On the Best Uniform Polynomial Approximation to the Checkmark Function
OON THE BEST UNIFORM POLYNOMIALAPPROXIMATION TO THE CHECKMARK FUNCTION
P. D. DRAGNEV † , A. R. LEGG, AND R. ORIVE ∗ Abstract.
The best uniform polynomial approximation of the check-mark function f ( x ) = | x − α | is considered, as α varies in ( − , n , the minimax error E n ( α ) is shown to be piecewiseanalytic in α . In addition, E n ( α ) is shown to feature n − α and theirdynamics are completely characterized. We also prove a conjecture ofShekhtman that for odd n , E n ( α ) has a local maximum at α = 0. Introduction
Our purpose is to study the best polynomial approximation (in the uni-form norm) to the so–called checkmark function, i.e.(1.1) f ( x ) = f ( x ; α ) = | x − α | , x ∈ [ − , , α ∈ ( − , . Given a nonnegative integer n and α ∈ ( − ,
1) we denote by p n ( x ) = p n ( x ; α )the best polynomial approximation of degree at most n to the function f ( x )in (1.1). This p n ( x ; α ) is known as the minimax polynomial and satisfies thecondition(1.2) E n ( α ) := (cid:107) f − p n (cid:107) = min q ∈ P n (cid:107) f − q (cid:107) , where P n denotes the linear space of all polynomials of degree at most n and (cid:107) · (cid:107) denotes the uniform (or Chebyshev) norm in the interval [ − , f ( x ; 0) = | x | is aclassical problem in Approximation Theory (see Bernstein [1, 2] and Stahl[8], among many others), since it is one of the simplest nontrivial functions tobe approximated. In the classical treatments of the problem, the asymptoticrates of approximation when the degrees of the polynomials (or rationalfunctions) tend to infinity were considered. Here we instead consider thedegree n fixed and focus on the evolution of the minimax polynomial asthe parameter α varies through ( − , Date : February 19, 2021. † The research of this author was supported in part by NSF grant DMS-1936543. ∗ The research of this author was supported in part by Ministerio de Ciencia e Innovaci´onunder grant MTM2015-71352-P, and the PFW Scholar-in-Residence program. a r X i v : . [ m a t h . C A ] F e b P. D. DRAGNEV † , A. R. LEGG, AND R. ORIVE ∗ It is well known that the best polynomial approximation in the uniformnorm to f of degree at most n is uniquely characterized by the followingequioscillation property. There exist at least n + 2 points z , z , · · · z N in[ − ,
1] such that the minimax error E n ( α ) defined in (1.2) is attained withalternating signs, that is,(1.3) e n ( z i ; α ) := f ( z i ) − p n ( z i ; α ) = (cid:15) ( − i (cid:107) f − p n ( · ; α ) (cid:107) = (cid:15) ( − i E n ( α ) , for i = 1 , , · · · , N , where (cid:15) = ± . The points z i are called alternationpoints. We call the set of all such z i the alternation set, and denote it by A n ( α ). This characterization is the basis for the numerical algorithm tocompute the minimax polynomial, which is commonly known as the Remezalgorithm (see [6] or [5, Ch. 1]). In [4, Lemma 3] it is proven that forthe checkmark function the exact number of alternation points is either n + 2 or n + 3, with α always belonging to A n ( α ), in such a way that p n ( α ; α ) = E n ( α ) . Moreover, it is shown there that for n = 2 and n = 3,the case where the set of alternation points has n + 3 points is exceptional ,and only occurs for finitely many values of α . At these finitely many valuesof α , p n ( x ; α ) coincides with p n +1 ( x ; α ) . Another important result previously established in [4, Theorem 2] assertsthat for values of α close to +1 (or similarly to −
1, by the symmetry ofthe problem), say α ∈ ( α , E n ( α ) can be described interms of the classical Chebyshev polynomials. And for a contiguous rangeof values α ∈ [ α , α ], the graph of E n ( α ) has a V-shape. This means thatfor some α ∈ ( α , α ), E n ( α ) is decreasing and linear on [ α , α ], increasingand linear on [ α , α ] , and continuous on [ α , α ] . The proof is based on thefact that for such range of values of α , the domain endpoint x = +1 is asoft–endpoint, that is, the minimax polynomial for f in [ − ,
1] is also theminimax polynomial for intervals of the form [ − , b ], for b ≥ b , with b < n + 3 points for some α then this α is a tip of a V-shape for E n ( α ). In fact, on p. 151 of [4] it is conjecturedthat the number of such V-shapes is exactly n −
1. Our Theorem 4.7 belowestablishes this conjecture.The simplest nontrivial cases are those of n = 2 and n = 3, which werestudied in [4]. It is illustrative to review these. It is easy to plot E ( α ) and E ( α ) numerically as functions of α , yielding Figure 1. Observe that thegraph of E ( α ) has a V-shape in a symmetric interval about α = 0, wherethe “tip” of the V-shape is located. Outside the V-shape, E ( α ) can bedetermined using Chebyshev polynomials as mentioned above. The graphof E ( α ) is slightly more involved, showing two symmetric V-shapes, andnow α = 0 is a local and in fact absolute maximum. At α = 0, note thatthe two graphs intersect, reflecting the fact that p ( x ; 0) = p ( x ; 0). HECKMARK FUNCTION 3
Figure 1.
The graphs of E ( α ) and E ( α ).In Figure 2, we numerically plot E n ( α ) for n = 1 , , · · · , . As n increases,it appears that more and more V-shapes appear, at whose tips the graphsof E n ( α ) and E n +1 ( α ) coincide. Figure 2.
