Gorin's problem for individual simple partial fractions
aa r X i v : . [ m a t h . C A ] J u l Gorin’s problem for individual simple partialfractions
Petr Chunaev ∗ and Vladimir Danchenko † Abstract
The main result of the paper is a lower estimate for the moduli of imaginaryparts of the poles of a simple partial fraction (i.e. the logarithmic derivativeof an algebraic polynomial) under the condition that the L ∞ ( R ) -norm of thefraction is unit (Gorin’s problem). In contrast to the preceding results, theestimate takes into account the residues associated with the poles.Moreover, a new estimate for the moduli is obtained in the case when the L ∞ ( R ) -norm of the derivative of the simple partial fraction is unit (Gelfond’sproblem). Keywords:
Gorin’s problem, Gelfond’s problem, logarithmic derivative of analgebraic polynomial, simple partial fraction, least deviation.
Gorin’s problem is formulated as follows. Find a lower estimate for d n ( R , p ) = inf (cid:8) Y ( ρ n ) : k ρ n k L p ( R ) ≤ (cid:9) , < p ≤ ∞ , (1) where Y ( ρ n ) := min k =1 ,...,m | Im ξ k | , and ρ n ( z ) := ln m Y k =1 ( z − ξ k ) n k ! ′ = m X k =1 n k z − ξ k (2) is a simple partial fraction (SPF), i.e. the logarithmic derivative of an algebraicpolynomial Q ( z ) = Q mk =1 ( z − ξ k ) n k of a given degree n = P mk =1 n k (the number n iscalled the order of the SPF). Note that SPFs, being discrete Cauchy potentials, havewide applications in Electrostatic Field Theory, Potential Theory (see [4, §1], [5, §3])and Theory of Partial Differential Equations [8].We start with a brief history of the problem (1). For p = ∞ (this is the most dif-ficult case) the problem was considered in [2,7–11]. The question about the principal ∗ [email protected]; National Center for Cognitive Research, ITMO University † [email protected]; Vladimir State University named after Alexander and Nikolay Stole-tovs d n ( R , ∞ ) ≥ c ( n ) > was stated and positively resolvedby Gorin [8]. Later, Nikolaev [10] obtained the estimate d n ( R , ∞ ) ≥ √ − n − , n ∈ N . In [10], the following problem was also stated:
Is it true that d n ( R , ∞ ) → as n → ∞ ? An essential improvement for Nikolaev’s estimate was done by Gelfond [7]: d n ( R , ∞ ) ≥ (17 ln n ) − , n ≥ n . Furthermore, Katsnelson [9] obtained a certain improvement of this estimate butwith the same logarithmic rate of minorant decrease. However, Nikolaev’s problemremained open. The final solution on the whole class of SPFs (2) was given in [2]: d n ( R , ∞ ) ≍ ln ln n ln n . (3) It was also shown there that for finite p the value d n ( R , p ) does not tend to zeroand is bounded from below by a positive constant depending only on p . Actuallythe following stronger result holds [2]: k ρ ± n k L ∞ ( R ) ≤ β p · k ρ n k qL p ( R ) , where β p ≤ p sin − q ( π/p ) , p − + q − = 1 , and ρ ± n denote partial sums from (2) containing all poles from the half-planes C ± ,correspondingly. This means that, in contrast to the uniform case, there is almostno compensation of partial sums ρ ± n in the integral metric on R . The value of β p wasrevisited in [1] but the question about the sharpness of the constant ˜ β p obtainedthere with respect to the rate of p still remains open.Later on, analogues of Gorin’s problem were considered for