EXPLICIT POLYNOMIAL PRESERVING TRACE LIFTINGS

ON A TRIANGLE

MARK AINSWORTH AND LESZEK DEMKOWICZ

Abstract. We give an explicit formula for a right inverse of the traceoperator from the Sobolev space H1(T ) on a triangle T to the trace space

H1/2(∂T ) on the boundary. The lifting preserves polynomials in thesense that if the boundary data are piecewise polynomial of degree N ,then the lifting is a polynomial of total degree at most N and the liftingis shown to be uniformly stable independently of the polynomial order.Moreover, the same operator is shown to provide a uniformly stablelifting from L2(∂T ) to H

1/2(T ). Finally, the lifting is used to constructa uniformly bounded right inverse for the normal trace operator fromthe space H(div; T ) to H−1/2(∂T ) which also preserves polynomials.Applications to the analysis of high order numerical methods for partialdifferential equations are indicated.

1. Introduction

Let T denote a triangle and let F ∈ H1/2(∂T ) be given boundary data.We say that EF is a stable, polynomial preserving trace lifting of F pro-vided that the trace of EF on the boundary ∂T coincides with F ; if F isa polynomial of degree N , then EF is a polynomial of total degree at mostN ; and, there exists a positive constant C (independent of F ) such that

‖EF‖H1(T ) ≤ C‖F‖H1/2(∂T ). Here, H1(T ) and H1/2(∂T ) denote the usualSobolev spaces [1, 21].

The existence of polynomial trace liftings is of fundamental importancein the design and analysis of high order numerical methods for partial differ-ential equations. The paper of Babuška and Suri [4] is ostensibly concernedwith the convergence analysis of high order finite element methods for secondorder elliptic problems, but contains a key result establishing the existenceof polynomial extensions which plays a vital role in the analysis and ap-propriate treatment of the non-homogeneous essential boundary data. A

function F defined on the real line may be extended to a function F̃ de-fined on the upper half plane in a bounded fashion using a convolution, oraveraging, type operator

F 7→ F̃ : (x, y) → 12

∫

R

χI(s) F (x + sy) ds =1

2y

∫ x+y

x−yF (s) ds (1)

1991 Mathematics Subject Classification. 46E35, 47B35 (65N55, 65N35).Key words and phrases. Trace lifting. Polynomial extension. Polynomial lifting. Do-

main decomposition. p-version finite element method. Spectral element method.Partial support of M.A. from the Leverhulme Trust through a Leverhulme Trust Fel-

lowship and the Institute for Computational Engineering and Sciences at the University ofTexas at Austin is gratefully acknowledged. The work of the second author was supportedby Air Force Contract F49620-98-1-0255 .

1

2 MARK AINSWORTH AND LESZEK DEMKOWICZ

where χI denotes the characteristic function for the interval I = (−1, 1).Operators of the form (1) and the study of trace and extension theorems haveenjoyed a close relationship (see, for instance, Gagliardo [15] and Lions [18]and the references therein).

The polynomial extension result of Babuška and Suri [4] uses a successionof applications of the operator (1) applied to carefully crafted choices of thedata F to construct a polynomial lifting on a triangle and a square. Asimilar procedure was subsequently used in the more widely cited work ofBabuška et. al. [3] and applied to the analysis of domain decompositionpreconditioning techniques for high order finite element methods.

Maday [20] shows that the lifting defined in [4] provides stable polyno-mial liftings in weighted Sobolev spaces on a triangle or a square. Theseresults are then used in the study Hilbertian interpolation between poly-nomial subspaces of the weighted Sobolev spaces, which are of importancein the analysis of spectral type methods for partial differential equations.Subsequently, polynomial liftings in weighted Sobolev spaces on a squarewere studied by Bernardi and Maday [9] and used to derive improved errorestimates for the accuracy of the pressure in spectral collocation methodsfor the approximation of Stokes’ equations. Polynomial liftings onto higherorder Sobolev spaces are treated by Bernardi, Dauge and Maday [7] capital-ising on the earlier approach in [4, 20]. Polynomial liftings on a tetrahedronwere developed by Muñoz-Sola [22] following an idea mentioned in [20], andused to derive convergence estimates for high order finite element schemesfor second order elliptic problems in three dimensions, while the case of acube is studied by Ben Belgacem [5].

The importance of polynomial liftings in the development and analysisof domain decomposition algorithms lies in their close relationship with dis-crete harmonic extensions [11, 14]. These algorithms rely, either directly orindirectly, on minimising the coupling between basis functions associatedwith the interfaces and the interior functions associated with the subdo-mains themselves. Indeed, minimising this coupling is often the decisivestep in developing effective preconditioning techniques but generally comesat a high cost in terms of complexity. This begs the question of whether itmight be possible to construct basis functions associated with the interfacesthat a priori have (nearly) minimal coupling with the interior degrees offreedom?

In principle, one could use any of the polynomial liftings described earlierto define suitable basis functions to minimise the coupling with interior func-tions. Unfortunately, those constructions are based on a sequence of liftingsof the type (1) as in Babuška and Suri [4], and as such do not lend them-selves to efficient practical implementation. The question naturally arisesof whether it is possible to replace the sequence of successive applicationsof extension (1) by a one-step extension consisting of three simultaneousapplications.

The rule

EMSF (x, y) =

(1 − x − y√

3

) (1 + x − y√

3

) √3

2y

∫ x+y/√3

x−y/√

3

F (s)

1 − s2 ds, (2)

EXPLICIT TRACE LIFTINGS 3

suggested in [20], constitutes a one-step extension of data supported on asingle edge, and may be shown to be uniformly bounded using argumentssimilar to those found in [22]. Unfortunately, this operator does not readilygeneralise to give a one-step extension for more general data, although itmay be used sequentially to construct a stable trace lifting.

A one-step extension formula that is applicable to general data, if feasi-ble, is attractive not only in the practical construction of basis functions butalso theoretically (we show how it may used to derive liftings in H(div;T )).Unfortunately, the data for each of the three independent extension fromthe individual edges of the triangle must be judiciously chosen in order thatthe combined effect is to produce a stable lifting of the original data. Thisleads to a coupled system of three integral equations determining the ad-missible choices of data. Furthermore, the system has non-trivial kernelcorresponding to non-uniqueness in the admissible choices of boundary datafor the one-step extension. Together, these features frustrate attempts toderive explicit, closed-form solutions. Nevertheless, by exploiting rotationalsymmetry properties of the triangle, we are able to decouple the systeminto three scalar integral equations involving the cube roots of unity. Ananalysis of the scalar integral equations reveals (as expected) non-trivial ker-nels, and in addition, that solvability requires the data satisfy appropriatecompatibility conditions. We exhibit explicit solutions of the equations andestablish continuous dependence with respect to certain subspaces of theSobolev space H1/2. Interpreting these results in the context of the originalcoupled system, we exhibit closed form solutions, show that the compatibil-ity conditions are automatically satisfied and deduce continuous dependencein natural norms.

