Part I, Chapter 1
Introduction to finite elements
This chapter introduces key notions about finite elements that are repeatedly
used in this book, i.e., degrees of freedom, shape functions, and interpolation
operator. These notions are illustrated on Lagrange finite elements, for which
the degrees of freedom are values at specific nodes, and modal finite elements,
for which the degrees of freedom are moments against specific test functions.
1.1 Introductory example
This section introduces the notion of finite element through a simple example
in one space dimension. Let K = [−1, 1] and consider a smooth function
v : K → R; to fix the ideas, let us assume that v is continuous, i.e., v ∈ C 0 (K).
Our objective is to devise an interpolation operator that approximates v in
a finite-dimensional functional space, say P . For simplicity, we assume that
P = Pk for some k ∈ N, where Pk is the real vector space composed of
univariate polynomial
P functions of degree at most k, i.e., a function p : R → R
is in Pk if p(t) = l∈{0: k} αl tl where αl ∈ R for all l ∈ {0: k}.
Let us consider (k+1) disctint points {ai }i∈{0: k} in K, and let us call these
points nodes. Then, we would like to construct an operator IK : C 0 (K) → Pk
such that IK (v) verifies
IK (v) ∈ Pk ,
IK (v)(ai ) = v(ai ), ∀i ∈ {0: k},
(1.1)
for any v ∈ C 0 (K). These conditions uniquely determine IK (v) since a polynomial in Pk is uniquely determined by the value it takes at (k + 1) distinct
points in R. For the same reason, Pk is pointwise invariant under IK , i.e.,
IK (p) = p for all p ∈ Pk . To obtain an explicit representation of IK (v), we
introduce the Lagrange interpolation polynomials defined as follows:
Q
i6=j∈{0: k} (t − aj )
[a0 ...ak ]
,
∀t ∈ R,
(1.2)
Li
(t) = Q
i6=j∈{0: k} (ai − aj )
♥♥
2
Chapter 1. Introduction to finite elements
[a ]
for all i ∈ {0: k}. We set L0 0 ≡ 1 if k = 0. By construction, the Lagrange
interpolation polynomials are such that
[a0 ...ak ]
Li
(aj ) = δij ,
∀i, j ∈ {0: k},
(1.3)
where δij is the Kronecker delta. The Lagrange interpolation polynomials of
degree k ∈ {1, 2, 3} using equidistant nodes in K (including both endpoints)
[a ...a ]
are shown in Figure 1.1. We observe that the family {Li 0 k }i∈{0: k} forms
a basis of Pk . Indeed, since dim(Pk ) = k + 1, we only need to show linear
Pk
[a ...a ]
independence. Assume that i=0 αi Li 0 k ≡ 0, then evaluating this linear
combination at the nodes {ai }i∈{0: k} implies that a0 = . . . = ak = 0, which
proves the result. The polynomial function IK (v) defined by (1.1) has the
[a ...a ]
following representation in the basis {Li 0 k }i∈{0: k} :
IK (v)(t) =
k
X
[a0 ...ak ]
v(ai )Li
(t).
(1.4)
i=0
1
1
1
0.5
0.5
0.5
0
0
0
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1
Fig. 1.1. Lagrange interpolation polynomials with equidistant nodes in the interval
K = [−1, 1] of degree k = 1 (left), 2 (center), and 3 (right).
Remark 1.1 (Key concepts). To sum up, we have used three important
ingredients to build the interpolation operator IK : the interval K = [−1, 1],
the finite-dimensional space P = Pk , and a set of degrees of freedom, i.e.,
linear mappings {σi }i∈{0: k} acting on continuous functions, which consist of
evaluations at the nodes {ai }i∈{0: k} , i.e., σi (v) = v(ai ). A key observation
concerning the degrees of freedom is that they uniquely determine functions
in P , i.e., if σi (p) = 0 for all i ∈ {0: k} for some p ∈ Pk then p = 0.
1.2 Finite element as a triple
Following Ciarlet [141, p. 93], a finite element is defined as a triple: (K, P, Σ).
Part I. Getting Started
3
Definition 1.2 (Finite element). Let d ≥ 1. Let nsh ≥ 1. A finite element
consists of a triple, (K, P, Σ), where:
(i) K is a compact, connected, Lipschitz set in Rd with non-empty interior.
