A criterion for the best uniform approximation by simple partial fractions in terms of alternance

We study the best uniform approximation of real-valued continuous functions on closed intervals of the real axis by real-valued simple partial fractions (s. f.). Recall that a real-valued simple partial fraction $R_n$ of degree $n$, $n=0,1,2,\dots$, is the logarithmic derivative of an algebraic polynomial of degree $n$ with real coefficients:

$$\begin{equation*} R_n=\frac{Q'}Q, \qquad Q(x)=a_0+a_1x+\dots+a_{n-1}x^{n-1}+x^n, \qquad Q\not\equiv 0, \quad x,a_j\in\mathbb{R}. \end{equation*}$$

Thus $R_0(x)\equiv 0$ and

$$\begin{equation} R_n(x)=\sum_{j=1}^k \frac{m_j}{x-z_j}\,, \qquad m_1+\dots+m_k=n, \qquad n\ge 1, \end{equation} \tag{ 1 }$$

where the poles $z_j\in\mathbb{C}$ are all pairwise- distinct zeros of the polynomial $Q$, and the numbers $m_j\in\mathbb{N}$ are their multiplicities. Clearly, the zeros are symmetric with respect to the real axis and the multiplicities of complex- conjugate zeros coincide.

Consider a simple partial fraction of order at most $n$ of the best uniform approximation of a function $f$ on a closed interval $S=[x_0-r,x_0+r]$, $r> 0$. We denote it by

$$\begin{equation*} R^*(x)=R^*(n,f,S;x), \qquad x\in S\subset\mathbb{R}, \end{equation*}$$

and sometimes call this fraction optimal.

Simple partial fractions were used by Macintyre and Fuchs (1940), Gonchar (1950s) and Dolzhenko (1960s) in connection with extremal problems in classes of rational functions. As an independent tool of approximation, simple partial fractions were first used by V. I. Danchenko and D. Ya. Danchenko (1999). The theory of approximation by simple partial fractions and their modifications was further developed by V. I. Danchenko, Borodin, Kosukhin, Protasov, Kayumov, Novak and others. Some of these papers are listed in the references. For example, analogues of the classical polynomial theorems of Mergelyan, Jackson, Bernshtein, Zygmund, Dzyadyk and Walsh were obtained for s. f. (see [1], [2]). On the other hand, there are a number of principal approximative properties that do not hold for polynomials. One such property of simple partial fractions is the possibility of approximation on unbounded sets (see [3]–[6]).

Another principal feature is that the fraction $R^*(n,f,S;x)$ is, generally speaking, non-unique and cannot be characterized by the number of alternance points ¹ for $f- R^*$. The non-uniqueness of $R^*$ even in the presence of an alternance of $n+ 1$ points (as well as the non-necessity of such an alternance for the best approximation) was first shown in [7] by an example for $n=2$. This example was completely generalized to $n=3$ in [8] and to arbitrary $n\ge 2$ in [9]. Thus for s. f. there is no exact analogue of Chebyshev's classical theorem on alternance (see [10], p. 69). Our main result (Theorem 1) is an analogue of Chebyshev's theorem under an additional restriction on the poles of the simple partial fraction that delivers the alternance. Many authors have sought such an analogue, and a number of general conjectures were stated in [11]–[14]. Some of them will be established here for the first time.

The following conjecture of V. I. Danchenko [12] is still open: for every continuous function $f$, the fraction $R^*(n,f,S;x)$ is unique and is characterized by an alternance of $n\,{+}\,1$ points whenever the degree $n\,{\ge}\,n_0(f,S)$ is sufficiently large. This conjecture has been proved only for constant functions $f(x)=\mathrm{const}$ (see [7], [15], [16]). We note that the problem of the best approximation of constants by simple partial fractions is one of the possible analogues of Chebyshev's problem on the monic polynomial of a given degree with least deviation from zero.

We now state our results. Let $f$ be a continuous real function on the closed interval $S=[x_0-r,x_0+r]$, and let $B(S)$ be the closed disc with diameter $S$:

$$\begin{equation*} B(S)=\{w\in\mathbb{C}\colon |w-x_0|\le r\}. \end{equation*}$$

Theorem 1.  If all poles of a real-valued s. f. $R_n$ of degree $n$ of the form (1) lie outside the disc $B(S)$, then $R_n(x)\equiv R^*(n,f,S;x)$ if and only if the difference $f- R_n$ admits an alternance of $n+1$ points on $S$. In this case $R_n$ is the unique fraction of best uniform approximation in the class of all s. f. of degree at most $n$. The condition on the position of the poles cannot be weakened.

The impossibility of weakening the condition on the poles means that if at least one pair of complex- conjugate poles of the fraction of best approximation lies in the disc $B(S)$, then, generally speaking, the best approximation is non-unique and is not characterized by an alternance of $n+1$ points in $S$. Examples are given in §2.

In Theorem 1, the degree of the s. f. of best approximation is fixed (and is equal to $n$). In the next theorem we get rid of this restriction as far as the sufficiency of the alternance is concerned. The condition also seems to be redundant as far as the necessity of the alternance and the uniqueness of the best approximation are concerned.

Theorem 2.  If a simple partial fraction $r_k$ of degree $k\le n$ has no poles in the disc $B(S)$ and the difference $f-r_k$ has an alternance of $n+1$ points on $S$, then $r_k(x)\equiv R^*(n,f,S;x)$.

Theorem 2 is proved in §4. Theorem 1 is proved in §3 (sufficiency of the alternance) and §5 (necessity of the alternance and uniqueness of the best approximation).

Remark 1.  In the case of $n$ distinct poles of $R_n$, Theorem 1 was stated and proved by the author in [14] (sufficiency of the alternance) and [17] (necessity of the alternance and uniqueness of the best approximation). We now do not require all poles to be distinct. This form of the criterion was announced by the author in [16]. Novak [11] had earlier proved an analogue of Theorem 1 in the case of $n$ distinct real poles except for the uniqueness assertion. The proofs in [11], [14], [17] make essential use of the determinant identity (8), which was conjectured (in other terminology) in [11] and verified only for small $n$. The author has learned that (8) was proved for all $n$ by Borchardt (see [18], Russian p. 18). This promotes the previous results from conjectures to fully fledged theorems. In this paper we also use Borchardt's identity.

Remark 2.  For $n\ge 3$ it is still an open question whether one can replace the disc in Theorem 1 by a smaller figure containing the closed interval $S$. Such a replacement is impossible for $n=2$ because of the non-uniqueness example in [7], where the poles of each s. f. of degree 2 of the best approximation to $f(x)=x+1$, $x\in [-1,1]$, lie precisely on the unit circle.

Remark 3.  The requirement of the continuity of $f$ in Theorem 1 is redundant in the following sense. Let $f$ and $R$ be bounded functions. We put $\rho=\sup_S |f-R|$ and say that points $t_1<t_2<\dots<t_{s}$ of the closed interval $S$ form a generalized alternance on $S$ for the difference $f-R$ if every point $t_j$ is contained in an arbitrarily small subinterval $U_j=U_j(t_j)\subset S$ such that at least one of the two following alternances holds:

$$\begin{equation*} \begin{aligned} \sup_{U_1} (f-R)=\rho, &\qquad\inf_{U_2} (f-R)=-\rho, & \qquad\sup_{U_3} (f-R)=\rho, &\qquad \dots, \\ \inf_{U_1} (f-R)=-\rho, &\qquad \sup_{U_2} (f-R)=\rho, &\qquad \inf_{U_3} (f-R)=-\rho, & \qquad \dotsc. \end{aligned} \end{equation*}$$

If we define the alternance in this way, then Theorem 1 holds for all bounded functions $f$. Only the slightest changes are needed in the proof.

