thus, if ^ ! = 0 on sfor each s = 1; ; N; we obtain (29).
Step 2. Analogous calculations, starting from (24), establishes the identity kd!k2+ k !k2 kr!k2 = N X s=1 Z s " h y !; !i y !; y@! @ n 1X l=1 Ely !; @ @El ( y !) + n 1X l=1 ljEly !j2 # :
Using that y ! = 0 on s for each s = 1; ; N; this yields (30). This …nishes the proof.
We …nally are ready to prove the theorem.
Proof (Theorem 20). By Theorem 28, the result is immediate since for a generalized polytope, the principal curvatures on every face are 0:
Theorem 28 also immediately implies the following.
Theorem 30 Let Rn be a bounded open Lipschitz set with piecewise C2 boundary, i.e. @ = [Ns=1 s; where the s are C2 and relatively open subset of @ and @ n [Ns=1 s has zero surface
measure. If the principal curvatures are all nonnegative at every point on [N
s=1 s; then
kr!k2 kd!k2+ k !k2; 8 ! 2 CT1 ; k [ CN1 ; k :
Remark 31 Note that unlike the case of smooth domains, here the hypothesis of all principal curvatures being non-negative does not imply that the domain is convex. For example, the domain given in polar coordinates by
= f(r; ) : 0< < 2 0; r 2 [0; 1)g R2;
for some 0 < 0 < =2; is a piecewise C2 Lipschitz domain which satis…es the property, but
is neither convex, nor can be approximated from the inside by smooth 1 convex domains since smooth 1 convex domains are necessarily convex. Hence, the result for such domains is not covered by the result in [27].
For k = 1; Theorem 28 also immediately implies, as a corollary, the following variant of Korn inequality with the precise constant, which was already observed in Bauer-Pauly [3].
Corollary 32 Let Rn be a bounded open Lipschitz set with piecewise C2 boundary, i.e.
@ = [N
surface measure. If the principal curvatures are all nonpositive at every point on [N
s=1 s; then
kruk2 2 krsymuk2; 8 u 2 CT1 ; 1 [ CN1 ; 1 ; where the symmetric gradient rsymu is de…ned by
(rsymu)ij= 1 2 @ui @xj +@u j @xi ; for i; j = 1; ; n: Proof Theorem 28 in this case implies,
kruk2 kcurl uk2+ kdiv uk2; 8 u 2 CT1 ; 1 [ CN1 ; 1 : Integrating the pointwise identity
jruj2= jrsymuj2+1 2jcurl uj
2
and combining with the inequality above gives, krsymuk2 12kruk2= 1
2 kruk
2
kcurl uk2 12kdiv uk2 0: This completes the proof.
7
Appendix: an integral identity for di¤erential forms
We recall that Rnis a bounded open smooth set with exterior unit normal : When we say thatE1; ; En 1is an orthonormal frame …eld of principal directions of @ with associated principal
curvatures 1; ; n 1; we mean that f ; E1; ; En 1g form an orthonormal frame …eld of Rn
and, for every 1 i n 1;
n X j=1 Eij l xj = iE l i ) i= n X j;l=1 El iE j i xlj :
The following proposition was used only implicitly and is another version of the identity obtained in [13] (see also Theorem 5.7 in [14]). We …rst state and prove it in the Euclidean setting and then for general Riemannian manifolds, using then the notation of di¤erential geometry and referring to [12].
Theorem 33 Let 0 k n: Let E1; ; En 1 be an orthonormal frame …eld of principal
di-rections of @ with associated principal curvatures 1; ; n 1: Then every ; 2 C1 ; k
where Ei1 ik = Ei1 ^ ^ Eik: In particular the following two identities hold. For every 2 WT1;2 ; k Z jd j2+ j j2 jr j2 = Z @ X 1 i1< <ik 1 n 1 ; ^ Ei1 ik 1 2 X j =2fi1; ;ik 1g j
and for every 2 WN1;2 ; k Z jd j2+ j j2 jr j2 = Z @ X 1 i1< <ik n 1 jh ; Ei1 ikij 2 X j2fi1; ;ikg j:
Remark 34 If k = 0 or k = n; then the the right hand side has to be understood as 0 by de…nition. If k = 1; the theorem reads as
Z (hd ; d i + h ; i hr ; r i) = Z @ (h ^ d ( y ) ; ^ i + h y ( ^ ) ; y i) + Z @ n 1 X i=1 ih ; Eii h ; Eii + Z @ h ; i h ; i n 1X j=1 j and, in particular, Z jd j2+ j j2 jr j2 = 8 > > > > < > > > > : Z @ ( 1+ + n 1) jh ; ij2 if 2 WT1;2 ; 1 n 1X i=1 Z @ ijh ; Eiij2 if 2 WN1;2 ; 1 :
Proof The theorem follows from Theorem 5.7 in [14] and Lemma 14. The last two identities also follow from (29), respectively (30), of Theorem 28 and the remark following it.