(ii) P is a finite-dimensional vector space of functions p : K → Rq for some
positive integer q (typically q = 1 or d); P is generally composed of polynomial functions, possibly composed with some smooth diffeomorphism.
(iii) Σ is a set of nsh linear forms from P to R, say Σ = {σi }i∈N with
N := {1: nsh }, such that the map ΦΣ : P → Rnsh acting as
ΦΣ (p) = σi (p) i∈N
(1.5)
is an isomorphism. The linear forms σi are called degrees of freedom.
Proposition 1.3 (Shape functions). dim(P ) = nsh and there exists a basis
{θi }i∈N of P such that
σi (θj ) = δij ,
∀i, j ∈ N ,
(1.6)
with δij the Kronecker symbol. The functions θi are called shape functions.
Proof. Direct consequence of ΦΣ : P → Rnsh being an isomorphism; observe
nsh
that {θi }i∈N is the image by Φ−1
.
⊓
⊔
Σ of the canonical basis of R
Another important consequence of (iii) is that Σ is a basis of L(P ; R), the
space of the linear forms from P to R. Indeed, dim(L(P ; R)) = dim(P ) = nsh ,
and
Plinear independence of the degrees of freedom follows from the fact that
if
P i∈N λi σi vanishes identically on P for some real numbers λi , then 0 =
i∈N λi σi (θj ) = λj for all j ∈ N .
Remark 1.4 (Unisolvence). Owing to the rank theorem (or rank-nullity
theorem), the bijectivity of the map ΦΣ is equivalent to
(
dim P ≥ nsh = card Σ,
∀p ∈ P, [ σi (p) = 0, ∀i ∈ N ] =⇒ [ p = 0 ].
This property is henceforth referred to as unisolvence.
⊓
⊔
Example 1.5 (Set K). A precise definition of Lipschitz sets can be found
in Definition B.1. At this stage, one can think of K as being a polyhedron
in Rd , e.g., a triangle or a square if d = 2 and a tetrahedron, a cube or a
prism if d = 3; see Figure 1.2. More generally, K can also be the image of a
polyhedron by some smooth diffeomorphism.
⊓
⊔
4
Chapter 1. Introduction to finite elements
Fig. 1.2. Simple examples of polyhedra in R2 and R3 ; dashed lines are used for
hidden parts of the polyhedron in R3 .
1.3 Interpolation: finite element as a quadruple
The notion of interpolation operator is central to the finite element theory;
this operator maps functions defined over K onto P using the shape functions
and the degrees of freedom of the finite element. The term “interpolation” is
used here in a broad sense, since the degrees of freedom are not necessarily
point evaluations. The construction of the interpolation operator requires to
extend the domain of the linear forms in Σ so that they can act on functions
in a space larger than P , which we denote V (K).
Definition 1.6 (Interpolation operator). Let (K, P, Σ) be a finite element. Assume that there exists a Banach space V (K) ⊂ L1 (K; Rq ) such that:
(i) P ⊂ V (K).
(ii) The linear forms {σ1 , . . . , σnsh } can be extended to L(V (K); R). (This
means that these linear forms are bounded on V (K), i.e., there exists
cΣ such that |σi (v)| ≤ cΣ kvkV (K) for all v ∈ V (K) and all i ∈ N .)
Then, we define IK : V (K) → P , the interpolation operator associated with
the finite element (K, P, Σ), by
X
IK (v)(x) =
σi (v)θi (x),
∀x ∈ K,
(1.7)
i∈N
for all v ∈ V (K). V (K) is the domain of IK , and P is its codomain. By
construction, IK ∈ L(V (K); V (K)).
Proposition 1.7 (P -invariance). P is pointwise invariant under IK , i.e.,
IK (p) = p for all p ∈ P . As a result, IK is a projection, i.e., IK ◦IK = IK
P
P
Proof. Letting p = j∈N αj θj yields IK (p) = i,j∈N αj σi (θj )θi = p owing
to (1.6). This shows that P is pointwise invariant under IK , whence it immediately follows that IK is a projection.