These results are then used to give an explicit formula for a right inverse ofthe trace operator from the Sobolev space H1(T ) on a triangle T to the trace

space H1/2(∂T ) on the boundary. The lifting preserves polynomials in thesense that if the boundary data are piecewise polynomial of degree N , thenthe lifting is a polynomial of total degree at most N and the lifting is shownto be uniformly stable independently of the polynomial order. Moreover, thesame operator is shown to provide a uniformly stable lifting from L2(∂T )

to H1/2(T )—a result that is of importance in the analysis of high orderboundary element methods [2]. Finally, the lifting is used to construct auniformly bounded right inverse for the normal trace operator from thespace H(div;T ) to H−1/2(∂T ) that also preserves polynomials.

2. Construction of the Trace Lifting

Let T denote the reference triangle with vertices located at A, B and Cas shown in Fig. 1 and let F ∈ H1/2(∂T ) be given real-valued data. Thegoal of the present work is to give an explicit construction for a stable,polynomial preserving trace lifting on the triangle. To this end, let ΦA, ΦBand ΦC be H

1/2-functions defined on the edges of the triangle whose valueswill ultimately be determined in terms of the data F ∈ H1/2(∂T ). For agiven point P ∈ T , we construct three lines passing through P parallel tothe edges of the triangle. The lines are used to define portions γA(P ), γB(P )and γC(P ) of the boundary as shown in Fig. 1. The value of the lifting EF

4 MARK AINSWORTH AND LESZEK DEMKOWICZ

B: (0, 3)

y

C: (−1,0) A: (0,1)xγ

γC

B

P: (x,y)

(P)

(P)Aγ

(P)

Figure 1. Reference triangle T showing portions of bound-ary γA(P ), γB(P ) and γC(P ) used in the definition (3) of thevalue of the extension at a point P ∈ T .

at a point P is defined to be the sum of the average values of the functionsΦA, ΦB and ΦC over their respective sub-portions of the boundary:

EF (P ) = –

∫

γA(P )ΦA + –

∫

γB(P )ΦB + –

∫

γC(P )ΦC (3)

where

–

∫

γΦ =

1

|γ|

∫

γΦ.

By introducing a parametrisation in terms of s ∈ (−1, 1) on each edge,oriented so that the boundary is traversed in an anticlockwise sense, wearrive at an explicit expression for the value of the lifting at the point Pwith coordinates (x, y) in the form:

EF (x, y) =1

1 + x − y/√

3

∫ 1−2y/√3

−(x+y/√

3)ΦA(s) ds +

√3

2y

∫ x+y/√3

x−y/√

3ΦB(s) ds

+1

1 − x − y/√

3

∫ −x+y/√3

2y/√

3−1ΦC(s) ds. (4)

By taking a sequence of points approaching a given point P located on theedge opposite to vertex A, we find that (in the sense of traces),

F|A(s) = ΦA(s) +1

1 − s

∫ 1

sΦB(−t) dt +

1

1 + s

∫ s

−1ΦC(−t) dt

where s ∈ (−1, 1) is the value of the parameter corresponding to the point Pand F|A denotes the restriction of F to the edge opposite vertex A. Likewise,by repeating this process for the remaining two edges, we obtain two furtherconditions on the functions Φ, resulting in the following system of integral

EXPLICIT TRACE LIFTINGS 5

equations:

F|A(s) = ΦA(s) +1

1 − s

∫ 1

sΦB(−t) dt +

1

1 + s

∫ s

−1ΦC(−t) dt

F|B(s) = ΦB(s) +1

1 − s

∫ 1

sΦC(−t) dt +

1

1 + s

∫ s

−1ΦA(−t) dt

F|C(s) = ΦC(s) +1

1 − s

∫ 1

sΦA(−t) dt +

1

1 + s

∫ s

−1ΦB(−t) dt

(5)

for all s ∈ (−1, 1). Thus, the task of constructing a stable, trace liftingis equivalent to establishing existence of solutions to the system (5) andcontinuous dependence on the data. Observe that we cannot expect a uniquesolution to the system since ΦA(s) = c1s

2 + 2c0s and ΦB(s) = ΦC(s) =c1s

2 − c0 satisfies the homogeneous equations for all c0 and c1.The study of (5) is facilitated by taking advantage of the invariance of the

reference triangle under rotation. Specifically, let ω ∈ {exp(im2π/3) : m =−1, 0, 1} denote a fixed, cube root of unity. Let us fix upon any particularedge (the edge opposite vertex A, say) of the triangle and consider theeffect of successively rotating the triangle through angles 0, 2π/3 and 4π/3,multiplying the equations on the edge by 1, ω and ω2(= ω) respectively, andsumming the resulting equations. The ensuing equation relates data

Fω = F|A + ωF|B + ωF|C (6)

to the new independent variable

Φω = ΦA + ωΦB + ωΦC (7)

through a scalar equation of the form:

Fω(s) = Φω(s) +ω

1 − s

∫ 1

sΦω(−t) dt +

ω

1 + s

∫ s

−1Φω(−t) dt. (8)

Alternatively, the same effect may be achieved purely algebraically by form-ing appropriate linear combinations of the equations appearing in the sys-tem (5).

By proceeding in this fashion for each distinct cube root of unity, weeffectively decouple the system (5) into three scalar equations involving newvariables Φ1, Φω and Φω. The fact that both complex cube roots of unity ωsatisfy 1 + ω + ω = 0 may be exploited to verify that the original variablesare uniquely determined in terms of the new variables by the relations

ΦA =13 (Φ1 + Φω + Φω)

ΦB =13 (Φ1 + ωΦω + ωΦω)

ΦC =13 (Φ1 + ωΦω + ωΦω) .

(9)

Hence, it suffices to study the scalar integral equations of the form (8) incases where ω is a cube root of unity.