The graphs of E n ( α ), for n = 1 , , . . . , E n ( α ) as α varies in ( − , n . In the next section, the monotonicity ofthe alternation points as functions of the parameter α is established. InSection 3, we show piecewise analyticity of the minimax error E n ( α ) andprove a conjecture of Shekhtman that E n ( α ) has a local maximum at α = 0when n is odd. In Section 4, the conjecture about the exact number ofV-shapes in the graph of E n ( α ), posed in [4], is resolved, which also allowsus to describe the phase transitions of the alternation points. These phasetransitions are illustrated by means of suitable numerical examples. P. D. DRAGNEV † , A. R. LEGG, AND R. ORIVE ∗ Monotonicity of the alternation points
We start our analysis by discovering monotonicity of the alternationpoints with respect to the parameter α , where α is not a tip of a V-shape.First note the following fact about continuity. Lemma 2.1.
Given a nonnegative integer n , The minimax polynomial p n ( x ; α ) and the minimax error E n ( α ) vary continuously in the parameter α ∈ [ − , . The proof follows from the continuity of the best approximation operatorfor subspaces of finite dimension (see e.g. [3, Th. 1.2, p. 60]).
Lemma 2.2.
Suppose f ( x ) is a nonconstant continuous function on [ a, b ] and differentiable on ( a, b ) and that f ( a ) ≥ ≥ f ( b ) ( resp. f ( a ) ≤ ≤ f ( b )) .Then there is a point z ∈ ( a, b ) such that f (cid:48) ( z ) < resp. f (cid:48) ( z ) > .Proof. This is a consequence of the Mean Value Theorem. (cid:3)
In what follows, it will be useful to have a notation for a normalized errorfunction. For each α ∈ ( − , g α ( x ) := p n ( x ; α ) − | x − α | E n ( α ) . Thus, g α is normalized in such a way that | g α ( x ) | ≤ , x ∈ [ − , . Theorem 2.3.
On any interval of α values not including a tip of a V-shape,the elements of A n ( α ) are all non-decreasing in α .Proof. For α in a linear part of a V-shape the monotonicity follows from alinear transformation (see [4, Remark on p. 154]).For α not in a V-shape we know from [4] that A n ( α ) contains exactly n + 2 alternation points, and {− , α, } ⊂ A n ( α ). So, we write A n ( α ) := {− < u k ( α ) < . . . < u ( α ) < α < v ( α ) < . . . < v l ( α ) < } , where k + l = n −
1. Hereafter, denote by u i the alternation points locatedto the left of α , and by v j , those to the right. The u i and v j are enumeratedby their distance to α . It is convenient to say u k +1 ( α ) := − v l +1 ( α ) := 1,and u ( α ) := α =: v ( α ).Now, let us take β > α sufficiently close to α such that A n ( β ) = {− < u k ( β ) < . . . < u ( β ) < β < v ( β ) < . . . < v l ( β ) < } , and consider the function(2.2) g ( x ) = g ( x ; α, β ) := g α ( x ) − g β ( x ) , x ∈ R , with g α and g β given by (2.1). As an illustration, Figure 3 depicts the case n = 6 and the graphs of g α , g β , and g for α = 0 . β = 0 . g is a continuous and piecewise polynomial function dif-ferentiable on R \ { α, β } , whose second derivative g (cid:48)(cid:48) ( x ) is a polynomial ofdegree at most n − α and β . By continuity,for β close to α we may assume that u i ( β ) is close to u i ( α ) and v j ( β ) is close HECKMARK FUNCTION 5
Figure 3.
Graphs of functions g α (dash-dot), g β (dash) and g (solid) for n = 6 , α = 0 .
4, and β = 0 . v j ( α ), so that sign e n ( u i ( α ); α ) = sign e n ( u i ( β ); β ) , i = 1 , . . . , k , as wellas sign e n ( v j ( α ); α ) = sign e n ( v j ( β ); β ) , j = 1 , . . . , l , where e n ( · ; · ) is givenby (1.3). Observe that g ( −
1) = g (1) = 0.We will account for all the zeroes of g (cid:48)(cid:48) ( x ) and show that they are sim-ple and located in ( − , g ( u ( α )) > g ( u ( α )) ≤
0, soby Lemma 2.2 we have a point z ∈ ( u ( α ) , u ( α )) at which g (cid:48) ( z ) > g ( u ( α )) ≥ g ( u ( α )) ≤
0, there is a point z ∈ ( u ( α ) , u ( α )) at which g (cid:48) ( z ) <
0. Therefore, applying the Mean ValueTheorem there is y ∈ ( z , z ) ⊂ ( u ( α ) , u ( α )) such that g (cid:48)(cid:48) ( y ) = g (cid:48) ( z ) − g (cid:48) ( z ) z − z > . Continuing the argument in a similar fashion we derive the existence ofpoints z i ∈ ( u i +1 ( α ) , u i ( α )), i = 0 , . . . , k , such that sign g (cid:48) ( z i ) = ( − i ,which yields the existence of points y i ∈ ( u i +1 ( α ) , u i − ( α )), i = 1 , . . . , k ,such that sign g (cid:48)(cid:48) ( y i ) = ( − i − . Observe that y i and y i +2 belong to distinctintervals and g (cid:48)(cid:48) ( y i ) and g (cid:48)(cid:48) ( y i +1 ) have opposite signs, hence all of the y i ’sare distinct. Therefore, by the Intermediate Value Theorem there are atleast k − g (cid:48)(cid:48) ( x ) in the interval ( y k , y ). We note that if l = 0, wehave accounted for all ( n −
