View of On Berman's phenomenon for (0,1,2) Hermite-Fejér interpolation

J. Numer. Anal. Approx. Theory, vol. 48 (2019) no. 1, pp. 3–15 ictp.acad.ro/jnaat

ON BERMAN’S PHENOMENON

FOR (0,1,2) HERMITE–FEJ ´ER INTERPOLATION

GRAEME J. BYRNE^∗and SIMON J. SMITH^†

Abstract. Given f ∈ C[−1,1] andn points (nodes) in [−1,1], the Hermite–

Fej´er interpolation(HFI) polynomial is the polynomial of degree at most 2n−1 which agrees withf and has zero derivative at each of the nodes. In 1916, L.

Fej´er showed that if the nodes are chosen to be the zeros ofTn(x), thenth Cheby- shev polynomial of the first kind, then the HFI polynomials converge uniformly to f as n → ∞. Later, D. L. Berman established the rather surprising result that this convergence property is no longer true for allfif the Chebyshev nodes are augmented by including the endpoints −1 and 1 as additional nodes. This behaviour has become known as Berman’s phenomenon. The aim of this pa- per is to investigate Berman’s phenomenon in the setting of (0,1,2) HFI, where the interpolation polynomial agrees with f and has vanishing first and second derivatives at each node. The principal result provides simple necessary and suf- ficient conditions, in terms of the (one-sided) derivatives offat±1, for pointwise and uniform convergence of (0,1,2) HFI on the augmented Chebyshev nodes if f∈C⁴[−1,1], and confirms that Berman’s phenomenon occurs for (0,1,2) HFI.

MSC 2010. Primary 41A05; Secondary 41A10.

Keywords. interpolation, polynomial interpolation, Hermite–Fej´er interpola- tion, Chebyshev nodes, Berman’s phenomenon.

1. INTRODUCTION

Supposef ∈C[−1,1] and let

(1) X={x_k,n :k= 0,1,2, . . . , n−1; n= 1,2,3, . . .}

be an infinite triangular matrix of nodes such that, for all n, (2) 1≥x0,n> x1,n > . . . > xn−1,n ≥ −1.

The well-known Lagrange interpolation polynomial of f is the polynomial Ln(X, f)(x) =Ln(X, f, x) of degree at mostn−1 which satisfies

Ln(X, f, x_k,n) =f(x_k,n), 0≤k≤n−1.

∗Department of Mathematics and Statistics, La Trobe University, P. O. Box 199, Bendigo, VIC 3552, Australia, e-mail: [email protected].

†54 Broad Parade, Bendigo, VIC 3550, Australia, e-mail: [email protected]. The work of this author has been supported by La Trobe University.

A classic result due to Faber [6] states that for any X there exists f ∈ C[−1,1] so that L_n(X, f) does not converge uniformly to f on [−1,1] as n → ∞. On the other hand, a more positive result occurs for the matrix of Chebyshev nodes

T =ⁿx_k,n= cos ^2k+1_2n π:k= 0,1,2, . . . , n−1; n= 1,2,3, . . .^o where, for each n, the x_k,n are the zeros of the nth Chebyshev polynomial Tn(x) = cos(narccosx),−1≤x≤1. This result states that if the modulus of continuityω(δ;f) of f is defined by

ω(δ;f) = max|f(s)−f(t)|:−1≤s, t≤1,|s−t| ≤δ , thenL_n(T, f) converges uniformly to f under the quite mild restriction

ω(_n¹;f) logn→0, asn→ ∞ (see Rivlin [11, Chapter 4] for details and references).

A generalization of Lagrange interpolation is provided by Hermite–Fej´er in- terpolation (HFI). Given a non-negative integermand nodesXdefined by (1) and (2), the (0,1, . . . , m) HFI polynomialH_m,n(X, f)(x) =H_m,n(X, f, x) off is the unique polynomial of degree at most (m+ 1)n−1 which satisfies the (m+ 1)n conditions







H_m,n(X, f, x_k,n) =f(x_k,n), 0≤k≤n−1,

Hm,n^(r) (X, f, x_k,n) = 0, 1≤r≤m, 0≤k≤n−1.