⊓
⊔
A finite element could actually be defined as a quadruple (K, P, Σ, V (K)),
where the triple (K, P, Σ) satisfies Definition 1.2 and V (K) satisfies Definition 1.6. The choice for V (K) plays a key role in the analysis of the approximation properties of IK , and many choices are possible. Under the mild assumption that P ⊂ L∞ (K; Rq ), it is possible to extend the degrees of freedom
so that V (K) = L1 (K; Rq ) (see §11.2). However, different (more restrictive)
Part I. Getting Started
5
choices for V (K) are interesting in practice. For instance, it is often useful to
build IK v by using the values of the function v (or of its derivatives) at some
points in K, or along the edges of K, or on the faces of K. In this case, it is natural to set V (K) = C l (K; Rq ) for some integer l ≥ 0 or V (K) = W s,p (K; Rq )
for some real numbers s ≥ 0 and p ∈ [1, ∞], see §B.6. (We abuse the notation
since we should write W s,p (int(K); Rq ), where int(K) denotes the interior of
the set K in Rd .)
1.4 Examples: Lagrange and modal finite elements
We consider in this section Lagrange and modal finite elements.
1.4.1 Lagrange finite elements
We focus our attention on scalar-valued Lagrange (or nodal) elements. The extension to vector-valued functions can be done by proceeding componentwise.
The degrees of freedom of Lagrange finite elements are point values.
Definition 1.8 (Lagrange finite element). Let (K, P, Σ) be a scalar-valued
finite element (q = 1 in Definition 1.2). If there is a set of points {ai }i∈N in
K such that, for all i ∈ N ,
σi (p) = p(ai ),
∀p ∈ P,
(1.8)
(K, P, Σ) is called a Lagrange finite element. The points {ai }i∈N are called
the nodes of the element. The shape functions {θi }i∈N , which are such that
θi (aj ) = δij ,
∀i, j ∈ N ,
(1.9)
form the so-called nodal basis of P associated with the nodes {ai }i∈N .
Examples are presented in Chapters 2 and 3. Following Definition 1.6, the
L
acts as follows:
Lagrange interpolation operator IK
X
L
IK
(v)(x) =
v(ai )θi (x),
∀x ∈ K.
(1.10)
i∈N
L
By construction, IK
(v) matches the values of v at the Lagrange nodes, i.e.,
L
L
IK (v)(aj ) = v(aj ) for all j ∈ N . A possible choice for the domain of IK
is
0
V (K) = C (K) equipped with the uniform norm kvkC 0 (K) = supx∈K |v(x)|
for all v ∈ C 0 (K) (recall that K is a closed set in Rd , so that functions
in C 0 (K) are continuous up to the boundary). An alternative choice using
Sobolev spaces is V (K) = W s,p (K) with p ∈ [1, ∞] and s > dp , since functions
in W s,p (K) have a continuous representative for s > dp ; see Theorem B.101.
6
Chapter 1. Introduction to finite elements
1.4.2 Modal finite elements
The degrees of freedom of modal finite elements are moments against test
functions using some measure over K. For simplicity, we consider the uniform
measure and work in L2 (K; Rq ) with q ≥ 1.
Proposition 1.9 (Modal finite element). Let K be as in Definition 1.2.
Let P be a finite-dimensional subspace of L2 (K; Rq ) and let {ζi }i∈N be a basis
of P . Let Σ = {σi }i∈N be composed of the following linear forms σi : P → R:
Z
1
(ζi , p)ℓ2 (Rq ) dx,
∀p ∈ P, ∀i ∈ N .
(1.11)
σi (p) =
|K| K
(The factor |K|−1 is meant to make σi independent of the size of K.) Then,
(K, P, Σ) is a finite element called modal finite element.
Proof. We verify unisolvence. By definition, dim(PP
) = card(Σ). Let p ∈ P
be such
that
σ
(p)
=
0
for
all
i
∈
N
.
Writing
p
=
i
i∈N αi ζi , we infer that
R
P
⊓
⊔
0 = j∈N αj σj (p) = |K|−1 K kpk2ℓ2 (Rq ) dx, so that p = 0.
Examples are presented in Chapter 2. Let us introduce the mass matrix
M of order nsh with entries
Z
Mij =
(ζi , ζj )ℓ2 (Rq ) dx,
∀i, j ∈ N .