6 MARK AINSWORTH AND LESZEK DEMKOWICZ

The study of these integral equations forms the topic of Section 4, whereit is shown that

Φ1 =1

4D

(1 − x2)∫ x

−1

F∆1 (s)

1 − s ds +1

1 + x

∫ x

−1(1 − s2)FΣ1 (s) ds

−(1 − x2)∫ 1

x

F∆1 (s)

1 + sds − 1

1 − x

∫ 1

x(1 − s2)FΣ1 (s) ds

(10)where D denotes the derivative and, FΣ1 and F

∆1 respectively denote the

even and odd parts of the data F1 = F|A + F|B + F|C . The correspondingexpression in the case of a complex cube root ω is slightly more involved.Let Fω = F|A + ωF|B + ωF|C , define

g±(x) = (1 − x)∫ x

−1

F∆ωR(s)

1 − s ds ∓ (1 + x)∫ 1

x

F∆ωL(s)

1 + sds (11)

where F∆ωL(s) = ωFω(s) − ωFω(−s) and F∆ωR(s) = ωFω(s) − ωFω(−s). Thesolution Φω is given by Φω =

12(Σω + ∆ω) where

∆ω(x) = (12)

−D{

g+(x) +1

2

∫ 1

−1g+(s) ds − 1

1 + x

∫ x

−1g+(s) ds − 1

1 − x

∫ 1

xg+(s) ds

}

and

(ω − ω)Σω(x) = (13)

D

{g−(x) − 1

1 + x

∫ x

−1g+(s) ds +

1

1 − x

∫ 1

xg+(s) ds +

1

2x

∫ 1

−1g+(s) ds

}.

Observe that it is convenient to express these quantities as derivativessince it is ultimately their primitives that are needed in the expression forthe extension formula (3). By inspection we see that if the data F arecontinuous piecewise polynomial of degree at most N , then Φ1 and Φω arepolynomials of degree at most N + 1. In turn, ΦA, ΦB and ΦC are alsopolynomials of degree at most N + 1 and it follows that the extension EFdefined in (4) preserves polynomials.

We now state our main result while postponing the proof until Section 5:

Theorem 1. Let F : ∂T → R be given data and let ω denote either of thecomplex cube roots of unity. Define Φ1 and Φω as above. If the functionsΦA, ΦB and ΦC appearing in (4) are taken to be

ΦA =13 {Φ1 + 2 Re(Φω)}

ΦB =13 {Φ1 + 2 Re(ωΦω)}

ΦC =13 {Φ1 + 2 Re(ωΦω)} ,

(14)

then the operator E defined in (4) formally defines a trace lifting of F . More-over, if the data F are continuous piecewise polynomials of degree at mostN on the boundary, then EF is a polynomial of total degree at most N . Thelifting is stable in the sense that

EXPLICIT TRACE LIFTINGS 7

2 4 6 8 10 12 14 16 18 20 220.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

Polynomial Degree

Ope

rato

r N

orm

M−S ExtensionA−D ExtensionOptimal Polynomial ExtensionOptimal Extension (Approx.)

Figure 2. Comparison of norm of various extension opera-tors for polynomial data supported on a single edge.

(1) if F ∈ L2(∂T ), then

‖EF‖H1/2(T ) ≤ C‖F‖L2(∂T ), (15)

(2) if F ∈ H1/2(∂T ), then

‖EF‖H1(T ) ≤ C‖F‖H1/2(∂T ) (16)

where C denotes positive constants independent of F and N .

In principle, bounds for the constants C could be extracted from theproof, although we shall not attempt this. At any rate the resulting boundsare likely to be overly pessimistic and we shall instead content ourselves withevaluating the norm of the extension operator for polynomial data defined ona single edge for increasing degree, giving the results shown in Fig. 2 for thecase of H1-extensions. In addition, we also present values of the norm of theextension operator defined in (2). For comparison, we also give values of thenorms for two other extension operators. The optimal extension operatorwould be obtained by solving the Dirichlet problem for the Laplacian usingpolynomial data, while the optimal polynomial extension may be obtainedby computing the Galerkin approximation of the Dirichlet problem usingpolynomials of total degree equal at most to the order of the boundarydata. By increasing the order of approximation, we are able to compute anapproximation to the optimal extension. The cost of evaluating the norms ofthese additional operators becomes prohibitive for large polynomial order,and only values for orders up to 13 have been given.

8 MARK AINSWORTH AND LESZEK DEMKOWICZ

2.1. An H(div;T )-Extension Operator. Let H(div;T ) denote the space

H(div;T ) = {v ∈ L2(T ) : div v ∈ L2(T )} (17)equipped with the graph norm. This function space arises naturally in thevariational formulation of any problem in continuum mechanics where aconservation principle is at work [12]. It is well-established that for any given

data G ∈ H−1/2(∂T ), there exists U ∈ H(div;T ) whose normal componenton the boundary of the triangle agrees with G in the sense of H−1/2(∂T )and such that

‖U‖H(div;T ) ≤ C‖G‖H−1/2(∂T ). (18)The existence and construction of stable, polynomial extensions is of consid-erable importance in the numerical analysis of high order numerical methodsfor such problems [13]. The fact that the boundary data G is an elementof a dual space obliges one to consider a single step lifting from the entireboundary, and the extension operator E of Theorem 1 proves useful in thisrespect.

Denote the average value of G by G0 = 〈G, 1〉/|∂T |, where 〈·, ·〉 de-notes the duality pairing on H−1/2(∂T ) × H1/2(∂T ), and note that G0 iswell-defined and bounded in absolute value by a multiple of ‖G‖H−1/2(∂T ).Clearly, G − G0 ∈ H−1/2(∂T ) has vanishing average value and as a con-sequence there exists F ∈ H1/2(∂T ), determined up to the addition of anarbitrary constant, such that

G − G0 =∂F

∂sin H−1/2(∂T ) (19)

where s denotes arc-length on the boundary ∂T . In particular, F may bechosen so that ‖F‖H1/2(∂T ) ≤ C‖G‖H−1/2(∂T ). Define an operator E by therule

EG =

[x√

3

y√

3 − 1

]G0 + grad

⊥(EF ). (20)

Observe that E is well-defined since E preserves constants meaning that EGis independent of the particular choice of F . Our main result concerningproperties of E is now an easy consequence of Theorem 1:

Corollary 1. Let G ∈ H−1/2(∂T ). Then, EG ∈ H(div;T ) satisfiesn · EG |∂T = G in H−1/2(∂T ) (21)

where n is the unit outward normal on ∂T , and

‖EG‖H(div;T ) ≤ C‖G‖H−1/2(∂T ) (22)where C is a positive constant independent of G. Moreover, if G can beidentified with a piecewise polynomial of degree at most N on each edge,then EG is polynomial of total degree at most N .