Note that H_0,n(X, f) =L_n(X, f). The original motivation for studying HFI was provided by Fej´er [7], who in 1916 showed that if f ∈ C[−1,1], then kH_1,n(T, f)−fk → 0 as n → ∞ (here and subsequently, k · k denotes the uniform norm on [−1,1]). Thus on the Chebyshev nodes, (0,1) HFI succeeds where Lagrange interpolation may fail.

In the years since Fej´er’s work, (0,1, . . . , m) HFI has been much studied by many authors. In this paper our focus is on an aspect of HFI that has become known as Berman’s phenomenon, which occurs if the Chebyshev nodes are augmented by the end points of the interval [−1,1]. In other words, we will be studying (0,1, . . . , m) HFI on the nodesTa={x_k,n+2 : 0≤k≤n+ 1, n= 1,2,3, . . .}, where

(3)







x_0,n+2 = 1, x_n+1,n+2=−1, x_k,n+2 = cos ^(2k−1)π_2n , 1≤k≤n.

Thus, for each n, thexk,n+2 are the zeros of (1−x²)Tn(x).

Initially it might be thought that augmenting the Chebyshev nodes with±1 will have little effect on the convergence behaviour of interpolation polynomials when compared with interpolation on the Chebyshev nodes alone. However, D. L. Berman [1] was able to show that if f(x) = |x|, then H_1,n(T_a, f,0) diverges, while later [2] he showed that for g(x) =x²,H_1,n(T_a, g, x) does not

converge to x² at any point of (−1,1). (This result for x² does not extend to [−1,1] because ±1 are interpolation nodes for all n.) An explanation for Berman’s phenomenon was provided by Bojanic, as follows.

Theorem 1 (Bojanic). [3]. If f ∈ C[−1,1] has left and right derivatives f_L⁰(1) and f_R⁰ (−1) at 1 and −1, respectively, then H1,n(Ta, f) converges uni- formly to f on[−1,1] if and only if f_L⁰(1) =f_R⁰(−1) = 0.

At this point it is natural to ask whether Berman’s phenomenon occurs for (0,1, . . . , m) HFI if m 6= 1. When m = 0 (Lagrange interpolation) it is straightforward to show that it does not occur. This follows from the repre- sentation

L_n+2(T_a, f, x) =L_n(T, f, x) + ¹₂T_n(x)(1+x)[f(1)−Ln−1(T, f,1)]

+ (−1)ⁿ(1−x)[f(−1)−Ln−1(T, f,−1)] , which can be verified by observing that both sides of the equation are poly- nomials of degree at mostn+ 1 which agree at the n+ 2 nodes x_k,n+2 given by (3). Thus Ln(Ta, f) → f uniformly on [−1,1] whenever Ln(T, f) → f uniformly on [−1,1].

The question of whether Berman’s phenomenon occurs for (0,1, . . . , m) HFI for any m >1 was answered in the affirmative by Cook and Mills [5] in 1975, who showed that ifh(x) = (1−x²)³, thenH_3,n(T_a, h,0) diverges. (Incidentally, it was in [5] that the termBerman’s phenomenonwas first used.) The result of Cook and Mills was later extended by Maky [9] who showed thatH_3,n(T_a, h, x) diverges at each point in (−1,1). These findings for (0,1,2,3) HFI on Ta

contrast with the earlier result of Krylov and Steuermann [8] that H_3,n(T, f) converges uniformly to f on [−1,1] forany f ∈C[−1,1].

In this paper our focus will be on (0,1,2) HFI. Here it was shown by Sz- abados and Varma [13] that, as with Lagrange interpolation, for any matrix of nodes X there exists f ∈ C[−1,1] so that H2,n(X, f) does not converge uniformly to f on [−1,1]. On the other hand, and again like Lagrange inter- polation, it follows from Byrne et al. [4, Theorem 1] that ifω(1/n;f) logn→0 asn→ ∞, thenH_2,n(T, f) converges uniformly to f.