(1.12)
K
One can verify that the matrix M is symmetric positive-definite
(see ExerP
cise 1.4). Then, the shape functions are given by θj = |K| m∈N (M−1 )mj ζm ,
for all j ∈ N , since
Z
Z
X
1
(ζi , θj )ℓ2 (Rq ) dx =
(M−1 )mj
σi (θj ) =
(ζi , ζm )ℓ2 (Rq ) dx
|K| K
K
m∈N
X
=
(M−1 )mj Mim = δij ,
m∈N
for all i, j ∈ N . If the basis {ζi }i∈N is L2 -orthogonal, the shape functions
are given by θi = (|K|/kζi k2L2 (K;Rq ) )ζi for all i ∈ N . Furthermore, following
m
Definition 1.6, the modal interpolation operator IK
acts as follows:
X 1 Z
m
IK
(v)(x) =
(ζi , v)ℓ2 (Rq ) dx θi (x),
∀x ∈ K.
(1.13)
|K| K
i∈N
The domain of IK can be defined to be V (K) = L2 (K; Rq ), or even V (K) =
L1 (K; Rq ) if P ⊂ L∞ (K; Rq ).
Part I. Getting Started
7
1.5 The Lebesgue constant(♦)
Recall from Definition 1.6 that the interpolation operator IK is a bounded
linear operator from V (K) onto P . Since P ⊂ V (K), we can equip P with the
norm of V (K) and view IK as an endomorphism in V (K), i.e., IK ∈ L(V (K)).
In this section, we study the quantity
kIK kL(V (K)) :=
kIK (v)kV (K)
,
kvkV (K)
v∈V (K)
sup
(1.14)
which is called the Lebesgue constant for IK . In this book, we abuse the
notation by writing the supremum over v ∈ V (K) instead of v ∈ V (K) \ {0}.
Lemma 1.10 (Lower bound). kIK kL(V (K)) ≥ 1.
Proof. Since P is non-trivial (i.e., P 6= {0}) and since IK (p) = p for all p ∈ P
owing to Proposition 1.7, we infer that
kIK (p)kV (K)
kIK (v)kV (K)
≥ sup
= 1.
kvk
kpkV (K)
p∈P
V (K)
v∈V (K)
sup
⊓
⊔
The Lebesgue constant arises naturally in the estimate of the interpolation
error in terms of the best approximation of V (K) by P . In particular, the next
result shows that a large value of the Lebesgue constant may lead to poor
approximation properties of IK .
Theorem 1.11 (Interpolation error). The following holds:
kv − IK (v)kV (K) ≤ (1 + kIK kL(V (K)) ) inf kv − pkV (K) ,
p∈P
(1.15)
for all v ∈ V (K), and kv − IK (v)kV (K) ≤ kIK kL(V (K)) inf p∈P kv − pkV (K) if
V (K) is a Hilbert space.
Proof. Since IK (p) = p for all p ∈ P , we infer that v − IK (v) = (I − IK )(v) =
(I − IK )(v − p) where I is the identity operator in V (K), so that
kv − IK (v)kV (K) ≤ kI − IK kL(V (K)) kv − pkV (K)
≤ (1 + kIK kL(V (K)) )kv − pkV (K) ,
where we have used the triangle inequality. We obtain (1.15) by taking the
infimum over p ∈ P . Assume now that V (K) is a Hilbert space. We use
the fact that in any Hilbert space H, any operator T ∈ L(H) such that
0 6= T ◦T = T 6= I verifies kT kL(H) = kI − T kL(H) , see Kato [305], Xu and
Zikatanov [482], Szyld [441]. We can apply this result with H = V (K) and
T = IK . Indeed, IK 6= 0 since P is non-trivial, IK 6= I since P is a proper
subset of V (K), and IK ◦IK = IK owing to Proposition 1.7. We infer that
kv − IK (v)kV (K) ≤ kI − IK kL(V (K)) kv − pkV (K) = kIK kL(V (K)) kv − pkV (K) ,
and conclude by taking the infimum over p ∈ P .
⊓
⊔
8
Chapter 1. Introduction to finite elements
Example 1.12 (Lagrange elements). The Lebesgue constant for the LaL
grange interpolation operator IK
with nodes {ai }i∈N and space V (K) =
0
N
L
C (K) is denoted Λ := kIK kL(C 0 (K)) , so that, owing to Theorem 1.11,
L
kv − IK
(v)kC 0 (K) ≤ (1 + ΛN ) inf p∈P kv − pkC 0 (K) . One can verify (see Exercise
1.5) that ΛN = kλN kC 0 (K) with the Lebesgue function λN (x) :=
P
⊓
⊔
i∈N |θi (x)|, for all x ∈ K.
