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BERNSTEIN’S INEQUALITY FOR MULTIVARIATE

POLYNOMIALS ON THE STANDARD SIMPLEX

LOZKO B. MILEV AND SZIL ´ARD GY. R ´EV ´ESZ

Received 31 July 2003

The classical Bernstein pointwise estimate of the (first) derivative of a univariate alge-braic polynomial on an interval has natural extensions to the multivariate setting. How-ever, in several variables the domain of boundedness, even if convex, has a considerable geometric variety. In 1990, Y. Sarantopoulos satisfactorily settled the case of a centrally symmetric convex body by a method we may call “the method of inscribed ellipses.” On the other hand, for the general case of nonsymmetric convex bodies we are only within a constant factor of an exact inequality. The best known results suggest relevance of the generalized Minkowski functional, and a natural conjecture for the exact Bernstein fac-tor was formulated with this geometric quantity. This work deals with the most natural and simple nonsymmetric case, that of a standard simplex inRd, and computes the exact yield of the method of inscribed ellipses. Although the known general estimates of the Bernstein factor are improved for the simplex here, we find that not even the exact yield of the inscribed ellipse method reaches the conjecture. However, we also show that for an arbitrary convex body the subset of ridge polynomials satisfies the conjecture.

1. Introduction

If a univariate algebraic polynomialpis given with degree at mostn, then by the classical Bernstein-Szeg˝o inequality (see [1,10,11]), we have

p(x)n

p2

C[a,b]−p2(x)

(b−x)(x−a) (a < x < b). (1.1)

This inequality is sharp for everynand every pointx∈(a,b), as

sup     

p(x)

p2

C[a,b]−p2(x)

: degp≤n,p(x)<pC[a,b]     =

n

(b−x)(x−a). (1.2)

We may say that the upper estimate (1.1) is exact, and the right-hand side is just the “true Bernstein factor” of the problem.

Copyright©2005 Hindawi Publishing Corporation

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In the multivariate setting a number of extensions were proved for this classical result. However, due to the geometric variety of possible convex sets replacing intervals ofR, our present knowledge is still not final. The exact Bernstein inequality is known only for symmetric convex bodies, and we are within a bound of some constant factor in the general, nonsymmetric case.

For more precise notation we may define formally for any topological vector spaceX, a subsetK⊂X, and a pointx∈Kthenth “Bernstein factor” as

Bn(K,x) :=n1sup     

D p(x)

p2

C(K)−p2(x)

: degp≤n,p(x)<pC(K)    

, (1.3)

whereD p(x) is the derivative ofpatx, and even for an arbitrary unit vectory∈X

Bn(K,x,y) :=1 nsup

    

D p(x),y

p2

C(K)−p2(x)

: degp≤n,p(x)<pC(K)    

. (1.4)

In the present paper, first we study the standard simplex of thed-dimensional Eu-clidean spaceRd. We find the exact yield of the possibly nicest available method—the method of inscribed ellipses, introduced into the subject by Sarantopoulos [9]—for ar-bitrary interior points of this nonsymmetric convex body. It will be seen that for this particular case this calculation improves upon the previously known general estimate. On the other hand, the perhaps most intriguing conjecture in the topic, which relates the “true Bernstein factor” to the “generalized Minkowski functional,” remains still open even for the standard simplex.

On the other hand, all known lower estimates use ridge polynomials some way. So it is of interest to test, whether ridge polynomials can be used to disprove the conjecture. It turns out that ridge polynomials always satisfy the above-mentioned conjecture, even for general convex bodies in arbitrary normed spaces.

2. A review of Sarantopoulos’ method of inscribed ellipses

Recall that aconvex bodyin a topological vector spaceX(e.g., inRd) is a bounded, closed convex set that has a nonempty interior. Polynomials and continuous polynomials are defined over topological vector spaces, see, for example, [2]. The set of continuous poly-nomials overX will be denoted byᏼ=ᏼ(X) and polynomials inᏼ with degree not exceedingnbyᏼn=n(X). In this section, we review the inscribed ellipse method. Al-though for the reader’s convenience we include short proofs, we emphasize that, unless otherwise stated, results in this section are due to Sarantopoulos [9].

Lemma2.1 (inscribed ellipse lemma). LetKbe any subset in a vector spaceX. Suppose that x∈Kand the ellipse

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lies insideK. Then for any polynomialpnof degree at mostnthe Bernstein-type inequality

D pn(x),yn b

pn 2

C(K)−pn2(x) (2.2)

holds true.

