Elementary Solutions of the Bernstein Problem on Two Intervals
First we note that the best polynomial approximation to jxj on the set, which consists of an interval on the positive half-axis and a point on the negative half-axis, can be given by means of the classical Chebyshev polynomials. Then we explore the cases when a solution of the related problem on two...
Gespeichert in:
| Datum: | 2012 |
|---|---|
| 1. Verfasser: | |
| Format: | Artikel |
| Sprache: | English |
| Veröffentlicht: |
Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
2012
|
| Schriftenreihe: | Журнал математической физики, анализа, геометрии |
| Online Zugang: | http://dspace.nbuv.gov.ua/handle/123456789/106708 |
| Tags: |
Tag hinzufügen
Keine Tags, Fügen Sie den ersten Tag hinzu!
|
| Назва журналу: | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| Zitieren: | Elementary Solutions of the Bernstein Problem on Two Intervals / F. Pausinger // Журнал математической физики, анализа, геометрии. — 2012. — Т. 8, № 1. — С. 63-78. — Бібліогр.: 7 назв. — англ. |
Institution
Digital Library of Periodicals of National Academy of Sciences of Ukraine| id |
irk-123456789-106708 |
|---|---|
| record_format |
dspace |
| spelling |
irk-123456789-1067082016-10-04T03:02:18Z Elementary Solutions of the Bernstein Problem on Two Intervals Pausinger, F. First we note that the best polynomial approximation to jxj on the set, which consists of an interval on the positive half-axis and a point on the negative half-axis, can be given by means of the classical Chebyshev polynomials. Then we explore the cases when a solution of the related problem on two intervals can be given in elementary functions. Вначале показываем, что решение задачи о наилучшей полиномиальной аппроксимации функции |x| на множестве, состоящем из интервала на положительной полуоси и точки на отрицательной полуоси, может быть выражено через классические полиномы Чебышева. Далее мы изучаем вопрос о том, в каких случаях решение аналогичной задачи на объединении двух интервалов может быть выражено в сходных терминах. 2012 Article Elementary Solutions of the Bernstein Problem on Two Intervals / F. Pausinger // Журнал математической физики, анализа, геометрии. — 2012. — Т. 8, № 1. — С. 63-78. — Бібліогр.: 7 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106708 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| institution |
Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| collection |
DSpace DC |
| language |
English |
| description |
First we note that the best polynomial approximation to jxj on the set, which consists of an interval on the positive half-axis and a point on the negative half-axis, can be given by means of the classical Chebyshev polynomials. Then we explore the cases when a solution of the related problem on two intervals can be given in elementary functions. |
| format |
Article |
| author |
Pausinger, F. |
| spellingShingle |
Pausinger, F. Elementary Solutions of the Bernstein Problem on Two Intervals Журнал математической физики, анализа, геометрии |
| author_facet |
Pausinger, F. |
| author_sort |
Pausinger, F. |
| title |
Elementary Solutions of the Bernstein Problem on Two Intervals |
| title_short |
Elementary Solutions of the Bernstein Problem on Two Intervals |
| title_full |
Elementary Solutions of the Bernstein Problem on Two Intervals |
| title_fullStr |
Elementary Solutions of the Bernstein Problem on Two Intervals |
| title_full_unstemmed |
Elementary Solutions of the Bernstein Problem on Two Intervals |
| title_sort |
elementary solutions of the bernstein problem on two intervals |
| publisher |
Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| publishDate |
2012 |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/106708 |
| citation_txt |
Elementary Solutions of the Bernstein Problem on Two Intervals / F. Pausinger // Журнал математической физики, анализа, геометрии. — 2012. — Т. 8, № 1. — С. 63-78. — Бібліогр.: 7 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT pausingerf elementarysolutionsofthebernsteinproblemontwointervals |
| first_indexed |
2025-07-07T18:52:59Z |
| last_indexed |
2025-07-07T18:52:59Z |
| _version_ |
1837015372587335680 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2012, vol. 8, No. 1, pp. 63–78
Elementary Solutions of the Bernstein Problem on Two
Intervals
F. Pausinger
I.S.T. Austria
Am Campus 1, A-3400 Klosterneuburg, Austria
E-mail: florian.pausinger@ist.ac.at
Received August 31, 2011
First we note that the best polynomial approximation to |x| on the set,
which consists of an interval on the positive half-axis and a point on the
negative half-axis, can be given by means of the classical Chebyshev poly-
nomials. Then we explore the cases when a solution of the related problem
on two intervals can be given in elementary functions.
Key words: Chebyshev polynomials, polynomial approximation of |x|,
Bernstein problem.
Mathematics Subject Classification 2000: 41A10.