To investigate (0,1,2) HFI on Ta it proves helpful to follow the approach of Bojanic [3], and introduce incremental modifications to the (0,1,2) HFI process on T. To this end, with nodes x_k = x_k,n+2 defined by (3) and f ∈ C[−1,1], define the polynomialQ2,n+2(Ta, f) of degree at most 3n+ 1 by the 3n+ 2 conditions

(4)







Q2,n+2(Ta, f, x_k) =f(x_k), 0≤k≤n+ 1, Q⁰_2,n+2(T_a, f, x_k) =Q⁰⁰_2,n+2(T_a, f, x_k) = 0, 1≤k≤n,

and define the polynomialR2,n+2(Ta, f) of degree at most 3n+ 3 by the 3n+ 4 conditions

(5)







R_2,n+2(T_a, f, x_k) =f(x_k), R⁰_2,n+2(T_a, f, x_k) = 0, 0≤k≤n+ 1,

R⁰⁰_2,n+2(T_a, f, x_k) = 0, 1≤k≤n.

Also, for future reference, recall that H_2,n+2(T_a, f) is defined by the 3n+ 6 conditions

(6)







H_2,n+2(T_a, f, x_k) =f(x_k),

H_2,n+2⁰ (T_a, f, x_k) =H_2,n+2⁰⁰ (T_a, f, x_k) = 0, 0≤k≤n+ 1.

Our principal results are presented in the following theorem. Note that throughout this paper we are concerned with functions defined on [−1,1], and so any derivative evaluated at 1 or −1 is assumed to be the appropriate one-sided derivative.

Theorem2. Suppose that the polynomialsQ_2,n+2(T_a, f),R_2,n+2(T_a, f)and H_2,n+2(T_a, f)are defined by(4)–(6), where the interpolation nodesx_k=x_k,n+2 are defined by (3).

a) If f ∈C[−1,1] and lim

n→∞kH_2,n(T, f)−fk= 0, then

n→∞lim kQ_2,n+2(Ta, f)−fk= 0.

b) If f ∈ C²[−1,1], then lim

n→∞kR_2,n+2(T_a, f)−fk= 0 if and only if f⁰(1) =f⁰(−1) = 0. Furthermore, if f⁰(1) and f⁰(−1)are not both 0, then R_2,n+2(T_a, f, x) is divergent if 0 < |x| < 1, and R_2,n+2(T_a, f,0) converges tof(0) if and only if f⁰(1) =f⁰(−1).

c) If f ∈C⁴[−1,1], then lim

n→∞kH_2,n+2(T_a, f)−fk= 0 if and only if (7) f⁰(1) =f⁰(−1) =f⁰⁰(1) =f⁰⁰(−1) = 0.

If0<|x|<1 and(7)does not hold, thenH2,n+2(Ta, f, x) is divergent;

in particular, iff⁰(1) and f⁰(−1)are not both 0, then

(8) lim sup

n→∞

1

n²|H_2,n+2(Ta, f, x)|>0.

Furthermore,H_2,n+2(T_a, f,0) converges to f(0) if and only if f⁰(1) = f⁰(−1)andf⁰⁰(1) =−f⁰⁰(−1) (which occurs, for example, iff is odd).

If f⁰(1)6=f⁰(−1), then

(9) lim sup

n→∞

1

n²|H_2,n+2(Ta, f,0)|>0.

Theorem 2 confirms that Berman’s phenomenon occurs for (0,1,2) HFI. In particular, the following results hold.

Corollary3. (0,1,2)HFI provides the following illustrations of Berman’s phenomenon.

a) If f(x) =x, then lim sup

n→∞

1

n²|H_2,n+2(Ta, f, x)|>0 for 0<|x|<1 and

n→∞lim H_2,n+2(T_a, f,0) = 0.

b) If g(x) =x², then lim sup

n→∞

1

n²|H_2,n+2(T_a, g, x)|>0 for all x in (−1,1).

The theorem will be proved viaa sequence of lemmas in the next section.

2. PROOF OF THE MAIN RESULT (THEOREM 2)

We begin by noting that the polynomials Q2,n+2(Ta, f), R2,n+2(Ta, f) and H_2,n+2(T_a, f), defined by (4)–(6), are related to each other and toH_2,n(T, f) according to the formulas

Q_2,n+2(T_a, f, x) = (10)

=H2,n(T, f, x) +¹₂(Tn(x))³

×(1 +x) (f(1)−H2,n(T, f,1)) + (−1)ⁿ(1−x) (f(−1)−H2,n(T, f,−1))], R_2,n+2(T_a, f, x) =

(11)

=Q_2,n+2(T_a, f, x) +¹₄(T_n(x))³(1−x²)

×(1 +x)Q⁰_2,n+2(T_a, f,1)−(−1)ⁿ(1−x)Q⁰_2,n+2(T_a, f,−1) and

H2,n+2(Ta, f, x) = (12)

=R_2,n+2(T_a, f, x)−₁₆¹ (T_n(x))³(1−x²)²

×(1 +x)R⁰⁰_2,n+2(Ta, f,1) + (−1)ⁿ(1−x)R⁰⁰_2,n+2(Ta, f,−1). Each of (10)–(12) can be verified by simply checking that the polynomial on the right-hand side satisfies the defining conditions (in terms of degree and values at the interpolation nodes (3)) of the polynomial on the left-hand side.

