UDC 517.52

MATHEMATICS

A. L. LEVIN, V. M. TIKHOMIROV

ON APPROXIMATIONS OF ANALYTIC FUNCTIONS BY RATIONAL FUNCTIONS

(Presented by Academician A. N. Kolmogorov on 8 VII 1966)

Let \(B\) be a Banach space of functions \(f(z)\) of a complex variable \(z\); \(\|f\|\) the norm of \(f\) in \(B\); \(\mathscr P_n\) the linear space of polynomials
\(P_n(z)=a_0+a_1z+\cdots+a_nz^n\) of degree \(\leq n\); \(\mathscr R_n\) the set of all rational functions
\(R_n(z)=P_{1k}(z)/P_{2s}(z)\), \(\max(k,s)\leq n\).

Put
\[ e_n(f)=\inf_{P_n\in\mathscr P_n}\|f-P_n\|,\qquad r_n(f)=\inf_{R_n\in\mathscr R_n}\|f-R_n\|. \]

In this paper we mainly take as \(B\) the space \(A_\infty^1\) of functions \(f(z)\), analytic for \(|z|<1\) and continuous in the closed disk \(|z|\leq 1\), with norm
\(\|f\|=\max_{|z|=1}|f(z)|\). Our concrete results follow from the following elementary lemma of functional analysis.

Lemma. Let \(B\) be any Banach space and \(L\) its subspace. Let \(\varphi_0\in L\) be an element of best approximation to a given \(f_0\in L\). Then we have:
\[ \|f_0-\varphi_0\|=\max_{\substack{F\in B^*\\ F\perp L}} |F(f_0)|/\|F\|, \]
and the maximum is attained.

Indeed, since
\[ |F(f_0)|=|F(f_0-\varphi)|\leq \|F\|\cdot\|f_0-\varphi\|, \]
we have
\[ \|f_0-\varphi\|\geq |F(f_0)|/\|F\| \quad \forall\,\varphi\in L. \]
At the same time there exists a functional \(F\) such that
\[ \|f_0-\varphi_0\|\cdot\|F\|=|F(f_0)| \]
(see \((^3)\)). This is an obvious consequence of the Hahn–Banach theorem. Conversely, if \(F\perp L\) and
\[ |F(f_0)|=\|F\|\cdot\|f_0-\varphi_0\|, \]
then we have
\[ \|f_0-\varphi_0\|=|F(f_0)|/\|F\|=|F(f_0-\varphi)|/\|F\|\leq \|f_0-\varphi\|,\qquad \forall\varphi\in L, \]
and \(\varphi_0\) is an element of best approximation.

We pass to approximations by rational functions. In this paper, continuing \((^1)\), we carry out a comparison of the apparatus of approximation by rational functions and by polynomials.

Theorem 1. Put
\[ f(z)=\sum_{m=0}^{\infty} c_m z^{\lambda_m}, \]
where \(c_m>0\),
\[ \sum_{m=0}^{\infty} c_m<\infty \]
and \(\lambda_{m+1}/\lambda_m\) is nonintegral. Let \(n\) be a natural number and let \(m_n\) be the first index such that
\[ \lambda_{m_n-1}-\lambda_{m_n-2}\leq n<\lambda_{m_n}-\lambda_{m_n-1}. \]
Then
\[ r_n(f)\geq \sum_{m>m_n} c_m. \]

We note that in \((^1)\) the same function is considered, and for approximation in the norm \(L_2\) the estimate
\[ r_n\geq c_{m_n}/2n \]
is obtained. From the theorem just stated it follows at once that if
\[ \lambda_{m_n-1}\leq n<\lambda_{m_n}-\lambda_{m_n-1}, \]
then
\[ r_n(f)=e_n(f), \]
since, obviously,
\[ e_n(f)\leq \sum_{m\geq m_n} c_m. \]

Proof of Theorem 1. Fix the poles \(a_1,\ldots,a_n\) of the rational function \(R_n(z)\). Put

\[ \mu_k=\exp(\pi i k)\prod_{l=1}^{n} \left(\exp\left(\frac{\pi i k}{\lambda_{m_n}}\right)-\frac{1}{\overline{a_l}}\right) \left(\exp\left(-\frac{\pi i k}{\lambda_{m_n}}\right)-\frac{1}{a_l}\right). \]

For any \(f\in A_\infty^1\) put

\[ F(f)=\sum_{k=0}^{2\lambda_{m_n}-1} f(z_k)\overline{\mu_k}, \quad \text{where } \quad z_k=\exp\left(\frac{\pi i k}{\lambda_{m_n}}\right). \]