1. Introduction
1.1. Setting of the Problem
In [1, 2] Eremenko and Yuditskii studied asymptotics of the error Ln of the
best polynomial approximation of the function sgn(x) on two intervals [−A,−1]∪
[1, B]. It was noted that in the case when one of the intervals degenerates to
a point, say A = 1, the extremal polynomial Pn(x) can be explicitly written
by means of the classical Chebyshev polynomials. Moreover, such a relation
holds true for a certain sufficiently small extension of the set until it reaches
the critical value A∗ > 1 (note that a further extension of the interval leads to
a representation of the extremal solution via the Zolotarev polynomials), (see
Figure 1).
With these results in mind, we study the best polynomial approximation to
the function |x| on the set [−D,−C]∪ [B, A] that consists of two intervals, one on
the positive and one on the negative half-axes, so that the problem also admits
an elementary solution. Recall that Bernstein [3, 4] found that for the error En
This work is supported by the Austrian Science Fund (FWF), Project P22025-N18.
c© F. Pausinger, 2012
F. Pausinger
Fig. 1. Best polynomial approximation of sgn(x) with B = 10, n = 3. Solution
on {−1} ∪ [1, B] (a). Maximal extension with Chebyshev polynomials (b).
of the best uniform approximation of |x| on [−1, 1] by polynomials of degree n
the following limit exists:
lim
n→∞nEn = µ > 0.
His question, whether µ can be expressed in terms of any known transcendental
functions, remains one of the most famous open problems in approximation theory
(for recent developments in this area see [5]).
Similarly to the ideas in [1], we start with the set {−C}∪ [B, A] that consists
of an interval on the positive half-axis and a point on the negative half-axis and
show that for arbitrary A, B,C > 0 these extremal polynomials can be described
in terms of Chebyshev polynomials.
The polynomials of the best approximation of sgn(x) possess the following
crucial property: all critical values, barring one possible exception, are on the
level 1 ± Ln on the positive half-axis and on the level −1 ± Ln on the negative
half-axis (see Figure 1). It allows in an easy way to relate some of them with the
Chebyshev and Zolotarev polynomials. The structure of the extremal polynomials
that approximate |x| is much more delicate. Nevertheless, we are able to find the
maximal interval bound D = D(A,B, C) such that for given A,B, C the extremal
polynomial on [−D,−C] ∪ [B, A] can still be described in terms of Chebyshev
polynomials.
Our considerations are based on the Chebyshev Alternation Theorem which
we would like to recall before stating our main result.
64 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 1
Elementary Solutions of the Bernstein Problem on Two Intervals
1.2. Cheybshev Polynomials and Alternation Theorem
We follow [6] (see also [7]) in this introductory subsection. Let K ⊂ R be an
arbitrary compact set consisting of at least n+2 points, and let s(x) and f(x) be
real continuous functions with s(x) > 0 for x ∈ K. Due to Chebyshev, we want
to determine a polynomial P (x) of degree at most n such that its deviation from
f(x) with the weight s(x), namely
L(P ) = ||(f − P )/s||C(K),
is as small as possible. Denote by
M(P ) = {x ∈ K : |f(x)− P (x)|/s(x) = L(P )}
the set of points of maximal deviation. A set of points of alternation is defined
to be a maximal subset {x1, . . . , xω} ⊂ M(P ) such that (f(x)−P (x))/s(x) takes
the values ±L(P ) with alternating signs at the points of this set. Note that in
general a set of points of alternation is not unique. However, it follows from the
definition that all sets of points of alternation have the same cardinality. Now we
can state the important theorem which is usually referred to as the Chebyshev
Alternation Theorem:
Theorem 1. There exists a unique polynomial P (x) of degree at most n de-
viating least from the function f(x) with the weight s(x) on the compact set K.
This polynomial is completely characterized by the property that a set of points of
alternation consists of at least n + 2 points.
Note that we can consider the polynomial of degree n−1 deviating least from
the function f(x) = xn with the weight s(x) = 1 on [−1, 1] or, in other words,
the polynomial of degree n with leading coefficient 1 deviating least from zero
on [−1, 1]. This is the classical Chebyshev problem and its solutions are known
as Chebyshev polynomials Tn(x) of degree n. The polynomials Tn have always
exactly n + 1 alternation points in [−1, 1].
1.3. Notations and Main Result
Let Pn(x) be the best approximation of |x| by polynomials of degree at most
n on the set I := [−D,−C] ∪ [B, A], with A,B,C, D > 0 and A > B and
D ≥ C. Our goal is to obtain a representation of the extremal polynomial on
I for given A, B,C and different values of D. We will study in which cases this
representation can be written in terms of Chebyshev polynomials Tn of degree n.
Our main result can be stated as follows:
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 1 65
F. Pausinger
Theorem 2. For given A, B,C > 0, there are values L,α and β such that
there exists D(A,B, C) with the property that for all values D with D ≥ D ≥ C
the extremal polynomial on I = [−D,−C] ∪ [B, A] can be written in the form
Pn(x) = L · (−1)n Tn(y(x, α, β)) + x, (1)
where L denotes the approximation error and y(x, α, β) = −1 + x−β
α−β · 2.
