Three-convex approximation by quadratic splines with arbitrary fixed knots

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EAST JOURNAL ON APPROXIMATIONS
Volume , Number (2002), 1-12
THREE–CONVEX APPROXIMATION BY QUADRATIC
SPLINES WITH ARBITRARY FIXED KNOTS
A. V. Prymak
Department of Mathematical Analysis,
Faculty of Mechanics and Mathematics,
Kyiv National Taras Shevchenko University,
Kyiv, 01033, UKRAINE
E-mail: prymak@mail.univ.kiev.ua
A function fis said to be 3-convex on [a, b] if it has a convex derivative
on (a, b). For every 3-convex function an estimate of approximation by
3-convex quadratic splines with arbitrary fixed knots is obtained in terms
of the modulus of smoothness of order 3 of the function.
1. Introduction
An approach to the construction of shape-preserving splines goes back to
the famous paper by DeVore [1], where he has constructed monotone splines
of order rthat provide proper estimates for the approximation of monotone
functions. For other constructions of comonotone and coconvex splines see
the papers by Hu [3], Yu, Leviatan and Hu [4], Kopotun [6], Shevchuk [10].
Recently, the interest to 3-convex (and in general, to q-convex, q3) ap-
proximation was intensified.
Let us first explain what do we mean by a generalized convexity (i.e., by
q-convexity). For a fixed positive integer q,
[x0, . . . , xq;f] :=
q
X
i=0
f(xi)
Qq
j=0,j6=i(xixj)
2Convex approximation
is the q-th divided difference of the real function fat the points x0, . . . , xq. We
shall denote by ∆qthe set of all q-convex functions on [1,1], that is, the set of
all functions such that for all choices of q+1 distinct points x0, . . . , xq[1,1]
we have [x0, . . . , xq;f]0. Thus, ∆1and ∆2are the sets of non-decreasing
and convex functions on [1,1], respectively, ∆3is the set of 3-convex func-
tions, or, in other words, the set of all functions, having convex first derivative
on (1,1). Remark that if fq,q2, then fis a continuous function on
(1,1).
For the free knots shape-preserving spline approximation Kopotun and
Shadrin have proved [7] the estimate
E(q)
N,r (f)pB·EN,r (f)p
for fqLp[1,1], where
EN,r (f)p:= inf{kfskLp[1,1] :sSN,r },
E(q)
N,r (f)p:= inf{kfskLp[1,1] :sSN,r q},
SN,r is the set of splines of degree r1 with Nfree knots, and the constants
Aand Bdepend on qand ronly. This estimate in turn generalizes the results
by Leviatan and Shadrin [8] for q= 1,2, and Petrov [9] for q= 3. So, the
case of free knots spline shape-preserving approximation is reduced to the
approximation without constrains, and hence, Jackson-type estimates hold in
the shape-preserving case.
In the present paper we consider splines with arbitrary, but fixed knots.
Let us denote by L(x) := L(x;f;x0, . . . , xn) the Lagrange polynomial of
degree nthat interpolates the function fat the points x0, . . . , xn. As
usual, ωk(f;t;I) is the k-th modulus of smoothness of the function fon the
interval I, and kfkI:= max
tI|f(t)|is the uniform norm of the continuous
function fon the closed interval I.
Konovalov and Leviatan kindly informed the author that for each
f3C(2)[1,1] they have constructed (see [5]) a 3-convex quadratic
spline swith equidistant knots, satisfying
(1) kfsk[1,1] c
n2ω1µf00;1
n; [1,1].
Also, the spline provides a proper estimate for the simultaneous approxima-
tion for the first and second derivatives.
They also raised a question: whether such a spline exists for arbitrary
fixed knots, as well as, whether for a given f3C[1,1] a spline s
A. V. Prymak 3
exists that provides ω3¡f;1
n; [1,1]¢in the right-hand side of (1) instead of
c
n2ω1¡f00;1
n; [1,1]¢.
We give an affirmative answer to both of these questions: our main result
is
Theorem 1. For every 3-convex and continuous on [a, b]function Fand
each partition a=x0< x1<··· < xn=b, there exists a quadratic 3-convex
spline Swith knots x0, x1, . . . , xnsatisfying
(2) kFSk[a,b]cΛω3,
where
Λ := max ©Λi¯¯i= 0, n 3ª,
Λi:= Ãxi+3 xi
min ©xi+j+1 xi+j¯¯j= 0,2ª!3
,
ω3:= max ©ω3(f; (xi+3 xi)/3; [xi, xi+3]) ¯¯i= 0, n 3ª,
and cis an absolute constant.