Also note (Byrneet al.[4, Section 1]) that H2,n(T, f) has the explicit formula (13) H2,n(T, f, x) = _n¹3 (Tn(x))³Sn(f, x)

where, withx_k=x_k,n+2 defined by (3), (14) Sn(f, x) =

n

X

k=1

(−1)^k−1q1−x²_k(x−x^1−xxk

k)³ −_2(x−x^xk

k)² +_2(x−xⁿ²⁻¹

k)

f(xk).

The results of the following two lemmas will be used to develop alternative representations for R_2,n+2(T_a, f) and H_2,n+2(T_a, f) to those in (11) and (12).

Lemma 4. For x_k =x_k,n+2 defined by (3) and f ∈C[−1,1], the following summation formulas hold.

(15) ^Tn_n^(x)

n

X

k=1

(−1)^k−1√

1−x²_k

x−x_k f(x_k) =L_n(T, f, x)

(16) ^Tn_n^(x)

n

X

k=1

(−1)^k−1√

1−x²

k

(x−x_k)² f(xk) = ^T_T^n0(x)

n(x)Ln(T, f, x)−L⁰_n(T, f, x)

Tn(x) n

n

X

k=1

(−1)^k−1√

1−x²_k

(x−x_k)³ f(x_k) = _T0

n(x) Tn(x)

2

−¹₂^T_T^n00(x)

n(x)

L_n(T, f, x) (17)

−^T_T^n0(x)

n(x)L⁰_n(T, f, x) +¹₂L⁰⁰_n(T, f, x)

Tn(x) n

n

X

k=1

(−1)^k−1√

1−x²_k

(x−x_k)⁴ f(xk) = (18)

=

1 6

T_n⁰⁰⁰(x)

Tn(x) −^Tn0_(T^(x)Tn00(x)

n(x))² +^T_T^n0(x)

n(x)

3

L_n(T, f, x) +

1 2

T_n⁰⁰(x)

Tn(x) −^T_T^n0(x)

n(x)

2

L⁰_n(T, f, x) +¹₂^T_T^n0(x)

n(x)L⁰⁰_n(T, f, x)−¹₆L⁰⁰⁰_n(T, f, x) Proof. The formula (15) for Lagrange interpolation on the Chebyshev nodes is well-known (see, for example, Rivlin [12, Section 1.3]). The remaining for- mulas (16)–(18) follow by successive differentiation of (15).

Lemma5. Forx_k=x_k,n+2 defined by(3), the following summation formu- las hold.

n

X

k=1

(−1)^k−1√

1−x²_k

1−x_k =n

(19)

n

X

k=1

(−1)^k−1√

1−x²_k (1−xk)² =n³ (20)

n

X

k=1

(−1)^k−1√

1−x²_k

(1−x_k)³ = ¹₆n³(5n²+ 1) (21)

n

X

k=1

(−1)^k−1√

1−x²_k

(1−x_k)⁴ = ₉₀¹n³(61n⁴+ 25n²+ 4) (22)

n

X

k=1

(−1)^k−1√

1−x²

k

(1+xk)^r = (−1)ⁿ⁺¹

n

X

k=1

(−1)^k−1√

1−x²

k

(1−x_k)^r , r = 1,2,3, . . . (23)

Proof. Iff(x)≡1, thenLn(T, f, x)≡1, so (15) gives

n

X

k=1

(−1)^k−1√

1−x²_k x−x_k = _Tⁿ

n(x).

Putting x = 1 gives (19). The formulas (20)–(22) follow from (16)–(18) in similar fashion after noting that if f(x) ≡1, then L^(r)n (T, f, x)≡0 forr ≥1, and (Rivlin [12, p. 38])

(24) T_n^(r)(1) = ^n2(n2−12₁⁾⁽ⁿ_.3².5...(2r−1)^{−22)...(n2−(r−1)2)}.