Our paper is structured as follows. In Section 2, we consider the problem of
finding a solution to our problem on I = {−Q}∪[B,A]. In this case α = A, β = B
and L = Ln(A,B, Q) can be given by the explicit Formula (5). In Section 3, we
apply the results of Section 2 and prove the first version of our main result.
In this case α = A, β = B and L = Ln(A,B,Q), where Q is an appropriately
chosen point in the second interval [−D,−C]. In Section 4, we study further
extensions of I by certain changes of the parameters α and β. We conclude
our considerations with the observation that any further extension of I involves
polynomials that are not Chebyshev polynomials any more. Section 5 is finally
concerned with the special case B = 0, where the set I might degenerate to an
interval.
2. Polynomials of Least Deviation from |x| on {−Q} ∪ [B,A]
In this section we will show that there always exists an extremal polynomial
of the above form for fixed but arbitrary A,B, Q on I = {−Q} ∪ [B, A]. Let
Hn(x, α, β, γ) = Ln(α, β, γ) · (−1)n Tn(y(x, α, β)) + x, (2)
where y(x, α, β) = −1 + x−β
α−β · 2.
In the following, we set α = A, β = B and γ = Q, and show that for the
given set I, Pn = Hn(x,A,B,Q), moreover, Ln is the approximation error (5).
Proposition 1. For given A,B, Q > 0, A,B, Q ∈ R, there exists Ln > 0 such
that Pn = Hn(x,A,B,Q) with y(x,A, B) = −1 + x−B
A−B · 2.
P r o o f. Note that for x ∈ [B, A] we have y(x,A, B) ∈ [−1, 1]. Hence we
have n + 1 alternation points in the interval [B, A]. In order to have an extremal
polynomial of degree n on I we need n+2 alternation points due to the Chebyshev
theorem. Therefore we introduce an additional relation, which already suffices
to compute Ln(A,B,Q) in a unique way, and solve the following set of equations
for Ln:
Hn(x,A, B, Q) = Ln(A,B, Q) · (−1)n Tn(y(x,A, B)) + x (3)
Hn(−Q,A, B, Q) = Q− Ln(A,B, Q). (4)
66 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 1
Elementary Solutions of the Bernstein Problem on Two Intervals
We get that
Ln(A, B,Q) =
2Q
1 + (−1)n Tn(y(−Q,A, B))
. (5)
Note that Ln > 0, and that (4) gives us the required (n + 2)-th alternation point
such that the polynomial Hn is indeed extremal on I.
These extremal polynomials have the following crucial properties:
Proposition 2. For given A,B, Q > 0, the first derivative H ′
n(x, A,B, Q)
of the extremal polynomial Pn on I is monotonously increasing in (−∞, B] and
H ′
n(B,A, B, Q) > −1.
P r o o f. Note that Tn(y(x,A, B)) has n + 1 points of alternation in [B, A].
Hence n zeros of Tn lie in [B,A], which means that all zeros of Tn lie in [B, A].
Consequently, [B, A] contains n − 1 critical points of Tn, which are exactly the
n − 1 zeros of T ′n. Therefore T ′n is monotonous outside of the interval [B, A].
Since we always have by definition that Hn(B, A, B, Q) = B +Ln, it follows that
T ′n is monotonously increasing in (−∞, B].
Now suppose H ′
n(B, A, B,Q) ≤ −1. Since Hn(B, A, B,Q) = B + Ln and
since the derivative is monotonic, we have Hn(x, A,B, Q) > |x| for all x ∈
(−∞, B). However, this is a contradiction to Hn(−Q,A,B,Q) = Q−Ln. Hence
H ′
n(B,A, B, Q) > −1.
Proposition 3. For given A, B and for all Q ∈ (0,∞), there exists x ∈ (Q,∞)
such that Hn(−x,A,B,Q) > x + Ln(A, B,Q).
P r o o f. Recall that
Hn(x,A, B, Q) = Ln(A,B, Q)(−1)nTn(y(x,A, B)) + x
and that Tn(x) behaves asymptotically like xn for x → −∞. Therefore, for
sufficiently large x ∈ (Q,∞), we get
Hn(−x, A,B, Q) = Ln(A,B, Q)(−1)nTn(y(−x,A, B))− x > x + Ln(A,B, Q)
since
1
x
((−1)nTn(y(−x,A,B))− 1) >
2
Ln(A,B,Q)
,
where the right-hand side is constant.
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 1 67
F. Pausinger
As a consequence of these three propositions we can state the following:
• If for a given Q, H ′
n(−Q,A, B, Q) > −1, then there is Q̃ ∈ (Q,∞) such that
Hn(−Q̃, A, B, Q) = Q̃ − Ln(A,B, Q). Indeed, there exists a point ξ1 > Q
such that Hn(−ξ1, A, B, Q) < ξ1 − Ln(A,B, Q) and due to Proposition 3
there also exists a point ξ2 > ξ1 with Pn(−ξ2, A, B, Q) > ξ2 + Ln(A, B,Q).