\(F\) is a linear continuous functional on \(A_\infty^1\). Let us clarify its properties.
\[ F(1)=\sum \overline{\mu_k}=\overline{\sum \mu_k}=0, \]
since the sum \(\sum \mu_k\) contains terms of the form

\[ \operatorname{const}(a_1,\ldots,a_n) \sum_{k=0}^{2\lambda_{m_n}-1} \exp\left(\frac{\pi i k}{\lambda_{m_n}}s\right)\exp(\pi i k), \tag{*} \]

which is equal to zero if
\[ \exp(\pi i)\exp\left(\frac{\pi i}{\lambda_{m_n}}s\right)\ne 1. \]
But this is so, because
\[ -n\leq s\leq n, \quad \text{and} \quad n<\lambda_{m_n}. \]
Further,

\[ F\left(\frac{1}{z-a_l}\right) =\sum_k \frac{\overline{\mu_k}}{z_k-a_l} =-\frac{1}{a_l^2}\sum \frac{\overline{\mu_k}}{z_k-1/\overline{a_l}} =0 \quad (l=1,\ldots,n), \]

because \(\mu_k\) contains the factor \(z_k-1/\overline{a_l}\), and after canceling it we again obtain terms of the form \((*)\).

Finally,

\[ F(z^p)=\sum_{k=0}^{2\lambda_{m_n}-1}\overline{\mu_k}z_k^p. \]

This sum contains terms of the form

\[ \sum_{k=0}^{2\lambda_{m_n}-1} \exp(-\pi i k)\exp\left(\frac{\pi i k}{\lambda_{m_n}}s\right) \exp\left(\frac{\pi i k}{\lambda_{m_n}}p\right). \]

We want \(F(z^p)=0\) for \(0\leq p\leq \lambda_{m_n-1}\). For this, as above, we must have
\[ s+p<\lambda_{m_n} \]
or
\[ s<\lambda_{m_n}-p\leq \lambda_{m_n}-\lambda_{m_n-1}. \]
But this is satisfied by assumption. It remains to note that

\[ F\left(z^{\lambda_{m_n}}\right) = \sum_{k=0}^{2\lambda_{m_n}-1} \prod_{l=1}^{n} \left|\exp\left(\frac{\pi i k}{\lambda_{m_n}}\right)-\frac{1}{\overline{a_l}}\right|^2 =\|F\| \]

and even
\[ F\left(z^{\lambda_m}\right)=\|F\| \]
for \(m\geq m_n\) (because of the evenness of \(\lambda_{m+1}/\lambda_m\)). By the lemma,

\[ r_n(f)\geq |F(f)|/\|F\|=\sum_{m\geq m_n} c_m, \]
as was required.

Corollary (see (1)). If the function considered \(f(z)\in A_R^1\), i.e.
\[ c_m\sim 1/R^m,\quad R>1, \]
and if
\[ \lambda_{m+1}/\lambda_m\to \infty, \]
then we have:

\[ \varlimsup_{n\to\infty} r_n(f)^{1/n} = \varlimsup_{n\to\infty} e_n(f)^{1/n} = 1/R. \]

Theorem 1, in a somewhat different formulation and by other methods, was obtained by E. P. Dolzhenko in (2). The question arises whether Theorem 1 can be strengthened, i.e. whether there exists a function \(f(z)\in A_\infty^1\) such that
\[ e_m(f)=r_m(f),\quad m=0,1,\ldots,n,\ldots \]
The answer to this question is given by

Theorem 2. The functions \(f(z)\equiv a_k z^k+b_k\), and only they, have the property that
\[ e_n(f)=r_n(f),\quad \forall n=0,1,\ldots \]

Theorem 2 follows from the following two theorems, which are of independent interest.

Theorem 3. Suppose it is known that the polynomial
\[ P_k^0(z)=a_0+a_1z+\cdots+a_kz^k,\quad a_k\ne 0, \]
is a polynomial of degree \(k>0\) of best approximation for \(f(z)\), and \(n+1\) is the first number exceeding \(k\) such that
\[ e_k(f)=e_n(f)>e_{n+1}(f). \]
Then
\[ e_n(f)>r_n(f). \]

Indeed, by the condition, the space \(\mathscr P_n\) is supporting to the sphere of the space \(A_\infty^1\) of radius \(\|f-P_k^0\|\) with center at \(f-P_k^0\), but the space \(\mathscr P_{n+1}\) is not supporting to it. This means that there exists a variation
\[ t\cdot Q_{n+1}(z)\equiv t\cdot(\varepsilon_0'+\varepsilon_1'z+\cdots+\varepsilon_{n+1}'z^{n+1}),\quad t>0, \]
such that
\[ \|f-P_k^0+t\cdot Q_{n+1}(z)\|=\|f-P_k^0\|-ta+o(t),\quad a>0. \]