Finally, (23) is a straightforward consequence of the symmetry of the Cheby- shev nodes (i.e. x_k,n+2 =−x_n−k+1,n+2 for 1≤k≤n).

We now obtain alternative representations for the interpolation polynomials R2,n+2(Ta, f) andH2,n+2(Ta, f) to those in (11) and (12).

Lemma 6. Forx_k=x_k,n+2 defined by(3) and f ∈C[−1,1], R_2,n+2(T_a, f, x) =H_2,n(T, f, x)

(25)

+ ¹₄(T_n(x))³ⁿ(1 +x)²(2−x) [f(1)−H_2,n(T, f,1)]

+ (−1)ⁿ(1−x)²(2 +x) [f(−1)−H2,n(T, f,−1)]

+ (1−x²) [(1 +x)A_1,n−(1−x)B_1,n]^o and

H2,n+2(Ta, f, x) =R2,n+2(Ta, f, x) +₁₆¹ (Tn(x))³(1−x²)²× (26)

×^h3x[f(1)−H_2,n(T, f,1)]−(−1)ⁿ[f(−1)−H_2,n(T, f,−1)]

+ (1 +x)(A_2,n+B_1,n)−(1−x)(B_2,n+A_1,n)ⁱ where

A1,n = _n¹₃

n

X

k=1

(−1)^k−1q1−x²_k(1−x³

k)² +_2(1−xⁿ²⁻¹

k)

_f(1)−f(x

k) 1−x_k , (27)

B1,n = _n¹₃

n

X

k=1

(−1)^k−1q1−x²_k(1+x³

k)² +_2(1+xⁿ²⁻¹

k)

_{f(−1)−f(x}

k) 1+xk , (28)

A_2,n = _n¹₃

n

X

k=1

(−1)^k−1q1−x²_k(1−x¹⁵

k)³ +_(1−xⁿ²⁺²

k)² +ⁿ_1−x²⁻¹

k

_f(1)−f(x

k) 1−x_k , (29)

B_2,n = _n¹₃

n

X

k=1

(−1)^k−1q1−x²_k(1+x¹⁵

k)³ +_(1+xⁿ²⁺²

k)² +ⁿ_1+x²⁻¹

k

_{f(−1)−f(x}

k) 1+xk . (30)

Proof. For convenience, denote the RHS of (25) byr(x). To establish (25) it is sufficient to show r(x) satisfies the 3n+ 4 defining conditions (5) of R_2,n+2(T_a, f). In fact, apart from r⁰(1) = r⁰(−1) = 0, the conditions are verified easily. To show r⁰(1) = 0, firstly observe from (5) that (upon using T_n⁰(1) =n²)

r⁰(1) =H_2,n⁰ (T, f,1) + 3n²(f(1)−H2,n(T, f,1))−A1,n. However, by (13) and (14),

H_2,n⁰ (T, f,1) = 3n²H2,n(T, f,1) +_n¹₃S_n⁰(f,1),

where

(31) S_n⁰(f,1) =−

n

X

k=1

(−1)^k−1q1−x²_k(1−x³

k)³ +_2(1−xⁿ²⁻¹

k)²

f(x_k).

Thus

r⁰(1) = 3n²f(1) +_n¹3S_n⁰(f,1)−A_1,n,

and this is zero by (20), (21) and (27). The result r⁰(−1) = 0 is established by similar means.

Again for convenience, denote the RHS of (26) by h(x). By the same ar- guments as above, it is evident that (26) will be proved if it can be shown that h⁰⁰(1) =h⁰⁰(−1) = 0. To showh⁰⁰(1) = 0, begin by noting that from (25) and (26),

h⁰⁰(1) =H_2,n⁰⁰ (T, f,1) + (7n⁴−n²) (f(1)−H2,n(T, f,1))−(6n²+ 2)A1,n+A2,n. (32)

However, by (13) and (14),

(33) H_2,n⁰⁰ (T, f,1) = (7n⁴−n²)H2,n(T, f,1) +_n⁶S⁰_n(f,1) + _n¹₃S_n⁰⁰(f,1), where

(34) S_n⁰⁰(f,1) =

n

X

k=1

(−1)^k−1q1−x²_k(1−x^3xk+12

k)⁴ +_(1−xⁿ²⁻¹

k)³

f(x_k).