As an application of Theorem 1, we consider the problem of the best approximation of continuous functions $f$ whose modulus is small on the interval $x\in [-1,1]$. We recall that Kosukhin [2] established, under very general assumptions, a connection between the least deviations from polynomials and simple partial fractions. To state his result in the case of approximation on $[-1,1]$, we put

$$\begin{equation*} F(x)=f(x)\exp\biggl(\int_{c}^x f(t)\,dt\biggr), \qquad c\in[-1,1] \end{equation*}$$

($c$ is fixed). Let $\delta_n(f)$ (resp. $E_{n-1}(F)$) be the least deviation on $[-1,1]$ of $f$ from simple partial functions of degree at most $n$ (resp. of $F$ from algebraic polynomials of degree at most $n-1$) and let $\|\cdot\|$ be the uniform norm on $[-1,1]$. Put $M= \|f\|$. Then the following estimate holds (see [2]):

$$\begin{equation} [2(1+M)e^M]^{-1} \delta_n(f)\le E_{n-1}(F) \le(1+2Me^M)e^M\delta_n(f), \qquad n\in\mathbb{N}. \end{equation} \tag{ 2 }$$

By definition we have $\delta_n(f)=\|R^*-f\|$, where $R^*(x)\equiv R^*(n,f,[-1,1];x)$. Therefore,

$$\begin{equation*} \|R^*\|\le \|f\|+\delta_n(f)\le M+E_{n-1}(F)\cdot 2(1+M)e^M. { } \end{equation*}$$

Furthermore, it is proved in [19] that the poles of every s. f. $R_n$ of degree $n$ satisfying the condition $\|R_n\|\le b^{-n-1}$ for some $b>2$ lie in the exterior of the ellipse

$$\begin{equation*} z=\biggl(\frac{b}{4}+\frac{1}{b}\biggr)\cos t+i\biggl(\frac{b}{4}-\frac{1}{b}\biggr)\sin t, \qquad t\in[0,2\pi]. \end{equation*}$$

The interior of this ellipse contains the unit disc if $b>2(1+\sqrt{2})$. Therefore we obtain the following corollary from Theorem 1 and the previous estimates.

Corollary 1.  Suppose that the following inequality holds for some constants $c\in[-1,1]$ (see the definition of $F)$, $b>2(1+\sqrt{2})$ and $n\in\mathbb{N}$:

$$\begin{equation*} M+E_{n-1}(F)\cdot 2(1+M)e^M\le b^{-n-1}, \qquad M=\|f\|, \end{equation*}$$

and the s. f. $R^*(n,f,[-1,1];x)$ has degree $n$. Then the best approximation is unique and the difference $f-R^*$ admits an alternance consisting of $n+1$ points of $[-1,1]$.

In §2 we apply this result to a concrete function $f$ (see the example in §2.4). Note that if this inequality holds, then the function itself has small modulus since we necessarily have $M\le b^{-n-1}$.

A lower bound for the least deviation can be obtained from the following analogue of de la Vallée-Poussin's polynomial theorem (see [10], pp. 63–64).

Theorem 3.  If the poles of a s. f. $r_k$ of degree $k\le n$ lie outside the disc $B(S)$ and there are $n+1$ points $t_1<t_2<\dots<t_{n+1}$ on the closed interval $S$ such that $f(t_j)-r_k(t_j)=\pm(-1)^j A_j$, $A_j> 0$, $j=1,\dots,n+1$, then

$$\begin{equation*} \max_{x\in S}|f(x)-R^*(n,f,S;x)|\ge \min \{A_1,A_2,\dots,A_{n+1}\}. \end{equation*}$$

Theorem 3 is proved in §4. Note that Theorem 3 with $k=n$ was obtained in [11] (see also [14]) under some additional assumptions (see Remark 1). Here we get rid of these restrictions.

The most important auxiliary tool in the proofs of Theorems 1–3 is the following Theorem 4, which can be deduced from Borchardt's identity. Theorem 4 is of independent interest. In particular, it asserts that the functions (3) below satisfy the Haar condition on $S$. We recall some terminology in this connection (see [10], pp. 79–88).

Continuous complex functions $f_1,\dots,f_n$ satisfy the Haar condition on a compact set $K\subset\mathbb{C}$ (containing more than $n$ points) if every generalized polynomial of the form $\alpha_1 f_1+\dots+\alpha_n f_n$ with constants $\alpha_j\in\mathbb{C}$ (not all equal to 0) has no more than $n-1$ distinct zeros on $K$. Clearly, this condition is equivalent to the inequality $\det(f_j(c_l))_{l,j=1}^n\ne 0$ for any $n$ distinct points $c_l\in K$. Haar showed that it is also equivalent to the uniqueness of the polynomial $\alpha_1^*f_1+\dots+\alpha_n^* f_n$ of least deviation for all continuous functions $f$ on $K$. In particular, if the functions $f_j$ are real and satisfy the Haar condition on a compact set $K\subset\mathbb{R}$, then the system $\{f_1,\dots,f_n\}$ is called a Chebyshev system on $K$. Chebyshev's alternance theorem and some other results in the theory of approximation by algebraic polynomials can be transferred verbatim to the case of Chebyshev systems.

We consider a system of $n$ functions of the form

$$\begin{equation} f_{j,s}(x)=(x-z_j)^{-s}, \qquad j=1,\dots,k, \quad s=2,\dots, m_j+1 \quad \biggl(m_j\ge 1, \ \sum_{j=1}^k m_j=n\biggr), \end{equation} \tag{ 3 }$$

where $z_1,\dots,z_k\in\mathbb{C}$ are distinct points which are situated symmetrically with respect to the real axis, and with every point $z_j$ we associate a number $m_j\in\mathbb{N}$ such that $\sum m_j=n$ and the numbers $m_j$ associated with complex- conjugate $z_j$ are equal to each other.

Theorem 4.  If the numbers $z_j$ in the definition of the system (3) of $n$ functions lie outside the disc $B(S)$, then the sum of the multiplicities of the zeros on $S$ of the generalized polynomial $F(\alpha;x)=\sum_{j,s} \alpha_{j,s}f_{j,s}(x)$ with arbitrary numerical coefficients $\alpha_{j,s}$, not all equal to $0$, does not exceed $n-1$.

Theorem 4 is proved in §6. We mention separately the case when the functions (3) are real-valued. In this case Theorem 4 yields the following corollary.

Corollary 2.  If the numbers $z_j$ in the definition of the system (3) are real and lie outside the closed interval $S$, then the functions (3) form a Chebyshev system on $S$.

This assertion was first proved in [11] (see also [14], [17]) in the case when $k= n$, that is, for the system of functions $\{(x-z_j)^{-2}\}_{j=1}^n$ with $n$ distinct real poles $z_j$, under the assumption that the identity (8) holds.

Unlike the Haar condition, the property of the system (3) stated in Theorem 4 restricts the sum of the multiplicities of the zeros of $F$. In what follows we call this property the strengthened Haar condition. Chebyshev systems possessing this property (see [20], Russian p. 58) are referred to as $ET$-systems, while arbitrary Chebyshev systems are alternatively referred to as $T$-systems.