Proof. Elementary computation may be used to verify that (21) holds thanksto (19) and, with the aid of the triangle inequality, that

‖EG‖H(div;T ) ≤ C |G0| + |EF |H1(T ).The bound (22) follows at once thanks to Theorem 1 and the estimates for|G0| and ‖F‖H1/2(∂T ) recorded earlier. The statement concerning polynomialdata is a trivial consequence of Theorem 1. ¤

EXPLICIT TRACE LIFTINGS 9

3. Some auxiliary operators and their properties

Let I denote the reference interval (−1, 1). For integer n, the usualSobolev space [1] is denoted by Hn(I). For non-integer s ∈ (n, n + 1),the space Hs(I) is defined by the K-method of interpolation [6] between thespaces Hn(I) and Hn+1(I).

Let I denote the identity, M denote the averaging operator

Mφ =1

2

∫

Iφ(s) ds, (23)

and formally define operators AL and AR by the rules

AL φ(s) =1

1 + s

∫ s

−1φ(t) dt (24)

and

AR φ(s) =1

1 − s

∫ 1

sφ(t) dt (25)

for s ∈ I. Operators of this type are sometimes referred to as Hardy av-eraging operators [10, 23, 24], and the following forms of Hardy’s inequal-ity [17, 21], valid for α > 0, will be useful in establishing certain of theirproperties:

∫

I

∣∣∣∣(1 + s)−α

∫ s

−1

F (t)

1 + tdt

∣∣∣∣2 ds

1 + s≤ 1

α2

∫

I

∣∣(1 + s)−αF (s)∣∣2 ds

1 + s. (26)

and∫

I

∣∣∣∣(1 + s)α

∫ 1

s

F (t)

1 + tdt

∣∣∣∣2

ds

1 + s≤ 1

α2

∫

I|(1 + s)αF (s)|2 ds

1 + s. (27)

The operators are related by the rule (AR φ)− = AL φ− where we use thenotation φ± to denote the functions s 7→ φ(±s). In the sequel, we shalltacitly exploit the relationship between AR and AL to deduce propertiesof AR from those of AL. Consequently, only properties of AL are givenexplicitly:

Lemma 1. Let AL be defined as above. Then,

(1) for n = 0, 1, . . ., and φ ∈ Hn(I),

(AL φ)(n)(s) =

1

(1 + s)n+1

∫ s

−1(1 + t)nφ(n)(t) dt (28)

and

‖(AL φ)(n)‖L2(I) ≤(

n +1

2

)−1‖φ(n)‖L2(I) (29)

(2) for n = 0, 1, . . ., the operator AL : Hn+1/2(I) → Hn+1/2(I) is

bounded and linear with

‖AL φ‖Hn+1/2(I) ≤ C‖φ‖Hn+1/2(I) (30)where C is a positive constant independent of φ

(3) ∫

I|φ(s) − AL φ(s)|2

ds

1 + s≤ 1

3‖φ‖2

H1/2(I). (31)

10 MARK AINSWORTH AND LESZEK DEMKOWICZ

Proof. (1) The statement is obviously true in the case n = 0. Proceed-ing by mathematical induction, suppose the assertion holds up to n, thendifferentiation gives

(AL φ)(n+1)(s) =

1

1 + sφ(n)(s) − n + 1

(1 + s)n+2

∫ s

−1(1 + t)nφ(n)(t) dt

and integrating by parts shows that the assertion holds up to n + 1. Usingexpression (28) and applying inequality (26) with α = n + 1/2 and F (t) =

(1 + t)n+1φ(n)(t) gives∫

I

∣∣∣(AL φ)(n)(s)∣∣∣2

ds =

∫

I

∣∣∣∣1

(1 + s)n+1

∫ s

−1(1 + t)nφ(n)(t)

∣∣∣∣2 ds

1 + s

≤(

n +1

2

)−2 ∫

I|φ(n)(t)|2 dt.

(2) Apply operator interpolation using cases n and n + 1 in part (1).(3) By writing

φ(s) − AL φ(s) =1

1 + s

∫ s

−1(φ(s) − φ(t)) dt

and using the Cauchy-Schwarz inequality, we deduce that

|φ(s) − AL φ(s)|2 ≤1

3(1 + s)

∫

I

∣∣∣∣φ(s) − φ(t)

s − t

∣∣∣∣2

dt

which, after rearranging and integrating over I, gives the claimed estimate.¤

Let BR and BL denote the operators formally defined by the rules

BR φ(s) = (1 − s)∫ s

−1

φ(t)

1 − t dt (32)

and

BL φ(s) = −(1 + s)∫ 1

s

φ(t)

1 + tdt. (33)

These operators are related to dual Hardy averaging operators [23]. We need

some further notation before recording some useful properties. Let H1/2L (I)

denote the subspace

H1/2L (I) =

{φ ∈ H1/2(I) : (1 + s)−1/2φ ∈ L2(I)

}(34)

equipped with the norm defined by

‖v‖2H

1/2L (I)

= ‖v‖2H1/2(I)

+

∫

I|v(s)|2 ds

1 + s. (35)

Likewise, H1/2R (I) denotes the subspace obtained by replacing 1+ s by 1− s

in the definition. Following [19], the intersection of these spaces is denoted

by H1/200 (I) and is equipped with the norm

‖v‖2H

1/200

(I)= ‖v‖2

H1/2(I)+

∫

I|v(s)|2 ds

1 − s2 . (36)

EXPLICIT TRACE LIFTINGS 11

As before, it is our intention to use the relationship (BL φ)− = BR φ− to de-duce properties of BR from those of BL without explicitly drawing attentionto the fact.

Lemma 2. Let D denote the derivative operator. Then

(1) For ψ ∈ L2(I),‖(1 + s)−1 BL ψ‖L2(I) ≤ 2‖ψ‖L2(I), (37)

‖BL ψ‖L2(I) ≤2

3‖(1 + s)ψ‖L2(I) (38)

and

‖D BL ψ‖L2(I) ≤ 3‖ψ‖L2(I) (39)(2) BL : L2(I) → H1/200 (I) is bounded with

‖BL ψ‖H1/200

(I)≤ C‖ψ‖L2(I) (40)

(3) D BL : H1/2L (I) → H1/2(I)/R is bounded with

‖D BL ψ‖H1/2(I) ≤ C‖ψ‖H1/2L (I). (41)

In each case, C is a positive constant independent of ψ.

Proof. (1) The first two estimates are simply inequality (27) with α chosento be 1/2 and 3/2 respectively. Applying the triangle inequality to theidentity

D BL ψ(s) = −∫ 1

sψ(t)

dt

1 + t+ ψ(s)

and using inequality (27) with α = 1/2, gives

‖DBL ψ‖L2(I) ≤ 2‖ψ‖L2(I) + ‖ψ‖L2(I). (42)(2) Operator interpolation using the results of part (1) gives boundedness

of BL : L2(I) → H1/2(I). It suffices to show that there exists a positiveconstant C such that if ψ ∈ L2(I), then

∫

I

|BL ψ(s)|21 − s2 ds ≤ C‖ψ‖

2L2(I)

.