2) zeros of g (cid:48)(cid:48) ( x ).Analogously, we derive the existence of ζ j ∈ ( v j ( β ) , v j +1 ( β )), j = 0 , . . . , l ,such that sign g (cid:48) ( ζ j ) = ( − j and points η j ∈ ( v j − ( β ) , v j +1 ( β )), j =1 , . . . , l , such that sign g (cid:48)(cid:48) ( η j ) = ( − j . This adds another l − g (cid:48)(cid:48) ( x ) in the interval ( η , η l ). Finally, since g (cid:48)(cid:48) ( y ) > > g (cid:48)(cid:48) ( η ) we accountfor one more zero of g (cid:48)(cid:48) ( x ) in ( y , η ).In summary, all the zeroes of g (cid:48)(cid:48) ( x ) are accounted for and belong to theinterval ( y k , η l ) ⊂ ( − , g (cid:48)(cid:48) ( x ) is a P. D. DRAGNEV † , A. R. LEGG, AND R. ORIVE ∗ polynomial of exact degree n −
2. Take into account also that g (cid:48) ( x ) doesnot change sign on ( −∞ , z k ) and on ( ζ l , ∞ ). Otherwise, as above we wouldbe able to derive an extra zero of g (cid:48)(cid:48) ( x ), which will lead to a contradiction.(This fact will be used again in the proofs of Proposition 2.5 and Theorem3.2.)From here we can prove the (non-strict) monotonicity of the alterna-tion points. By symmetry it suffices to consider only the u i , i = 1 , . . . , k .First, let us establish that u k ( α ) ≤ u k ( β ). If this were not the case, as sign [ g ( u k ( β )) g ( u k ( α )) ] < z k +1 ∈ ( − , u k ( β )) suchthat g (cid:48) ( z k +1 ) g (cid:48) ( z k ) < z k ∈ ( u k ( β ) , u k ( α ))). Hence, we add onemore zero of g (cid:48)(cid:48) ( x ), which is a contradiction.If monotonicity of any of u i ( α ) , i = 1 , . . . , k − i such that u i +1 ( α ) ≤ u i +1 ( β ) < u i ( β ) < u i ( α ). But then g (cid:48) willchange sign on the interval ( u i +1 ( α ) , u i ( α )), which adds an extra zero to g (cid:48)(cid:48) ( x ), which leads again to a contradiction. This completes the proof of thetheorem. (cid:3) Remark . We reiterate for future use the fact arising from the proof that g (cid:48) cannot change sign on either of the the intervals ( −∞ , z k ) , ( ζ l , ∞ ).Even more, we can derive existence and monotonicity of an additionalextremum of the function g α which lies outside the interval [ − , Proposition 2.5.
Suppose that {± } ⊂ A n ( α ) and α is neither the tip ofa V-shape for E n − ( α ) nor part of a V-shape of E n ( α ) . Then, there exists w = w ( α ) ∈ R \ [ − , such that g (cid:48) α ( w ) = 0 . Moreover, in the intervalsof values of α where such w exists, it is also monotonically increasing withrespect to α .Proof. Since α is not a tip of a V-shape for E n − ( α ) or E n ( α ), we know p n ( x ; α ) is of degree n and A n ( α ) has n + 2 points. We also know that g α ( x )alternates monotonicity on the intervals ( u i +1 ( α ) , u i ( α )), i = 0 , . . . , k , and( v i ( α ) , v i +1 ( α )), i = 0 , . . . , l , which implies g (cid:48) α ( x ) changes sign at least k times on ( − , α ) and at least l times on ( α,
1) (in particular, sign g (cid:48) α ( x ) =( − k on ( − , u k ( α ))). Consequently g (cid:48)(cid:48) α ( x ) has at least k − − , α )and at least l − α, {± } ⊂ A n ( α ), we have that g α ( −
1) =( − n +1 g α (+1). Thus, there must be an additional extremum w = w ( α )outside [ − , g α behaves as a polynomial of degree n (even orodd) at ±∞ .To derive the monotonicity of w ( α ), let β > α be as in the proof of Theo-rem 2.3 with the additional assumption that w ( α ) and w ( β ) are close to eachother and outside [ − , g (cid:48) ( x ) preserves signon ( −∞ , z k ) and on ( ζ l , ∞ ). This implies that g ( x ) preserves monotonicityon these subintervals.We first consider the case w ( α ) , w ( β ) ∈ ( −∞ , −
1) (see Figure 4 for illus-tration). Observe that g α ( x ) and g β ( x ) share alternating monotonicity onthe intervals ( u i +1 ( β ) , u i ( α )), i = 0 , . . . , k . Recall also that sign g (cid:48) ( z i ) = HECKMARK FUNCTION 7
Figure 4. { g α , g β , g ( x ) } for n = 6 , α = 0 .