Using Theorem 4, we prove in §3 the following theorem on the uniqueness of the solution of the multiple interpolation problem. We recall that an interpolating s. f. need not be unique, in contrast to the interpolating polynomial. This was first observed for the simple interpolation problem in [21] and [11], where one can also find some uniqueness conditions in terms of the table of interpolation nodes. The issue of the unique solubility of the interpolation problem was further treated in [13], [14], [22] and elsewhere.

Theorem 5.  If a real-valued s. f. $R_n$ of degree $n$ of the form (1) with all poles lying outside the disc $B(S)$ interpolates a table with $q$ pairwise- distinct nodes $c_l\in S$ of total multiplicity $n$, that is,

$$\begin{equation} \begin{gathered} (R_n)^{(s)}(c_l)=y_{l,s+1}, \qquad y_{l,s+1}\in\mathbb{R}, \\ l=1,\dots, q, \quad s=0,\dots, p_l-1, \quad p_l\ge 1, \quad p_1+\dots+p_q=n, \end{gathered} \end{equation} \tag{ 4 }$$

then $R_n$ is the unique interpolating fraction in the class of all s. f. of degree $\le n$ for all possible nodes $c_l$, values $y_{l,s+1}$ and multiplicities $p_l$.

In §§2.1–2.3 we establish the unimprovability of the condition on poles in Theorem 1. In §2.4 we give an example relating to Corollary 1. The example in §2.5 concerns the case when the best approximation is attained at a s. f. with one pole.

2.1. 

We show by an example that if at least one pair of complex- conjugate poles of a s. f. possessing an $(n+1)$-point alternance lies in the disc $B(S)$, then the fraction need not be optimal.

Consider a s. f. $R(x)$ of degree $n=4$ with poles $0.2\pm 0.2i$ and $-1\pm 0.3i$, and take the interval $[-1,1]$ for $S$. Clearly, the second pair of poles lies outside the closed unit disc, and the first pair lies inside.

The function $R(x)$ has exactly five points of local extremum on $[-1,1]$:

$$\begin{equation*} w_1=-1, \quad w_2\in(-0.8,-0.7), \quad w_3\in(0,0.1), \quad w_4\in(0.3,0.5), \quad w_5=1 \end{equation*}$$

(alternately minima and maxima) and takes the following values $u_j=R(w_j)$ at these points:

$$\begin{equation*} \begin{gathered} u_1=-1.6216216\dots, \quad u_2=1.2642412\dots, \\ u_3=-3.1738633\dots, \quad u_4=6.3689408\dots, \quad u_5=3.3309362\dotsc. \end{gathered} \end{equation*}$$

We put $d=(u_2-u_1)/2=1.4429314\dots$ and define a continuous piecewise-linear function $f(x)$ by letting it be constant on the intervals

$$\begin{equation*} f(x) = \begin{cases} u_1+d, & \quad x\in[-1,-0.4], \\ u_3+d, & \quad x\in[-0.2,0.1], \\ u_4-d, & \quad x\in[0.3,0.6], \\ u_5+d, & \quad x\in[0.7,1] , \end{cases} \end{equation*}$$

and connecting the neighbouring endpoints of these intervals by straight lines on the remaining part of $[-1,1]$.

By construction, the $n+1=5$ points $w_1,\dots,w_5$ form an alternance for $f- R$ on $[-1,1]$ and we have $\max_{x\in[-1,1]}|f(x)-R(x)|=d$. Nevertheless, $R$ is not a fraction of best approximation for $f$. For example, we easily verify that the s. f. $\widetilde{R}(x)$ with poles $0.20007\pm 0.20002i$ and $-1.00001\pm 0.30003i$ is a better approximation to $f$: $\max_{x\in[-1,1]}|f(x)-\widetilde{R}(x)|=1.4427864\ldots<d$.

2.2. 

2.3. 

Following [13], [8], we construct for every $n=2,3,\dots$ an example showing that if all poles of the fraction lie in the disc $B(S)$, then even the presence of an alternance of $2n-1$ points does not guarantee that the approximation is optimal.

It is proved in [13] that for all $x_0$ and $r> 0$ there are real polynomials $Q$ and $P$ of degree $n$ which are positive on $S=[x_0-r,x_0+r]$ and such that the polynomial $H=P'Q-Q'P$ has exactly $2n-2$ simple zeros on the interval $(x_0-r,x_0+r)$.

Indeed, suppose for example that $n=2m$ is even, choose points $x_1<x_2<\cdots \cdots<x_m<y_1<y_2<\dots<y_m$ on the interval $(x_0-r,x_0+r)$ and put

$$\begin{equation*} q(x)=[(x-x_1)\dotsb(x-x_m)]^2, \qquad p(x)=[(x-y_1)\dotsb(x-y_m)]^2, \qquad H_0=p'q-q'p. \end{equation*}$$

The polynomial $H_0$ has at least $n-1$ distinct zeros on the closed interval $[x_1,x_m]$. These are the $m$ zeros $x_k$ and at least one zero on each of the $m-1$ intervals $(x_k,x_{k+1})$ (since the signs of the derivatives $H_0'(x_k)=-q''(x_k)p(x_k)\ne 0$ coincide). We similarly see that $H_0$ has at least $n-1$ distinct zeros on $[y_1,y_m]$. Since the polynomial $H_0$ is of degree $2n-2$, it has no zeros other than those listed, and all the listed zeros are simple. Putting $Q(x)=q(x)+\varepsilon$ and $P(x)=p(x)+\varepsilon$, where $\varepsilon> 0$ is arbitrarily small, we see that the polynomials $Q$ and $P$ are positive and, by continuity, the polynomials $H=P'Q-Q'P$ and $H_0$ have the same number of zeros on $S$, that is, $2n-2$ zeros. (A similar, but more involved, argument works for odd $n$.)

Since $\varepsilon$ is small, it follows by continuity that all zeros of $Q$ and $P$ lie inside the disc $B(S)$.

We now repeat an argument in [8]. Let $c_l$, $l=1,\dots, 2n-2$, be the zeros of $H$ enumerated in ascending order. Putting also $c_0=x_0-r$ and $c_{2n-1}=x_0+r$, we have $c_0<c_1<\dots<c_{2n-2}<c_{2n-1}$.

By construction, all the poles of the simple partial fractions $R_n=Q'/Q$ and $L=P'/P$ lie inside the disc $B(S)$ but outside the closed interval $S$, and the difference $L-R_n\equiv H/(PQ)$ has exactly $2n-2$ simple zeros on $S$ and preserves sign on each of the $2n-1$ intervals $I_l=(c_l,c_{l+1})$, $l=0,\dots,2n-2$. This sign alternates from one interval to another.

There is no loss of generality in assuming that $\operatorname{sign} (L(x)-R_n(x))=(-1)^{l+1}$ for $x\in I_l$, $l=0,\dots, 2n-2$. We fix an arbitrary number $A>\max_{x\in S}|L(x)-R_n(x)|$ and define a continuous real-valued function $f$ by the following (easily achievable) conditions:

1) $f(c_l)=R_n(c_l)=L(c_l)$, $l=1,\dots,2n-2$;

2) $(L(x)-f(x))\cdot(-1)^l> 0$ for $x\in I_l$, $l=0,\dots,2n-2$;

3) $\max_{x\in I_l}|f(x)-R_n(x)|=A$, $l=0,\dots,2n-2$.