Hence there is Q̃ with ξ2 > Q̃ > ξ1 for which the claim holds true.
• If for a given Q, H ′
n(−Q,A, B, Q) < −1, then there is Q̃ ∈ (0, Q) with
Hn(−Q̃, A, B, Q) = Q̃ − Ln(A,B, Q). Indeed, there exists ξ < Q with
Hn(−ξ, A, B, Q) < ξ−Ln(A, B,Q). Due to the monotonicity of the deriva-
tive and since Hn(B, A, B,Q) = B + Ln(A,B, Q), there is Q̃ ∈ (0, Q) for
which the claim holds true.
Note that since there is a unique extremal polynomial for a given Q, the corre-
sponding points Q̃ are also unique (see Figure 2).
Fig. 2. Solutions of the point-interval problem for n = 3 and A = 6, B = 1 resp.
Q = 0.5 in (a) and Q = Q∗(6, 1) = 1.5 in (b).
Moreover, for different values of Q the corresponding polynomials Hn only
differ in the factor Ln with which Tn is multiplied. The polynomial Tn has all its
zeros in [B,A], which means that the only points of intersection of the graphs of
two extremal polynomials for different values of Q can lie in [B, A]. Therefore we
can conclude that for the given values Q1 > Q2 with H ′
n(−Q1, A, B, Q1) < −1
and H ′
n(−Q2, A, B, Q2) < −1 we always have Q̃2 > Q̃1.
Consequently, since we already know that there is a unique solution to our
problem for every Q ∈ (0,∞), it is easy to see that there exists a unique point
Q∗ = Q∗(A,B) which is the fixed point of the above described involution Q 7→ Q̃.
For this point it holds that
Hn(−Q∗, A,B,Q∗) = Q∗ − Ln(A,B,Q∗), (6)
H ′
n(−Q∗, A,B,Q∗) = −1. (7)
68 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 1
Elementary Solutions of the Bernstein Problem on Two Intervals
Furthermore, we can define the values C∗(A,B) < D∗(A,B)
Hn(−D∗, A, B,Q∗) = D∗ + Ln(A, B,Q∗), (8)
Hn(−C∗, A, B,Q∗) = C∗ + Ln(A,B, Q∗). (9)
The value D∗(A,B) always exists due to Proposition 3, whereas C∗(A, B)
exists because
Hn(0, A, B,Q∗(A,B)) = Ln(A,B, Q∗)(−1)nTn
(
−1 + 2
0−B
A−B
)
+ 0
= Ln(A,B, Q∗)(−1)nTn(−1− ε) > Ln(A,B,Q∗). (10)
The values C∗, Q∗ and D∗ play a crucial role in the first extension of the set (see
next section).
3. First Extension of I
We have seen that we can always solve our problem for arbitrary A,B,Q on
I = {−Q} ∪ [B,A]. The main idea of our first extension is to move, if possible,
the leftmost alternation point (which initially is at C = Q). Thus we consider
a new point-interval problem with the additional constraint Hn(−C) ≤ C + Ln.
Formally, we change the parameter γ in the solution of the initial point-interval
problem.
We will see that if C ≥ Q∗, then no non-trivial extension of the interval
is possible. Note that in the following we always study the maximal possible
extension of I to the left, in other words, we always fix A,B, C and compute the
biggest possible interval bound D = D(A,B, C) > C such that there exists Hn
of the form (2) on [−D,−C] ∪ [B, A].
In order to simplify the proof of Theorem 3, it is useful to have the following
property:
Proposition 4. For given A,B and arbitrary Q, Ln(A,B, Q) is increasing
for Q ∈ (0, Q∗) and decreasing for Q ∈ (Q∗,∞).
P r o o f. Recall that for different values of Q the corresponding polynomials
Hn only differ in the factor Ln with which Tn is multiplied. Moreover, Tn has all
its zeros in [B, A], which means that the only points of intersection of the graphs
of two extremal polynomials for different values of Q can be in [B, A].
Suppose there exists Q ∈ (0, Q∗) such that Ln(A,B, Q) = Ln(A, B,Q∗),
which means we have only one polynomial to consider. For this polynomial it
holds that Hn(−Q∗) = Q∗ − Ln as well as Hn(−Q) = Q − Ln. However, this is
a contradiction to the monotonicity of H ′
n since we know that H ′
n(−Q∗) = −1.
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 1 69
F. Pausinger
Now suppose there exists Q ∈ (0, Q∗) such that Ln(A,B, Q) > Ln(A,B,Q∗).
This implies that Hn(B, A, B, Q) > Hn(B, A,B, Q∗) and also that H(−Q,A, B, Q)
< H(−Q∗, A, B, Q∗). This means that there exists a point of intersection of the
two graphs in (−Q∗, B). However, this is a contradiction to the fact that Tn has
all its zeros in [B,A]. Hence, for all Q ∈ (0, Q∗) it holds that Ln(A,B, Q) <
Ln(A,B, Q∗).