Applying the Cauchy-Schwarz inequality gives∫

I

|BL ψ(s)|21 − s2 ds =

∫

I

1 + s

1 − s

∣∣∣∣∫ 1

s

ψ(t)

1 + tdt

∣∣∣∣2

ds ≤∫

I(1 + s)

∫ 1

s

|ψ(t)|2(1 + t)2

dt ds.

Supposing for the moment that the order of integration may be interchanged,the right hand side may be rewritten as

∫

I

|ψ(t)|2(1 + t)2

∫ t

−1(1 + s) dsdt =

1

2‖ψ‖2L2(I).

The interchange of order is now justified using Fubini’s theorem and the factthat the above integral is bounded whenever ψ ∈ L2(I).

(3) Observing that

D2BL ψ(s) = Dψ(s) −

ψ(s)

1 + s

12 MARK AINSWORTH AND LESZEK DEMKOWICZ

and applying the triangle inequality and using (42) gives the estimate

‖D BL ψ‖H1(I) ≤ ‖ψ‖H1(I) + ‖(1 + s)−1ψ‖L2(I) ≤ C‖ψ‖H1(I).Interpolating using the above estimate and (42) gives

‖D BL ψ‖H1/2(I) ≤ C‖ψ‖H1/200

(I)(43)

as claimed. The proof is concluded by noting that D BL ψ has vanishing

mean I since BL ψ ∈ H1/200 (I). ¤The relationship between AL and BL is the subject of the following result.

Lemma 3. Let I−AL be formally defined by the rule

(I−AL)φ(s) = φ(s) −1

1 + s

∫ s

−1φ(t) dt, s ∈ I. (44)

Then, the following operators are bounded, linear and bijective

(1) I−AL : L2(I)/R → L2(I) with (I−AL)D BL = I on L2(I).(2) I−AL : H1/2(I)/R → H1/2L (I) with (I−AL)D BL = I on H

1/2L (I).

Proof. We use the same notation for both operators described above sincethis is unlikely to cause confusion. Lemma 1(1) shows that the first operatoris bounded while parts (2) and (3) of Lemma 1 show that the second operatoris bounded. Suppose that (I−AL)φ vanishes for φ ∈ L2(I),

(1 + s)φ(s) =

∫ s

−1φ(t) dt, s ∈ I,

then φ must be constant and as a consequence both operators are injective.Let ψ ∈ L2(I) then, thanks to Lemma 2(1), DBL ψ ∈ L2(I)/R, since BL ψ ∈H

1/200 (I) by Lemma 2(2). The same result also shows that

(I−AL)D BL ψ = −D(

(1 + s)

∫ 1

s

ψ(t)

1 + tdt

)+

∫ 1

s

ψ(t)

1 + tdt = ψ

and we conclude that the first operator is surjective and that (I−AL)D BLreduces to the identity on L2(I)/R. The argument for the second operatorfollows along the same lines using Lemma 2(3) in place of Lemma 2(1). ¤

4. Analysis of the Integral Equations

Throughout this section, let ω denote a fixed, cube root of unity and, forsuitable data f : I → C, consider the integral equation

φ(x) +ω

1 − x

∫ 1

xφ(−t) dt + ω

1 + x

∫ x

−1φ(−t) dt = f(x), x ∈ I. (45)

4.1. Compatibility Conditions. We begin by deriving conditions on thedata f that are necessary for the existence of a solution φ ∈ L2(I) or φ ∈H1/2(I). The conditions in the latter case will be expressed in terms offunctions f∆L = ωf+−ωf− and f∆R = ωf+−ωf−. For simplicity, we suppressthe dependence on ω in these notations since this will be clear from the

context. The relationship (f∆R )− = −(f∆L )+ shows that f∆L ∈ H1/2L (I) is

equivalent to f∆R ∈ H1/2R (I). Although this fact could be used to streamline

the statement of the next result, we prefer instead to highlight symmetry:

EXPLICIT TRACE LIFTINGS 13

Lemma 4. Suppose that φ satisfies equation (45). Then,

(1) if φ ∈ L2(I), then f ∈ L2(I) and ‖f‖L2(I) ≤ C‖φ‖L2(I)(2) if φ ∈ H1/2(I), then f ∈ H1/2(I) must satisfy f∆L ∈ H

1/2L (I), or

equivalently f∆R ∈ H1/2R (I), and

‖f∆L ‖H1/2L (I) + ‖f∆R ‖H1/2R (I) ≤ C‖φ‖H1/2(I). (46)

In each case, C is a positive constant that only depends on ω.

Proof. The first statement follows at once by writing

f = φ + ω AL φ− + ω AR φ−

and then applying the triangle inequality along with estimates from Lemma 1(1).The proof in the second case is less straightforward. Write

f∆L (s) = ω(I−AL)φ+(s) − ω(I−AL)φ−(s) − AR(φ+ − φ−)(s).With the aid of parts (2) and (3) of Lemma 1, we see that the first termmay be bounded as

‖ω(I−AL)φ+‖H1/2L (I) ≤ C‖φ‖H1/2(I)and observe that the second term may be bounded in the same way. Lemma 1may also be used to obtain

‖AR φ−‖H1/2(I) + ‖AR φ+‖H1/2(I) ≤ C‖φ‖H1/2(I).

It only remains to bound ‖(1 + s)−1/2 AR(φ+ − φ−)‖L2(I). Now,

AR(φ+ − φ−)(s) =1

1 − s

∫ 1

s(φ(t) − φ(−t)) dt

and by noting that the integrand is an odd function, we may rewrite this inthe form

AR(φ+ − φ−)(s) =1

1 − s

∫ 1

|s|(φ(t) − φ(−t)) dt.