Clearly, the deviation of the s. f. $R_n$ from $f$ on $S$ is equal to $A$ and the difference $f-R_n$ has an alternance on $S$ consisting of the $2n-1$ points $t_k\in I_k$. However, $R_n$ is not optimal because, by construction, the deviation of the s. f. $L$ from $f$ is strictly less than $A$.

2.4. 

Consider the family of functions

$$\begin{equation} f(x)=f(a,p\,;x)=\frac{ae^{ax}}{1+e^{ax}-e^{ap}}, \qquad ap\le 0, \qquad x,a\in\mathbb{R}, \quad p\in[-1,1] \end{equation} \tag{ 5 }$$

(where $a$ and $p$ are parameters). The condition $ap\le 0$ guarantees that these functions are continuous for all $x$ ($1+e^{ax}-e^{ap}> 0$) and monotone increasing ($f\,'(x)= a^2e^{ax}(1- e^{ap})\times [1+ e^{ax}-e^{ap}]^{-2}\ge 0$). For example, the family (5) contains all constants: $f(x)\equiv a$ for $p= 0$. If $a\ne 0$, then $f$ and $a$ have the same sign. Therefore, by monotonicity,

$$\begin{equation*} M=\|f\|\equiv\max_{x\in[-1,1]}|f(x)|= \frac{|a|e^{|a|}}{1+e^{|a|}-e^{ap}}\le |a|. \end{equation*}$$

We fix a parameter $p\in[-1,1]$ and put $c=p$ in the definition of $F(x)$ (see §1). Then it can easily be seen that $F(x)=ae^{ax}$. We use the following estimate for $E_{n-1}(F)$ (see [23], Russian pp. 229–230). If $F^{(n)}(x)\ge 0$ $(\le0)$ everywhere on $[-1,1]$, then

$$\begin{equation*} \frac{1}{2^{n-1\,}n!}\min_{x\in[-1,1]}|F^{(n)}(x)|\le E_{n-1}(F)\le \frac{1}{2^{n-1\,}n!}\max_{x\in[-1,1]}|F^{(n)}(x)|. \end{equation*}$$

Since in our case $F^{(n)}(x)=a^{n+1}e^{ax}$, we obtain

$$\begin{equation*} \frac{|a|^{n+1}e^{-|a|}}{2^{n-1\,}n!}\le E_{n-1}(F)\le \frac{|a|^{n+1}e^{|a|}}{2^{n-1\,}n!}. \end{equation*}$$

Using this and (2), we find a bound for the least deviation from $f(a,p;\cdot)$:

$$\begin{equation} \frac{e^{-M-|a|}}{1+2Me^M}\frac{|a|^{n+1}}{2^{n-1\,}n!}\le \delta_n(f)\le 2(1+M)e^{M+|a|}\frac{|a|^{n+1}}{2^{n-1\,}n!}, \qquad M=\|f\|. \end{equation} \tag{ 6 }$$

This estimate has exact order in $n$. In particular, we obtain an estimate of exact order for the least deviations in the problem of the best approximation of constants.

Thus we have the following assertion whose last part follows from Corollary 1, the inequality $M\le |a|$ and the upper bound in (6).

Proposition 1.  For all $p\in[-1,1]$ and $a\in\mathbb{R}$ with $ap\le 0$, the least deviation of a s. f. of degree at most $n$ from a function $f$ of the form (5) satisfies the estimate (6), and this estimate is of exact order. If the s. f. $R^*$ of the best approximation of $f$ has degree $n$ for $|a|\le 5^{-n-2}$, then the best approximation is unique and the difference $f-R^*$ has an alternance of $n+1$ points on $[-1,1]$.

2.5. 

Suppose that $f=P_{n-1}$ is an algebraic polynomial of degree at most $n-1$ and the poles of a real-valued s. f. $R_n$ of the form (1) lie outside the disc $B(S)$. By Theorem 1 and Chebyshev's alternance theorem, the equality $R_n(x)\equiv R^*(n,P_{n-1},S;x)$ holds if and only if $P_{n-1}$ has the least deviation from $R_n$ on $S$ among all polynomials of degree at most $n-1$. Then $\max_S |P_{n-1}-R_n|$ coincides with $\delta_n(P_{n-1},S)$ (the least deviation of s. f. of degree at most $n$ from $P_{n-1}$) and $E_{n-1}(R_n,S)$ (the least deviation of polynomials of degree at most $n-1$ from $R_n$); compare with [13].

Theorem 2 enables us to strengthen this assertion. Namely, suppose that the poles of a s. f. $r_k$ of degree $k\le n$ lie outside the disc $B(S)$. If $P_{n-1}$ is a polynomial of least deviation from $r_k$ on $S$, then $r_k$ is a s. f. of degree $\le n$ of least deviation from $P_{n-1}$ on $S$, and the magnitude of this deviation is $\delta_n(P_{n-1},S)=E_{n-1}(r_k,S)$.

We now calculate the deviation in the case of the interval $[-1,1]$ and a unipolar s. f. $R_k(x)=k/(x-z_0)$, $z_0> 1$. For real $A$ and $a> 1$ we have $E_{n-1}(A/(x- a),[-1,1])=|A|(a-\sqrt{a^2-1})^{n-1}/(a^2-1)$ (this follows from [10], pp. 70–72, where this quantity is found for $A= 1$). Using this and what was said above, we obtain the following result for the polynomial $P_{n-1}$ of least deviation from the fraction $R_k$ of degree $k\le n$:

$$\begin{equation*} \delta_n(P_{n-1},[-1,1])= k\Bigl(z_0-\sqrt{z_0^2-1}\,\Bigr)^{n-1}(z_0^2-1)^{-1}, \qquad z_0>1. \end{equation*}$$

In this section we use Theorem 4 to prove the sufficiency of the alternance in Theorem 1. We also prove Theorem 5.

Theorem 4 is applied to simple partial fractions by means of the following lemma, which was obtained by the author [13] in the case of $n$ distinct points $z_j$. We state the lemma for all simple partial fractions, although it will be used only for real-valued ones.

Lemma 1.  Suppose that $R_n(x)=(\ln Q(x))'$, where $Q$ is a polynomial of degree $n$ of the form

$$\begin{equation*} Q(x)=(x-z_1)^{m_1}\dotsb(x-z_k)^{m_k}, \qquad m_1+\dots+m_k=n, \end{equation*}$$

$z_1,\dots,z_k\in\mathbb{C}$ are distinct points and $m_j$ are positive integers. Then for every s. f. $L(x)\,{=}\,(\ln P(x))'$ of degree at most $n$ there are $n$ uniquely determined numbers $\alpha_{j,s}$ such that

$$\begin{equation} L(x)-R_n(x)\equiv\frac{Q(x)}{P(x)}\sum_{j=1}^k\sum_{s=2}^{m_j+1} \frac{\alpha_{j,s}}{(x-z_j)^s}. \end{equation} \tag{ 7 }$$

Proof.  We have $L-R_n=H/(QP)$, where $H=P'Q-Q'P$. Regard the formula $P'Q-Q'P=H$ as a differential equation for an unknown polynomial $P$ of degree ${\le}\,n$ with known polynomials $Q$ (of degree $n$) and $H$ (of degree ${\le}\,2n{ }{-}{ }2$). Integrating this equation, we obtain

$$\begin{equation*} P(x)=Q(x)C+Q(x)\int\frac{H(x)}{Q^2(x)}\,dx, \qquad C=\mathrm{const}. \end{equation*}$$