Consider the numbers \(\varepsilon,\varepsilon_0,\varepsilon_1,\ldots,\varepsilon_n\), defined by the equalities
\[ \varepsilon a_k=\varepsilon_{n+1}',\quad (\varepsilon a_{k-1}+\varepsilon_n)=\varepsilon_n',\ldots,\quad (\varepsilon a_0)+\varepsilon_{n-k+1}=\varepsilon_{n-k+1}',\quad \varepsilon_{n-k}=\varepsilon_{n-k}',\ldots,\quad \varepsilon_0=\varepsilon_0'. \]
Then we have
\[ R_n(z;t)\equiv \frac{P_k^0(z)-t\cdot(\varepsilon_0+\varepsilon_1z+\cdots+\varepsilon_nz^n)} {1-\varepsilon t z^{\,n-k+1}} =P_k^0(z)-t\cdot Q_{n+1}(z)+o(t). \]

Hence, as is easy to see, there is a \(t_0>0\) such that
\[ \|f(z)-R_n(z;t_0)\| =\|f(z)-P_k^0(z)+t_0Q_{n+1}(z)-o(t_0)\| < \|f(z)-P_k^0(z)\|. \]
The theorem is proved.

Corollary. Let \(e_0(f)>e_k(f)=\cdots=e_n(f)>e_{n+1}(f)\). By Theorem 3 we have \(r_n(f)<e_n(f)\). Hence, if \(r_n(f)=e_n(f)\ \forall n\), then one must have
\[ e_0(f)=\cdots=e_{k-1}>e_k(f)=\cdots=e_n(f)=\cdots . \]
But since \(e_n(f)\to0\) as \(n\to\infty\), we obtain \(e_k(f)=0\), and hence \(f(z)\) is a polynomial of degree \(k\), and moreover such that \(e_{k-1}(f)=e_0(f)\).

Theorem 4. Among all polynomials \(P_k(z)\) of degree \(k\), only for polynomials \(P_k(z)\equiv a_kz^k\) do we have
\[ e_{k-1}\bigl(P_k(z)\bigr)=\|P_k(z)\|=|a_k|. \tag{**} \]

Indeed, for these polynomials condition \((**)\) is satisfied. Let us prove this. Fix the poles \(\alpha_1,\ldots,\alpha_{k-1}\). Put, for any \(f(z)\in A_\infty^1\),
\[ F(f)=\int_{|z|=1} f(z)\overline{\mu}(z)|dz|,\quad \text{where}\quad \mu(z)=z^k\prod_{i=1}^{k-1}(z-\alpha_i)\left(\frac{1-\overline{\alpha_i}z}{z}\right). \]
\(F\) is obviously a linear continuous functional on \(A_\infty^1\). We have:

a) \[ \overline{z-\alpha_i}=\frac1z-\overline{\alpha_i} \]
on the circle \(|z|=1\), i.e.
\[ \prod_{i=1}^{k-1}(z-\alpha_i)\left(\frac1z-\overline{\alpha_i}\right) \]
is real and positive.

b) \(F\perp 1,\ 1/(z-\alpha_i),\ i=1,\ldots,k-1\), by Cauchy’s theorem, since \(\mu(z)/z\) and \(\mu(z)/(\overline z-\overline{\alpha_i})=z\mu(z)/(1-\alpha_i z)\) are analytic. (We note that \(|dz|=dz/z\) if \(|z|=1\).)

c)
\[ |F(a_kz^k)| =|a_k|\int_{|z|=1}\prod_{i=1}^{k-1}(z-\alpha_i)\left(\frac{1-\overline{\alpha_i}z}{z}\right)|dz| =\|F\|\,\|a_kz^k\|; \]
\(F\) satisfies the conditions of the lemma with \(\varphi_0=0\), whence
\[ \|a_kz^k\|\le \left\|a_kz^k-c_0-\sum_{i=1}^{k-1}\frac{c_i}{z-\alpha_i}\right\|. \]
In view of the arbitrariness of \(\alpha_i\) and \(c_i\), we obtain \(r_n(a_kz^k)\ge |a_k|\), as was required.

This result was obtained by another method by V. M. Tikhomirov in 1962. We shall now prove that if the polynomial \(P_k(z)\not\equiv a_kz^k\), then condition \((**)\) is not satisfied for it.

If for the polynomial \(P_k(z)\not\equiv a_kz^k\) condition \((**)\) is satisfied, then, by the lemma, there exists a functional
\[ F(f)=\int_{|z|=1} f(z)\,d\mu(z) \]
such that
\[ \left|F\bigl(P_k(z)\bigr)\right| =\left|\int_{|z|=1}P_k(z)\,d\mu(z)\right| =\max |P_k(z)|\int_{|z|=1}|d\mu(z)|, \tag{1} \]
\[ F\perp z^s,\quad s=0,1,\ldots,k-1. \tag{2} \]

In view of condition (1), the measure \(\mu\) is concentrated on the set of points at which \(|P_k(z)|\) attains its maximum. But this set consists of \(m\) points \((m\leq k)\). Hence
\[ F(f)=\sum_{i=1}^{m} f(\xi_i)\mu(\xi_i). \]
Condition (2) gives:
\[ \sum_{i=1}^{m}\mu_i=0,\qquad \sum_{i=1}^{m}\mu_i\xi_i^s=0,\qquad s=1,2,\ldots,k-1. \]

This system of \(k\) equations with \(m\) \((m\leq k)\) unknowns has only the trivial solution. The contradiction obtained proves Theorem 4. Theorem 4 can be strengthened in the following direction.