Corollary 1. For the equidistant knots xj=a+j(ba)/n we have
kFSk[a,b]c ω3µf;ba
n; [a, b].
In particular, if FC(1)[a, b], then
kFSk[a,b]cµba
nω2µf0;ba
n; [a, b],
if FC(2)[a, b], then
kFSk[a,b]cµba
n2
ω1µf00;ba
n; [a, b],
and if FC(3)[a, b], then
kFSk[a,b]cµba
n3°
°
°F(3)°
°
°[a,b].
Corollary 2. For the interval [1,1] and the Chebyshev partition
xj=cos ³
n´,j= 0, n, we have
kFSk[1,1] c ω3
ϕµf;1
n,
4Convex approximation
where ω3
ϕis the Ditzian – Totik [2] modulus of smoothness of order 3 and
ϕ(x) := 1x2. In particular, if FC(3)(1,1), then
kFSk[1,1] c
n3°
°
°ϕ3F(3)°
°
°[1,1].
Remark. The spline Sconstructed in Theorem 1 satisfies
°
°F0S0°
°[a,b]cpΛω3Var,
|F00(x+) S00 (x+)| ≤ cVar, x [a, b),
where cis an absolute constant, and
Var := max ©Var(F00,(xi1, xi))¯¯i= 1, nª.
To prove Theorem 1, we first prove Theorem 10and then, for interior
intervals we reduce Theorem 1 to Theorem 10and provide some additional
arguments for subintervals near the end-points aand b.
Theorem 10.For every 3-convex function Fwith continuous derivative
on [a, b]and each partition a=x0< x1<··· < xn=b, there exists a
quadratic 3-convex spline Swith knots x0, x1, . . . , xnsatisfying
(3) kFSk[a,b]cmax ©mi¯¯i= 1, nª,
where cis an absolute constant and
mi:= Zxi
xi1
(L(x;F0;xi1, xi)F0(x)) dx, i = 1, n.
In Section 2 we prove some lemmas and present an auxiliary construction
for the proof of Theorem 10. Section 3 contains the proofs of Theorem 10and
Theorem 1.
2. An auxiliary construction and lemmas
Let fC[a, b] be a given convex function. Then f0exists a.e. in (a, b) and
is monotone. For briefness, we shall write f0(x) instead of f0(x+), if x6=b,
and f0(b) instead of f0(b). Let a=x0< x1<···< xn=bbe an arbitrary
fixed partition of [a, b]. Set
M := max ½Zxi
xi1
(L(x;f;xi1, xi)f(x)) dx ¯¯¯i= 1, n¾.
A. V. Prymak 5
We shall write sAi,j to note that sis a convex piecewise linear function on
[xi, xj] with knots xi, . . . , xj, satisfying
(4) f0(xl1)s0(θ)f0(xl), l =i+ 1, j, θ (xl1, xl),
and
s(xi) = f(xi), s(xj) = f(xj).
Notice that s0(θ) = const for θ(xl1, xl) since swas supposed linear on the
subintervals (xl1, xl). For each k= 0, n we introduce an auxiliary function
sk. To do this, we set
(5) gk(t) := ½f0(xi1), t [xi1, xi), i =k+ 1, n
f0(xi), t (xi1, xi], i = 1, k,
and then define
sk(x) := f(xk) + Zx
xk
gk(t)dt.
Evidently, skis a convex linear spline on [a, b] with knots x0, . . . , xnsatisfy-
ing (4), and
sk(x)f(x), x [a, b], sk(xk) = f(xk).
Let 0 i < j nbe a fixed pair of integers. Next we construct a function
si,j Ai,j as follows. Since sj(xi)si(xi), sj(xj)si(xj), and sjsi
is a continuous function, then there exists a point θ(xi, xj) such that
si(θ) = sj(θ). In view of notation (5),
gi(t)gj(t), t [a, b],
and hence sjsiis a non-decreasing function. This yields
max{si(x), sj(x)}=½si(x), x θ
sj(x), x > θ,
and we set si,j(x) := max{si(x), sj(x)}. Clearly, θ[xm1, xm] for some
integer m,i+ 1 mj. Using this m, we finally set
si,j (x) := ½si,j (x), x [xm1, xm]
L(x;si,j ;xm1, xm), x [xm1, xm].
It is easy to see that si,j Ai,j .
This auxiliary construction of splines si,j plays an important role in the
proof of Theorem 10. It will be applied in the forthcoming Lemma 2.
6Convex approximation
Evidently,
(6) si,j (x)f(x), x [xm1, xm],
and
(7) si,j (x)L(x;f;xm1, xm), x [xm1, xm].