Clearly, each zero of $Q$ of multiplicity $m_0$ is also a zero of $H=P'Q-Q'P$ of multiplicity at least $m_0-1$. Therefore $H(x)=(x-z_1)^{m_1-1}\dotsb(x-z_k)^{m_k-1}h(x)$ for some polynomial $h$, and the (unique) partial fraction expansion of the regular fraction $H/Q^2$ is given by

$$\begin{equation*} \frac{H(x)}{Q^2(x)}=\frac{h(x)}{(x-z_1)^{m_1+1}\dotsb(x-z_k)^{m_k+1}}= \sum_{j=1}^k\sum_{s=1}^{m_j+1}\frac{\alpha_{j,s}}{(x-z_j)^s}. \end{equation*}$$

Note that the coefficients $\alpha_{j,1}$ of the fractions $(x-z_j)^{-1}$ must necessarily be equal to zero (otherwise integration gives an uncancellable logarithm, contrary to the requirement that `$P$ is a polynomial'). Thus we have the identity

$$\begin{equation*} \frac{H(x)}{Q^2(x)}=\sum_{j=1}^k\sum_{s=2}^{m_j+1}\frac{\alpha_{j,s}}{(x-z_j)^s}\,, \end{equation*}$$

which proves the lemma. $\square$

Here are some corollaries of Theorem 4. In their statements we assume that the s. f. $R_n$ are real-valued. Hence the points $z_j$ in the definition of $Q(x)$ (see Lemma 1) must be situated symmetrically with respect to the real axis, and the positive integers $m_j$ must be equal for conjugate $z_j$.

Corollary 3.  If all poles of a s. f. $R_n$ of degree $n$ of the form (1) lie outside the disc $B(S)$, and if $L\not\equiv R_n$ is another real-valued s. f. of degree at most $n$, then the sum of the multiplicities of the zeros of $L-R_n$ on $S$ does not exceed $n-1$. In particular, the difference $L-R_n$ has at most $n-1$ distinct zeros on $S$.

Proof.  We have $Q(x)\ne 0$ for $x\in S$. Hence it follows from Lemma 1 that each zero of multiplicity $p$ of $L-R_n$ is also a zero of multiplicity $p$ of the sum on the right-hand side of (7). But this sum is a polynomial in the system of functions (3) and, by Theorem 4, cannot have more than $n-1$ roots on $S$ (counting multiplicities). $\square$

Theorem 5 is merely a restatement of Corollary 3.

Corollary 4.  If all poles of a s. f. $R_n$ of degree $n$ of the form (1) lie outside the disc $B(S)$, then the sum of the multiplicities of the zeros of the first derivative of $R_n$ on $S$ does not exceed $n-1$. In particular, $(R_n)'$ has no more than $n-1$ distinct zeros on $S$.

Proof.  It follows from Corollary 3 that the polynomial $H=P'Q-Q'P$ has at most $n-1$ zeros on $S$ (counting multiplicities) if $Q$ is a real polynomial of degree $n$ all of whose roots lie outside $B(S)$ and $P$ is another real polynomial of degree at most $n$. Putting $P=Q'$ in the definition of $H$, we conclude that the polynomial $\widetilde{H}=Q''Q-(Q')^2$ has at most $n-1$ roots on $S$ (counting multiplicities). It remains to note that $\widetilde{H}/Q^2\equiv (Q\,'/Q)\,' \equiv(R_n)'$. $\square$

Corollary 5.  (sufficiency of the alternance) Suppose that $f$ is a continuous function on the closed interval $S$, the poles of a s. f. $R_n$ of degree $n$ of the form (1) lie outside the disc $B(S)$, and the difference $f-R_n$ has an alternance of $n+1$ points on $S$. Then $R_n(x)\equiv R^*(n,f,S;x)$.

Proof.  Suppose that $f-R_n$ admits an alternance $t_1<t_2<\dots<t_{n+1}$ of $n+ 1$ points on the interval $S\,{=}\,[x_0-r,x_0+r]$. Assume that there is a s. f. $L$ of degree ${\le}\,n$ providing a better approximation to $f$ on $S$ than $R_n$: $\max_S |f-R_n|\,{>}{ }\max_S |f-L|$. At the points of alternance we have

$$\begin{equation*} \begin{aligned} \operatorname{sign}\bigl[L(t_j)-R_n(t_j)\bigr] &\equiv\operatorname{sign} \bigl[(L(t_j)-f(t_j))-(R_n(t_j)-f(t_j))\bigr] \\ &=\pm (-1)^j,\qquad j=1,\dots,n+1. \end{aligned} \end{equation*}$$

In particular, the function $L-R_n$ is continuous on $S$ and changes sign on each of the $n$ closed intervals $[t_j,t_{j+1}]$. Hence it has at least $n$ distinct zeros on the interval $(x_0-r,x_0+r)$. By Corollary 3 it follows that $L\equiv R_n$. This contradicts the assumption that $L$ is a better approximation to $f$. $\square$

In this section we prove Theorems 2, 3. Since a straightforward application of the method used in §3 becomes cumbersome when the order of the approximating s. f. is less than $n$, we prove Theorems 2, 3 in another way.

Proof. of Theorem 2 When $k=n$ the theorem coincides with Corollary 5 (see §3). Let us prove that an alternance of $n+1$ points guarantees the best approximation of $f$ by $r_k$ even for $k<n$.

Assume the opposite: there is a s. f. $R$ of degree at most $n$ which provides a better approximation, that is, $\|R-f\|=\|r_k-f\|-\varepsilon$, where $\varepsilon> 0$ (here and in what follows, $\|\cdot\|$ is the uniform norm for the closed interval $S$). We introduce the notation

$$\begin{equation*} R_n(x)=r_k(x)+\frac{n-k}{x+c}, \qquad f_n(x)=f(x)+\frac{n-k}{x+c}, \end{equation*}$$

where $c\in\mathbb{R}\setminus S$, $c> 0$. For every such $c$, the simple partial fraction $R_n$ is a best approximation to $f_n$. This follows from Corollary 5 since the degree of $R_n$ is equal to $n$, all its poles lie outside $B(S)$, and the difference $R_n-f_n$ has an alternance of $n+1$ points on $S$ because of the identity $R_n(x)-f_n(x)\equiv r_k(x)-f(x)$ and the presence of such an alternance for $r_k-f$.

By the triangle inequality we have the following chain:

$$\begin{equation*} \begin{aligned} \|R-f_n\| &\le \|R-f\|+\|f-f_n\|=\|r_k-f\|-\varepsilon+\|f-f_n\| \\ &\le \|r_k-R_n\|+\|R_n-f_n\|+2\|f-f_n\|-\varepsilon. \end{aligned} \end{equation*}$$

Clearly, $\|r_k-R_n\|$ and $\|f-f_n\|$ decrease to zero as $c\to +\infty$. Hence there is a $C_0=C_0(\varepsilon)> 0$ such that $\|r_k-R_n\|+2\|f- f_n\|< \varepsilon$ for all $c>C_0$. Thus for every $c>C_0$ we have

$$\begin{equation*} \|R-f_n\|<\|R_n-f_n\|, \end{equation*}$$

contrary to the assumption that $R_n$ is a best approximation of $f_n$. Therefore our assumption is false and $r_k$ is indeed a best approximation to $f$. $\square$

Proof. of Theorem 3 When $k=n$ the theorem is proved by contradiction in the same way as Corollary 5. We claim that the case $k<n$ can easily be reduced to the case $k=n$ as in the proof of Theorem 2. Indeed, given $r_k$ and $f$, we construct a fraction $R_n$ and a function $f_n$ as above. It follows from the hypothesis of Theorem 3 and the identity $R_n(x)-f_n(x)\equiv r_k(x)-f(x)$ that we have $f_n(t_j)- R_n(t_j)=\pm(-1)^jA_j$, $A_j> 0$, $j=1,\dots,n+1$, at the points $t_1<t_2<\dots<t_{n+1}$ of the interval $S$. Since all the poles of the s. f. $R_n$ of degree $n$ lie outside the disc $B(S)$ and, since the theorem is already known for fractions of degree $n$, the following bound holds for all $c$ (see the proof of Theorem 2):

$$\begin{equation*} \delta_n(f_n,S)\ge \mu=\min \{A_1,A_2,\dots,A_{n+1}\}. \end{equation*}$$

Here and in what follows we write $\delta_n(\widetilde{f},S)$ for the least deviation of simple partial fractions of degree at most $n$ from the function $\widetilde{f}$ on $S$.