Proof. Consider the trigonometric polynomialT(t) :=pn(r(t)) of degree at mostn. Since r(t)⊂K we clearly have T ≤ pnC(K). According to the Bernstein-Szeg˝o inequality [10] (see also [11]) for trigonometric polynomials,

T(t)nT2T(t)2npn2 C(K)−pn

r(t)2 (∀t∈R). (2.3)

In particular, fort=0, we get

T(0)np

n 2

C(K)−pn2(x). (2.4)

By the chain rule

T(0)=D pnr(0),r(0)=D pn(x),by, (2.5)

which completes the proof.

The Minkowski functional [5] of a convex bodyKis defined by

ϕ(x) :=ϕ(K,x) :=inf{λ >0 :x∈λK}. (2.6)

Clearly,xK:=ϕ(x) is a norm onXif and only ifKis centrally symmetric with respect to the origin. IfKis not centrally symmetric, the same functional can be used; however, there is another extension, the “generalized Minkowski functional”α(K,x), which also goes back to Minkowski [5] and Radon [6], see also [3,7]. For our present purposes we defineα(K,x) forxintKas follows. First, let

γ(K,x) :=inf

2

x−axb

a−b : a,b∈∂K, such thatx[a,b]

. (2.7)

Then we set

α(K,x) :=1−γ2(K,x). (2.8)

For many other equivalent formulations, geometric properties, and applications in ap-proximation theory, see [7] and the references therein.

Lemma2.2. LetKbe a centrally symmetric convex body in a vector spaceXand letx∈K. The ellipser(t)=costx+bsinty(t[−π,π))lies inKwhenever

yK=1, b=1− x2

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Proof. The assertion is equivalent tor(t)K≤1 for everyt. By the triangle and Cauchy inequalities,

r(t)

K≤ |cost|xK+b|sint|yK≤

cos2t+ sin2tx2

K+b2y2K=1. (2.10)

Lemma 2.2is proved.

The maximal chord ofKin directionv=0is

τ(K,v) :=supλ >0 :∃y,z∈Ksuch thatz=y+λv, (2.11)

see, for example, [12]. Note that in normed spaces thewidth of Kisw(K) :=inf{τ(K,v) : v =1}, see, for example, [7]. Mutatis mutandis to the previous lemma we can deduce also the following variant.

Lemma2.3. LetKbe a centrally symmetric body inX, where(X, · )is a normed space. LetϕK= · Kbe the Minkowski functional (norm) generated byK. Then for every nonzero vectory∈Xthe ellipser(t)=costx+bsinty(t[−π,π))lies inKwith

b:=

1−ϕ2(K,x)

ϕ(K,y) . (2.12)

Theorem2.4. Letpnbe any polynomial of degree at mostnover the normed spaceX. Then for any unit vectory∈Xthe following Bernstein-type inequality holds:

D pn(x),ynpn2C(K)−p2n(x) 1− x2

K

. (2.13)

Proof. The proof follows from combining Lemmas2.1and2.2. Theorem2.5. LetKbe a symmetric convex body andya unit vector in the normed space X. Letpnbe any polynomial of degree at mostn. Then,

D pn(x),y2npn2C(K)−pn2(x)

τ(K,y)1−ϕ2(K,x). (2.14)

In particular,

D pn(x)2npn2C(K)−p2n(x)

w(K)1−ϕ2(K,x) , (2.15)

wherew(K)stands for the width ofK.

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Definition 2.6. For arbitraryK⊂X andx∈K,y∈X the corresponding “best ellipse constants” are the extremal quantities

E(K,x,y) :=supb:r⊂Kwithras given in (2.1), (2.16) E(K,x) :=infE(K,x,y) :y∈X,y =1. (2.17)

Clearly, the inscribed ellipse method yields Bernstein-type estimates whenever we can derive some estimate of the ellipse constants. In case of symmetric convex bodies, Saran-topoulos’s Theorems2.4and2.5are sharp; for the nonsymmetric case we know only the following result.

Theorem2.7 (Kro ´o-R´ev´esz [4]). LetKbe an arbitrary convex body,xintKandy = 1, whereXcan be an arbitrary normed space. Then,

D pn(x),y2npn2C(K)−pn2(x)

τ(K,y)1−α(K,x) , (2.18)

for any polynomialpnof degree at mostn. Moreover,

D pn(x)2npn 2

C(K)−pn2(x) w(K)1−α(K,x)

22npn 2

C(K)−pn2(x)

w(K)1−α2(K,x) . (2.19)

Note that in [4] the best ellipse is not found; the construction there gives only a good estimate, but not an exact value of (2.16) or (2.17). In fact, here we quoted [4] in a strengthened form: the original paper contains a somewhat weaker formulation only.