Note that for each Q 6= Q∗ there exists a unique Q̃ such that Ln(A,B, Q) =
Ln(A,B, Q̃). Therefore if Ln is increasing for Q ∈ (0, Q∗), it is decreasing for
Q ∈ (Q∗,∞).
Theorem 3. For given A, B let C∗, Q∗, D∗ be as defined above. For arbitrary
C > 0 there exists D1 ≥ C such that for any D with D1 ≥ D ≥ C there
exists an extremal polynomial Hn(x, α, β, γ) with y(x, α, β) = −1 + x−β
α−β · 2 on
I = [−D,−C] ∪ [B, A].
More precisely, set α = A, β = B. Then:
(I) if C ∈ (0, C∗), D1 is such that Hn(−C, A,B,D1) = C + Ln(A,B,D1). For
any D with D1 ≥ D ≥ C, we set γ = D, and the extremal polynomial on I
is given by Hn(x, A,B, D).
(II) if C ∈ [C∗, Q∗], D1 = D∗.
(II.a) For Q∗ > D ≥ C, we set γ = D, and the extremal polynomial on I is
given by Hn(x,A, B, D).
(II.b) For D∗ ≥ D ≥ Q∗, we set γ = Q∗, and the extremal polynomial on I
is given by Hn(x,A, B,Q∗).
(III) if C ∈ (Q∗,∞), D1 is such that Hn(−D1, A, B,C) = D1 + Ln(A,B, C).
For any D with D1 ≥ D ≥ C, we set γ = C, and the extremal polynomial
on I is given by Hn(x,A, B, C).
Remark 1. For an illustration of Case 1, see Figure 3 and Figure 4. For
an example of a diagram of the maximal first extensions for given A,B, n and
different values of C see Figure 5(a).
P r o o f. Case 1. Let C ∈ (0, C∗). Note that C < C∗ implies D1 < Q∗.
We have to show that for any D with D1 ≥ D ≥ C
Hn(−C, A,B, D) ≤ C + Ln(A,B,D).
Due to Proposition 4, we have
Hn(−x,A, B,D1) ≥ Hn(−x,A, B, C)
for x ∈ (0, D1).
70 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 1
Elementary Solutions of the Bernstein Problem on Two Intervals
Fig. 3. Illustration of Case 1 for A = 6, B = 1, C = 0.5. Solution on {−C}∪[B, A]
(a). Extension to [−D1,−C] ∪ [B,A] (b).
Fig. 4. Illustration of Case 1 for A = 6, B = 1, C = 0.5 . Zoom (Dashed Box) of
Figure 3.
Suppose that there exists γ with D1 ≥ γ ≥ C such that
Hn(−C, A,B, γ) > C + Ln(A,B, γ),
then
(−1)n Tn(y(−C,A, B))(Ln(A,B,D1)− Ln(A,B, γ)) =
= Hn(−C, A,B,D1)−Hn(−C, A, B, γ)
< C + (Ln(A, B,D1)− C − (Ln(A,B, γ)
= Ln(A,B,D1)− Ln(A,B, γ).
This is a contradiction to the fact that (−1)nTn(y(−C,A, B)) > 1 since −C < B.
Therefore we have shown
Hn(−C, A,B,D) ≤ C + Ln(A,B,D).
Case 2. Let C ∈ [C∗, Q∗]. We have to show that for γ = D and Q∗ > D ≥ C
Hn(−C, A,B,D) ≤ C + Ln(A,B,D).
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 1 71
F. Pausinger
We can use a similar argument as in the first case. Due to Proposition 4, we
know that
Hn(−x,A,B,Q∗) ≥ Hn(−x,A, B, C)
for x ∈ (0, Q∗). Suppose there exists γ with Q∗ > γ ≥ C such that
Hn(−C, A,B, γ) > C + Ln(A,B, γ).
Then we also have that
Hn(−C∗, A, B, γ) > C∗ + Ln(A,B, γ)
due to the monotonicity of H ′
n. Hence,
(−1)n Tn(y(−C∗, A, B))(Ln(A,B, Q∗)− Ln(A,B, γ)) =
Hn(−C∗, A, B, Q∗)−Hn(−C∗, A, B, γ)
< C∗ + (Ln(A, B,Q∗)− C∗ − (Ln(A,B, γ)
= Ln(A,B, Q∗)− Ln(A,B, γ).
This yields the same contradiction as before. Note that for γ = Q∗ we always
have
x− Ln(A,B, Q∗) ≤ Hn(x,A, B, Q∗) ≤ x + Ln(A,B, Q∗)
for x ∈ (C, D∗). Hence, Hn(x,A, B, Q∗) is maximal for any D with D∗ ≥ D ≥ Q∗
on I.