Let $r^*$ be a s. f. of degree $\le n$ of best uniform approximation to $f$ on $S$. Assume that $\delta_n(f,S)=\mu-\varepsilon$, where $\varepsilon> 0$. Let $C_1$ be a positive number such that $\|f_n-f\|<\varepsilon$ for all $c>C_1$. Then, by our assumption,

$$\begin{equation*} \delta_n(f_n,S)\le \|f_n-r^*\|\le \|f_n-f\|+\|f-r^*\|=\|f_n-f\|+ \delta_n(f,S)<\mu \end{equation*}$$

for $c>C_1$. This contradicts the inequality $\delta_n(f_n,S)\ge \mu$. Hence $\delta_n(f,S)\ge \mu$. $\square$

Here we prove the remaining assertions in Theorem 1. To simplify the notation, we assume that $S=[-1,1]$ (the case of an arbitrary closed real interval $S=[x_0-r,x_0+r]$ is reduced to this by the linear transformation $v= (w-x_0)/r$).

Necessity of the alternance of $n+1$ points. Suppose that the poles of the s. f. $R_n$ of degree $n$ of the best approximation to a continuous function $f$ on $[-1,1]$ lie outside the disc $B([-1,1])$, and let $d=\max_{x\in[-1,1]}|f(x)-R_n(x)|$ be the least deviation.

Assume that $f-R_n$ has an alternance of only $m\le n$ points on $[-1,1]$. Then there are points $x_1=-1<x_2<\dots<x_m<x_{m+1}= 1$ such that

1) $(f-R_n)(x_j)= 0$ for all $j=2,\dots, m$;

2) for every $j=1,\dots,m$ the difference $f-R_n$ attains an extremal value $\pm (-1)^j d$ on the closed interval $[x_j,x_{j+1}]$.

We note that the extremum on $[x_j,x_{j+1}]$, $j\,{=}\,2,\dots,m-1$, is necessarily attained at an interior point while the extremum on $[x_1,x_2]$ (resp. $[x_m,x_{m+1}]$) may also be attained at the endpoint $x_1$ (resp. $x_{m+1}$).

We first consider the case when $m=n$. Define the following $n$-tuple of points $c_j$:

$$\begin{equation*} c_1=-1-\varepsilon, \qquad c_j=x_j, \quad j=2,\dots,n, \end{equation*}$$

where $\varepsilon> 0$ is small so that the poles of the s. f. $R_n$ lie outside the disc $B(\widetilde{S})$ with diameter $\widetilde{S}=[-1-\varepsilon,1+\varepsilon]$. For all $\tau\in\mathbb{R}$ of sufficiently small modulus, the problem of interpolating the $n$-node table

$$\begin{equation*} \{(c_j,y_j)\}_{j=1}^n, \qquad y_1=R_n(c_1)+\tau, \qquad y_j=R_n(c_j), \quad j=2,\dots,n, \end{equation*}$$

by simple partial fractions of degree $\le n$ has a unique solution $L(x;\tau)$. The degree of $L$ is equal to $n$, and the poles of $L$ lie outside the disc $B(\widetilde{S})$. This follows by continuity from Theorem 5 (see §1) since the only solution of the interpolation problem for $\tau= 0$ is the s. f. $R_n(x)\equiv L(x;0)$.

Furthermore, by Theorem 5, the graphs of $L$ and $R_n$ for small $\tau\ne 0$ have no intersections on $\widetilde{S}$ except for the nodes $c_2,\dots,c_n$, and all of them are simple zeros of the difference $\Delta(x)=L(x;\cdot)-R_n(x)$. Hence the function $\Delta(x)$, whose modulus can be made arbitrary small on $\widetilde{S}$ by an appropriate choice of $\tau$, preserves its sign on each of the intervals $[c_1,c_2)$, $(c_j,c_{j+1})$, $j=2,\dots,n-1$, $(c_n,1+\varepsilon]$, and this sign alternates as we pass from any interval to the next. We can choose the sign of $\tau$ so that $\max_{x\in[-1,1]}|f(x)-L(x;\tau)|<d$. This contradicts the optimality of $R_n$ and hence proves the necessity of the existence of an alternance of $n+1$ points.

It remains to consider the case when $m<n$. We reduce it to the previous case by complementing the $m$-tuple (defined as above) of the points $c_j$, $j=1,\dots,m$, with points $c_{m+1}<\dots<c_n$ in the interval $(1,1+\varepsilon)$. This proves the necessity of the alternance.

The uniqueness of the best approximation is proved in the same way as Corollary 5 (see §3) with the following difference. If the s. f. $R_n$ of degree $n$ with poles outside $B(S)$ and another s. f. $L$ of degree $\le n$ are fractions of least deviation from $f$ on $S$, then the existence of an alternance of $n+1$ points for $f-R_n$ on $S$ (which has been just proved) means that the difference $L-R_n=(L-f)-(R_n-f)$ has at least $n$ zeros on $S$ (counting multiplicities). More precisely, the sum of the number of simple zeros and twice the number of multiple zeros is not smaller than $n$ (compare [13]). By Theorem 5 we have $L\equiv R_n$. This completes the proof of Theorem 1.

We divide the proof of Theorem 4 into three parts (cases). In the first and second parts we show that the system (3) satisfies the Haar condition on $S$ for $k=n$ and $k< n$ respectively. In the third part we study the general case of the strengthened Haar condition. Note that the Haar condition for $k=n$ was verified in [14], but we reproduce this proof (that uses the identity (8)) for the sake of completeness.

It is convenient to prove the Haar condition (a particular case of Theorem 4) in terms of certain determinants. The fixed numbers $z_j$ will be referred to as poles (not points) in contrast with the variable points $c_l$ appearing in the statement of the Haar condition (see §1).

Theorem 4'.  Suppose that for each of $k$ distinct poles $z_1,\dots,z_k\in\mathbb{C}$ of the system of functions (3), which are situated symmetrically with respect to the real axis, we are given a number $m_j\in\mathbb{N}$ such that $\sum m_j=n$, and conjugate $z_j$ correspond to equal $m_j$. If all the poles $z_j$ lie outside the disc $B(S)$, then the determinant

$$\begin{equation*} \widetilde{D}=\det\bigl((c_l-z_1)^{-2},\dots,(c_l-z_1)^{-m_1-1};\dots; (c_l-z_k)^{-2},\dots,(c_l-z_k)^{-m_k-1}\bigr)_{l=1}^n \end{equation*}$$

is different from zero for any $n$ distinct points $c_l\in S$.