As mentioned above, one of the most intriguing questions of the topic is the following conjecture, formulated first in [7].

Conjecture2.8 (R´ev´esz and Sarantopoulos). LetXbe a topological vector space, and let K be a convex body inX. For every pointxintK and every (bounded) polynomial pof degree at mostnoverX, we have

D p(x)2np2

C(K)−p2(x)

w(K)1−α2(K,x), (2.20)

wherew(K)stands for the width ofK.

3. The inscribed ellipse method for the standard simplex

We denote byᏼd

nthe space of polynomials ofdvariables and total degree≤n. LetSd−1:= {x∈Rd:|x|

2=1}be the unit sphere inRd, where|x|2:=(di=1x2i)1/2is the Euclidean norm ofx=(x1,. . .,xd)Rd. The derivative of pn∈dnin direction ySd−1 will be denoted byDypn.

Let

∆:=d:=

x1,. . .,xd:xi≥0,i=1,. . .,d, d

i=1 xi≤1

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be the standard simplex in Rd. For fixedx=(x1,. . .,x

d)int∆, andy=(y1,. . .,yd), |y|2=1, we consider the set of ellipses (2.1), where

a=a1,. . .,ad

Rd, bR. (3.2)

Clearly the best ellipse constant of∆is the extremal quantity

E(∆,x,y)=maxb:a1,. . .,ad,b

Ω, (3.3)

where

Ω:=a1,. . .,ad,b

:r(t)∆for everyt∈[−π,π)⊂Rd+1. (3.4)

Lemma3.1. The explicit formula

E(∆,x,y)=

y2 1 x1 +···+

y2 d xd +

y1+···+yd 2

1−x1− ··· −xd 1/2

(3.5)

holds true.

Proof. In coordinate form

Ω:=

a1,. . .,ad,b:aicost+byisint+xi−ai≥0 (i=1,. . .,d),

cost d

i=1

ai+bsint d

i=1 yi+

d

i=1 xi−

d

i=1

ai≤1∀t∈[−π,π)

.

(3.6)

We need the following simple lemma. Lemma3.2. The sets

=(a,b,c) :acost+bsint≤c∀t∈[−π,π),

=(a,b,c) :c≥0,a2+b2c2 (3.7)

coincide.

Proof. If (a,b,c)ᏹ, thenchas to be nonnegative. The casea=b=0 is trivial, so we can assume that at least one ofaandbis nonzero. Letξbe defined to satisfy

sinξ=√ a

a2+b2, cosξ= b

a2+b2. (3.8)

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Conversely, let (a,b,c)ᏺ. The Cauchy-Schwartz inequality gives (acost+bsint)2 a2+b2c2, that is, (a,b,c), which concludes the proof.

UsingLemma 3.2and settingZ:=b2, we arrive at the problem

maxZ :z:=a1,. . .,ad,Z

Θ, (3.9)

where

Θ:=  

z:Z≥0,Z y2i ≤x2i−2xiai(i=1,. . .,d),xi−ai≥0 (i=1,. . .,d),

Z d

i=1 yi

2

1

d

i=1 xi

2 + 2

1

d

i=1 xi

d

i=1 ai, 1

d

i=1 xi+

d

i=1 ai≥0

  .

(3.10)

Note thatΘRd+1 is bounded, hence compact. Indeed,xand the inequalities ai≤xi,i=1,. . .,d,di=1ai≥di=1xi−1 imply that|ai| ≤1,i=1,. . .,d. At least one ofyi, i=1,. . .,dis different from zero. Ifyj=0, then

Z≤x 2 j−2xjaj

y2j

3 y2j

. (3.11)

It follows that the optimal value of the problem (3.9) is finite.

Next, we quote the Kuhn-Tucker Theorem, as given in [8, Corollary 28.3.1].

Lemma3.3 (Kuhn-Tucker theorem). Let the functions fi:Rn→R,i=0,. . .,m, be diff er-entiable, where fi is convex fori=0,. . .,rand fiis affine fori=r+ 1,. . .,m. Consider the extremal problem

(P)

minf0(z) (3.12)

subject to constrains f1(z)0,. . .,fr(z)0, fr+1(z)=0,. . .,fm(z)=0. Suppose that the optimal value in (P) is not−∞and (P) has at least one feasible solution, which satisfies all the inequality constrains fori=1,. . .,rwith strict inequality. Thenzis a solution of (P) if and only if there exists numbersλ:=(λ1,. . .,λm)which together withzsatisfy the following:

(a)λi≥0, fi(z)0andλifi(z)=0,i=1,. . .,r, (b) fi(z)=0fori=r+ 1,. . .,m,

(c) gradf0(z) +mi=1λigradfi(z)=0.