4, and β = 0 . − i , i = 0 , . . . , k , z i ∈ ( u i +1 ( α ) , u i ( α )). As α is not part of a V-shape for E n ( α ), we must have g (cid:48) α ( − (cid:54) = 0. We may assume β > α is close enoughthat g (cid:48) β ( − (cid:54) = 0 as well. Then the functions g, g α , and g β have the samemonotonicity in neighborhoods of − sign g (cid:48) ( z k ) = sign g (cid:48) α ( z k ) =( − k ).Without loss of generality we assume that the function g ( x ) is decreasingon ( −∞ , z k ), or g (cid:48) ( x ) < g α ( x ) is decreasingon ( w ( α ) , −
1) and increasing on ( −∞ , w ( α )). Similarly, we have that g β ( x )is decreasing on ( w ( β ) , −
1) and increasing on ( −∞ , w ( β )). If we assumethat w ( β ) < w ( α ), as g (cid:48) α ( w ( α )) = 0 one derives g (cid:48) α ( w ( α )) − g (cid:48) β ( w ( α )) >
0, acontradiction with g ( x ) being decreasing on ( −∞ , − w ( α ) , w ( β ) ∈ (1 , ∞ ) follows by considering the approximationof the checkmark functions f ( x ; − α ) and f ( x ; − β ) and the symmetry of theminimax problem (see [4]).This proves the proposition. (cid:3) Analyticity of E n ( α ) and Shekhtman’s Conjecture. Throughout this section we will show that the minimax error E n ( α ) is areal analytic function of the parameter α ∈ ( − , p n ( x ; α ), as well as the alternation points { u j } and { v k } willbe established. At the end of the section we prove Shekhtman’s Conjecturethat for odd n , E n ( α ) has a local maximum at α = 0. Theorem 3.1.
The error function E n ( α ) is real analytic on ( − , , withthe exception of the endpoints and the tips of V -shapes. The same holds forthe alternation points and the coefficients of the minimax polynomial. P. D. DRAGNEV † , A. R. LEGG, AND R. ORIVE ∗ Proof.
Suppose n ≥ E ( α ) is quadratic). Con-sider α in an open interval excluding the tips and endpoints on the V -shapesof E n ( α ). If the interval is within a V -shape, then E n ( α ) is linear and theminimax polynomial can be found by linear transformation from the min-imax polynomial of one of the endpoints, which yields the theorem in thiscase.For the rest of the proof we assume that α does not belong to a V -shape.Then {− , α, } ⊂ A n ( α ), and the number of alternation points to the leftof α is the same for all α in the interval (so too for the right of α ). Weagain call the alternation points − < u k < u k − < · · · < u < α 1. It is understood that the u j and v i depend on α , and that k + l = n − 1. Consider now for each fixed α the function p n ( x ; α ) − | x − α | . It is smooth away from x = α and has local extrema ateach x = u j and x = v i . Thus the derivative is 0 at each u j and v i . Let usdenote p n ( x ; α ) = c n x n + · · · + c x + c , where the coefficients c j depend on α. We now phrase the solution to the minimax problem in terms of a systemof 2 n +1 equations in 2 n +1 unknowns, with α as an independent parameter.We have that p n ( x ; α ), E n ( α ), and the u j , v i solve the following equations: c n ( − n + c n − ( − n − + · · · + c − α − − k +2 E n = 0 c n u nk + c n − u n − k + · · · + c − α + u k + ( − k +1 E n = 0... ... ... c n u n + c n − u n − + · · · + c − α + u + ( − E n = 0 c n α n + c n − α n − + · · · + c + ( − E n = 0 c n v n + c n − v n − + · · · + c + α − v + ( − E n = 0... ... ... c n v nl + c n − v n − l + · · · + c + α − v l + ( − l +1 E n = 0 c n + c n − + · · · + c + α − − l +2 E n = 0 nc n u n − k + ( n − c n − u n − k + · · · + 0 + 0 + 1 = 0... ... ... nc n u n − + ( n − c n − u n − + · · · + 0 + 0 + 1 = 0 nc n v n − + ( n − c n − v n − + · · · + 0 + 0 − nc n v n − l + ( n − c n − v n − l + · · · + 0 + 0 − HECKMARK FUNCTION 9 Note that the first n + 2 equations describe the equioscillation at thealternation points, and the last n − p n ( x ; α ) − | x − α | has vanishing derivative at the u j and v i . Thus, we do have a total of n + 2 + ( n − 1) = 2 n + 1 equations, and the unknowns are the c j , the u j ,the v i and E n , amounting to ( n + 1) + ( n − 1) + 1 = 2 n + 1 unknowns, with α as a parameter.To determine the analyticity of E n ( α ) as a function of α , we appeal tothe implicit function theorem. We need to differentiate each of the aboveequations with respect to each of the unknowns and gather these partialderivatives in a matrix, where the first n + 1 columns contain the derivativeswith respect to the c j ; column n + 2 will contain the derivatives with respectto E n ; and the next n − u k , u k − , · · · , u , v , v , · · · , v l . The rows of the matrix will keep thesame order as the equations above, so for instance the first row will containall the partial derivatives from the first equation in the order just specified.So denote by F = ( F , . . . , F n +1 ) the vector function F : R n +2 → R n +1 and y = ( x , α ) = (cid:0) { c i } ni =0 , E n , { u i } ki =1 , { v i } li =1 , α (cid:1) ∈ R n +2 in such a waythat the above system has