Finally, using the fact that (1 + s)(1 − s)2 ≥ (1 − |s|)2 for s ∈ I, we obtain

‖(1 + s)−1/2 AR(φ+ − φ−)‖2L2(I)

≤∫

I

∣∣∣∣∣1

1 − |s|

∫ 1

|s|(φ(t) − φ(−t)) dt

∣∣∣∣∣

2

ds

then, noting that the integrand is an even function, this may be written inthe form

2

∫ 1

0

∣∣∣∣1

1 − s

∫ 1

s(φ(t) − φ(−t)) dt

∣∣∣∣2

ds

and, applying Hardy’s inequality, this may be bounded by 8‖φ(t) − φ(−t)‖2L2(I)which is in turn bounded in terms of ‖φ‖2L2(I). This completes the proof ofthe estimate for ‖f∆L ‖H1/2L (I). The estimate for the remaining term followsdirectly from the relationship between f∆R and f

∆L alluded to earlier. ¤

14 MARK AINSWORTH AND LESZEK DEMKOWICZ

The remainder of this section is concerned with proving these conditionsgiven in Lemma 4 are sufficient for existence of solutions, obtaining closedform expressions and establishing continuous dependence on the data. Webegin by deriving a pair of equations for the odd and even parts of φ givenby ∆ = φ+ −φ− and Σ = φ+ +φ− respectively. Substituting for φ+ and φ−in terms of ∆ and Σ in equation (45) gives

(I +ω AL +ω AR)Σ + (I−ω AL −ω AR)∆ = 2f+and replacing x by −x and simplifying gives

(I+ω AL +ω AR)Σ − (I−ω AL −ω AR)∆ = 2f−.Adding and subtracting these equations gives the following system

(2 I−(ω + ω)(AL + AR)) ∆ + (ω − ω)(AL −AR)Σ = 2f∆

(ω − ω)(AL −AR)∆ + (2 I+(ω + ω)(AL + AR)) Σ = 2fΣ(47)

where f∆ and fΣ denote the odd and even parts of f . Observe that theexistence of a solution of this system having the correct parity may be usedto infer the existence of a solution of the scalar equation (45) by retracingthe steps leading to (47).

4.2. Analysis in the case ω = 1. If ω is taken to be unity, the system (47)decouples to give

(I−AL −AR)∆ = f∆ (48)and

(I+ AL + AR)Σ = fΣ. (49)

Lemma 4 gives necessary conditions for the existence of a solution in L2(I)

or H1/2(I) to (45) in terms of the data f . The following result shows thatthese conditions are also sufficient.

Lemma 5. Suppose ω = 1. Equation (45) has a formal solution φ whoseodd and even parts are given by

∆ =1

2D

{(1 − x2)

∫ x

−1

f∆(s)

1 − s ds − (1 − x2)

∫ 1

x

f∆(s)

1 + sds

}(50)

and

Σ =1

2D

{1

1 + x

∫ x

−1(1 − s2)fΣ(s) ds − 1

1 − x

∫ 1

x(1 − s2)fΣ(s) ds

}. (51)

Moreover, if the data f satisfy

(1) f ∈ L2(I), then‖∆‖L2(I) ≤ C‖f∆‖L2(I) (52)

and

‖Σ‖L2(I) ≤ C‖fΣ‖L2(I), (53)(2) fΣ ∈ H1/2(I) and f∆ ∈ H1/200 (I), then

‖∆‖H1/2(I) ≤ C‖f∆‖H1/200

(I)(54)

and

‖Σ‖H1/2(I) ≤ C‖fΣ‖H1/2(I). (55)

EXPLICIT TRACE LIFTINGS 15

In each case, C is a positive constant independent of f .

Proof. It is not difficult to check that the expressions for ∆ and Σ indeedhave the correct parities. Therefore, thanks to the foregoing arguments,it suffices to check that ∆ and Σ satisfy equations (48) and (49). Simplecomputation reveals that

(AL + AR)∆ = −x∫ x

−1

f∆(s)

1 − s ds + x∫ 1

x

f∆(s)

1 + sds,

while expanding the derivative gives

∆ = f∆ − x∫ x

−1

f∆(s)

1 − s ds + x∫ 1

x

f∆(s)

1 + sds.

Together, these show that the expression for ∆ satisfies (48). Writing

∆ = f∆ − x(

1

1 − x BR f∆ +

1

1 + xBL f

∆

)

and using Lemma 2(1) gives ‖∆‖L2(I) ≤ C‖f∆‖L2(I) for f∆ ∈ L2(I). Fur-thermore, for f∆ ∈ H10 (I), we have

D∆ = D f∆ −(

1

1 − x BR f∆ +

1

1 + xBL f

∆

)+ x(AR −AL)D f∆,

and hence, using part (1) of Lemma 1 to bound the final term by 4‖D f∆‖L2(I)and we conclude that

‖∆‖H1(I) ≤ C‖f∆‖H1(I) ≤ C‖D f∆‖L2(I)for f∆ ∈ H10 (I). Interpolation then gives the bound for ‖∆‖H1/2(I).

Turning to the even part, expanding the derivative gives

Σ = fΣ − 12(1 + x)2

∫ x

−1(1 − s2)fΣ(s) ds − 1

2(1 − x)2∫ 1

x(1 − s2)fΣ(s) ds.

(56)Meanwhile, elementary manipulations give

AL Σ =1

2(1 + x)2

∫ x

−1(1 − s2)fΣ(s) ds

− 12(1 − x2)

∫ 1

x(1 − s2)fΣ(s) ds + 1

4(1 + x)

∫ 1

−1(1 − s2)fΣ(s) ds

with a similar expression for AR Σ. Simplifying leads to the conclusion thatΣ satisfies (49). Let F± be a primitive of (1 ± s)fΣ, then we may useLemma 1(1) to rewrite (56) in the form

Σ = fΣ − 12

DAL F− − 1

2DAR F

+.

Once again thanks to Lemma 1, we have

1

2‖D AL F−‖L2(I) ≤

1

3‖DF−‖L2(I) ≤ C‖fΣ‖L2(I)

and1

2‖D2 AL F−‖L2(I) ≤

1

5‖D2 F−‖L2(I) ≤ C‖fΣ‖H1(I)

16 MARK AINSWORTH AND LESZEK DEMKOWICZ

with similar bounds for derivatives of AR F+. Thus, we arrive at ‖Σ‖Hr(I) ≤

C‖fΣ‖Hr(I) for r = 0 and r = 1, and the bound in the case r = 1/2 is againobtained by interpolation. ¤

4.3. Analysis in the case ω 6= 1. If ω is a complex cube root of unity,then ω + ω = −1 and the system (47) takes the form

(2 I+ AL + AR)∆ + (ω − ω)(AL −AR)Σ = 2f∆

(ω − ω)(AL −AR)∆ + (2 I−AL −AR)Σ = 2fΣ.(57)

A necessary condition for the existence of a solution is that the data satisfy

f∆L ∈ H1/2L (I), or equivalently, f

∆R ∈ H

1/2R (I). The coupled nature of (57)

frustrates efforts to obtain closed form solutions for the odd and even parts.Nevertheless, by analogy with Lemma 5, we have the following result:

