Part I, Chapter 3 Simplicial Finite Elements 3.1 Simplices and Barycentric Coordinates

Home , Affine transformation, Degrees of freedom

Part I, Chapter 3 Simplicial ﬁnite elements

This chapter deals with finite elements (K,P,Σ) where K is a simplex in ♥♥ Rd with d 2, the degrees of freedom Σ are either nodal values in K or ≥ integrals over the faces (or edges) of K, and P is the space Pk,d composed of multivariate polynomials of total degree at most k 0. We focus our attention on scalar-valued finite elements, but most of the results≥ in this chapter can be extended to the vector-valued case by reasoning componentwise. Examples of vector-valued elements where the normal and the tangential components at the boundary of K play specific roles are studied in Chapters 8 and 9.

3.1 Simplices and barycentric coordinates

Simplices in Rd, d 1, are ubiquitous objects in the ﬁnite element theory. We review important properties≥ of simplices in this section.

3.1.1 Simplices

Definition 3.1 (Simplex, vertices). Let d 1. Let zi i 0: d be a set of d ≥ { } ∈{ } points in R such that the vectors z1 z0,..., zd z0 are linearly independent. Then, the convex hull of these{ − points is called− a}simplex in Rd, say K = conv(z0,..., zd). By definition, K is a closed set. If d = 1, a simplex is an interval [xl,xr] with real numbers xl < xr; if d = 2, a simplex is a triangle; and if d = 3, a simplex is a tetrahedron. The points zi i 0: d are the vertices of K. { } ∈{ } Example 3.2 (Unit simplex). The unit simplex in Rd corresponds to the d choice x R 0 xi 1, i 1:d , i 1: d xi 1 . This set has { ∈ | ≤ ≤ ∀ ∈ { } ∈{ } ≤ } d-dimensional measure equal to 1 . d! P Definition 3.3 (Faces, normals). For all i 0:d , the convex hull of the set z ,..., z z is denoted F and is called∈ { the face} of K opposite to the { 0 d} \ { i} i 30 Chapter 3. Simplicial finite elements

d vertex zi. By construction, Fi is a subset of an aﬃne hyperplane in R , and the unit normal to Fi pointing outward K is denoted nK,i. The unit outward normal vector on ∂K is denoted nK .

Deﬁnition 3.4 (l-faces). For all l 0:d 1 , an l-face of K is the convex ∈ { − } hull of a subset of zi i 0: d of cardinality (l + 1). By deﬁnition the l-faces are closed. The 0-faces{ } ∈{ of K} are the vertices of K. The 1-faces of K are called edges. In dimension d = 2, the notions of edge and face coincide. In dimension d = 1, the notions of vertex, edge, and face coincide.

Example 3.5 (Number of faces and edges). The number of l-faces in a d d+1 simplex in R is l+1 since an l-face is deﬁned by selecting a subset of vertices with cardinality (l +1). In particular, there are (d +1) faces and vertices, and d(d+1) there are 2 edges.

3.1.2 Barycentric coordinates in a simplex d d Let K be a simplex in R with vertices zi i 0: d . For all x R , we denote { } ∈{ } ∈ by λi(x), for all i 1:d , the components of the vector x z0 in the basis (z z ,..., z ∈z {), i.e.,} − 1 − 0 d − 0 x z = λ (x)(z z ). (3.1) − 0 i i − 0 i 1: d ∈{X }

Differentiating twice (3.1), we infer that the second Fréchet derivative of λi 2 d vanishes identically, i.e., D λi(x)(h1, h2) = 0 for all h1, h2 R . This implies ∈ d that λi is an affine function of x, i.e., there exist γi R and gi R such d ∈ ∈ that λi(x)= γi + gi x for all x R . Note that Dλi is independent of x and · d ∈ Dλi(h)= gi h for all h R ; in other words, λi = gi. To allow· all the vertices∈ of K to play a symmetric∇ role, we introduce the additional function λ0(x) := 1 i 1: d λi(x), so that the following holds − ∈{ } for all x Rd: ∈ P

λi(x)=1 and x = λi(x)zi. (3.2) i 0: d i 0: d ∈{X } ∈{X } A consequence of the above deﬁnitions is that