Case 3. Let C ∈ (Q∗,∞). In this case we can only trivially extend the
solution of {−C} ∪ [B,A] to the point D1 that is defined via
Hn(−D1, A,B, C) = D1 + Ln(A,B, C).
Due to Proposition 3, this point always exists and, since H ′
n(−C, A, B,C) < −1,
we always have that
x− Ln(A,B, C) ≤ Hn(x,A, B, C) ≤ x + Ln(A,B,C)
for x ∈ (C, D1).
Remark 2. We can see that it is possible to extend I = {−C} ∪ [B,A] if
we change the parameter γ and, therefore, the factor Ln in the representation of
the extremal polynomial. Since Ln also denotes the approximation error and it is
increasing for increasing C ∈ (0, Q∗) and decreasing for increasing C ∈ (Q∗,∞),
it is not surprising that we can only trivially extend our interval (to the left) for
any C ∈ (Q∗,∞).
72 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 1
Elementary Solutions of the Bernstein Problem on Two Intervals
4. Further Extension of I
In the previous section we extended I by changing the parameter γ of the
extremal polynomial. At the end of the first extension step we obtained a new
alternation point in each of the three cases, so that we had 2 in the subset of the
negative half-axes and still (n + 1) in the subset of the positive half-axes. In this
section we show that we can extend I further by changing the parameters α and β
accordingly. That is, we change the interval on which the Chebyshev polynomial
Tn is extremal. The idea is that since we have one alternation point more after
the first extension than we need for applying the alternation theorem, we will try
to extend our interval by shifting the rightmost resp. the leftmost alternation
point on the positive half-axes. Of course, there is a natural limitation in this
shift, namely, we can only shift as long as we do not lose a second alternation
point in I. Formally, we define the right limit point Ā > A for given A,B,C in
the following way:
Hn(A, Ā, B,C) = A− (−1)nLn(Ā, B, C), (11)
H ′
n(A, Ā, B,C) = 1, (12)
and H ′
n(x, Ā, B,C) > 1 or < 1 for all x ∈ (A,∞). This means we shift the
rightmost critical point of the polynomial Hn(x,A, B,C) to the position A and
obtain a new polynomial Hn(x, Ā, B,C). Note that for any α with A ≤ α ≤ Ā
the graph of the corresponding polynomial Hn(x, α,B,C) has n − 1 points of
intersection with the graph of Hn(x, A,B, C) in the interval [B, A]. Analogously,
we can define, if possible, a point B̄ > 0 by shifting the leftmost critical point to
the position B. In the case when this point does not exist, we consider β with
0 < β ≤ B instead of B̄ ≤ β ≤ B.
Proposition 5. For given A,B, α and β with A ≤ α ≤ Ā and B̄ ≤ β ≤ B,
we have
Q∗(A,B) ≤ Q∗(α, B) and D∗(A,B) ≤ D∗(α, B), (13)
C∗(A, β) ≤ C∗(A,B). (14)
P r o o f. To prove the first part, note that Ln(A,B, Q∗(A,B)) <
Ln(α, B, Q∗(α, B)) and hence
Hn(B, α, B, Q∗(α,B)) > Hn(B, A,B, Q∗(A,B)),
Hn(−Q∗(α,B), α,B,Q∗(α, B)) < Hn(−Q∗(α, B), A, B, Q∗(A,B)),
which means that the two graphs intersect in the interval (−Q∗(α, B), B). More-
over, there are n−1 points of intersection in [B, A], and since Hn(x, α,B,Q∗(α,B))
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 1 73
F. Pausinger
−Hn(x,A, B, Q∗(A,B)) is a polynomial of degree n, these are all intersections.
Furthermore, we have for all x ∈ (−∞, B) that
(−1)nT ′n(y(x,A,B)) ≤ (−1)nT ′n(y(x, α,B))
and therefore
Ln(A,B, Q∗(A, B))(−1)nT ′n(y(x, A,B))≤Ln(α, B, Q∗(α, B))(−1)nT ′n(y(x, α, B)).
This implies
−1 = H ′
n(−Q∗(A,B), A, B,Q∗(A,B)) ≤ H ′
n(−Q∗(A,B), α,B, Q∗(α, B)),
and due to the monotonicity of the derivative,
Q∗(A,B) ≤ Q∗(α,B).
Moreover, since Hn(x, α, B,C) < Hn(x, A,B, C) for all x ∈ (−∞,−Q∗(α,B)),
it follows that
D∗(A,B) < D∗(α, B).
For the second part, note that Ln(A,B, Q∗(A,B)) < Ln(A, β, Q∗(A, β)) and
that
Hn(B, A, β, Q∗(A, β)) < Hn(B, A, B, Q∗(A,B)),
Hn(−Q∗(A, β), A, β,Q∗(A, β)) < Hn(−Q∗(A, β), A,B,−Q∗(A,B)).