other sets (semi-axes, segments, rectifiable compacts, etc.) and with a different normalisation of (2).Detailed history of the problems and related results in this direction are summarisedin the survey [4]. Here we only mention Gelfond’s problem that we consider below.In [10], estimates of type (1) but with the normalisation of the derivative of SPFare considered, d ′ n ( R , p ) = inf (cid:8) Y ( ρ n ) : k ρ ′ n k L p ( R ) ≤ (cid:9) , < p ≤ ∞ . (4) The following estimate was obtained in [10]: d ′ n ( R , ∞ ) ≥ const · − n/ . Later, Nikolaevgeneralised and improved some of Gelfond’s results. In particular, he showed in [11]that d ′ n ( R , ∞ ) ≥ const · n − / . Much later, the following weak equivalence was proved in [2]: inf (cid:8) Y ( ρ n ) : k ρ ′ n k L ∞ ( R ) ≤ , ρ n = ρ + n (cid:9) ≍ ln n √ n , (5) The expression α ≍ β means that there exist absolute positive constants c and C such that cβ ≤ α ≤ Cβ . ρ n = ρ + n belongto the half-plane C + . It is plausible that the same holds in the general case, for d ′ n ( R , ∞ ) .Note that the exchange ̺ ( z ) = cρ n ( c z ) , c = k ρ n k − qL p ( R ) saves the form of a SPF,and k ̺ k L p ( R ) = 1 and Y ( ̺ ) = c − Y ( ρ n ) . Consequently, (1) may be rewritten as d n ( R , p ) = inf ρ n n Y ( ρ n ) k ρ n k qL p ( R ) o , p − + q − = 1 , (6) where the infimum is taken over all SPFs (2) with no poles on R . Analogously,considering the SPF ̺ ( z ) = cρ n ( c z )) , c := k ρ ′ n k − qq +1 L p ( R ) , one gets k ̺ ′ k L p ( R ) = 1 and Y ( ̺ ) = c − Y ( ρ n ) , and therefore d ′ n ( R , p ) = inf ρ n n Y ( ρ n ) k ρ ′ n k qq +1 L p ( R ) o , p − + q − = 1 . (7) Thus Gorin’s and Gelfond’s problems can be thought as finding the least de-viation from zero in L p ( R ) of SPFs (2) and their derivatives under the condition Y ( ρ n ) = 1 , or, which is the same, under the condition that all SPFs (2) have acommon fixed pole, say, ξ = i . In this sense, the problems are analogues of classicalChebyshev’s problem on the least deviation from zero of a unitary polynomial ofa fixed degree. This circumstance, in particular, leads to more general approxima-tion problems for SPFs (2) and their derivatives on R and other sets, making theestimates for (1) and (4) still topical (see [4]).Recall that the two-sided estimate (3) is valid for the class of all SPFs (2), withno attention to the multiplicity of the roots of Q . A natural question about theestimation of Y ( ρ n ) for an individual normalised SPF ρ n , taking into account n k ,arises. It is answered in the following theorem. Theorem 1.
There is an absolute c > such that for any pole ξ k it holds that | Im ξ k | · k ρ n k L ∞ ( R ) ≥ c (ln n ) /n k + 1(ln n ) /n k − · ln ln n ln n > c n k ln n , n ≥ . (8) Note that the second inequality in (8) follows from the simple inequality µ t + 1 µ t − > t ln µ , where µ = ln n, t = 1 n k > . Thus Theorem 1 provides a continuous scale of additional factors in (3). Forexample, if n k ≤ ln ln n , then the first inequality in (8) has the same rate as in (3).For n k , satisfying the opposite inequality, the second inequality in (8) is more precisethan (3).As for the estimates for d ′ n ( R , ∞ ) , we prove the following theorem in the generalcase, i.e. without any assumptions on the location of poles. Theorem 2.