By substituting (27), (29), (31), (33) and (34) into (32), and employing sum- mation formulas (20)–(22), it follows (after somewhat tedious calculations) that h⁰⁰(1) = 0. The result h⁰⁰(−1) = 0 is proved in similar fashion.

The next result characterizes the quantities A_i,n and B_i,n of Lemma 6 in terms of values of Lagrange interpolation polynomials and their derivatives.

Lemma 7. For i = 1,2, the quantities A_i,n and B_i,n that are defined by (27)–(30)can be written as

A_1,n = 3n²[f(1)−L_n(T, f,1)] + ⁷ⁿ_2n²⁻¹2 L⁰_n(T, f,1)−_2n³2L⁰⁰_n(T, f,1), (35)

B1,n =(−1)ⁿ⁺¹3n²[f(−1)−Ln(T, f,−1)]

(36)

−⁷ⁿ_2n²⁻¹₂ L⁰_n(T, f,−1)−_2n³₂L⁰⁰_n(T, f,−1),

A2,n=(11n⁴+ 7n²) [f(1)−Ln(T, f,1)] + ^27n4+11n_2n2 ²⁻²L⁰_n(T, f,1) (37)

− ⁸ⁿ_n²₂⁺¹L⁰⁰_n(T, f,1) + _2n⁵₂L⁰⁰⁰_n(T, f,1), B_2,n=(−1)ⁿ⁺¹(11n⁴+ 7n²) [f(−1)−L_n(T, f,−1)]

(38)

−^27n4+11n_2n2 ²⁻²L⁰_n(T, f,−1)−⁸ⁿ_n²₂⁺¹L⁰⁰_n(T, f,−1)−_2n⁵₂L⁰⁰⁰_n(T, f,−1).

Proof. To obtain (35), begin by writing (27) as A1,n= _n³₃

n

X

k=1

(−1)^k−1√

1−x²

k

(1−x_k)³ g(x_k) +ⁿ_2n²⁻¹₃

n

X

k=1

(−1)^k−1√

1−x²

k

(1−x_k)² g(x_k),

whereg(x) =f(1)−f(x). Now apply the identities (16) and (17) atx= 1, not- ing that Ln(T, g, x) =f(1)−Ln(T, f, x) and using the result (24) forTn^(r)(1).

To obtain (36), apply (16) and (17) atx=−1 to (28) withg(x) =f(−1)−f(x), and use Tn^(r)(−1) = (−1)^n+rTn^(r)(1). The remaining expressions (37) and (38)

are proved in a similar fashion.

For a function f that has continuous derivatives on [−1,1], the following lemma characterizes the limiting behaviour of the A_i,n and B_i,n asn→ ∞ in terms of these derivatives. The proof of the lemma uses the following result which is a special case of a more general theorem concerning Lagrange inter- polation on Jacobi nodes that is due to Neckermann and Runck [10, Satz 2, p. 168]:

(39) Ifm >0 and f^(2m)∈C[−1,1], then lim

n→∞kL^(m)_n (T, f)−f^(m)k= 0.

Lemma8. SupposeAi,nandBi,nare defined by(27)–(30). Iff ∈C²[−1,1], then

n→∞lim A_1,n = ⁷₂f⁰(1), (40)

n→∞lim(−1)ⁿB_1,n = ⁷₂f⁰(−1).

(41)

If f ∈C⁴[−1,1], then

n→∞lim

A_2,n−²⁷ⁿ₂²f⁰(1)= ¹¹₂f⁰(1)−8f⁰⁰(1), (42)

n→∞lim

(−1)ⁿB2,n−²⁷ⁿ₂²f⁰(−1)= ¹¹₂f⁰(−1) + 8f⁰⁰(−1).

(43)

Proof. Supposef ∈C²[−1,1]. We work with the expression (35) for A_1,n. Firstly, with x1 = cos(π/(2n)), it follows from the Mean Value Theorem that there existsα∈(x₁,1) so that

f(1)−L_n(T, f,1)=(f(1)−L_n(T, f,1))−(f(x₁)−L_n(T, f, x₁))

=(1−x1) f⁰(α)−L⁰_n(T, f, α)

≤2 sin² _4n^π kf⁰−L⁰_n(T, f)k.