Hence, there is no point of intersection of the two graphs in (−Q∗(A, β), B),
(if there were one, this would imply that there has to be another one to satisfy
these relations) and
Hn(−C∗(A,B), A, β, Q∗(A, β)) < Hn(−C∗(A,B), A, B, Q∗(A,B))
= C∗(A,B) + Ln(A,B, Q∗(A,B)),
such that
C∗(A, β) < C∗(A,B).
This technical proposition is the base of the following considerations. We im-
prove each case of Theorem 3 in a separate lemma from which our main theorem,
as stated in the introduction, immediately follows.
Lemma 1. For given A,B and C ∈ (Q∗(Ā, B),∞), there exists D2 > D1
such that Hn(x, Ā, B,C) is maximal on [−D2,−C] ∪ [B,A].
74 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 1
Elementary Solutions of the Bernstein Problem on Two Intervals
P r o o f. As in the previous proposition, we can see that Hn(x, α,B,C) −
Hn(x,A, B, C) for A ≤ α ≤ Ā is a polynomial of degree n with n − 1 zeros in
[B,A] and an additional zero in [−C, B]. This implies
Hn(x, α,B, C) ≤ Hn(x,A, B,C)
for all x ∈ (−∞,−C), and
Hn(−D1, α, B, C) ≤ Hn(−D1, A,B, C) = Ln(A,B, C) + D1 < Ln(α, B, C) + D1.
Consequently, due to Proposition 3, there exists D2 > D1 such that
Hn(−D2, α,B,C) = Ln(α, B, C) + D2.
To get a polynomial which is maximal for any D with D1 ≤ D ≤ D2, one has to
chose an appropriate α with A ≤ α ≤ Ā.
Lemma 2. For given A,B and C ∈ (C∗(A,B), Q∗(Ā, B)), there exists α,
with A ≤ α ≤ Ā and D2 > D1 such that Hn(x, α, B,Q∗(α,B)) is maximal on
[−D2,−C] ∪ [B, A].
P r o o f. Case 1. Let C∗(A,B) ≤ C∗(Ā, B). For A, B,C we know from
Theorem 3 that D1 = D∗(A,B).
However, for all C ∈ (C∗(A, B), C∗(Ā, B)), there exists α, with A ≤ α ≤ Ā,
such that
Hn(−C,α, B, Q∗(α, B)) = C + Ln(α, B, Q∗(α, B)).
Due to Proposition 5, D2 = D∗(α, B) ≥ D∗(A,B) = D1.
Now consider C ∈ (C∗(Ā, B), Q∗(Ā, B)). In this case we can again apply
Case 2 of Theorem 3 and see due to Proposition 5 that the maximal extension
D2 = D∗(Ā, B) for Hn(x, Ā, B,Q∗(Ā, B)).
Case 2. Let C∗(Ā, B) ≤ C∗(A, B). Like in the second subcase of the first
case, we can apply Case 2 of Theorem 3 and see due to Proposition 5 that the
maximal extension D2 = D∗(Ā, B) for Hn(x, Ā, B, Q∗(Ā, B)).
Lemma 3. For given A,B and C ∈ (0, C∗(A,B)), there exists β, with B̄ ≤
β ≤ B, and D2 > D1 such that Hn(x,A, β, Q∗(A, β)) is maximal on [−D2,−C]∪
[B,A].
P r o o f. Case 1. Let C ∈ (C∗(A, B̄), C∗(A,B)). For all C there exists β,
with B̄ ≤ β ≤ B, such that
Hn(−C, A, β, Q∗(A, β)) = C + Ln(A, β, Q∗(A, β)).
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 1 75
F. Pausinger
Note that C = C∗(A, β) and, therefore, D2 = D∗(A, β) due to Case 2 of Theorem
3. From Case 1 of Theorem 3 we know that D1 < Q∗(A,B). So we have to show
that D2 = D∗(A, β) ≥ Q∗(A,B). Suppose D∗(A, β) < Q∗(A,B). This would
mean that the point of intersection of the graphs of the two polynomials is in
(0, Q∗(A, B)). However, as we can see from Proposition 5, this point lies in
(Q∗(A, B),∞). Hence, D∗(A, β) ≥ Q∗(A,B) and, therefore, D2 > D1, and the
extremal polynomial is given by Hn(x,A, β,Q∗(A, β)).
Case 2. Let C ∈ (0, C∗(A, B̄)). In this case we have to apply Case 1 of
Theorem 3 twice. First, we get the relation Hn(−C, A, B,D1) = C+Ln(A,B,D1)
for D1. Secondly, we get the relation Hn(−C, A, B̄, D2) = C + Ln(A, B̄,D2) for
D2. Suppose that D2 < D1, due to the relations for these two values this would
again yield a contradiction to the fact that the point of intersection of the two
graphs is in the interval (Q∗(A,B),∞).
If B̄ does not exist, we can apply Case 1 for all C.
These three lemmata can be combined to obtain Theorem 2. According to
our observations, we can classify C for given A,B as shown in Table 1. For an
example of a diagram of the maximal second extensions for given A,B, n and
different values of C see Figure 5(b).