There exists an absolute c > such that d ′ n ( R , ∞ ) ≥ c r ln nn , n ≥ n . (9) Proof of the estimate (8)
It is sufficient to prove (8) in the case when k ρ n k L ∞ ( R ) = 1 (see (6)). For determinacy, we obtain a lower estimate for y = Im z assuming that z = iy is one of the poles of ρ n belonging to the upper half-plane C + .First we get estimates under the following additional assumptions.1) The poles of ρ n and corresponding residues are symmetric with respect to thereal and imaginary axes so that the poles on the imaginary axis have even residues.This happens e.g. if the symmetrisation from Section 2.4 is applied.By z k , k = 1 , . . . , m , we denote the poles of SPF ρ n belonging to the upperhalf-plane C + , and by n k the corresponding residues so that the order of SPF equals n = 2 P mk =1 n k . We aim to estimate the imaginary part y > of the pole z = iy .Let B ( z ) := m Y k =1 ( z − z k ) n k ( z − z k ) n k , µ ( x ) = 12 i B ′ ( x ) B ( x ) = m X k =1 n k y k ( x − x k ) + y k , x ∈ R . (10) We use one more assumption.2) For real x and x | µ ( x ) − µ ( x ) | ≤ (cid:18) r y (cid:19) , r := | x − x | . (11) Below we show that the assumptions 1)–2) do not limit the generality of the problem.In the Blaschke product (10), the products of factors with poles z k and − z k , beingsymmetric with respect to the imaginary axis, are non-negative on the imaginaryaxis as ( iy − z k )( iy − z k ) ( iy + z k )( iy + z k ) = ( iy − z k )( iy − z k ) ( − iy + z k )( − iy + z k ) = (cid:12)(cid:12)(cid:12)(cid:12) iy − z k iy − z k (cid:12)(cid:12)(cid:12)(cid:12) . Therefore the symmetry assumption 1) implies that B ( iy ) ≥ for all y ∈ R and < B ( iy ) < for y > . The assumption 1) also implies that µ is an even positivefunction on the real axis. Furthermore, the following partial fraction decompositionholds: − B ( z )1 + B ( z ) = i η X k =1 µ ( t k ) 1 z − t k , η := n/ m X k =1 n k , (12) with pairwise distinct finite real t k being the roots of the equation B ( x ) = − . Thepoints t k locate on the real axis symmetrically with respect to the origin (it followsfrom the equality B ( x ) = B ( − x ) ). For determinacy, let t k < t k +1 ( k = 1 , . . . , η − )and let t k < for k = 1 , . . . , η and t k > for k = η + 1 , . . . , η . Set r k = t η + k , k = 1 , . . . , η. On each segment [ t k , t k +1 ] the argument of the Blaschke product B ( x ) has incrementof π , in particular, Z r k µ ( x ) dx = 12 i Z r k (ln( B ( x ))) ′ dx = 12 Z r k (arg( B ( x ))) ′ dx = π k − . (13) θ ∈ (0 , and y = y θ ( y = Im z ). Since < B ( iy ) = (1 − θ ) n (1 + θ ) n m Y k =2 ( iy − z k ) n k ( iy − z k ) n k < ε, ε := (1 − θ ) n (1 + θ ) n , (14) the decomposition (12) leads to − ε ε < − B ( iy )1 + B ( iy ) = i η X k =1 µ ( t k ) 1 iy − t k = η X k =1 µ ( r k ) y y + r k , and thus η X k =1 µ ( r k ) y y + r k ≥ − δ, δ := 2 ε ε = 2(1 − θ ) n (1 − θ ) n + (1 + θ ) n . (15) . Let τ > and µ ( τ ) = min [0 ,τ ] µ ( x ) , µ ( τ ) = max [0 ,τ ] µ ( x ) . Fix r > and divide the sum in (15) into the two: S ( r ) + S ( r ) := X r k ≤ r + X r k >r ! µ ( r k ) y y + r k . To estimate S , take into account (13): r k µ ( r k ) ≥ Z r k µ ( x ) dx = π k − , r k ≥ π (2 k − µ ( r k ) , which implies that S ( r ) ≤ X r k ≤ r µ ( r ) y y + π (2 k − µ ( r ) ≤ µ ( r ) µ ( r ) ∞ X k =1 µ ( r ) y (2 µ ( r ) y ) + π (2 k − = µ ( r ) µ ( r ) e µ ( r ) y − e µ ( r ) y + 1 . The sum of the series is known, see e.g. [6, Chapter 2, §3].To estimate S , consider (12). By Cauchy’s integral formula, π Z i ∞− i ∞ ξ + r ) − B ( ξ )1 + B ( ξ ) dξ = η X k =1 µ ( r k ) 1( r + r k ) . Furthermore, recall that B ( iy ) ≥ for all y ∈ R . By this reason, the modulus of theleft hand side of (12) is at most and therefore π (cid:12)(cid:12)(cid:12)(cid:12)Z i ∞− i ∞ ξ + r ) − B ( ξ )1 + B ( ξ ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ π Z i ∞− i ∞ | ξ + r | = 12 r . S ( r ) ≤ X r k >r y µ ( r k ) r k ≤ X r k >r µ ( r k ) y ( r + r k ) ≤ y r . The sum of the estimates for S and S and (15) give the inequity µ ( r ) µ ( r ) e µ ( r ) y − e µ ( r ) y + 1 + 4 y r ≥ − δ, y = y θ. Thus the following lemma is true.