Let us denote
i,j (x) := Zx
xi
(si,j (t)f(t)) dt.
Lemma 1. The function i,j is continuous on [xi, xj]. It has at most 3
intervals of monotonicity with at most one interval where the function is in-
creasing. Moreover, on this interval the oscillation of i,j does not exceed M.
Proof. It is sufficient to show that there is at most one interval where
si,j (t)> f(t). Indeed, (6) implies that if si,j (t)> f (t), then t[xm1, xm].
Since si,j is linear on [xm1, xm], then si,j fis concave on [xm1, xm] and
hence, there is at most one interval where it is positive. The estimate of the
oscillation follows from (7) and the definition of M. 2
Remark. Lemma 1 yields the inequality
(8) ki,j k[xi,xj]≤ |i,j (xj)|+ M.
Lemma 2. Let i,1in1, be a fixed integer. Then there exist an
integer j,i+ 1 jnand a spline s?
i,j Ai,j satisfying
(9) °
°?
i,j °
°[xi,xj]5 M,
and if j < n, then
(10) ∆?
i,j (xj)0
where ?
i,j (x) := Zx
xi
(s?
i,j (t)f(t)) dt.
Proof. Consider the numbers ∆i,i+1(xi+1),i,i+2(xi+2), . . . , i,n(xn). Lem-
ma 1 implies that any of them is bounded by M. If for some k, 1 kni,
the inequalities 2 M i,i+k(xi+k)0 hold, then (8) yields that if one
takes j:= i+kand s?
i,j := si,j , then (9) and (10) are true. If ∆i,i+k(xi+k)>0
for all k,k= 1, n i, then we take j:= n,s?
i,j := si,j and (9) holds. Other-
wise, there exists k, 1 kni, such that ∆i,i+k(xi+k)<2 M. Assume
A. V. Prymak 7
that kis the smallest number satisfying this inequality. Evidently k2. Set
j:= i+kand
˜si,j (x) := ½si,j1(x), x [xi, xj1)
L(x;f;xj1, xj), x [xj1, xj].
It is easy to see that ˜si,j Ai,j . Finally, we define s?
i,j as follows
s?
i,j := λsi,j + (1 λsi,j ,
where
λ:= 2 M
|i,j (xj)|.
Note that λ(0,1). Then, since si,j ,˜si,j Ai,j , the function s?
i,j also belongs
to Ai,j . Now we prove that s?
i,j satisfies (10). The minimality of kyields
0<i,j1(xj1)M. Thus, Lemma 1 implies ki,j1k[xi,xj1]M and
hence
(11) °
°
°
°Zx
xi
si,j (t)f(t)) dt°
°
°
°[xi,xj]2 M.
We obtain
?
i,j (xj) = λi,j (xj) + (1 λ)Zxj
xi
si,j (t)f(t)) dt
≤ −2 M + (1 λ) 2 M <0,
so, the inequality (10) holds. Applying (8) and (11), we get
°
°?
i,j °
°[xi,xj]λki,j k[xi,xj]+ (1 λ)°
°
°
°Zx
xi
si,j (t)f(t)) dt°
°
°
°[xi,xj]
λ(|i,j (xj)|+ M) + 2 M = 2 M + λM + 2 M 5 M,
and thus (9) holds, which completes the proof. 2
Lemma 3. Let Fbe a 3-convex function on [z0, z3]and z0< z1< z2< z3
be some real numbers. With
l0(x) := F00(z1+)(xz1) + F0(z1),
l1(x) := L(x;F0;z1, z2),
l2(x) := F00(z2)(xz2) + F0(z2),
the inequality
(12) ¯¯¯¯Zzi+1
zi
(F0(x)li(x)) dx¯¯¯¯cΛω3(F; (z3z0)/3; [z0, z3]), i = 0,2
8Convex approximation
holds, where
Λ := Ãz3z0
min ©zj+1 zj¯¯j= 0,2ª!3
,
and cis an absolute constant.
Proof. Note that
F0(x)li(x), x (zi, zi+1), i = 0, i = 2,
and
F0(x)l1(x), x [z1, z2].
Therefore the sign of Rzi+1
zi(F0(x)li(x)) dx equals (1)i.