Theorem \(4'\). If \(f(z)\in A_\infty^1\), \(|f(e^{i\theta})|=\|f\|\), \(0\leq\theta\leq 2\pi\), and in the disk \(|z|<1\) \(f(z)\) has more than \(k-1\) zeros, then \(r_{k-1}(f)=e_{k-1}(f)=\|f\|\).

We note that, by the maximum principle, the indicated class of functions is exhausted by Blaschke products.

Let \(K\) be a subset of \(A_\infty^1\). Denote
\[ \varepsilon_n(K)=\sup_{f\in K} e_n(f),\quad \rho_n(K)=\sup_{f\in K} r_n(f). \]

Let \(B_r\) be the class considered in work \((^4)\).
\[ B_r=\{f(z): f(z)\in A_\infty^1,\quad |f^{(r)}(z)|\leq 1\ \text{for}\ |z|\leq 1\}. \]
From \((^4)\) it follows that
\[ \varepsilon_n(B_r)=1/(n+1)\cdots(n-r+2). \]
From Theorem 4,
\[ \rho_n(B_r)\geq r_n\left(\frac{z^{n+1}}{(n+1)\cdots(n-r+2)}\right) =1/(n+1)\cdots(n-r+2), \]
whence \(\varepsilon_n(B_r)=\rho_n(B_r)\).

Theorem 4 also holds in the multidimensional case. If we consider approximations of functions analytic in the polydisc \(T\) \((|z_i|<1,\ i=1,\ldots,k)\) and continuous in \(\overline{T}\), by rational functions of the form
\[ R_n(z_1,\ldots,z_k)=Q_n/P_n, \]
where \(Q_n\) and \(P_n\) are polynomials in \(z_1,\ldots,z_k\) of degree \(\leq n\) (in the totality of the variables), and the polynomial \(P_n\) has no zeros in \(\overline{T}\), and take as norm
\[ \|f\|=\max_{|z_1|=\cdots=|z_k|=1}|f(z_1,\ldots,z_k)|, \]
then for the function
\[ f_0(z)=(z_1z_2\cdots z_k)^{n+1} \]
we have \(r_n(f_0)=\|f_0\|\).

Theorem \(4'\) is also generalized to the case of an annulus.

Consider the space \(A_\infty^{1,r}\) of functions analytic in the annulus \(K\) \((r<|z|<1)\), continuous in \(\overline{K}\), with norm
\[ \|f\|=\max_{|z|=1,\ |z|=r}|f(z)|. \]
Consider approximations of functions from \(A_\infty^{1,r}\) by rational functions
\[ R_n(z)=(b_0+b_1z+\cdots+b_nz^n)/(a_0+a_1z+\cdots+a_nz^n), \]
where the zeros of the denominator lie outside \(\overline{K}\). Then, if \(f_0\in A_\infty^{1,r}\) has the properties:

a) \(f_0(z)\) is analytic in the disk \(|z|<1\) (respectively, outside the disk \(|z|\leq r\));

b) \(|f_0(e^{i\theta})|=\|f\|\), \(0\leq\theta\leq 2\pi\) (respectively, \(|f(re^{i\theta})|=\|f\|\));

c) \(f_0(z)\) has more than \(n\) zeros in the disk \(|z|<1\) (respectively, outside the disk \(|z|\leq r\)),

then we have \(r_n(f_0)=e_n(f_0)=\|f_0\|\).

If, instead of condition a), one requires only that \(f_0(z)\) be meromorphic in the corresponding domain, then c) must be strengthened and it must be required that \(f_0(z)\) have, in addition, more than \(p-1\) zeros in the disk \(|r|<1\). (Here \(p\) denotes the number of poles of \(f_0(z)\) in the corresponding domain.) The proofs of these assertions are almost a verbatim repetition of the proof of Theorem \(4'\).

Moscow State University
named after M. V. Lomonosov

Received
24 VI 1966

REFERENCES

\(^{1}\) V. D. Erokhin, DAN, 128, No. 1, 32 (1959).
\(^{2}\) E. P. Dolzhenko, DAN, 166, No. 3 (1966).
\(^{3}\) V. S. Videnskii, UMN, 11, 5, 174 (1956).
\(^{4}\) K. I. Babenko, Izv. AN SSSR, Ser. Mat., 22, 4 (1958).