the vector form(3.1) F ( y ) = , ∈ R n +1 . To prove that (3.1) implies the existence of an implicit function G : R → R n +1 , G ( α ) = x , we require that the Jacobian matrix(3.2) J = (cid:20) ∂ F ∂ x (cid:21) is nonsingular. In such case, we would have that(3.3) (cid:20) d G dα (cid:21) (2 n +1) × = − J − n +1) × (2 n +1) · (cid:20) ∂ F ∂α (cid:21) (2 n +1) × The Jacobian matrix J given in (3.2) may be written in block form: J = (cid:20) A ( n +2) × ( n +2) B ( n +2) × ( n − C ( n − × ( n +2) D ( n − × ( n − (cid:21) The matrix B contains the derivatives of the equations for the alternationpoints with respect to the u j , v i . Since each such equation features at mostone of the u j or v i , and by the ordering chosen for listing the equations andtaking the derivatives, we see that B is a diagonal matrix. The first andlast rows of B correspond to equations that feature none of the u j or v i ,so these are 0-rows. For the second row of B , the main diagonal entry is p (cid:48) n ( u k ; α ) + 1, but by equation n + 2 in the system, this is 0. Similarly, allmain diagonal entries of B are zero. Thus, B is the null matrix.Next consider D. This matrix contains derivatives of the last n − u j , v i . Here again, each equation features just one of the u j or v i , so all off-diagonal entries of D are 0. The first entry of row 1 is p (cid:48)(cid:48) n ( u k ; α ), † , A. R. LEGG, AND R. ORIVE ∗ and it is simple to determine that actually D is a diagonal matrix comprisedof the second derivatives of p n ( x ; α ) evaluated at the various u j , v i . At thispoint, we conclude that(3.4) det J = det A · k (cid:89) j =1 p (cid:48)(cid:48) n ( u j ; α ) · l (cid:89) i =1 p (cid:48)(cid:48) n ( v i ; α ) . Since the local extrema at { w ( α ) , u k , · · · , u , v , · · · , v l } alternate n − p n ( x ; α ) − | x − α | , the zeros of p (cid:48)(cid:48) n ( x ; α ) areaway from the alternation points by Rolle’s Theorem. Hence, the productof second derivatives in (3.4) is nonzero.Our final task is to show that det A (cid:54) = 0, where A = ( − n ( − n − · · · − − k +2 u nk u n − k · · · u k − k +1 ... ... . . . ... ... ... u n u n − · · · u − α n α n − · · · α − v n v n − · · · v − ... ... . . . ... ... ... v nl v n − l · · · v l − l +1 · · · − l +2 . By permuting columns, let us denoteˆ A := ( − k +2 − · · · ( − n − ( − n ( − k +1 u k · · · u n − k u nk ... ... ... . . . ... ...( − u · · · u n − u n ( − α · · · α n − α n ( − v · · · v n − v n ... ... ... . . . ... ...( − l +1 v l · · · v n − l v nl ( − l +2 · · · , and we derivedet A = ( − k · ε · k +1 (cid:88) j =1 V ( u j ) + V ( α ) + l +1 (cid:88) j =1 V ( v j ) , where ε is the sign ( ± 1) of the permutation of columns taking A to ˆ A , andwhere V ( u j ) , j = 1 , . . . , k + 1 , V ( α ) and V ( v j ) , j = 1 , . . . , l + 1 , denote theVandermonde determinants obtained as minors by deleting the first columnand the row corresponding to u j , α or v j respectively in ˆ A. Since each ofthese V ( · ) is strictly positive, det A is non-zero.Thus J is nonsingular and the theorem is proved. (cid:3) HECKMARK FUNCTION 11 The Implicit Function Theorem also allows us to find an expression for E (cid:48) n ( α ) in terms of the alternation points. From (3.3) we have(3.5) E (cid:48) n ( α ) = − R n +2 (cid:20) ∂ F ∂α (cid:21) , where R n +2 stands for the ( n + 2)–row of matrix J − and (cid:20) ∂ F ∂α (cid:21) T = (cid:2) − · · · − p (cid:48) n ( α ; α ) 1 · · · · · · (cid:3) , where − 1, 1 and 0 are repeated k + 1, l + 1 and n − E (cid:48) n ( α ) = det A n +2 / det A, where A n +2 = ( − n ( − n − · · · − u nk u n − k · · · u k u n u n − · · · u α n α n − · · · α − p (cid:48) n ( α ; α ) v n v n − l · · · v l − v nl v n − l · · · v l − 11 1 · · · − But, similarly as we did above, we can permute the columns of A n +2 toobtain ˆ A n +2 := − · · · ( − n − ( − n u k · · · u n − k u nk ... ... ... . . . ... ...1 1 u · · · u n − u n − p (cid:48) n ( α ; α ) 1 α · · · α n − α n − v · · · v n − v n ... ... ... . . . ... ... − v l · · · v n − l v nl − · · · Since this uses the same permutation that took A to ˆ A , we can just as wellrepresent E (cid:48) n ( α ) as E (cid:48) n ( α ) = det ˆ A n +2 det ˆ A . But notice now that by taking an appropriate linear combination ofcolumns 2 through n + 1 in ˆ A n +2 , we can produce values of the deriva-tive of the minimax polynomial. Adding this linear combination of columnsto the first column will not change the determinant of the matrix, but thenew first column will be (cid:2) ( p (cid:48) n ( − α ) + 1) 0 0 · · · p (cid:48) n (1; α ) − (cid:3) T . † , A. R. LEGG, AND R. ORIVE ∗ Then by expansion on this column,det ˆ A n +2 = ( p (cid:48) n ( − α ) + 1) · V ( − 1) + ( − n +3 ( p (cid:48) n (1; α ) − · V (1)So at this point we have the following formula for the derivative of theerror function:(3.6) E (cid:48) n ( α ) = ( p (cid:48) n ( − α ) + 1) · V ( − 1) + ( − n +1 ( p (cid:48) n (1; α ) − · V (1)( − k (cid:16)(cid:80) k +1 j =1 V ( u j ) + V ( α ) + (cid:80) l +1 j =1 V ( v j ) (cid:17) Observe that V (1) and V ( − 1) share as a common factor the productof mutual distances among the u j , α, v j , which is itself the Vandermondedeterminant µ ( α ) := det u k · · · u n − k ... ... . . . ...1 u · · · u n − α · · · α n − v · · · v n − ... ... . . . ...1 v l · · · v n − l . The remaining factors of V (1) and V ( − 1) are the products of the distancesof each alternation point to either − δ − ( α ) := ( α + 1) · k (cid:89) j =1 ( u j + 1) · l (cid:89) j =1 ( v j + 1) ,δ ( α ) := (1 − α ) · k (cid:89) j =1 (1 − u j ) · l (cid:89) j =1 (1 − v j ) . In summary, we obtain the two identities V ( − 1) = µ ( α ) · δ ( α ) , V (1) = µ ( α ) · δ − ( α ) . For convenience, we introduce one last symbol. Let ∆ denote the quantity∆( α ) = µ ( α ) (cid:80) k +1 j =1 V ( u j ) + V ( α ) + (cid:80) l +1 j =1 V ( v j ) . In order to prove Shekhtman’s conjecture, it will be advantageous todivide (3.6) through by E n ( α ), and then observe that we have the followingexpression for the logarithmic derivative of the error function:(3.7) E (cid:48) n ( α ) E n ( α ) = ( − k · ∆( α ) · (cid:2) g (cid:48) α ( − δ ( α ) + ( − n +3 g (cid:48) α (1) δ − ( α ) (cid:3) Here g α is as in (2.1). With this formula we are ready to prove the conjecture. Theorem 3.2. For each odd n , the function E n ( α ) has a local maximum at α = 0 . HECKMARK FUNCTION 13 Proof. For n odd, (3.7) takes the form E (cid:48) n ( α ) E n ( α ) = ( − k · ∆( α ) · (cid:2) g (cid:48) α ( − δ ( α ) + g (cid:48) α (1) δ − ( α ) (cid:3) . As Lemma 4.3 below will indicate, since α = 0 is a tip of a V-shape for E n − ( α ) (cf. [4]), it follows that α = 0 is not within a V-shape of E n ( α ).If α = 0 were the endpoint of a V-shape for E n ( α ), then by symmetry itwould be a tip of a V-shape, which was just excluded. So by Theorem 3.1, E n ( α ) is analytic at α = 0 and the use of (3.7) is justified.We note that ∆( α ) is always positive, and by symmetry E (cid:48) n (0) = 0 ,g (cid:48) ( − 1) = − g (cid:48) (1) (cid:54) = 0, and δ (0) = δ − (0) . Furthermore, by symmetry at α = 0 we have k = l , and this persists for α sufficiently close to 0.Now we use the first derivative test. We have just mentioned that E (cid:48) n (0) =0, so if we can establish that E (cid:48) n ( α ) < α , we will befinished. The symmetry of the problem will ensure E (cid:48) n ( α ) > α < 0, and then the critical point α = 0 will necessarily be a local maximumof E n ( α ). So we only need to establish that for small α > 0, we have( − k (cid:2) g (cid:48) α ( − δ ( α ) + g (cid:48) α (1) δ − ( α ) (cid:3) < . To demonstrate this, for a fixed α > g ( x ) = g ( x ) − g α ( x ) as in (2.2). In the proof of Theorem 2.3, we demon-strated the existence of points z i ∈ ( u i +1 (0) , u i (0)), i = 0 , , · · · , k where sign g (cid:48) ( z i ) = ( − i , and points ζ i ∈ ( v i ( α ) , v i +1 ( α )) , i = 0 , , · · · , l where sign g (cid:48) ( ζ i ) = ( − i . These in turn led to the existence of points y i ∈ ( z i +1 , z i )and η i ∈ ( ζ i , ζ i +1 ) where sign g (cid:48)(cid:48) ( y i ) = ( − i − and sign g (cid:48)(cid:48) ( η i ) = ( − i . Wesaw also that all the roots of g (cid:48)(cid:48) ( x ) are in the interval ( y k , η l ) . We have that sign g (cid:48) ( − 1) = sign g (cid:48) ( z k ) = ( − k (see Remark 2.4).In other words sign ( − k ( g (cid:48) ( − − g (cid:48) α ( − , or ( − k g (cid:48) ( − > ( − k g (cid:48) α ( − . And since k = l , we similarly get (again see Remark 2.4)( − k g (cid:48) (1) > ( − k g (cid:48) α (1) . By continuity sign g (cid:48) ( − 1) = sign g (cid:48) α ( − 1) = ( − k and sign g (cid:48) (1) = sign g (cid:48) α (1) = ( − k +1 . At this point we have | g (cid:48) α ( − | = ( − k g (cid:48) α ( − < ( − k g (cid:48) ( − 1) = | g (cid:48) ( − | as well as −| g (cid:48) α (1) | = ( − k g (cid:48) α (1) < ( − k g (cid:48) (1) = −| g (cid:48) (1) | . † , A. R. LEGG, AND R. ORIVE ∗ We have also seen by symmetry that | g (cid:48) ( − | = | g (cid:48) (1) | , and the monotonic-ity of alternation points in Theorem 2.3 makes it clear that δ ( α ) < δ (0) = δ − (0) < δ − ( α ) . So finally, combining this all together we have( − k (cid:2) g (cid:48) α ( − δ ( α ) + g (cid:48) α (1) δ − ( α ) (cid:3) < | g ( − | · δ ( α ) − | g (1) | · δ − ( α ) , and the right hand side of this inequality is clearly negative. (cid:3) Having established Shekhtman’s conjecture, we formulate the followingcomplementary one. Conjecture. For all natural numbers n , the function E n ( α ) is concaveoutside of its V-shapes. For odd n , E n ( α ) has an absolute maximum at α = 0. 