Lemma 6. Suppose ω is a complex cube root of unity. Equation (45) hasa formal solution φ whose odd and even parts are given by ∆ and Σ respec-tively, where

∆ = −D (I+ M−AL −AR) (BR f∆R + BL f∆L ) (58)

and

(ω − ω)Σ =D

{(BR f

∆R − BL f∆L ) + (AR −AL +xM)(BR f∆R + BL f∆L )

}. (59)

Moreover, if the data f satisfy

(1) f ∈ L2(I), then

‖∆‖L2(I) + ‖Σ‖L2(I) ≤ C‖f‖L2(I), (60)

(2) f ∈ H1/2(I) satisfies the compatibility condition f∆L ∈ H1/2L (I), or

equivalently f∆R ∈ H1/2R (I), then

‖∆‖H1/2(I) + ‖Σ‖H1/2(I) ≤ C{‖f∆L ‖H1/2L (I) + ‖f

∆R ‖H1/2R (I)

}, (61)

where C is a positive constant independent of f .

Proof. It is not difficult to check that BR f∆R +BL f

∆L is an even function and

in turn that the right hand side of (58) defines an odd function. Likewise,BR f

∆R −BL f∆L is an odd function and the right hand side of (59) defines an

even function. Thus, the expressions for ∆ and Σ have the correct parities.Denote g = BR f

∆R + BL f

∆L and note that if f

∆R and f

∆L ∈ L2(I), then

Lemma 2(2) implies that g ∈ H1/200 (I). Let Φ = (I + M−AL −AR)g, thenwe claim that Φ satisfies

Φ + (BR AL + BL AR)D Φ = g. (62)

Elementary manipulation shows that

(BR AL ±BL AR)DΦ = (1 − x)∫ x

−1

Φ(s)

1 − s2 ds ± (1 + x)∫ 1

x

Φ(s)

1 − s2 ds.

EXPLICIT TRACE LIFTINGS 17

Now, noting that

1

2D(M + AL −AR)g

=1

1 − x2{

g(x) − 1 + x2(1 − x)

∫ 1

xg(s) ds − 1 − x

2(1 + x)

∫ x

−1g(s) ds

}

=1

1 − x2 (I + M−AL −AR)g =Φ(x)

1 − x2 .

and observing that (M + AL −AR)g(−1) vanishes thanks to g ∈ H1/200 (I),we deduce that ∫ x

−1

Φ(s)

1 − s2 ds =1

2(M + AL −AR)g. (63)

Furthermore, (M + AL −AR)g(1) = M g for g ∈ H1/200 (I) and, using (63), itfollows that ∫ 1

x

Φ(s)

1 − s2 ds =1

2(M−AL + AR)g.

Thus,

(BR AL + BL AR)DΦ = M g − x(AL −AR)g,and as a consequence

Φ + (BR AL + BL AR)DΦ = g + 2M g − (1 + x)AL g − (1 − x)AR g = gwhich establishes (62). By differentiating and substituting DΦ = −∆ wefind that ∆ satisfies

∆ = −D(BR f∆R + BL f∆L ) − D(BR AL + BL AR)∆. (64)Similar arguments reveal that

(BR AL −BL AR)∆ = −(BR AL −BL AR)DΦ = (AR −AL +xM)gand we deduce that Σ satisfies

(ω − ω)Σ = D(BR f∆R − BL f∆L ) + D(BR AL −BL AR)∆. (65)Adding and subtracting equations (64) and (65) gives

∆ + (ω − ω)Σ = −2 D BL(f∆L + AR ∆)∆ − (ω − ω)Σ = −2 D BR(f∆R + AL ∆).

Now, if the data f belong to L2(I), then the right hand side of the firstequation also belongs to L2(I). Moreover, AR ∆(−1) = M ∆ = 0 since ∆is odd, and if f∆L ∈ H

1/2L (I), then the right hand side belongs to H

1/2L (I).

Hence, applying the appropriate operator I−AL defined in Lemma 3, weobtain

(I−AL)∆ + (ω − ω)(I−AL)Σ = −2(f∆L + AR ∆).A similar treatment of the remaining equation using the hypotheses on f∆Rleads to

(I−AR)∆ − (ω − ω)(I−AR)Σ = −2(f∆R + AL ∆).Finally, adding these equations we arrive at

(2 I + AL + AR)∆ + (ω − ω)(AL −AR)Σ = 2f∆

18 MARK AINSWORTH AND LESZEK DEMKOWICZ

while subtraction gives

−3(AL −AR)∆ + (ω − ω)(2 I−AL −AR)Σ = 2(ω − ω)fΣ,and so, multiplying by (ω − ω)/3 and noting that |ω − ω| =

√3, we arrive

at

(ω − ω)(AL −AR)∆ + (2 I−AL −AR)Σ = 2fΣ.Thus, ∆ and Σ satisfy (57) and the formal solvability of (45) follows.

It remains to demonstrate the boundedness of ∆ and Σ. With the aid ofLemma 1(1), we find that

‖D AR BL f∆L ‖L2(I) + ‖D AL BL f∆L ‖L2(I) ≤ C‖BL f∆L ‖L2(I)and applying Lemma 1(2) gives

‖D AR BL f∆L ‖H1/2(I) + ‖DAL BL f∆L ‖H1/2(I)≤ C

{‖BL f∆L ‖L2(I) + ‖D BL f∆L ‖H1/2(I)

}.

It is trivial to see that the same bounds hold for D(I−M)BL f∆L . Byparts (1) and (3) of Lemma 2, we have

‖BL f∆L ‖L2(I) ≤ C‖f∆L ‖L2(I)and

‖BL f∆L ‖L2(I) + ‖D BL f∆L ‖H1/2(I) ≤ C‖f∆L ‖H1/2L (I).

By applying the triangle inequality, we obtain

‖D(I+ M−AR −AL)BL f∆L ‖L2(I) ≤ C‖f∆L ‖L2(I)and

‖D(I + M−AR −AL)BL f∆L ‖H1/2(I) ≤ C‖f∆L ‖H1/2L (I).