Example 1.13 (Modal elements). Consider a modal finite element, see
Proposition 1.9, with an L2 -orthogonal basis {ζi }i∈N of P and V (K) =
L2 (K; Rq ). Then, the L2 -orthogonal projection from L2 (K; Rq ) onto P , say
ΠP , is such that
!
Z
X
1
(ζi , v)ℓ2 (Rq ) dx ζi ,
∀v ∈ L2 (K; Rq ).
ΠP (v) =
kζi k2L2 (K;Rq ) K
i∈N
m
Since θi = (|K|/kζi k2L2 (K;Rq ) )ζi , we infer from (1.13) that IK
= ΠP . The
2
m
2
m
Pythagoras identity kvkL2 (K;Rq ) = kIK (v)kL2 (K;Rq ) + kv − IK
(v)k2L2 (K;Rq )
m
m
implies that kIK
kL(L2 (K;Rq )) ≤ 1, which in turn gives kIK
kL(L2 (K;Rq )) = 1
owing to Lemma 1.10.
⊓
⊔
In the rest of this section, we assume that V (K) is a Hilbert space with
inner product (·, ·)V (K) . Following ideas developed in Maday et al. [341], we
show that the Lebesgue constant can be related to the stability of an oblique
projector. Let qi be the Riesz–Fréchet representative in V (K) of the linear
form σi for all i ∈ N , i.e., σi (v) = (v, qi )V (K) for all v ∈ V (K) and all i ∈ N ,
see Theorem A.44, and set Q := span{(qi )i∈N }.
Lemma 1.14 (Oblique projection). Let IK be defined by (1.7). Then,
IK is the oblique projector onto P along Q⊥ , and the Lebesgue constant is
(p,q)V (K)
−1
kIK kL(V (K)) = αP
Q with αP Q = inf p∈P supq∈Q kpkV (K) kqkV (K) .
Proof. (1) Unisolvence implies that P ∩ Q⊥ = {0}. Let now v ∈ V (K). We
observe that IK (v) ∈ P and that
(qi , IK (v) − v)V (K) = σi (IK (v)) − σi (v) = 0,
∀i ∈ N .
Hence, IK (v) − v ∈ Q⊥ . From the decomposition v = IK (v) + (v − IK (v)),
we infer that V (K) = P + Q⊥ . Therefore, the sum is direct, and IK (v) is the
oblique projection of v onto P along Q⊥ . Furthermore, for all v ∈ V (K),
(IK (v), q)V (K)
(v, q)V (K)
= sup
≤ kvkV (K) ,
kqkV (K)
q∈Q
q∈Q kqkV (K)
αP Q kIK (v)kV (K) ≤ sup
−1
showing that kIK kL(V (K)) ≤ αP
Q.
(2) To prove the reverse bound, let us first show that IK (ΠQ (p)) = p for all
Part I. Getting Started
9
p ∈ P , where ΠQ denotes the orthogonal projector onto Q. We first observe
that
(IK (ΠQ (p)), q)V (K) = (ΠQ (p), q)V (K) = (p, q)V (K) ,
for all q ∈ Q, where we have used the fact that both IK and ΠQ are projectors
along Q⊥ . The above identity implies that IK (ΠQ (p)) − p ∈ P ∩ Q⊥ = {0}.
Hence, IK (ΠQ (p)) = p. Since P is a finite-dimensional space, a compactness
argument shows that there is p∗ ∈ P with kp∗ kV (K) = 1 such that αP Q =
supq∈Q
(p∗ ,q)V (K)
kqkV (K)
= kΠQ (p∗ )kV (K) . We conclude that
kIK kL(V (K)) ≥
kp∗ kV (K)
kIK (ΠQ (p∗ ))kV (K)
1
=
=
.
kΠQ (p∗ )kV (K)
kΠQ (p∗ )kV (K)
αP Q
⊓
⊔
Further results on the Lebesgue constant for one-dimensional Lagrange
elements can be found in §2.2.1.
Exercises
Exercise 1.1 (Linear combination). Let S ∈ Rnsh ×nsh be an invertible
matrix. Let (K,
PP, Σ) be a finite element. Let Σ̃ = {σ̃i }i∈N with degrees of
freedom σ̃i = i′ ∈N Sii′ σi′ for all i ∈ N .
(i) Prove that (K, P, Σ̃) is a finite element.