λ (z )= δ , i,j 0:d , (3.3) i j ij ∀ ∈ { }

This implies that the functions λi i 0: d are linearly independent: if the { } ∈{ } linear combination i 0: d βiλi(x) vanishes identically, evaluating it at a vertex z yields β = 0∈{ for all} j 0:d . Moreover, since K is the convex hull j j P ∈ { } of zi i 0: d , we infer that { } ∈{ } 0 λ (x) 1, x K, i 0:d . (3.4) ≤ i ≤ ∀ ∈ ∀ ∈ { } Part I. Getting Started 31

Deﬁnition 3.6 (Barycentric coordinates). The functions λi i 0: d are called the barycentric coordinates in K. { } ∈{ }

Proposition 3.7 (Geometric expression). The barycentric coordinates are such that F λ (x) = 1 | i| n (x z ), (3.5) i − d K K,i· − i | | Fi for all x K and all i 0:d , whence λi = d| K| nK,i. ∈ ∈ { } ∇ − | |

Proof. Since λi is constant on Fi, its gradient is collinear to nK,i. Since λi is aﬃne, we can write λi(x) = 1 cinK,i (x zi). We obtain ci by using that K = 1 F (n (z z )) for− all j = i·. − | | d | i| K,i· j − i 6 ⊓⊔ Remark 3.8 (Geometric identities). The following holds:

F n = 0, F n (c c )= K I , (3.6) | i| K,i | i| K,i ⊗ Fi − K | | d i 0: d i 0: d ∈{X } ∈{X } I where cFi is the barycenter of Fi, cK that of K, and d the identity matrix in d d R × . These identities hold in any polyhedron, see Exercise 3.1 for proofs. ⊓⊔

(♦) 3.1.3 Geometric map TA The following construction is important for finite elements having degrees of freedom over edges or faces, see 3.4 and Chapters 8 and 9 for examples. Let A be an affine subspace of Rd of§ dimension l 1:d 1 . We can build an l ∈ { − } A affine geometric map TA : R A as follows: Fix a simplex S in A with A → l l vertices zi i 0: l and fix a simplex S in R with vertices zi i 0: l and ∈{ } ∈{ } { } l { } l barycentric coordinates λi i 0: l . Typically, S is the unit simplex in R . { } ∈{ } A b l b l A Then, set TA(y) = i 0: l λi(y)zi for all y R . Note that TA(S ) = S ∈{ } ∈ and T (z ) = zA for all ib 0:l ; we often considerb the restriction of T to A i i P A l ∈ {b } b S and abuse the notation by still writing TA instead of TA Sl . b | b Propositionb 3.9 (Diffeomorphism). The map TA is a smooth diffeomorphism.

Proof. Since the barycentric coordinates λi i 0: l are aﬃne functions, the { } ∈{ } map TA is of class C∞ and DTA(y) is independent of y. Thus, we only need l l to verify that the linear map DTA : R Rb is invertible, which will follow by → l verifying that DTA is injective. Let h R be such that DTA(h) = 0. This A A ∈ yields i 1: l Dλi(h)(zi z0 ) = 0, so that Dλi(h) = 0 for all i 0:l ∈{ } − ∈ { } since the vectors (zA zA,..., zA zA) are linearly independent. We conclude P 1 0 l 0 using Exercise 3.2(iv).b − − b ⊓⊔ 32 Chapter 3. Simplicial ﬁnite elements

3.2 The polynomial space Pk,d

The real vector space Pk,d is composed of d-variate polynomial functions p : Rd R of total degree at most k. Thus, → P = span xα1 ...xαd , 0 α ,...,α k, α + ... + α k . (3.7) k,d { 1 d ≤ 1 d ≤ 1 d ≤ }

The importance of the polynomial space Pk,d is rooted in the fact that the Taylor expansion of order k of a d-variate function belongs to Pk,d. Another important fact is that, for any smooth function v : Rd R, → [ v P ] [ Dk+1v(x) = 0, x Rd ]. (3.8) ∈ k,d ⇐⇒ ∀ ∈

The vector space Pk,d has dimension

k +1 if d = 1, k + d 1 dim Pk,d = =  2 (k + 1)(k +2) if d = 2, (3.9) d   1 6 (k + 1)(k + 2)(k +3) if d = 3.  See Exercise 3.3 for the proof. We omit the subscript d and simply write Pk when the context is unambiguous. d An element α = (α1,...,αd) of N is called a multi-index, and its length is deﬁned to be α = α1 + ... + αd. A generic polynomial function p Pk,d can be written in| the| form ∈