Thanks to symmetry, there holds

‖D(I+ M−AR −AL)BR f∆R ‖L2(I) ≤ C‖f∆R ‖L2(I)and

‖D(I + M−AR −AL)BR f∆R ‖H1/2(I) ≤ C‖f∆R ‖H1/2R (I)and the bounds for ∆ follow at once. Concerning the bound for the evenpart Σ, we observe that the only term that has not already been accountedfor is the mean value M(BR f

∆L +BL f

∆R ). In view of the fact that the mean

value is constant, it suffices to note that the L2-norm of the argument ofM may be bounded in terms of ‖f∆L ‖L2(I) and ‖f∆R ‖L2(I) thanks to the firstpart of Lemma 2. ¤

5. Proof of Theorem 1

Let F : ∂T → R be given data and let F|A, F|B and F|C denote therestrictions of F to the edges opposite vertices A, B and C respectively.We begin by verifying that the data satisfy the compatibility conditionsdescribed in Lemma 4. The compatibility conditions given in Lemma 4(1)are trivially satisfied for data F ∈ L2(I), and so we turn our attention to

EXPLICIT TRACE LIFTINGS 19

the case where F ∈ H1/2(∂T ). It may be shown (e.g. [16]) that ‖F‖2H1/2(∂T )

is equivalent to

‖FA‖2H1/2(γA) + ‖FB‖2H1/2(γB)

+ ‖FC‖2H1/2(γC)+‖(1 + s)−1/2(FA(s) − FC(−s))‖2L2(I)+‖(1 + s)−1/2(FB(s) − FA(−s))‖2L2(I)+‖(1 + s)−1/2(FC(s) − FB(−s))‖2L2(I),

and it follows that FA(s)−FC(−s), FB(s)−FA(−s), FC(s)−FB(−s) belongto H

1/2L (I). To begin with, let ω denote an arbitrary cube root of unity. The

data Fω for the decoupled problem (8) are given by Fω = F|A +ωF|B +ωF|C .By writing

F∆ωL(s) = ωFω(s) − ωFω(−s) =ω (FA(s) − FC(−s)) + ω (FB(s) − FA(−s)) + (FC(s) − FB(−s)) ,

we see that F∆ωL ∈ H1/2L (I), or equivalently, F

∆ωR ∈ H

1/2R (I). Furthermore,

by applying the triangle inequality, we have

‖F∆ωL‖L2(I) + ‖F∆ωR‖L2(I) ≤ C‖F‖L2(∂T ) (66)and

‖F∆ωL‖H1/2L (I) + ‖F∆ωR‖H1/2R (I) ≤ C‖F‖H1/2(∂T ) (67)

where C is a positive constant independent of F . If ω = 1, then F∆ωL =F∆ωR = F

∆ and the above estimates specialise to

‖FΣ‖L2(I) + ‖F∆‖L2(I) ≤ C‖F‖L2(∂T ) (68)and

‖FΣ‖H1/2(I) + ‖F∆‖H1/200

(I)≤ C‖F‖H1/2(∂T ). (69)

The compatibility conditions of Lemma 4(2) are therefore satisfied automat-

ically for all values of ω provided that F ∈ H1/2(∂T ). Thanks to Lemmas 5and 6, solutions of the decoupled equation (8) exist for ω a cube root ofunity.

Suppose now that ω is a complex cube root of unity. The data Fω andFω satisfy the relation Fω = Fω. In turn, it is not difficult to verify that

F∆ωL = F∆ωL and F

∆ωR = F

∆ωR. The expressions for ∆ω and Σω given in

Lemma 6 depend on BR F∆ωR and BL F

∆ωL. However, in view of the fact

that BR F∆ωR = BR F∆ωR = BR F

∆ωR, along with a similar identity for BL F

∆ωL,

we conclude that ∆ω = ∆ω and Σω = Σω, and hence that Φω = Φω.Consequently, equations (9) reduce to the form claimed in Theorem 1.

Taking the average of the expressions for ∆ and Σ given in Lemma 5,we arrive at the formula (10) quoted earlier for Φ1. Moreover, again byLemma 5, we have

‖Φ1‖L2(I) ≤ C{‖FΣ‖L2(I) + ‖F∆‖L2(I)

}≤ C‖F‖L2(∂T )

and

‖Φ1‖H1/2(I) ≤ C{‖FΣ‖H1/2(I) + ‖F∆‖H1/2

00(I)

}≤ C‖F‖H1/2(∂T )

20 MARK AINSWORTH AND LESZEK DEMKOWICZ

thanks to (68) and (69). Observe that the data BR F∆ωR ±BL F∆ωL appearing

in Lemma 6 coincide with the quantities g± defined in (11). Consequently,expressions (12) and (13) agree with the ones quoted in Lemma 6, and henceΦω satisfies (8). Furthermore, using (66), (67) and Lemma 6, we have

‖Φω‖L2(I) ≤ C{‖F∆ωL‖L2(I) + ‖F∆ωR‖L2(I)

}≤ C‖F‖L2(∂T )

and

‖Φω‖H1/2(I) ≤ C{‖F∆ωL‖H1/2L (I) + ‖F

∆ωR‖H1/2R (I)

}≤ C‖F‖H1/2(∂T ).

Therefore, using equation (9) and the triangle inequality, we conclude that

‖ΦA‖L2(I) + ‖ΦB‖L2(I) + ‖ΦC‖L2(I) ≤ C‖F‖L2(∂T ). (70)and

‖ΦA‖H1/2(I) + ‖ΦB‖H1/2(I) + ‖ΦC‖H1/2(I) ≤ C‖F‖H1/2(∂T ). (71)By Lemma 7.1 of [3] or Theorem 2.2 of [8], there exists a positive constantC such that

‖Φ̃A‖H1/2(T ) ≤ C‖ΦA‖L2(I)and

‖Φ̃A‖H1(T ) ≤ C‖ΦA‖H1/2(I)where Φ̃A denotes the first term appearing in equation (4). The same esti-mates apply to the remaining terms giving

‖EF‖H1(T ) ≤ C{‖ΦA‖H1/2(I) + ‖ΦB‖H1/2(I) + ‖ΦC‖H1/2(I)

}

and

‖EF‖H1(T ) ≤ C{‖ΦA‖H1/2(I) + ‖ΦB‖H1/2(I) + ‖ΦC‖H1/2(I)

},

and, thanks to estimates (70) and (71), we conclude that the extension hasthe required stability properties. This completes the proof of Theorem 1since the claims concerning polynomial preserving properties of the exten-sions were discussed prior to Theorem 1.

References

[1] R. A. Adams, Sobolev spaces, Academic Press, New York-London, 1975. Pure andApplied Mathematics, Vol. 65.

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Mark Ainsworth, Department of Mathematics, Strathclyde University, 26

Richmond St., Glasgow G1 1XH, Scotland.

E-mail address: M.Ainsworth@strath.ac.uk

Leszek Demkowicz, Institute for Computational Engineering and Sciences,

201 E. 24th Street, The University of Texas at Austin, Austin TX 78712, USA.

E-mail address: leszek@ices.utexas.edu