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Now we turn to the problem (3.9). In the notations ofLemma 3.3we haven=d+ 1, m=r=2d+ 3 (and thus, in particular, (b) becomes void),z=(a,Z),

f0(a,Z)= −Z, f1(a,Z)= −Z,

fi+1(a,Z)=2xiai+y2iZ−x2i (i=1,. . .,d),

fd+2(a,Z)= −2

1 d

i=1 xi

d

i=1 ai+Z

d

i=1 yi 2 1 d

i=1 xi

2 ,

fd+2+j(a,Z)=aj−xj (j=1,. . .,d),

f2d+3(a,Z)= d

i=1 xi−1

d

i=1 ai.

(3.13)

Differentiation with respect tozgives

gradf0=gradf1=(0,. . ., 0,−1), gradf2=2x1, 0,. . ., 0,y21

, ..

.

gradfd+1=0,. . ., 0, 2xd,yd2

,

gradfd+2= 2 1 d

i=1 xi

,. . .,2

1

d

i=1 xi

,

d

i=1 yi

2 ,

gradfd+3=(1, 0,. . ., 0, 0), ..

.

gradf2d+2=(0,. . ., 0, 1, 0), gradf2d+3=(1,1,. . .,−1, 0).

(3.14)

Thus in problem (3.9) the system (a), (b), and (c) in (z,λ)=(a1,. . .,ad,Z,λ1,. . .,λ2d+3) becomes

λi≥0, i=1,. . ., 2d+ 3, (3.15)

Z≥0, (3.16)

2xiai+yi2Z−x2i 0, i=1,. . .,d, (3.17)

2

1 d

i=1 xi

d

i=1 ai+Z

d

i=1 yi 2 1 d

i=1 xi

2

0, (3.18)

ai−xi≤0, i=1,. . .,d, (3.19) d

i=1 xi−

d

i=1

ai−10, (3.20)

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λi+1

2xiai+y2iZ−x2i

=0, i=1,. . .,d, (3.22)

λd+2 2 1 d

i=1 xi

d

i=1 ai+Z

d

i=1 yi 2 1 d

i=1 xi

2

=0, (3.23)

λd+2+i

ai−xi

=0, i=1,. . .,d, (3.24)

λ2d+3 d

i=1 xi−

d

i=1 ai−1

=0, (3.25)

i+1xi−d+2

1 d

i=1 xi

+λd+2+i−λ2d+3=0, i=1,. . .,d, (3.26)

1−λ1+ d

i=1

λi+1yi2+λd+2 d

i=1 yi

2

=0. (3.27)

Ifxj=ajfor somej∈ {1,. . .,d}, then from (3.17) we getx2j+y2jZ≤0, hencexj=0, a contradiction in view ofxint∆.

Similarly, ifdi=1ai= d

i=1xi−1 it follows from (3.18) that 1 d

i=1xi=0. Therefore, from (3.24) and (3.25),λd+2+i=0 fori=1,. . .,d+ 1.

Now from these we will show thatλ2>0,. . .,λd+2>0. Indeed, ifλd+2were 0, then from (3.26) (due toxi>0,i=1,. . .,d) it follows thatλi+1=0,i=1,. . .,dand then from (3.27) λ1= −1, which contradicts (3.15). So,λd+2>0. Hence (3.26) leads to

λi+1=1 d

i=1xi

xi λd+2>0, fori=1,. . .,d, (3.28)

as stated. Then from (3.22)

ai=x 2 i−yi2Z

2xi , fori=1,. . .,d. (3.29)

We substitute these expressions in (3.23) and derive

Z=d 1

i=1

yi2/xi

+di=1yi 2

/1−d i=1xi

. (3.30)

Note thatZ >0 sincexint∆and|y| =1. Hence, from (3.21),λ1=0. Substituting (3.28) in (3.27) yields

λd+2=d 1

i=1yi 2

+1−d i=1xi

d i=1

yi2/xi

. (3.31)

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So, if the system (3.15)–(3.27) has a solution, it is unique. We have to check whether the so determineda1,. . .,ad,Z,λ1,. . .,λ2d+3really solve (3.15)–(3.27).

From the considerations until now it is clear that (3.15)–(3.18) and (3.21)–(3.27) are satisfied. It remains to consider (3.19) and (3.20).

By (3.29), inequalities (3.19) are equivalent to

xi−ai= x2

i +yi2Z

2xi 0, i=1,. . .,d, (3.32)

which hold true, becauseZ >0.