α α α1 αd p(x)= aαx , x := x1 ...xd , (3.10) α ∈AXk,d with real coefficients a and multi-index set = α Nd α k . Note α Ak,d { ∈ || |≤ } that card( ) = dim(P )= k+d . Ak,d k,d d d Lemma 3.10 (Trace). Let Hbe an affine hyperplane in R and let TH : d 1 R − H be an affine bijective map. Then, p TH Pk,d 1 for all p Pk,d. → ◦ ∈ − ∈ k+1 d 1 Proof. Observe first that D (p TH )(y) =0 for all y R − by using the chain rule and the fact that T is◦ affine, then apply (3.8).∈ H ⊓⊔

3.3 Lagrange (nodal) ﬁnite elements

We begin with a simple example, see Table 3.1 for k = 1.

Proposition 3.11 (Simplicial Lagrange, k = 1). Let K be a simplex in d R with vertices zi i 0: d . Let P = P1,d. Let Σ = σi i 0: d be the linear { } ∈{ } { } ∈{ } forms on P such that σi(p) = p(zi) for all i 0:d . Then, (K,P,Σ) is a Lagrange ﬁnite element. ∈ { } Part I. Getting Started 33

Proof. We verify unisolvence. First, we notice that dim P = d +1 = card Σ. Moreover, we observe that

p(x)= p(z )+ Dp(x z )= p(z )+ λ (x)Dp(z z ) 0 − 0 0 i i − 0 i 1: d ∈{X } = λ (x)(p(z )+ Dp(z z )) = λ (x)p(z ), i 0 i − 0 i i i 0: d i 0: d ∈{X } ∈{X } for all p P and all x Rd, where we have used that p is affine, the definition (3.1) of∈ barycentric coordinates,∈ the linearity of Dp, and the first identity in (3.2). This proves that a polynomial in P vanishing at the (d + 1) vertices of K vanishes identically. ⊓⊔ We now extend the above construction to arbitrary polynomial order k 1 using equi-distributed nodes, see Remark 3.13 for alternative choices. ≥

Proposition 3.12 (Simplicial Lagrange). Let K be a simplex in Rd. Let k+d k 1 and let P = Pk,d. Set nsh = d and consider the set of nodes ≥ αi aα α such that aα z0 = (zi z0). Let Σ = σα α { } ∈Ak,d − i 1:d k − { } ∈Ak,d be the linear forms on P such that σ∈{ (p)} = p(a ) for all α . Then, P α α k,d (K,P,Σ) is a Lagrange ﬁnite element. ∈ A

k+d Proof. We verify unisolvence. Since card Σ = d = dim Pk,d, we need to prove the following assertion which we call [Ak,d]: A polynomial p Pk,d d ∈ vanishing at all the Lagrange nodes aα α k,d of any simplex in R vanishes identically. Assertion [A ] holds for{ all} ∈Ak 1 owing to Proposition 2.4. k,1 ≥ Assume now that d 2 and that [Ak,d 1] holds for all k 1 and let us prove that [A ] holds≥ for all k 1. Assume− that p P ≥vanishes at all k,d ≥ ∈ k,d the Lagrange nodes of a simplex K. Let F0 be the face of K opposite to d 1 the vertex z and consider an affine bijective map T : R − H , where 0 H0 → 0 H0 is the affine hyperplane supporting F0. Then, p TH0 is in Pk,d 1 owing to Lemma 3.10, and this polynomial vanishes at all◦ the Lagrange− nodes of 1 d 1 the simplex T − (F ) in R − . Owing to [Ak,d 1], p TH 0, and since H0 − ◦ 0 ≡ TH0 is bijective, p F 0. Denoting by λ0 P1,d the barycentric coordinate | 0 ≡ ∈ associated with z0, this implies that there is q Pk 1,d such that p = λ0q (see Exercise 3.3(iv)). Let us now prove by induction∈ on− k that q 0. For k = 1, q is in P and q(z ) = 0 (since λ (z ) = 1 = 0 and p(z ) = 0);≡ hence, q 0. 0,d 0 0 0 6 0 ≡ This proves [A1,d]. Now, let us assume that [Ak 1,d] holds for k 2. Since − ≥ k 2, q vanishes at all the Lagrange nodes aα such that α1 + ... + αd < k (since≥ λ (a ) = 0 at these nodes), i.e., α +... +α k 1. Hence q vanishes 0 α 6 1 d ≤ − at all the Lagrange nodes aα, α k 1,d. Since these nodes belong again to ∈A − a simplex, [Ak 1,d] implies q 0. − ≡ ⊓⊔ Table 3.1 presents examples of node location and shape functions for k 1, 2, 3 in dimensions d 2, 3 . The bullets conventionally indicate the∈ {location} of the nodes; see∈ Exercise { } 3.4 for some properties of these nodes. 34 Chapter 3. Simplicial finite elements