Thus, by (39),

(44) lim

n→∞n²[f(1)−Ln(T, f,1)] = 0.

Also, if P_n denotes the set of polynomials of degree at most n, let q be the best uniform approximation to f⁰ inP_n−3. Then, by Markov’s inequality,

|L⁰⁰_n(T, f,1)| ≤ kL⁰⁰_n(T, f)−q⁰k+kq⁰k

≤(n−2)²kL⁰_n(T, f)−qk+kq⁰k

≤(n−2)² kL⁰_n(T, f)−f⁰k+kf⁰−qk+kq⁰k.

Now, there exists an absolute constant c so that kq⁰k ≤ c(n−3)²ω(_n−3¹ ;f⁰) (see Szabados and V´ertesi [14, p. 284]), so

1

n²|L⁰⁰_n(T, f,1)| ≤ kL⁰_n(T, f)−f⁰k+kf⁰−qk+c ω(_n−3¹ ;f⁰).

Since f ∈ C²[−1,1], it follows from (39) and the Weierstrass approximation theorem that

(45) lim

n→∞

1

n²L⁰⁰_n(T, f,1) = 0.

Substituting (44) and (45) into (35) and using (39) then gives the result (40).

The result (41) is established from (36) in near-identical fashion.

Now suppose f ∈ C⁴[−1,1]. Here we work with the expression (37) for A2,n. With xk = cos((2k−1)π/(2n)), by the Mean Value Theorem there exists β ∈(x₂, x₁) so that f⁰(β)−L⁰_n(T, f, β) = 0. Thus, with α∈ (x₁,1) as above, there existsγ ∈(β, α) such that

|f(1)−L_n(T, f,1)|=(1−x₁) f⁰(α)−L⁰_n(T, f, α)

=(1−x1) f⁰(α)−L⁰_n(T, f, α)− f⁰(β)−L⁰_n(T, f, β)

=(1−x1)(α−β) f⁰⁰(γ)−L⁰⁰_n(T, f, γ)

≤4 sin² _4n^π sin^23π_4nkf⁰⁰−L⁰⁰_n(T, f)k, and so, by (39),

(46) lim

n→∞n⁴[f(1)−Ln(T, f,1)] = 0.

Similarly,

f⁰(1)−L⁰_n(T, f,1)= f⁰(1)−L⁰_n(T, f,1)− f⁰(β)−L⁰_n(T, f, β)

≤2 sin^23π_4nkf⁰⁰−L⁰⁰_n(T, f)k, so

(47) lim

n→∞n²f⁰(1)−L⁰_n(T, f,1)= 0.

Now let r be the best uniform approximation to f⁰⁰ in P_n−4. By the method used above to derive (45) it follows that

1

n²|L⁰⁰⁰_n(T, f,1)| ≤ kL⁰⁰_n(T, f)−f⁰⁰k+kf⁰⁰−rk+c ω(_n−4¹ ;f⁰⁰), and so

(48) lim

n→∞

1

n²L⁰⁰⁰_n(T, f,1) = 0.

Substituting (46)–(48) into (37) and using (39) then gives (42). The result (43)

is established similarly.

We have now developed all the preliminary results needed to establish our theorem.

Proof of Theorem 2. Firstly observe that a) is an immediate consequence of the representation (10) for Q_2,n+2(T_a, f).

b) Suppose f ∈ C²[−1,1] and x ∈ (−1,1). Then H_2,n(T, f) converges uniformly tof, so by (25),R2,n+2(Ta, f, x) converges (pointwise or uniformly) tof(x) if and only if

n→∞lim(T_n(x))³[(1 +x)A_1,n−(1−x)B_1,n] = 0, which (by (40) and (41)) is equivalent to

(49) lim

n→∞(Tn(x))^3h(1 +x)f⁰(1) + (−1)ⁿ⁺¹(1−x)f⁰(−1)ⁱ= 0.

Clearly, then, R2,n+2(Ta, f) converges uniformly to f iff⁰(1) =f⁰(−1) = 0.

On the other hand, since T_n(0) = 0 for n odd and |T_n(0)|= 1 for neven, (49) holds true atx= 0 if and only iff⁰(1) =f⁰(−1). ThusR_2,n+2(T_a, f,0)→ f(0) if and only iff⁰(1) =f⁰(−1).