If fis 3-convex on [z0, z3], then
[z0, z1, z2, z3;f] = 1
z3z0
([z3, z2, z1;f][z2, z1, z0;f])
=1
z3z0µ[z3, z2;f][z2, z1;f]
z3z1[z2, z1;f][z1, z0;f]
z2z0
=α1I1+α2I2+α3I3,
where
Ii:= Zzi
zi1
f0(x)dx =f(zi)f(zi1), i = 1,3,
and
α1=1
(z3z0)(z3z1)(z3z2), α3=1
(z3z0)(z2z0)(z1z0),
α2=1
z3z0µ1
z2z0
+1
z3z1.
Thus,
(13) 0 [z0, z1, z2, z3;f] = α1I1+α2I2+α3I3.
We shall make use of the Whitney inequality. Let Pbe the quadratic poly-
nomial of best uniform approximation of Fon [z0, z3]. Then
kFPk[z0,z3]c0ω3(F; (z3z0)/3; [z0, z3]).
where c0is an absolute constant. We have
[z0, z1, z2, z3;F] = [z0, z1, z2, z3;FP] =
3
X
i=0
F(zi)P(zi)
Q3
j=0,j6=i(zizj)
4c0
ω3(F; (z3z0)/3; [z0, z3])
(min ©zj+1 zj¯¯j= 0,2ª)3.
A. V. Prymak 9
Thus
(14) [z0, z1, z2, z3;F]4c0
ω3(F; (z3z0)/3; [z0, z3])
(min ©zj+1 zj¯¯j= 0,2ª)3.
Let us set
I?
i:= Zzi
zi1
F0(x)dx, i = 1,3,
and define
f0(x) := ½l0(x), x [z0, z1)
F0(x), x [z1, z3],
f1(x) := ½l1(x), x [z1, z2]
F0(x), x /[z1, z2],
f2(x) := ½F0(x), x [z0, z2)
l2(x), x [z2, z3].
It is easy to see that the functions f0,f1and f2are convex on [z0, z3]. This
implies that the functions Fj(x) := Rx
z0fj(t)dt,j= 0,1,2, are 3-convex on
[z0, z3]. We shall prove (12) in the case i= 0 only. The cases i= 1 and i= 2
are similar. Recall that
sign αi= sign µZzi+1
zi
(F0(x)li(x)) dx= (1)i.
Taking into account (13) and (14), we obtain
0[z0, z1, z2, z3;F0]
=
3
X
i=1
αiI?
iα1Zz1
z0
(F0(x)l0(x)) dx
= [z0, z1, z2, z3;F]α1Zz1
z0
(F0(x)l0(x)) dx
4c0
ω3(F; (z3z0)/3; [z0, z3])
(min ©zj+1 zj¯¯j= 0,2ª)3Rz1
z0(F0(x)l0(x)) dx
(z3z0)3,
hence (12) holds for i= 0. 2
3. Positive results
Proof of Theorem 10.Assume that F(0) = 0, set f:= F0, and apply the
arguments of Section 2 to this function. Then
M = max ©mi¯¯i= 1, nª,
10 Convex approximation
and
(15) 0 Zx
xi1
(L(t;f;xi1, xi)f(t)) dt M, x [xi1, xi], i = 1, n.
We set S(x) := Rx
x0s(t)dt, where sA0,n is a linear convex spline, constructed
consequently as follows.
On the first step we put
s(x) := L(x;f;x0, x1), x [x0, x1].
Remark that L(x;f;xi1, xi)Ai1,i,i= 1, n. Besides, if 0 i < j < k n,
then the assumptions hAi,j and hAj,k imply hAi,k. On each step we
assume that sis already defined on [x0, xi], and for x[x0, xi] it satisfies the
conditions
(16) ¯¯¯¯Zx
x0
(s(t)f(t)) dt¯¯¯¯10 M,
and
(17) ¯¯¯¯Zxi
x0
(s(t)f(t)) dt¯¯¯¯5 M.
Then we extend sfurther on [xi, xj], for some j,i < j n, such that (16)
remains true for x[x0, xj] and the inequality (17), with xireplaced by xj,
holds whenever j < n. If j=n, then we need (16) only, for x[x0, xn].
Since there is a finite number of intervals, then we finish our procedure in a
finite number of steps. Further, taking into account (16) and the fact that
sA0,n, we obtain (3).
Let us describe our procedure in detail. Suppose (16) and (17) hold for
some i, 1 i < n. If Rxi
x0(s(t)f(t)) dt 0, then we take j:= i+ 1 and put
s(x) := L(x;f;xi, xj), x [xi, xj].