P1 P2 P3

1 λi λi(2λi 1) 2 λi(3λi 1)(3λi 2) − 9 − − 4λiλj 2 λi(3λi 1)λj 9 − λi(3λi 2)λj − 2 − 27λiλj λk

Table 3.1. Two- and three-dimensional P1, P2, and P3 Lagrange ﬁnite elements. Visible degrees of freedom are shown in black, hidden degrees of freedom are in white, and hidden edges are represented with dashed lines. The shape functions are expressed in terms of the barycentric coordinates; the ﬁrst line lists shape functions attached to nodes for all i 0:d , the second and third lines those attached to ∈ { } edges for all i,j 0:d , i

Possible choices for the domain of the interpolation operator are V (K) = C0(K) or V (K)= W s,p(K) with p [1, ] and s> d , see 1.4.1. ∈ ∞ p § Remark 3.13 (High-order). Finding Fekete points on simplices is delicate, see Chen and Babuˇska [132] and Taylor et al. [444] for results on triangles with polynomial order up to k = 13 and k = 19, respectively, see also Canuto et al. [120, p. 112]. A comparison of various nodal sets on triangles and tetrahedra can be found in Blyth et al. [55]. ⊓⊔ Remark 3.14 (Modal and hybrid simplicial elements). A hierarchic basis of the polynomial space Pk,d can be built by combining a hierarchical univariate basis of Pk,1 with the barycentric coordinates, see Ainsworth and Coyle [7] and Exercise 3.5. Another possibility is to introduce a nonlinear transformation mapping the simplex to a cuboid and to use tensor products of one-dimensional basis functions in the cuboid; see Proriol [382] for an early work, Dubiner [199], Owens [370], Karniadakis and Sherwin [304, 3.2]. § ⊓⊔ Part I. Getting Started 35

Remark 3.15 (Prismatic Lagrange elements). Let d 3 and set x′ = d d 1≥ + (x1,...,xd 1) for all x R . Let K′ be a simplex in R − and let [zd−,zd ] be a compact− interval with∈ non-empty interior. Then, the set K = x d + d { ∈ R x′ K′, x [z−,z ] is called a prism in R . Let k 1 and let | ∈ d ∈ d d } ≥ PRk = span p(x) = p1(x′) p2(xd) p1 Pk,d 1, p2 Pk,1 . Examples of { | ∈ − ∈ } prismatic Lagrange elements based on K and PRk with equidistributed nodes are shown in Table 3.2 for k 1, 2, 3 . ∈ { } ⊓⊔

PR1 PR2 PR3

Table 3.2. Nodes for prismatic Lagrange ﬁnite elements of degree 1, 2, and 3. The bullets indicate the location of the nodes; only visible nodes are shown.

( ) 3.4 Canonical hybrid ﬁnite elements ♦

We present a finite element (K,P,Σ) based on the polynomial space Pk,d whose degrees of freedom combine values at the vertices of the simplex K with integrals on the l-faces of K for l 1 (hence the name hybrid). This finite element is used in Chapter 9 to build≥ potentials of curl-free functions associated with the simplicial Nédélec finite element (hence the name canonical). For simplicity, we focus on the case d = 3; the construction is possible in any space dimension. d Let K be a simplex in R with vertices zi i 0: d (d = 3). The edges ∈{ } { } 1 of K are denoted Ei i 1:6 and its faces Fi i 0: d . Let TEi : S Ei ∈{ } ∈{ } 2 { } { } 1 → 2 and TFi : S Fi be affine bijective maps (see 3.1.2), where S and S are the unit simplices→ in R and R2. Let k 1§ be the polynomialb order. ≥ The canonicalb hybrid finite element involves vertex degrees of freedom,b edgeb degrees of freedom if k 2, surface (or face) degrees of freedom if k 3, and volume (or cell) degrees≥ of freedom if k 4. We consider the following≥ degrees of freedom: ≥ 36 Chapter 3. Simplicial finite elements