Lemma 1.
Under the assumptions and , for any r > it holds that e θ µ ( r ) y ≥ µ ( r ) + µ ( r ) − δµ ( r ) − y µ ( r ) /rµ ( r ) − µ ( r ) + δµ ( r ) + 4 y µ ( r ) /r , (16) where r > , θ ∈ (0 , , δ = 2(1 − θ ) n (1 − θ ) n + (1 + θ ) n . r and δ ∈ (0 , . Since µ ( r ) ≥ µ ( r ) ≥ , (16) gives e θ µ ( r ) y ≥ µ ( r ) − y µ ( r ) /rµ ( r ) − µ ( r ) + δµ ( r ) + 4 y µ ( r ) /r . From now on we think that y ≤ n / (in the otherwise case the inequality (8)is obvious), therefore µ ( r ) > µ (0) ≥ n /y > . Choose r > and then δ from theconditions r = 4 µ ( r ) y , µ ( r ) δ = 1 . Such r s obviously exist, possibly they are multiple. Solving the latter equation withrespect to θ gives θ = (2 µ ( r ) − /n − µ ( r ) − /n + 1 . (17) For the chosen r and θ the following inequalities are valid: e θ µ ( r ) y ≥ µ ( r ) − µ ( r ) − µ ( r ) + 2 , y ≥ θ µ ( r ) ln (cid:18) µ ( r ) − µ ( r ) − µ ( r ) + 2 (cid:19) ,y ≥ µ ( r ) (2 µ ( r ) − /n + 1(2 µ ( r ) − /n − (cid:18) µ ( r ) − µ ( r ) − µ ( r ) + 2 (cid:19) . By (11), the choice of r = 4 µ ( r ) y and the inequality µ ( r ) > , µ ( r ) − µ ( r ) ≤ (cid:18) r y (cid:19) ≤ µ ( r )) < µ ( r ) . Thus we have proved the following lemma.6 emma 2. If r = 4 µ ( r ) y and y ≤ n / , then y ≥ µ ( r ) (2 µ ( r ) − /n + 1(2 µ ( r ) − /n − (cid:18) µ ( r ) −
12 + 4 ln µ ( r ) (cid:19) ≥ c µ ( r ) /n + 1 µ ( r ) /n − · ln µ ( r ) µ ( r ) . (18) Now recall that k ρ n k L ∞ ( R ) ≤ and use the following estimate from [3]: k µ k L ∞ ( R ) ≤ const ln n. Taking into account that the minorant in (18) is decreasing as a function of µ ( r ) (it can be easily checked) and that µ ( r ) ≤ k µ k L ∞ ( R ) , we come to the first inequalityin (8). Thus the inequality (8) is proved under the assumptions 1)–2). . The general case of SPF (2) (of a given order n ) can bereduced to Lemma 2 as follows. Considering z = iy and y > , we make twosymmetrisations of the form s ( z ) = ρ n ( z ) + ρ n (¯ z ) , σ ( z ) = s ( z ) − s ( − ¯ z ) . These give a SPF of the form σ ( z ) = σ ( z ) + σ (¯ z ) of order n with poles symmetricwith respect to the real and imaginary axes so that the poles of σ belong to C + andthe poles of σ (¯ z ) to C − . Obviously, the maximum of | σ ( x ) | is at most four timesmore that the maximum of | ρ n ( x ) | and therefore k σ k L ∞ ( R ) ≤ . One of the poles of σ is still z = iy , y > , with the residue ≥ n . Furthermore, we exchange σ ( z ) for SPF ρ ( z ) = σ ( z − iy ) + σ ( z + iy ) , i.e. move the poles of σ ( z ) from the realaxis by y so that one of the poles of ρ is z = 2 iy , with the residue ≥ n . Thevalue of k ρ k