4. The V-shapes and the Phase diagram We now turn to proving the conjecture of [4] that n − E n ( α ). This will follow from a string of lemmasbased on studying the dynamics of the external extremum w ( α ) of g α . Lemma 4.3. A tip of a V-shape of E n − ( α ) cannot be within a V-shape of E n ( α ) .Proof. At the tip of a V-shape of E n − ( α ), ( p n − ( x ; α ) − | x − α | ) will have( n − − n − − , . This is because incase only n − E n ( α ), there mustbe n turning points for ( p n ( x ; α ) − | x − α | ) in the interval [ − , (cid:3) Lemma 4.4. Let α be a tip of a V-shape from E n − ( α ) . Let w ( α ) bethe extremum of g α ( x ) = p n ( x,α ) −| x − α | E n ( α ) which lies outside [ − , , when suchexists. Then lim α → α +0 w ( α ) = −∞ and lim α → α − w ( α ) = + ∞ .Proof. Since from above α is not within a V-shape of E n ( α ), we know that w ( α ) exists for all α in a punctured neighborhood of α , and that it ismonotone increasing in α on either side of α .Assume now that w ( α ) > 1, and let q α ( x ) = p (cid:48) n ( x ; α ) − E n ( α ) . For x > α we have g (cid:48) α ( x ) = q α ( x ) , and in particular q α ( w ( α )) = 0.Since the coefficients of the minimax polynomial along with the minimaxerror are analytic in α in a neighborhood of α , we know that for someanalytic functions b j ( α ), j = 0 , , · · · , n − 1, we have for all x :(4.8) q α ( x ) = n − (cid:88) j =0 b j ( α ) x j . HECKMARK FUNCTION 15 Since g (cid:48)(cid:48) α has no roots outside [ − , w ( α ) is analytic in α when it exists.As w ( α ) is monotone, we also know that ω = lim α → α − w ( α ) exists. As-sume for contradiction that ω were finite. In that case, we should have: q α ( ω ) = n − (cid:88) j =1 b j ( α ) ω j = lim α → α − n − (cid:88) j =1 b j ( α ) w ( α ) j = lim α → α − q α ( w ( α )) = lim α → α − . Hence g α has a critical point outside [ − , α → α − w ( α ) = + ∞ . The remainingcases are handled similarly. (cid:3) Lemma 4.5. Given a V-shape of E n ( α ) with left endpoint γ and right end-point β , we have lim α → β + w ( α ) = 1 , and lim α → γ − w ( α ) = − . Proof. Note that g β ( x ) must have a local extremum at x = 1 . For if | g β (1) | < 1, then by either shrinking or expanding the domain slightly by a lineartransformation we could keep the needed number of alternation points withinthe interval [ − , . Then β would be an interior point of a linear part of aV-shape, instead of an endpoint.Thus, without loss of generality, assume g β (1) = 1. Observe that thereis (cid:15) > α ∈ ( β − (cid:15), β ) we have g (cid:48) α ( x ) = q α ( x ) < x ∈ [1 , ( β + 1 + (cid:15) ) / ( β + 1 − (cid:15) )]. Continuity of the coefficients b j ( α ) in (4.8)implies g (cid:48) β ( x ) = q β ( x ) ≤ g (cid:48) β (1) = 0 and x = 1 is a local maximum for g β ( x ).Now since w ( α ) is monotone in the interval ( β, β + (cid:15) ) for any small (cid:15) , wehave that lim α → β + w ( α ) exists. By continuity of the minimax polynomialcoefficients, and since g α has only one external extremum for such α , we mustconclude that lim α → β + w ( α ) = 1, to match the location of the extremum of g β at x = 1.The other limit in the statement of the Lemma may be analyzed similarly. (cid:3) Next we derive an interlacing property of the V-shapes of E n − ( α ) and E n ( α ). Lemma 4.6. The following hold: • Between any two consecutive tips of V-shapes of E n ( α ) , there is atip of a V-shape of E n − ( α ) . † , A. R. LEGG, AND R. ORIVE ∗ • Between any two consecutive tips of V-shapes of E n − ( α ) , there is atip of a V-shape of E n ( α ) . • Between the final V-shape of E n − ( α ) and α = 1 , there is a V-shapeof E n ( α ) . • Between α = − and the first V-shape of E n − ( α ) , there is a V-shapeof E n ( α ) . Proof. The proof hinges on tracking the progress of w ( α ), the external ex-tremum of g α (using degree n ), as α increases.For the first statement, let α and α be consecutive tips of V-shapes of E n ( α ). As we let α grow from α , it will eventually exit the V-shape of α ,and then we will observe w ( α ) growing monotonically from a value of 1. Butas α approaches α , it will enter the V-shape of α from the left side, andso we observe w ( α ) growing monotonically to a value of − 1. Hence, theremust have been a discontinuity of w ( α ) somewhere in between. But w ( α )is analytic wherever it exists, and it exists everywhere outside the V-shapesof E n ( α ), except at tips of V-shapes of E n − ( α ) . So, we must conclude thatthere occurred a tip of a V-shape of E n − ( α ), where w ( α ) jumped from + ∞ to −∞ .For the second statement, let α and α now be two consecutive tips ofV-shapes of E n − ( α ) . As α grows from α to α , we see from Lemma 4.5that w ( α ) starts from −∞ at α , and tends to + ∞ at α . Somewhere inbetween, we see by continuity that w ( α ) = − 1, and we enter a V-shape of E n ( α ).The third statement follows directly from [4, Theorem 2] where the loca-tion of the final V-shape of E n ( α ) is determined explicitly and the fourthstatement follows by the even