Case 1 (the main case): $k=n$. We denote the $n$-tuples of the points $c_l$ and poles $z_j$ by

$$\begin{equation*} c=(c_1,\dots,c_n), \qquad z=(z_1,\dots,z_n). \end{equation*}$$

Since $m_j= 1$ for all $j$, the system of functions (3) takes the form $\{(x-z_j)^{-2}\}_{j=1}^n$, and the Haar condition is written as the inequality

$$\begin{equation*} D=D(c;z)=\det((c_l-z_j)^{-2})_{l,j=1}^n\ne 0 \end{equation*}$$

for all $n$-tuples of distinct points $c_l$ on the closed interval $S$. To prove this inequality, we use the following identity of Borchardt (1855), which was stated in the monograph [18] on permanents of matrices. (The permanent of a square matrix is the function calculated by the rule of expanding the determinant but taking the sign `plus' of every product of entries independently of the parity of the corresponding permutation.)

Borchardt's identity.  For any disjoint $n$-tuples $c$ and $z$ of complex numbers we have

$$\begin{equation} D(c;z)=W(c;z) T(c;z), \end{equation} \tag{ 8 }$$

where $W$ and $T$ are the determinant and permanent of the matrix $((c_l-z_j)^{-1})_{l,j=1}^n$ respectively.

The following expansion for the determinant $W(c;z)$ is known ([10], p. 29):

$$\begin{equation} W(c;z)=\det((c_l-z_j)^{-1})_{l,j=1}^n=\pm\frac{\prod_{n\ge j>l\ge1}(c_j-c_l)(z_j-z_l)}{g(z_1)\dotsb g(z_n)}, \end{equation} \tag{ 9 }$$

where the sign $\pm$ depends only on $n$, and $g(t)=(c_1-t)\dotsb(c_n-t)$. In our case, the points $c_l$ and the poles $z_j$ are pairwise distinct, whence the factor $W$ in (8) is different from 0.

The inequality $T\ne 0$ is established in Lemma 2 below. It holds for more general $n$-tuples $c$ and $z$ than in the inequality $W\ne 0$. The lemma will be proved in §7 (compare [14]). Thus we arrive at the desired inequality $D\ne 0$, which expresses the Haar condition for $n$ distinct poles.

Lemma 2.  The permanent $T(c;z)$ of the matrix $((c_l-z_j)^{-1})_{l,j=1}^n$ is non-zero for any $n$-tuple $c$ of points in $S$ (not necessarily distinct) and any $n$-tuple $z$ of poles lying outside the disc $B(S)$ symmetrically with respect to the real axis (not necessarily distinct).

Case 2 (the general case in Theorem 4'): $k<n$. We consider two basic subcases in detail.

Subcase 2.1: a real pole $z_1$ is equipped with a number $m=m_1\ge 2$, and the poles $z_2,\dots,z_k$ are equipped with numbers $m_j= 1$. (We note that here the number $k$ of distinct poles is equal to $n-m+1$ since $\sum m_j=n$.) The functions in the system (3) take the form

$$\begin{equation*} \bigl\{(x-z_1)^{-2}, (x-z_1)^{-3},\dots, (x-z_1)^{-m-1};(x-z_2)^{-2};(x-z_3)^{-2};\dots;(x-z_k)^{-2}\bigr\}. \end{equation*}$$

The determinant in the Haar condition is correspondingly written as

$$\begin{equation*} D_1=\operatorname{det}\bigl((c_l-z_1)^{-2}, (c_l-z_1)^{-3},\dots, (c_l-z_1)^{-m-1};(c_l-z_2)^{-2};\dots; (c_l-z_k)^{-2}\bigr)_{l=1}^n. \end{equation*}$$

We claim that $D_1\ne 0$ for all pairwise- distinct points $c_1,\dots,c_n\in S$.

Indeed, take $n$ pairwise- distinct points $v_1,\dots,v_n\in\mathbb{C}$ situated symmetrically with respect to the real axis and satisfying the equalities $v_1=z_1$, $v_{m+1}= z_2$, ... , $v_n=z_k$. Using them and the points $c_l$, we construct an auxiliary determinant $D(c;v)=\det((c_l-v_j)^{-2})_{l,j=1}^n$, where $v=(v_1,\dots,v_n)$. It is easily seen that

$$\begin{equation*} D_1=\mu\lim_{v_m,\dots,v_3,v_2\to v_1} \frac{\partial^{(m-1)+\dots+2+1} D(c;v)}{\partial v_m^{m-1} \dotsb \partial v_3^{2} \partial v_2}, \qquad \mu^{-1}=\prod_{s=1}^m s!\ne 0. \end{equation*}$$

The identity (8) enables us to replace $D(c;v)$ in this expression by $W(c;v)T(c;v)$. We expand the derivative of this product into a sum by Leibniz' formula. Each summand is a multiple of some derivative of $W(c;v)$. We easily see from the definition of $W(c;v)=\det((c_l-v_j)^{-1})_{l,j=1}^n$ that the limiting values of all relevant derivatives of $W$ (except possibly for the highest) are equal to zero because each of these limiting values is a determinant with at least two equal columns. Hence,

$$\begin{equation*} D_1=\mu\lim_{v_m,\dots,v_3,v_2\to v_1} \frac{\partial^{(m-1)+\dots+2+1} W(c;v)}{\partial v_m^{m-1} \dotsb\partial v_3^{2} \partial v_2} \lim_{v_m,\dots,v_3,v_2\to v_1}T(c;v). \end{equation*}$$

But the limit of the permanent $T$ is different from zero by Lemma 2 (whose hypothesis allows for coincidence of some points inside each of the two $n$-tuples of arguments). It remains to prove that the limit $A$ of the derivative of the determinant $W$ is non-zero. This limit can be written more conveniently as

$$\begin{equation} A=\lim_{v_m\to v_1}\frac{\partial^{m-1}}{\partial v_m^{m-1}}\biggl(\dotsb \biggl(\lim_{v_3\to v_1} \frac{\partial^2}{\partial v_3^2}\biggl(\lim_{v_2\to v_1} \frac{\partial W(c;v)}{\partial v_2}\biggr)\biggr)\dotsb\biggr). \end{equation} \tag{ 10 }$$

We see from (9) that

$$\begin{equation*} W(c;v)=\frac{v_1-v_2}{g(v_1)}\prod_{j=3}^n (v_2-v_j)\prod_{\substack{n\ge j>l\ge1\\ l,j\ne 2}}(v_l-v_j)\frac{Y(c)}{g(v_2)\dotsb g(v_n)}, { } \end{equation*}$$

where $Y(c)=\pm\prod_{n\ge j>l\ge1}(c_j-c_l)$. We find the value of the innermost bracket in the limiting equality (10):

$$\begin{equation*} \lim_{v_2\to v_1}\frac{\partial W(c;v)}{\partial v_2}= -\biggl[\frac{v_1-v_3}{g(v_1)}\biggr]^2\prod_{j=4}^n (v_1-v_j)^2\prod_{\substack {n\ge j>l\ge1\\ l,j\ne 2,\ne 1}}(v_l-v_j)\frac{Y(c)}{g(v_3)\dotsb g(v_n)}\ne 0. { } \end{equation*}$$

Then we calculate the limit of the second derivative of the resulting expression with respect to $v_3$ and so on. Each differentiation gives rise to a non-zero quantity since one can easily see that the limit (as $v_l\to v_1$) of the derivative of order $l- 1$ with respect to $v_l$ is taken from a function of the form $(v_1-v_l)^{l-1} Y_1$, where $\lim_{v_l\to v_1}Y_1\ne 0$. Hence $A\ne 0$, as required.