L ∞ ( R ) is then at most four times more than the maximum modulus of theinitial SPF (2). This follows from the maximum modulus principle for subharmonicfunctions: | ρ ( x ) | = 2 | Re σ ( x − iy ) | ≤ k Re σ k L ∞ ( R ) = k σ k L ∞ ( R ) ≤ . Now let us show that the SPF R defined by R ( z ) := 4 − ρ (4 − z ) = ̺ ( z ) + ̺ ( z ) , ̺ ( z ) := 14 σ (cid:18) z − iy (cid:19) , (19) satisfies the assumptions 1)–2). Indeed, its sup -norm on R is at most , its poles andresidues are symmetric with respect to the coordinate axes and moreover the residueof its pole z = 8 y i is at least n . What is more, Cauchy’s integral formula for z ∈ C − gives | σ ′ ( z ) | ≤ π Z R | σ ( x ) || x − z | dx ≤ | Im z | and therefore by (19), | ̺ ′ ( z ) | ≤ y + | Im z | .
7y integrating this estimate against the right angle γ with equal sides that belongsto the lower half-plane and is based on the segment [ x , x ] , r = x − x > , we get | ̺ ( x ) − ̺ ( x ) | ≤ Z γ | dz | y + | Im z | = 2 √ Z r/ dx − x + y + r/ < (cid:18) r y (cid:19) . Thus the inequality (11) holds and the SPF R satisfies the assumptions 1)–2).Consequently, the estimate (8) is valid for it. The difference between R and the initialSPF ρ n by means of n , n k and ξ k does not influence the rate in the estimate (8) butonly changes the absolute constant c . Let ρ n ( z ) = ρ + ( z ) + ρ − ( z ) , where ρ ± are SPFs whose poles lie in C ± . Let σ ( z ) = ρ ′ n ( z ) , σ ( z ) = ( ρ + ( z )) ′ , σ ( z ) = ( ρ − ( z )) ′ , so that σ ( z ) = σ ( z ) + σ ( z ) . For simplicity, suppose that k σ k L ∞ ( R ) = 1 . Lemma 3 . Given a fixed n ≥ and k σ k L ∞ ( R ) = 1 , k σ ( · − ih ) k L ∞ ( R ) ≤ n, h = 1 n . (20) Proof.
Cauchy’s integral formula for z ∈ C − gives | σ ′ ( z ) | ≤ π Z R | σ ( x ) || x − z | dx ≤ k σ k L ∞ ( R ) | Im z | = 1 | Im z | . For a fixed x ∈ R and h = 1 /n , this implies that | σ ( x − ih ) − σ ( x − i/h ) | ≤ Z /hh dyy = 4 ln n. Obviously, | σ ( x − i/h ) | ≤ n − and therefore | σ ( x − ih ) | ≤ n − + 4 ln n < n, n ≥ . The inequality (20) is proved.Furthermore, the estimate (5) and definition (7) for p = ∞ , q = 1 , leads to ( h + Y ( ρ + )) k σ ( · + ih ) k L ∞ ( R ) ≥ c ln n √ n , so that Lemma 3 implies h + Y ( ρ + ) ≥ c √ √ ln n √ n , Y ( ρ + ) ≥ c √ √ ln n √ n − n > c √ ln n √ n , n ≥ n ( c ) . Analogous inequalities hold for Y ( ρ − ) , too, thus the required inequality (9) follows.8 Acknowledgements
The reported study was funded by Russian Ministry of Education and Science (tasknumber 1.574.2016/1.4) and RFBR (project number 18-01-00744).
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