To prove that (3.20) holds, we substitute the values (3.29) forai,i=1,. . .,d, which gives

Z d

i=1 yi2 xi 2

d

i=1

xi. (3.33)

Substituting formula (3.30) forZ, the left-hand side is seen to remain below 1, while the right-hand side exceeds 1, providedxint∆. Thus also (3.20) holds true.

So, the optimal solution of the problem (3.9) isZas given in (3.30). Clearly,E(∆,x,y)=

Z, andLemma 3.1follows.

Theorem3.4. Let pn∈dn. Then for everyxint∆andySd−1,

Dypn(x)npn2C(∆)−pn2(x)

E(∆,x,y) , (3.34)

whereE(∆,x,y)is as given in (3.5).

Proof. According toLemma 3.1, the ellipse (2.1) with the parameterb=E(∆,x,y) from (3.5) belongs to∆. Hence the method of inscribed ellipses (cf.Lemma 2.1) gives (3.34).

Theorem 3.4is proved.

4. Comparisons and improvements for the standard triangle

In this section, we restrict ourselves to the cased=2. First, we see that estimate (3.34) is better than (2.18) whenK=∆.

We denote the vertices of∆byO=(0, 0),A=(1, 0),B=(0, 1), and the centroid (i.e., mass point) of∆byM=(1/3, 1/3).

A calculation shows that 1−α(∆,x)=2r(x), with

r:=r(x) :=minx1,x2, 1−x1−x2=         

x1, x∈ OMB,

x2, x∈ OMA,

1−x1−x2, x∈ AMB,

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and ify=(cosϕ, sinϕ) (0≤ϕ≤π), then

τ(∆,y)=                      1

y1+y2, ϕ∈ 0,π 2 , 1

y2, ϕ∈ π 2, 3π 4 ,

y11, ϕ∈

4 ,π

.

(4.2)

Theorem4.1. The inequality

1 E(∆,x,y)

2

τ(∆,y)1−α(∆,x) (4.3)

holds true for everyxint∆andyS1. The equality occurs if and only if (a)x(OM),y=(−√2/2,2/2);

(b)x(AM),y=(0, 1); (c)x(BM),y=(1, 0);

(d)x=M,y∈ {(0, 1), (1, 0), (−√2/2,2/2)}.

Proof. Combining (3.5) and (4.1), inequality (4.3) can be written in the equivalent form:

3

i=1

pi(x)qi(y) 2

τ2(∆,y), (4.4)

where

p1(x),p2(x),p3(x):=r(x) 1 x1, 1 x2, 1 1−x1−x2

,

q1(y),q2(y),q3(y):=y21,y22,

y1+y22 .

(4.5)

Note thatpi(x)1 for allxint∆andi=1, 2, 3. Hence,

3

i=1

pi(x)qi(y) 3

i=1

qi(y). (4.6)

Now the validity of (4.4) will follow from

3

i=1

qi(y) 2

τ2(∆,y). (4.7)

Depending onϕ, (4.7) is equivalent to y1y2≥0 ifϕ∈[0,π/2] ory1(y1+y2)≤0 ifϕ∈ (π/2, 3π/4] ory2(y1+y2)≤0 ifϕ∈(3π/4,π]. It is easily seen that these inequalities hold true. Moreover, (4.7) is an equality if and only ify∈ {(0, 1), (1, 0), (−√2/2,2/2)}.

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Ifxis an interior point for some of the trianglesOMA,OMB, andAMB, thenpi(x)<1 for alli, hence (4.6) is a strict inequality.

Ifx(OM), thenp1(x)=p2(x)=1 and p3(x)<1. So, (4.6) can be equality only if q3(y)=0, that is,y=(−√2/2,2/2). In the latter case also (4.7) is an equality. Similarly, one can prove statements (b) and (c). Finally, ifx=M, then p1=p2=p3=1 and (4.4) is an equality if and only ify∈ {(0, 1), (1, 0), (−√2/2,2/2)}.Theorem 4.1is proved.

In the next theorem we give a new estimation forD pn(x). Theorem4.2. Let pn∈ᏼ2n. Then for everyxint∆,

D pn(x)

2≤nE(x)

pn2C(∆)−p2n(x), (4.8)

where

E(x)= ! " "

#x11−x1+x21−x2+D(x)

2x1x21−x1−x2 (4.9)

with

D(x)=$x11−x1+x21−x2%24x1x21−x1−x2. (4.10)

Remark 4.3. The inequality $

x11−x1+x21−x2%24x1x21−x1−x2>$x11−x1−x21−x2%2 (4.11)

holds true forxint∆, henceD(x)>0.