σx(p)= p(z ), i 0:d , (3.11a) i i ∈ { } e 1 1 e σ (p)= (µm T − ) p dl, i 1:6 , m 1:n , (3.11b) i,m E ◦ Ei ∈ { } ∈ { sh} | i| ZEi s 1 1 s σ (p)= (ζm T − ) p ds, i 0:d , m 1:n , (3.11c) i,m F ◦ Fi ∈ { } ∈ { sh} | i| ZFi 1 σv (p)= ψ p dx, m 1:nv (3.11d) m K m ∈ { sh} | | ZK e k 1 s k 1 where nsh = − and µm m 1: ne is a basis of Pk 2,1, nsh = − and 1 ∈{ sh} − 2 { } v k 1 ζm m 1: ns is a basis of Pk 3,2, and nsh = − and ψm m 1: nv is a { } ∈{ sh} − 3 { } ∈{ sh} basis of Pk 4,3. Observe that the total number of degrees of freedom is − d k 1 d + 1 k + d n = − = = dim(P ), (3.12) sh j d j d k,d j=0 X − owing to Vandermonde’s convolution identity.

Proposition 3.16 (Canonical hybrid finite element). Let k 1. Let K d ≥ be a simplex in R , let P = Pk,d, and let Σ = σi i be the degrees of freedom defined in (3.11). Then, (K,P,Σ) is a finite{ element.} ∈N

Proof. Owing to (3.12), we only need to prove that if p P is such that ∈ k,d σi(p)=0 for all i , then p vanishes identically. First, p vanishes at all the vertices of K. If k∈N= 1, this concludes the proof. If k 2, ﬁx an edge E of K. ≥ Since p TE vanishes at the two endpoints of E, p TE = λ0λ1q where λ0,λ1 ◦ ◦1 ∈ P1,1 are the local barycentric coordinates over S and q Pk 2,1. Since the ∈ − 2 degrees of freedom of p attached to E vanish, we infer that S1 λ0λ1q dl = 0, which implies q = 0. This implies that p is identicallyb zero on all edges of K. If k = 2, this completes the proof since all the Lagrange nodesR b for k = 2 are located at the edges of K. If k 3, we proceed similarly by ﬁxing a face F of K and showing that p is identically≥ zero on all faces of K. If k = 3, this completes the proof since all the Lagrange nodes for k = 3 are located at the faces of K. Finally, for k 4, we infer that p =(λ0 ...λd)qK where λi i 0: d ≥ { } ∈{ } are the barycentric coordinates of K and qK Pk 4,d. Since the degrees of ∈ − 2 freedom of p attached to K vanish, we infer that K (λ0 ...λd)qK dx = 0, which implies q = 0. The proof is complete. K R ⊓⊔ The shape functions associated with the vertices, the edges, the faces, and ˜ K are denoted ξi i 0: d , µ˜i,m i 1:6 ,m 1: ne , ζi,m i 0: d ,m 1: ns , and { } ∈{ } { } ∈{ } ∈{ sh} { } ∈{ } ∈{ sh} ψ˜ v , respectively. All these functions are in P and form a basis m m 1: nsh k,d thereof.{ } ∈{ The interpolation} operator acts as follows: Part I. Getting Started 37

e 6 nsh g 1 K (v)(x)= v(zi)ξi(x)+ v(µm TEi )dl µ˜i,m(x) I Ei ◦ i 0: d i=1 m=1 ZEi ∈{X } X X | | s v nsh nsh 1 ˜ 1 ˜ + v(ζm TFi )ds ζi,m(x)+ vψm dx ψm(x), Fi ◦ K i 0: d m=1 ZFi m=1 ZK ∈{X } X | | X | | and its domain can be chosen to be C0(K) or W s,p(K) with s > d for p p ∈ (1, ] or with s = d for p = 1. ∞