Thus Theorem 4' holds under the hypotheses of Subcase 2.1.

Subcase 2.2: complex-conjugate poles $z_1$ and $z_2$ are equipped with equal numbers $m=m_1=m_2\ge 2$, and the remaining poles $z_3,\dots,z_k$ are equipped with numbers $m_j= 1$. Here $k=n-2m+2$. As above, we denote the determinant in the Haar condition for the corresponding system of functions by $D_2$ and consider an auxiliary determinant $D(c;v)$ with pairwise- distinct points $v_1,\dots,v_n\in\mathbb{C}$ situated symmetrically with respect to the real axis and satisfying the equalities $v_1=z_1$, $v_{m+1}=z_2$, $v_{2m+1}=z_3$, $v_{2m+2}=z_4$, ... , $v_n=z_k$. Note that the determinant $D_2$ coincides (up to a non-zero factor $\mu^2$) with the limit

$$\begin{equation*} \lim_{\substack{v_m,\dots,v_2\to v_1\\ v_{2m},\dots,v_{m+2}\to v_{m+1}}} \frac{\partial^{2[(m-1)+\dots+2+1]} D(c;v)}{\partial v_m^{m-1} \dotsb \partial v_3^{2} \partial v_2\partial v_{2m}^{m-1} \dotsb \partial v_{m+3}^{2} \partial v_{m+2}}. \end{equation*}$$

Using (8), (9) and Lemma 2, we now see that $D_2\ne 0$.

Thus Theorem 4' holds under the hypotheses of Subcase 2.2. In the general case 2, the theorem is proved by successive application of the scheme used in the two basic subcases to each pole with $m_j> 1$. This completes the proof of Theorem 4'.

Case 3 (the general case of Theorem 4). Here the polynomial in the system of functions (3) may have multiple zeros. Consider the homogeneous system of equations for the coefficients $\alpha_{j,s}$ obtained from (7) in view of the multiple interpolation conditions (4). It suffices to verify that the determinant $\widehat{D}$ of this system is non-zero. We easily see that $\widehat{D}$ is equal to the limit of a certain derivative of the determinant $\widetilde{D}$ (see the statement of Theorem 4') with respect to the variables $c_l$. For example, if the nodes $c_1,\dots,c_{n-2}$ in the conditions (4) are simple and the node $c_{n-1}$ is double, then $\widehat{D}=\lim_{c_n\to c_{n-1}} (\widetilde{D})'_{c_n}$. Arguing as above, we see that $\widehat{D}\ne 0$. This proves Theorem 4 (the strengthened Haar condition).

Here we reproduce the corresponding proof from [14] almost verbatim. As in [14], we only consider the case when $S=[-1,1]$. Let $m=0,1,2,\dots$ be the number of those real numbers $z_j$ that are smaller than $-1$ (if some of the $z_j$ coincide, then we count each of them).

When all the numbers $z_j$ are real, the signs of all summands in the expansion of $T$ are equal to $(-1)^m$ and, therefore, $T\ne 0$ (this was observed in [11]).

Suppose that there is at least one pair of complex- conjugate numbers $z_j$. Reorder the $z_j$ in such a way that their list begins with all pairs of complex- conjugate numbers and ends with real numbers. Take $n=2$. Under our assumptions we have $z_{1,2}= a\pm ib$, $b\ne 0$, and the numbers $c_1$ and $c_2$ range over $[-1,1]$. We calculate the permanent $T=T_2$ directly:

$$\begin{equation} T_2=\frac{1}{c_{1}-z_1}\,\frac{1}{c_{2}-z_2} +\frac{1}{c_{2}-z_1}\,\frac{1}{c_{1}-{z_2}}= \frac{2(a-c_2)(a-c_1)+2b^2}{[(a-c_1)^2+b^2][(a-c_2)^2+b^2]}. \end{equation} \tag{ 11 }$$

Since $(a-c)^2+b^2> 0$, it follows that the inequality $T_2\ne 0$ holds for all $c_1$, $c_2$ in $[-1,1]$ if and only if

$$\begin{equation*} \min_{-1\le c_1, c_2\le 1} \bigl[(a-c_2)(a-c_1)+b^2\bigr]>0. \end{equation*}$$

We easily see that the minimum is attained at the points $c_1=-c_2=\pm 1$ and is equal to $a^2+b^2-1$. Under the hypotheses of the lemma with $S=[-1,1]$ we have $a^2+b^2> 1$ (the points $z_1$ and $z_2$ lie outside the closed unit disc). Therefore $T_2> 0$. This proves the lemma for $n=2$.

We now suppose that $n>2$. The determinant $W=\det((c_l-z_j)^{-1})_{l,j=1}^n$ can be written by Laplace's theorem (see, for example, [24], p. 129) in the form

$$\begin{equation*} W=\sum_{1\le l_1<l_2\le n} W_2\begin{pmatrix} 1 & 2 \\ l_1 & l_2 \end{pmatrix} (-1)^{1+2+l_1+l_2} \widetilde{W}_{n-2} \begin{pmatrix} 1 & 2 \\ l_1 & l_2 \end{pmatrix}, \end{equation*}$$

where $W_2$ is the minor of order 2 formed by the entries of $((c_l-z_j)^{-1})_{l,j=1}^n$ situated at the intersection of the first and second columns and $l_1$th and $l_2$th rows, and $\widetilde{W}_{n-2}$ is the complementary minor of order $n{ }{-}{ }2$. Thus, by the definition of the permanent, we have

$$\begin{equation*} T=\sum_{1\le l_1<l_2\le n} T_2 \begin{pmatrix} 1 & 2 \\ l_1 & l_2 \end{pmatrix}\widetilde{T}_{n-2} \begin{pmatrix} 1 & 2 \\ l_1 & l_2 \end{pmatrix}, { } \end{equation*}$$

where $T_2$ and $\widetilde{T}_{n-2}$ are the permanents of those submatrices whose determinants are equal to $W_2$ and $\widetilde{W}_{n-2}$ respectively.

We claim that in our situation all the products $T_2\widetilde{T}_{n-2}$ have the same sign $(-1)^m$ and, therefore, the sum $T$ is non-zero. Indeed, by what was said above, we have

$$\begin{equation*} T_2 \begin{pmatrix} 1 & 2 \\ l_1 & l_2 \end{pmatrix} =\frac{1}{c_{l_1}-z_1}\,\frac{1}{c_{l_2}-z_2}+ \frac{1}{c_{l_2}-z_1}\,\frac{1}{c_{l_1}-z_2}>0 \end{equation*}$$

for all $c_{l_1},c_{l_2}\in [-1,1]$. Therefore the sign of the product $T_2 \widetilde{T}_{n-2}$ is equal to the sign of $\widetilde{T}_{n-2}$. If all $z_j$ are real except for $z_1$ and $z_2$, then we have seen that $\operatorname{sign}\widetilde{T}_{n-2}=(-1)^m$. If $z_3$, $z_4$ is the next pair of complex- conjugate poles, then the previous argument works for the permanent $\widetilde{T}_{n-2}$. After several iterations, factoring out only positive factors at every step, we exhaust all pairs of complex- conjugate $z_j$. Hence only sums with real $z_j$ remain, and the total sign is $(-1)^m$. The lemma is proved.