Fig. 5. Maximal interval extensions. Maximal first extension for A = 6, B = 1
(a). Maximal first and second extension for A = 6, B = 1 (b).
To sum up, we have proven that we can further extend our first maximal set
for each case of Theorem 3 using this classification of C and applying Lemmas
1,2,3.
76 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 1
Elementary Solutions of the Bernstein Problem on Two Intervals
Theorem 3 # Alt. Pts Lemma 1/2/3 # Alt. Pts
[−D,−C] [B, A] [−D,−C] [B, A]
C∈ (0, C∗(A, B)) 2 (n + 1) (0, C∗(A, B̄)) 2 n
(C∗(A, B̄), C∗(A, B)) 3 n
C∈ (C∗(A, B), Q∗(A, B)) 2 (n + 1) (C∗(A, B), C∗(Ā, B)) 3 n
(C∗(Ā, B), Q∗(A, B)) 2 n
C∈ (Q∗(A, B),∞) 2 (n + 1) (Q∗(A, B), Q∗(Ā, B)) 2 n
(Q∗(Ā, B),∞) 2 n
Table 1. Classification of C
Remark 3. Note that C∗(A,B) ≤ C∗(α,B) and D∗(A, β) ≤ D∗(A,B) for all
A ≤ α ≤ Ā and B̄ ≤ β ≤ B immediately imply that our second extension step is
the maximal possible extension such that Pn can be written in terms of Chebyshev
polynomials. We have computational evidence, but no proof that these relations
hold in general. If these relations do not hold (this can be computationally checked
in an easy way for arbitrary values), we have to be (a little) more careful with
the values C in the interval (C∗(α, B), C∗(A,B)). In this case we can, for given
A,B, C, change either the parameter α or β and apply the ideas from Lemmas
1, 2, 3. The maximal extension is then given by the maximum of the two solutions
we get. Note that if B̄ exists, we can change both parameters for the unique value
C = C∗(Ā, B̄) and therefore get a third case to consider when (computationally)
looking for the maximal extension.
5. The Special Case B = 0
With respect to the original Bernstein problem, it is interesting to consider
the special case B = 0. Note that our considerations of Section 2 still hold true
if we set B = 0. In this case we have C∗(A, 0) = 0 for any A > 0 since
Hn(0, A, 0, Q∗(A, 0)) = Ln(A, 0, Q∗(A, 0))(−1)nTn(−1)+0 = Ln(A, 0, Q∗(A, 0))+0.
Hence, we can reduce our set I to one interval if we choose C = C∗ = 0 and we
can apply Case 2 of Theorem 3 to see that D1 = D∗. In this case we have n + 3
alternation points and can further extend our interval. It follows from above that
C∗(A, 0) = C∗(Ā, 0) = 0 and, therefore, we can apply the ideas of Lemma 2 to
obtain D2 = D∗(Ā, 0) > D1. It is easy to see that this is the maximal possible
extension for Pn to be written in terms of Chebyshev polynomials on I.
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 1 77
F. Pausinger
Fig. 6. Solution of the point-interval problem for n = 3, A = 1, B = 0 and
Q = Q∗(1, 0) (a). Maximal extension for C = 0 (b).
Let us discuss the special example with A = 1, B = 0, n = 3. We get Q∗ =
1/4(−1+
√
3) = 0.183 . . . and D∗ = 0.3509 . . . (see Figure 6). More computations
show that the values Q∗ and D∗ decrease for increasing n. For our example we get
Ā = 4/3, Q∗(Ā, 0) = 1/3(−1 +
√
3) = 0.244 . . . and D2 = D∗(Ā, 0) = 0.4678 . . . .
References
[1] A. Eremenko and P. Yuditskii, Polynomials of the Best Uniform Approximation to
sgn(x) on Two Intervals. To appear in J. Anal. Math.
[2] A. Eremenko and P. Yuditskii, Uniform Approximation of sgn(x) by Polynomials
and Entire Functions. — J. Anal. Math. 101 (2007), 313–324.
[3] S. Bernstein, Sur la meilleure approximation de |x| par des polynomes des degrés
donnés. — Acta Math. 37 (1914), No. 1, 1–57.
[4] S. Bernstein, On the Best Approximation of |x|p by Polynomials of Very High
Degree. — Izv. Akad. Nauk SSSR (1938), 169–180.
[5] D. Lubinsky, On the Bernstein Constants of Polynomial Approximation. — Constr.
Approx. 25 (2007), No. 3, 303–366.
[6] N.I. Achieser, Vorlesungen über Approximationstheorie. Math. Lehrbücher, 2. Au-
flage, Band II Akademie Verlag, Berlin (1967).
[7] M.L. Sodin and P. Yuditskii, Functions Deviating Least from Zero on Closed Subsets
of the Real Axis. — St. Petersburg Math. J. 4 (1993), No. 2, 201–249.
78 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 1
|