For fixedxwith 0<|x|<1, it is well-known that there exists a subsequence {n_r}^∞_r=1 of natural numbers so that |T_nr(x)| →1. Since |T_nr+1(x)| → |x|, it follows that there is a subsequence {m_r}^∞_r=1 of natural numbers that con- tains infinitely many odd values and infinitely many even values and satisfies

|T_mr(x)| ≥ |x|/2 for all r. Thus (49) holds true if and only if (1 +x)f⁰(1) + (1−x)f⁰(−1) = 0 and (1 +x)f⁰(1)−(1−x)f⁰(−1) = 0, which is equivalent tof⁰(1) =f⁰(−1) = 0.

c) Suppose f ∈ C⁴[−1,1] and x ∈ (−1,1). Again H_2,n(T, f) converges uniformly to f, so by (25) and (26),

H_2,n+2(T_a, f, x) =

=f(x) +₁₆¹(Tn(x))³(1−x²)(1 +x)²(3−x)A1,n

−(1−x)²(3 +x)B_1,n+ (1 +x)²(1−x)A_2,n−(1−x)²(1 +x)B_2,n+o(1) where, here and subsequently, theo(1) term is uniform inx. Then by (40)-(43),

H_2,n+2(T_a, f, x) = (50)

=f(x) + ₃₂¹(T_n(x))³(1−x²)^h7(1 +x)²(3−x)f⁰(1)

−7(−1)ⁿ(1−x)²(3+x)f⁰(−1)+(1 +x)²(1−x)(27n²+11)f⁰(1)−16f⁰⁰(1)

−(−1)ⁿ(1−x)²(1 +x)(27n²+ 11)f⁰(−1) + 16f⁰⁰(−1)ⁱ+o(1).

It is obvious from this expression thatH2,n+2(Ta, f) converges uniformly tof iff⁰(1) =f⁰(−1) =f⁰⁰(1) =f⁰⁰(−1) = 0.

On the other hand, (50) shows that H2,n+2(Ta, f,0)→f(0) if and only if

n→∞lim

neven

(27n²+ 32)f⁰(1)−f⁰(−1)−16f⁰⁰(1) +f⁰⁰(−1)= 0,

which is equivalent tof⁰(1) =f⁰(−1) and f⁰⁰(1) =−f⁰⁰(−1). Furthermore, if f⁰(1)6=f⁰(−1), (9) follows from (50).

For fixedx with 0<|x|<1, supposeH_2,n+2(T_a, f, x) →f(x). Then, with {m_r} defined as in (b) above, it follows from (50) that

(51) lim

r→∞

h

7(1 +x)²(3−x)f⁰(1)−7(−1)^mr(1−x)²(3 +x)f⁰(−1) + (1 +x)²(1−x)(27m²_r+ 11)f⁰(1)−16f⁰⁰(1)

−(−1)^mr(1−x)²(1 +x)(27m²_r+ 11)f⁰(−1) + 16f⁰⁰(−1)ⁱ= 0.

Because {m_r} has infinitely many values of each parity, (51) implies that (1 +x)f⁰(1) + (1−x)f⁰(−1) = 0 and (1 +x)f⁰(1)−(1−x)f⁰(−1) = 0, which means f⁰(1) = f⁰(−1) = 0. A similar argument then shows that f⁰⁰(1) = f⁰⁰(−1) = 0. Finally, if f⁰(1) andf⁰(−1) are not both 0, (8) is a consequence

of (50) and the definition of m_r.

Remark. Theorem 1 for (0,1) HFI does not place any conditions on the derivatives of f, other than the existence of f⁰ at±1. By contrast, our The- orem 2 for (0,1,2) HFI imposes quite stringent conditions on the derivatives of f. However, it seems difficult to fully extend Bojanic’s methods to (0,1,2) HFI, essentially because the positivity of terms that occur when working with (0,1) HFI no longer applies to the corresponding terms in (0,1,2) HFI. Never- theless, it would be of interest to know whether the conditions of our theorem can be weakened.

Acknowledgements. The authors thank Professor Terry Mills for helpful discussions during the preparation of this paper.

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Received by the editors: October 7, 2018. Accepted: February 4, 2019. Published online:

October 10, 2019.