The inequality (15) yields (16) for x[x0, xj], and also (17), with xjinstead
of xi. Otherwise (that is, if Rxi
x0(s(t)f(t)) dt > 0), we apply Lemma 2. It
gives us some integer j,i+ 1 jn, and a spline s?
i,j , satisfying (10) if
j < n, and (9). We put s(x) := s?
i,j (x), x[xi, xj]. So, if j=n, then (9)
implies (16) for x[x0, xn] and the procedure is finished. Otherwise, (9),
(17), and (16) for x[x0, xi] imply (16) for x[x0, xj]. The inequality (10)
gives Rxi
x0(s(t)f(t)) dt Rxj
x0(s(t)f(t)) dt. Now, taking into account that
Rxi
x0(s(t)f(t)) dt > 0 and the estimate (9), we obtain (17) with xjinstead
of xi. Theorem 10is proven. 2
A. V. Prymak 11
Remark. The spline Sconstructed in the proof of Theorem 10satisfies
F(x0) = S(x0),
F0(x0) = S0(x0), F 0(xn) = S0(xn),
F00(x0+) = S00 (x0+), F 00 (xn) = S00(xn).
Proof of Theorem 1. Take y0:= x1, y1:= x2, . . . , yn2:= xn1. The first
derivative of Fexists on (x0, xn) and is continuous. Hence, it is continuous
on [y0, yn2] and we can apply Theorem 10to the function Fand the segment
[y0, yn2] with the partition y0, y1, . . . , yn2. We get some quadratic spline S1
with knots y0, . . . , yn2satisfying
kFS1k[y0,yn2]=kFS1k[x1,xn1]
cmax (Zxi
xi1
(L(x;F0;xi1, xi)F0(x)) dx¯¯¯¯¯
i= 2, n 1),
(18) F0(x1) = S0
1(x1), F 0(xn1) = S0
1(xn1),
(19) F00(x1+) = S00
1(x1+), F 00 (xn1) = S00
1(xn1).
Put
s(x) :=
F00(x1+)(xx1) + F0(x1), x [x0, x1)
S0
1(x), x [x1, xn1]
F00(xn1)(xxn1) + F0(xn1), x (xn1, xn],
and
S(x) := Zx
x1
s(t)dt +F(x1).
We shall prove that Ssatisfies (2). Indeed, note that the definition of S, the
convexity of S0
1, (18) and (19) imply that s0is a non-decreasing step function
on [x0, xn]. Hence, sis convex and consequently Sis 3-convex on [x0, xn]. In
order to prove (2) it is sufficient to show that, for i= 0, n 3,
Zxi+2
xi+1
(L(x;F0;xi+1, xi+2)F0(x)) dx cΛiω3(F; (xi+3 xi)/3; [xi, xi+3]),
Zx1
x0
(F0(x)s(x)) dx cΛ0ω3(F; (x3x0)/3; [x0, x3]),
12 Convex approximation
and
Zxn
xn1
(F0(x)s(x)) dx cΛn3ω3(F; (xnxn3)/3; [xn3, xn]).
These inequalities are evident corollaries of Lemma 3, and thus Theorem 1 is
proven. 2
Acknowledgement. The author thanks Professor I. A. Shevchuk for the
useful discussions on the paper.
References
[1] Ronald A. DeVore, Monotone approximation by splines, SIAM J. Math.
Anal. 8, 5 (1977), 891–905.
[2] Z. Ditzian and V. Totik, Moduli of Smoothness, Springer Verlag, Berlin,
1987.
[3] Yingkang Hu, Convex approximation by quadratic splines, J. Approx. The-
ory 74, 1 (1993), 69–82.
[4] Y. Hu, D. Leviatan and X. M. Yu, Convex polynomial and spline approxi-
mation in C[1,1], Constr. Approx. 10, 1 (1994), 31–64.
[5] V. N. Konovalov and D. Leviatan, Estimates on the approximation of 3-
monotone function by 3-monotone quadratic splines, East J. Approx. 7, 3 (2001),
333–349.
[6] K. A. Kopotun, Pointwise and uniform estimates for convex approximation of
functions by algebraic polynomials, Constr. Approx. 10, 2 (1994), 153–178.
[7] K. Kopotun and A. Shadrin, Shape-preserving approximation of k-monotone
functions by splines with free knots, to appear.
[8] D. Leviatan and A. Shadrin, On monotone and convex approximation by
splines with free knots, Annals of Numer. Math. 4, 1–4(1997), 415–434.
[9] P. P. Petrov, Three-convex approximation by free knot splines in C[a, b], Con-
str. Approx. 14, 2 (1998), 247–258.
[10] I. A. Shevchuk, One construction of cubic convex spline, Approximation and
Optimization, Proceedings of ICAOR, vol. 1, 357–368.
Received December 6, 2001
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