symmetry of the problem. (cid:3) Now we may finally establish the number of V-shapes of E n . Theorem 4.7. For each n ∈ N , the number of V-shapes in the graph of E n ( α ) is exactly n − .Proof. For each j ∈ N , define N ( j ) to be the number of V-shapes in thegraph of E j ( α ). The case N (1) = 0 is immediate since E ( α ) is quadratic,and the cases N (2) = 1 and N (3) = 2 are established in [4].Now consider an integer n ≥ 4. For induction, assume that N ( n − 1) = n − , and let the corresponding tips of V-shapes of E n − ( α ) be called α < α < · · · < α n − . By Lemma 4.6 there is at least one V-shape of E n ( α )in each of the intervals ( − , α ) , ( α , α ) , · · · , ( α n − , N ( n ) ≥ n − E n ( α ) must occur at least oneof the α k . By the pigeon-hole principle N ( n ) ≤ n − 1, which finishes theinduction. (cid:3) To finish, we use Theorem 2.3 and Proposition 2.5 to completely describethe phase transitions of the alternation points as α decreases from +1 to HECKMARK FUNCTION 17 − 1. For each of the V-shapes whose existence is guaranteed by Theorem4.7, we denote by β i,n the right endpoint, by α i,n the tip and by γ i,n the leftendpoint, i = 1 , . . . , n − E n ( α ).Throughout a V-shape, an external extremum w ( α ) will cross x = 1, andreplace 1 as the rightmost alternation point. Then at the tip of the V-shape,1 will again become an alternation point, making for n +3 alternation points.Then − − w ( α ) emerges from x = − 1. After this process, the left most u j will haveexited the interval, and we will have gained one more v j .Here is how this looks for the particular case of n = 5 (Phase 1) α is close to +1. As α decreases, w ( α ) (cid:38) 1. The alternationset is given by A ( α ) = {− < u < . . . < u < α < } . Figure 5. The graph of g α ( x ), when α = . (1st Transition) α = β , ≈ . . At this point, w ( α ) = 1 and α = β , is the right endpoint of a V–shape. (Phase 2) α ∈ ( α , , β , ). We pass through the right linear piece of theV–shape. The alternation set has cardinality 7 and is given by A ( α ) = {− < u < . . . < u < α < v } , for some α < v < (2nd Transition) α = α , ≈ . A ( α , ) and the graph of E ( α ) has a tip of a V-shape at α = α , . Thealternation set A ( α , ) has cardinality 8 A ( α , ) = {− < u < . . . < u < α < v < } and hence E ( α , ) = E ( α , ). † , A. R. LEGG, AND R. ORIVE ∗ Figure 6. The graph of E ( . (Phase 3) α ∈ ( γ , , α , ). We are now moving through the left linearpiece of the V–shape. The alternation set is given by A ( α ) = { u < . . . < u < α < v < } . (3rd Transition) α = γ , ≈ . − A ( γ , ). We have reached the left endpoint of the V–shape and w ( α ) = − (Phase 4) α ∈ ( β , , γ , ). The alternation set is now given by A ( α ) = {− < u < . . . < u < α < v < } . The external extremum w ( α ) is now decreasing from − −∞ (as α decreases). During this phase, at α = α , ≈ . E ( α ) has a tipof a V-shape. Here w ( α ) = −∞ and a degree loss takes place, that is, E ( α , ) = E ( α , ) . When α passes through this value, the sign of theleading coefficient of the minimax polynomial changes sign, and w ( α ) willdecrease from + ∞ to w = 1. HECKMARK FUNCTION 19 Figure 7. The graph of E ( . (4th Transition) α = β , ≈ . w ( α ) = 1 and we start thesecond (from the right to the left) V–shape of the graph of E ( α ).The phases above take place again, completing the second V–shape. Tak-ing into account the symmetry of E , two more symmetric V–shapes of E will lie in ( − , References [1] S. N. Bernstein: Sur le meilleure approximation de | x | par des polynomes de degr´esdonn´es, Acta Math. 37 (1913), 1–57.[2] S. N. Bernstein: On the best approximation of | x − c | p , Dokl. Acad. Nauka SSSR 18(1938), 379–384.[3] R. A. DeVore, G. G. Lorentz: Constructive approximation. Grundlehren der Mathe-matischen Wissenschaften, 303. Springer-Verlag, Berlin, 1993.[4] P. D. Dragnev, D. A. Legg, D. W. Townsend: Polynomial approximation of thecheckmark function. Advances in constructive approximation: Vanderbilt 2003, 149–164, Mod. Methods Math., Nashboro Press, Brentwood, TN, 2004.[5] P. P. Petrushev, V. A. Popov: Rational Approximation of real functions. Encyclope-dia of Mathematics and its Applications 28. Cambridge University Press, Cambridge,1987.[6] E. Ya. Remez: Sur le calcul effectif des polynomes d’approximation de Tschebyscheff.C. R. 199 (1934), 337–340.[7] B. Shekhtman: Private communication.[8] H. Stahl: Poles and zeros of best rational approximants of | x | . Constr. Approx. 10(1994), no. 4, 469–22. † , A. R. LEGG, AND R. ORIVE ∗ Department of Mathematical Sciences, Purdue University Fort Wayne, Ft.Wayne, IN 46805 Email address : [email protected] Department of Mathematical Sciences, Purdue University Fort Wayne, Ft.Wayne, IN 46805 Email address : [email protected] Departmento de An´alisis Matem´atico, Universidad de La Laguna, 38200,The Canary Islands, Spain Email address ::