Proof ofTheorem 4.2. Clearly, the inscribed ellipse method fromSection 2implies D pn(x)

2≤nE(x)

pn 2

C(∆)−p2n(x), (4.12) where

E(x) := 1

E(∆,x). (4.13)

It remains to prove the explicit representation (4.9) forE(x). From (3.5),

1 E2(∆,x,y)=

y21 x1 +

y22 x2 +

y1+y22 1−x1−x2 =

yTPy

x1x21−x1−x2, (4.14) where

P=

x21−x2 x1x2 x1x2 x11−x1

. (4.15)

SincePis a positive definite symmetric matrix, it follows that

maxyTPy:|y| =1=λ1, (4.16)

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Now, a computation gives

λ1=1 2 $

x11−x1+x21−x2+D(x)%, (4.17)

which implies (4.9).Theorem 4.2is proved.

In the next theorem we improveTheorem 2.7forK=∆.

Theorem4.4. Let pn∈ᏼ2nandpnC(∆)=1. Then for everyxint∆,

D pn(x) 2

3npn2C(∆)−p2n(x)

w(∆)1−α(∆,x) . (4.18)

Proof. According toTheorem 4.2, it suffices to prove

E(x)

3

w(∆)1−α(∆,x), xint∆. (4.19)

Sincew(∆)=√2/2, this is equivalent to

E2(x)1α(∆,x)6, forxint∆. (4.20)

We set besides (4.1) also

u=x11−x1, v=x21−x2, w=x1x2

1−x1−x2

r . (4.21)

Then (4.20) can be rewritten as

u+v+

(u+v)24rw6w. (4.22)

Since 6w(u+v)≥0, we arrive at the inequality

w$9w3(u+v) +r%0. (4.23)

To prove (4.23), we define the numberspandqby

{p,q,w}:=x11−x1−x2,x21−x1−x2,x1x2. (4.24)

Note thatw≥pandw≥q. It is seen thatu+v≤2w+p+qandrw=pq/r≥3pq. Hence,

w[9w−3(u+v) +r]≥3(w−p)(w−q)≥0.Theorem 4.4is proved.

Remark 4.5. The constant 6 in inequality (4.20) cannot be replaced by smaller one, since E2(M)(1α(∆,M))=6.

In the next theorem we present the estimation corresponding to 1−α2(∆,x). Theorem4.6. Let pn∈ᏼ2nandpnC(∆)=1. Then for everyxint∆,

D pn(x) 2

3 +5npn 2

C(∆)−pn2(x)

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Remark 4.7. This improves the constant inTheorem 2.7but falls short ofConjecture 2.8, since 22=2.8284···>3 +5=2.2882···>2.

Proof. Arguing as inTheorem 4.4, it suffices to prove that

E2(x)1α2(,x)2(3 +5), forxint. (4.26)

Note first that

1−α2(,x)=         

4x11−x1, x∈ OMB,

4x21−x2, x∈ OMA,

4x1+x21−x1−x2, x∈ AMB.

(4.27)

Letu,v,w, andrbe as inTheorem 4.4. Setc:=3 +5. Then (4.26) is equivalent to

u+v+

(u+v)24rw (1r)cw⇐⇒f(x) :=c2w2c(u+v)(1r)+4r(1r)20. (4.28)

Forβ∈[0, 1/3] we consider the line segments $

,Bβ%=(β,t) :β≤t≤12β, $

,Aβ %

=(t,β) :β≤t≤12β, $

,Bβ %

=(t, 1−β−t) :β≤t≤1.

(4.29)

We will prove that for everyβ∈[0, 1/3], f(x)0 on these segments. (1) Letx=(β,t)∈[Oβ,]. We have

u=β(1−β), v=t(1−t), r=β, w=t(1−β−t). (4.30)

Hence

f(x)=g(t) :=$2c(1−β)−c2%t2+c(c2)(1β)t+β(1β)2(42c). (4.31)

The leading coefficient ofgis negative, so it is sufficient to show thatg(β)0 andg(1 2β)0. We haveg(β)=β[c−2(1−β)][(1−β)(c−2)−cβ]. Since

c−2 c

β

1−β for everyβ∈

0,1 3

, (4.32)

it follows thatg(β)0. Next we compute

g(1−2β)=(410c)β32c28c+ 4β2+c26c+ 4β. (4.33)

Usingc2=6c4 we get

g(12β)=2β2$(25c)β+ 2c%0 for everyβ

0,1 3

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(2) Letx=(t,β)[Oβ,]. Then

u=t(1−t), v=β(1−β), r=β, w=t(1−t−β) (4.35)

and f(x)=g(t) is the same as in case (1). (3) Letx=(t, 1−β−t)∈[Aβ,]. Then

u=t(1−t), v=(t+β)(1−t−β), r=β, w=t(1−β−t),

f(x)=h(t) :=$4c(1−β)−c2%t2+c(1−β)$c−4(1−β)%t+ 2β(1−β)2(2−c). (4.36)

Since the leading coefficient is negative and, by continuity from cases (1) and (2),h(β)= f(Bβ)0 andh(1−2β)=f(Aβ)0, we conclude that f(x)0 on [Aβ,Bβ].Theorem 4.6

is proved.