(ii) Write the shape functions {θ̃j }j∈N and verify that the interpolation operator does not depend on the matrix S, i.e., ĨK (v)(x) = IK (v)(x) for
all v ∈ V (K) and all x ∈ K.
Exercise 1.2 (Hermite). Let K = [0, 1] and let P = P3 . Let Σ =
{σ1 , σ2 , σ3 , σ4 } be the linear forms on P such that, for all p ∈ P ,
σ1 (p) = p(0),
σ2 (p) = p′ (0),
σ3 (p) = p(1),
σ4 (p) = p′ (1).
(i) Show that (K, P, Σ) is a finite element.
(ii) Compute the shape functions.
(iii) Indicate possible choices for the domain of the interpolation operator in
the form V (K) = C l (K) or H s (K) for suitable l and s.
Exercise 1.3 (Powell–Sabin). Consider the unit segment K = [0, 1] and
let PK be composed of the functions that are piecewise quadratic over the
intervals [0, 12 ] ∪ [ 12 , 1] and are of class C 1 over K, i.e., functions in P are
continuous and their first derivative is continuous at 21 . Let Σ = {σ1 , . . . , σ4 }
be the linear forms on P such that σ1 (p) = p(0), σ2 (p) = p′ (0), σ3 (p) =
p(1), σ4 (p) = p′ (1). (Note: a two-dimensional version of this finite element on
triangles has been developed in [380].)
10
Chapter 1. Introduction to finite elements
(i) Prove that the triple (K, P, Σ) is a finite element.
(ii) Verify that the first two shape functions are
(
(
t(1 − 23 t),
1 − 2t2 ,
for t ∈ [0, 12 ],
θ
(t)
=
θ1 (t) =
2
1
2
2(1 − t)2 for t ∈ [ 12 , 1].
2 (1 − t)
for t ∈ [0, 21 ],
for t ∈ [ 21 , 1],
and compute the two other shape functions.
Exercise 1.4 (Mass matrix). Prove that the mass matrix M defined by
(1.12) is symmetric positive-definite.
Exercise 1.5 (Lebesgue constant for Lagrange element). Prove that
L
the Lebesgue constant ΛN defined in Example 1.12 is equal to kIK
kL(C 0 (K)) .
L
N
(Hint: To prove the bound kIK kL(C 0 (K)) ≥ ΛP , consider a set of functions
{ψi }i∈N taking values in [0, 1] and such that i∈N ψi (x) = 1 for all x ∈ K
and ψi (aj ) = δij for all i, j ∈ N .)
Exercise 1.6 (Lagrange interpolation). Let K = [a, b] and let p ∈ [1, ∞).
1
1
(i) Prove that kvkL∞ (K) ≤ (b − a)− p kvkLp (K) + (b − a)1− p kv ′ kLp (K) for all
Ry
v ∈ W 1,p (K) (Hint: for v ∈ C 1 (K), use v(x) − v(y) = x v ′ (t) dt, where
|v(y)| = minz∈K |v(z)|, then use the density of C 1 (K) in W 1,p (K)).
(ii) Prove that W 1,p (K) embeds continuously in C 0 (K).
L
(iii) Let IK
be the interpolation operator based on the linear Lagrange finite
element using nodes a and b. Determine the two shape functions and
L
prove that IK
can be extended to W 1,p (K).
(iv) Assuming that w ∈ W 1,p (K) is zero at some point in K, show that
kwkLp (K) ≤ (b − a)kw′ kLp (K) .
L
(v) Prove the following estimates: k(v − IK
v)′ kLp (K) ≤ (b − a)kv ′′ kLp (K) ,
L
L
′
L
v)′ kLp (K) ≤ kv ′ kLp (K) ,
kv − IK vkLp (K) ≤ (b − a)k(v − IK v) kLp (K) , k(IK
2,p
for all p ∈ (1, ∞] and all v ∈ W (K).
Exercise 1.7 (Variation on P2 ). Let K = [0, 1], P = P2 and Σ =
{σ1 , σ2 , σ3 } such that σ1 (p) = p(0), σ2 (p) = 2p( 12 ) − p(0) − p(1), σ3 (p) = p1).
(i) Show that (K, P, Σ) is a finite element.
(ii) Compute the shape functions.
(iii) Indicate possible choices for the domain of the interpolation operator in
the form V (K) = C l (K) or H s (K) for suitable l and s.