We now discuss the more general version of the identity. We assume that is an n dimensional compact orientable smooth Riemannian manifold with boundary @ : We also adopt the following abbreviations Tn 1k = I = (i1; ; ik) 2 Nk: 1 i1< < ik n 1 EI = (Ei1; ; Eik) for I 2 T k n 1 Ic is the complement of I in f1; ; n 1g :
In the next theorem the quantity Fk is the linear 0 th order (not di¤erential) operator given
by the Bochner-Weitzenböck formula, see [12] Section 2.2 for more details. On a ‡at manifold, for instance Rn; the operator F
k is equal to 0 (which explains its absence in Theorem 33).
Theorem 35 Let 0 k n: Then every ; 2 C1 ; k satisfy the equation
Remark 36 In the case = a similar form of this identity is known as Reilly formula, see Theorem 3 in [32] and the references there. In that case the identity simpli…es: using partial integration twice one obtains, see Remark 3.8 (iv) in [12],
Z
@ (h ^ d ( y ) ; ^ i + h y ( ^ ) ; y i) = 2
Z
@ h y ( ^ ) ; y i :
ProofThe theorem follows from [12] Theorem 3.7 and Remark 3.8 (ii) and from Lemma 37 below. The following Lemma is the analogue of Lemma 14. N ( T) denotes the normal (respectively
tangential) component of (see [12]).
Lemma 37 Let Sk be de…ned as in De…nition 3.3 in [12]. Then the following identities hold
(i) hSk ; Ni = X I2Tnk 11 2 4 ( ; EI) ( ; EI) X j2Ic j 3 5 (ii) hSn k( ); ( T)i = X I2Tk n 1 2 4 (EI) (EI) X j2I j 3 5 :
Proof Step 1. We …rst prove (i). Since E1; ; En 1 are principal directions of @ they satisfy
that
rEi = iEi and hEi; Eji = ij:
We now choose ( ; E1; ; En 1) as an orthonormal basis of Rn to evaluate the scalar product
hSk ; Ni : Tuples (Ei1; ; Eik) have no contribution in hSk ; Ni ; because Nis normal (actually
Sk too). We therefore obtain
hSk ; Ni =
X
1 i1< <ik 1 n 1
Sk ; Ei1; ; Eik 1 ; Ei1; ; Eik 1 ; (33)
where we have used that N ; Ei1; ; Eik 1 = ; Ei1; ; Eik 1 by de…nition of the normal
component. By de…nition of Sk; we …nd Sk ; Ei1; ; Eik 1 = n 1X j=1 k 1 X l=1 Ej; Ei1; ; Eil 1; II (Ej; Eil) ; Eil+1; Eik 1 n 1X j=1 II (Ej; Ej) ; Ei1; ; Eik 1 = A(i1; ;ik 1)+ B(i1; ;ik 1);
where II is the second fundamental form. We have also used that (Er)T = Er for any r =
1; ; n 1; since they are tangent vectors. We therefore get for any r; s II (Es; Er) = hrEsEr; i = hEr; rEs i = r rs :
This shows that
B(i1; ;ik 1)= ; Ei1; ; Eik 1
n 1X
In the same way we get A(i1; ;ik 1)= k 1 X l=1 Eil; Ei1; ; Eil 1; II (Eil; Eil) ; Eil+1; Eik 1 = k 1 X l=1 il Eil; Ei1; ; Eil 1; ; Eil+1; Eik 1 = ; Ei1; ; Eik 1 k 1X l=1 il: So we obtain that Sk ; Ei1; ; Eik 1 = ; Ei1; ; Eik 1 0 @ n 1 X j=1 j k 1X l=1 il 1 A = ; Ei1; ; Eik 1 X j2Ic j:
From this last equation and (33) the …rst identity (i) follows. Step 2. We now deduce (ii) from (i) in the following way
hSn k( ) ; ( T)i = X I2Tnn1k 1 2 4( ) ( ; EI) ( T) ( ; EI) X j2Ic j 3 5 : Note that the Hodge operator computes on k forms ! as
( !) (Xk+1; ; Xn) = ! (X1; ; Xk)
whenever (X1; ; Xn) is an orthonormal basis of Rn: We apply this to X = ( ; EI; EIc) : Thus
we obtain that for each I 2 Tn k 1
( ) ( ; EI) ( T) ( ; EI) = (EIc) (EIc) :
This leads to, renaming the summation index I ! Ic;
hSn k( ) ; ( T)i = X I2Tn(n11) k 2 4 (EIc) (EIc) X j2Ic j 3 5 = X I2Tk n 1 2 4 (EI) (EI) X j2I j 3 5 ; which proves (ii).
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