Remarks 4.8. (1) The constant in (4.26) cannot be replaced by a smaller one. To prove this, we consider

(BM)= &

(t, 12t) : 0< t <1 3 '

. (4.37)

Ifx(BM), then

E2(x)=35t+

25t222t+ 5 2t(12t) , 1−α2(∆,x)=4t(1t).

(4.38)

Hence

E2(x)1−α2(∆,x)=2(1−t)

35t+25t222t+ 5

12t ,

lim

x→B,x(BM)E

2(x)1α2(,x)=2(3 +5). (4.39)

(2) limx→BE2(x)(1−α2(∆,x)) does not exist. This can be seen calculating limits on the intervals (B,Ek), whereEk=(1/k, 0)(OA) (k >1). We skip the details.

(3) Observe that Theorems4.2,4.4, and4.6show that not even the exact yield of the inscribed ellipse method can reachConjecture 2.8.

5. Lower estimations by ridge functions

We consider the following question. All known lower estimates for the Bernstein factors used some kind of ridge polynomials, that is, polynomials composed from a linear form and some (in fact, a Chebyshev) polynomial. Can one sharpen these lower estimates to the extent thatConjecture 2.8will be disproved?

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w(K,v) :=h(K,v) +h(K,−v∗). Ridge polynomialsare defined as

n:=

p∈ᏼ:p(x)=PL(x),L∈X∗,P∈ᏼ1 n

, ᏾:=

(

n=1

n. (5.1)

We denote for anyv∗∈S∗the linear expression

t(x) :=tK,v,x:=2

v,x−hK,v+hK,−v∗

wK,v . (5.2)

In the following we assume, as we may, that ridge polynomials are expressed by using someL(x)=t(K,v,x).

Definition 5.1. For anyn∈Nthe corresponding “ridge Bernstein constant” is

Cn(K,x,y) :=1

nRn,|Rsup(x)|<RC(K)

DR(x),y

R2

C(K)−R2(x)

. (5.3)

Theorem5.2. For every convex bodyKandxintK,y∈S∗, it holds that

Cn(K,x,y) 2 τ(K,y)

1

1−α2(K,x). (5.4)

Proof. By the chain rule we have for anyR∈n,R=P(t(x)), the formula DR(x),y=Pt(x) 2

wK,v

v,yτ(K,2 y)P

t(x). (5.5)

Applying the Bernstein-Szeg˝o inequality fors∈(1, 1), we get P(s)

P2

C[1,1]−P2(s) ≤√ n

1−s2. (5.6)

Note that forTn, the classical Chebyshev polynomial of degreen, (and only for that) this last inequality is sharp. Puttings:=t(x) and combining the previous two inequalities, we are led to

1 n

DR(x),y

R2

C(K)−R(x) 2

τ(K,y)

1

1−t2K,v,x. (5.7)

Taking supremum with respect tov∗∈S∗on the right-hand side, we obtain a bound in-dependent ofv. In fact, according to [7, Proposition 4.1], the supremum is a maximum and is equal to (2/τ(K,y))(1/1−α2(K,x)). Thus taking supremum also on the left-hand

side,Theorem 5.2follows.

It follows from the definitions andLemma 2.1that

Cn(K,x,y)≤Bn(K,x,y)E(K1,x,y). (5.8)

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Theorem5.3. For everyxint∆andySd−1, the following inequality holds: 1

E(∆,x,y)

dCn(∆,x,y). (5.9)

Proof. Let

vi:=(0,. . ., 0, 1, 0,. . ., 0), i=1,. . .,d, (5.10)

be theith unit vector inRdand

vd+1:=√1

d(1, 1,. . ., 1). (5.11)

We denote

TnK,v,x:=Tnt(x) (5.12)

and define

C(i)

n (∆,x,y) := 1 n

DyTn

K,vi ,x

1−T2 n

K,v∗i ,x

, i=1,. . .,d+ 1. (5.13)

A computation gives

C(i)

n (∆,x,y)=

yi

xi

1−xi

, i=1,. . .,d,

C(d+1)

n (∆,x,y)=

d

i=1yi

d

i=1xi1−di=1xi .