Exercises

Exercise 3.1 (Geometric identities). Prove (3.6). (Hint: differentiate (3.2) and use (3.5).) Devise another proof valid in any polyhedron using the diver- gence theorem. Exercise 3.2 (Barycentric coordinates). Let K be a simplex in Rd. (i) Give the explicit expression of the barycentric coordinates over K in dimension d 1 when K is the unit simplex. ≥ (ii) For any x K, let Ki(x) be the simplex obtained by joining x to the d ∈ Ki(x) vertices zj with j = i. Show that λi(x)= | K | . 6 1 | | (iii) Prove that λi dx = K for all i 0:d , and that λi ds = K d+1 Fj 1 | | ∈ { } Fj for all j 0:d with j = i, and λi ds =0. (Hint: consider an d | | R ∈ { } 6 Fi R affine transformation mapping K to the unit simplex.) d R (iv) Prove that if h R satisfies Dλi(h) = 0 for all i 1:d , then h = 0. ∈ d∈ { } Prove that the family nK,i i 1: d is a basis of R . { } ∈{ } Exercise 3.3 (Space Pk,d).

(i) Give a basis for the polynomial space P2 in dimensions d 1, 2, 3 . (ii) Show that any polynomial p P can be written in the∈ form { } ∈ k,d

p(x1,...,xd)= r(x1,...,xd 1)+ xdq(x1,...,xd), −

with unique polynomials r Pk,d 1 and q Pk 1,d. ∈ − ∈ − (iii) Determine the dimension of Pk,d. (Hint: by induction on d.) d (iv) Let K be a simplex in R . Let F0 be the face of K opposite to the vertex z0. Prove that if p Pk,d is such that p F 0, then there is q Pk 1,d ∈ | 0 ≡ ∈ − such that p = λ0q. (Hint: write the Taylor expansion of p at zd and use (3.1) with zd playing the role of z0.) Exercise 3.4 (Nodes of simplicial Lagrange FE). Let K be a simplex d in R , and consider the set of nodes ai 1 i nsh with barycentric coordinates i { } ≤ ≤ i0 ,..., d , i ,...,i 0:k with i + ... + i = k. k k ∀ 0 d ∈ { } 0 d 38 Chapter 3. Simplicial ﬁnite elements

(i) Prove that the number of nodes located on any one-dimensional edge of K is (k + 1) in any dimension d 2. (ii) Prove that the number of nodes≥ located on any (d 1)-dimensional face − of K is the dimension of Pk in dimension (d 1). (iii) Prove that if k d, all the nodes are located− on the boundary of K. ≤ Exercise 3.5 (Hierarchical basis). Let k 1 and let θ ,...,θ be a ≥ { 0 k} hierarchical basis of Pk,1. Let λ0,...,λd be a basis of P1,d and assume that λ : Rd R is surjective for all{ i 0:d }(i.e., λ is not constant). i → ∈ { } i d (i) Show that the functions (mapping R to R) θ0(λi),...,θk(λi) are linearly independent, for all i 0:d .(Hint: consider{ a linear combination} α θ (λ ) P and∈ { prove} that the polynomial α θ l 0: k l l i ∈ k,d l 0: k l l ∈ P ∈{vanishes} at (k + 1) distinct points.) ∈{ } Pk,1 P (ii) Show that the functions (mapping Rd to R) from the set

S := θ (λ ) ...θ (λ ) (α ,...α ) Nd, α k k,d { α1 1 αd d | 1 d ∈ | |≤ } are linearly independent. (Hint: by induction on d.) (iii) Show that (Sk,d)k 0 is a hierarchical polynomial basis, i.e., Sk,d Sk+1,d ≥ ⊂ and Sk,d is basis of Pk,d. (Note: the (d +1) vertices of K do not play the same role; in fact, the vertex z0 plays a diﬀerent role since its barycentric coordinate λ0 is not used in the construction of the basis.)

Exercise 3.6 (Cubic Hermite triangle). Let K be a (non-degenerate) triangle with vertices z , z , z . Set P = P and { 0 1 2} 3,2

Σ = p(zi),∂x p(zi),∂x p(zi) 0 i 2 p(aK ) , { 1 2 } ≤ ≤ ∪ { } where aK is a point inside K. Show that (K,P,Σ) is a ﬁnite element. (Hint: show that any p P3,2 for which all the degrees of freedom vanish is identically zero on the three∈ edges of K and infer that p = cλ λ λ for some c R.) 0 1 2 ∈