(5.14)

Hence, from (3.5),

1 E(∆,x,y)=

! " " " #d

i=1

1−xi )

Cn(i)(∆,x,y) *2

+ d

i=1 xi

)

C(nd+1)(∆,x,y) *2

≤d max i=1,...,d+1C

(i)

n (∆,x,y)

dCn(∆,x,y).

(5.15)

Theorem 5.3is proved.

Corollary5.4. For everyxint∆andySd−1, it holds that

1≤Bn(∆,x,y) Cn(∆,x,y)

d, 1≤Bn(∆,x) Cn(∆,x)

d. (5.16)

Note that in [4] it is proved that for everyx0intKthere is a directiony0and a ridge polynomialTn(K,v0,x) such that

1 n

Dy0Tn

K,v0,x

1−Tn

K,v0,x

2 = 2

τK,y0

1−α2K,x 0

(18)

Consequently,

CnK,x0,y0

2 τK,y0

1−α2K,x 0

. (5.18)

Hence, for everyx0intKthere is ay0such that

Bn

K,x0,y0

Cn

K,x0,y0

≤√2. (5.19)

Comparing this withCorollary 5.4, we see that (for the case of the simplex) the latter ratio remains uniformly bounded for allxandy.

Acknowledgments

This joint work was done during the postdoctoral fellowship of the first author at the Alfr´ed R´enyi Institute of Mathematics, Hungarian Academy of Sciences, supported by the “Mathematics in Information Society” project in the framework of the European Com-munity’s “Confirming the International Role of Community Research” program. The first author was supported in part by the Bulgarian Ministry of Science under Project MM802/98. The work of the second author was supported by the French-Hungarian Scientific and Technological Governmental Cooperation, Project no. F-10/04 and by the Hungarian-Spanish Scientific and Technological Governmental Cooperation, Project no. E-38/04.

References

[1] B. Bojanov,Markov-type inequalities for polynomials and splines, Approximation Theory, X: Abstract and Classical Analysis (Missouri, 2001) (L. L. Schumaker C. K. Chui and J. Stoekler, eds.), Innov. Appl. Math., Vanderbilt University Press, Tennessee, 2002, pp. 31–90. [2] S. Dineen,Complex Analysis on Infinite-Dimensional Spaces, Springer Monographs in

Mathe-matics, Springer, London, 1999.

[3] B. Gr¨unbaum,Measures of symmetry for convex sets, Proceedings of Symposia in Pure Mathe-matics, Vol. VII, American Mathematical Society, Rhode Island, 1963, pp. 233–270. [4] A. Kro ´o and Sz. R´ev´esz,On Bernstein and Markov-type inequalities for multivariate polynomials

on convex bodies, J. Approx. Theory99(1999), no. 1, 134–152.

[5] H. Minkowski,Allgemeine Lehrs¨atze ¨uber die konvexe polyeder, Nachr. Ges. Wiss. G¨ottingen (1897), 198–219 (German), Ges. Abh.2, pp. 103–121, Leipzig–Berlin, 1911.

[6] J. Radon,Uber eine Erweiterung des Begri¨ ffes der konvexen Funktionen mit einer Anwendung auf die Theorie der konvexen K¨orper, S.-B. Akad. Wiss. Wien125(1916), 241–258 (German). [7] Sz. Gy. R´ev´esz and Y. Sarantopoulos,A generalized Minkowski functional with applications in

approximation theory, preprint, 2001.

[8] R. T. Rockafellar,Convex Analysis, Princeton Mathematical Series, no. 28, Princeton University Press, New Jersey, 1970.

[9] Y. Sarantopoulos,Bounds on the derivatives of polynomials on Banach spaces, Math. Proc. Cam-bridge Philos. Soc.110(1991), no. 2, 307–312.

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[11] J. G. van der Corput and G. Schaake,Ungleichungen f¨ur Polynome und trigonometrische Poly-nome, Compositio Math.2(1935), 321–361 (German), Correction ibid.,3(1936), 128. [12] R. Webster,Convexity, Oxford Science Publications, The Clarendon Press; Oxford University

Press, New York, 1994.

Lozko B. Milev: Department of Mathematics, University of Sofia, Boulevard James Boucher 5, 1164 Sofia, Bulgaria

E-mail address:milev@fmi.uni-sofia.bg

Szil´ard Gy. R´ev´esz: A. R´enyi Institute of Mathematics, Hungarian Academy of Sciences, P.O. Box 127 1364, Budapest, Hungary

References

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