MATHEMATICS

BLAGOVEST SENDOV

ON THE QUESTION OF THE EXPANSION OF REGULARLY MONOTONE FUNCTIONS IN A GONCHAROV SERIES

(Presented by Academician S. N. Bernstein, 15 VII 1957)

Recently we proved \((^1)\), using the method of Ya. A. Tagamlitskii \((^2)\), a theorem on the expansion of a certain class of regularly monotone functions in a Goncharov series. In the present note we shall prove this theorem by classical means and at the same time generalize it.

If one takes an arbitrary sequence \(\varepsilon\) \((\varepsilon_0, \varepsilon_1,\ldots,\varepsilon_n,\ldots;\ \varepsilon_n=1\) or \(-1)\), then there exist functions satisfying the conditions

\[ \varepsilon_n f^{(n)}(x)\geq 0,\quad 0\leq x\leq 1,\quad n=0,1,2,\ldots . \tag{1} \]

Such are, for example, the Goncharov polynomials corresponding to the nodes

\[ x_0,\ x_1,\ x_2,\ldots,\ x_n,\ldots;\quad x_n=\frac{|\varepsilon_n-\varepsilon_{n+1}|}{2}, \tag{2} \]

which we shall denote by \(P_n(x)\), so that

\[ P_0(x)=\varepsilon_0,\quad P_n(x)=\int_{x_0}^{x}\tau_0\,dt_1\int_{x_1}^{t_1}\tau_1\,dt_2\cdots \int_{x_{n-1}}^{t_{n-1}}\tau_{n-1}\,dt_n,\quad n=1,2,\ldots, \tag{3} \]

where \(\tau_k=\varepsilon_k\varepsilon_{k+1}\).

Denote by \(K_\varepsilon\) the set of regularly monotone functions satisfying conditions (1), and by \(K_\varepsilon^*\) the set of those functions \(f(x)\in K_\varepsilon\) for which \(f^{(n)}(x_n)=0\) \((n=0,1,2,\ldots)\).

It is not difficult to see that, if \(f(x)\in K_\varepsilon\), then

\[ f(x)=\sum_{\nu=0}^{\infty}\varepsilon_\nu f^{(\nu)}(x_\nu)P_\nu(x)+R(x), \]

where \(R(x)\in K_\varepsilon^*\).

From the works of S. N. Bernstein \((^{3,4})\) it is known that in the case of absolutely monotone functions \(R(x)\equiv 0\), while in the case of cyclically monotone functions of sine (cosine) type,
\[ R(x)=A\sin\frac{\pi}{2}x \quad \left(R(x)=A\cos\frac{\pi}{2}x\right). \]
In the work \((^1)\) we considered the case when the sequence \(\varepsilon\) is periodic, and established that \(R(x)=AR_\varepsilon(x)\), where \(A\geq 0\) and \(R_\varepsilon(x)\) does not depend on \(f(x)\).

Theorem 1. The set \(K_\varepsilon^*\) contains only one function, up to a nonnegative constant factor, if the sequence \(\varepsilon\) contains, an infinite number of times, one and the same combination consisting of \(p\geq 4\) terms of the form

\[ -\varepsilon,\ \varepsilon,\ \varepsilon,\ldots,\varepsilon,\ -\varepsilon,\quad \text{where } \varepsilon=\pm 1. \tag{4} \]

This condition will be called condition (A).

In view of the possibility of replacing \(x\) by \(1-x\) and \(f(x)\) by \(-f(x)\), we may assume, for definiteness, that \(\varepsilon_0=\varepsilon_1=1,\ x_0=0\). We restrict ourselves to the case \(\varepsilon=1\) (the case \(\varepsilon=-1\) is considered analogously).

The proof* of the theorem is based on the following lemmas.

Lemma 1. If condition (A) is satisfied, then the sequence (2) contains an unbounded number of zeros and ones.

Lemma 2. If

\[ f_1(x),\, f_2(x),\, \ldots,\, f_n(x),\, \ldots \tag{5} \]

is a sequence of functions belonging to \(K_\varepsilon\), and \(f_n(1)\le A,\ n=1,2,\ldots\), then from (5) one can extract a subsequence

\[ f_{n_1},\, f_{n_2},\, \ldots,\, f_{n_m},\, \ldots, \]

for which there exists the limit

\[ \lim_{m\to\infty} f_{n_m}^{(k)}(x)=f^{(k)}(x),\qquad k=1,2,\ldots,\quad x\in[0,1]. \]

We do not give the proof.

It follows from Lemma 2 that \(K_\varepsilon^*\) contains a nonvanishing function \(R(x)\), normalized by the condition \(R(1)=1\). Indeed, in place of (5) it suffices to take the sequence of polynomials

\[ \widetilde P_n(x)=\frac{1}{P_n(1)}P_n(x),\qquad \widetilde P_n(1)=1. \]

With the aid of Rolle’s theorem, Lemmas 3 and 4 are proved.

Lemma 3. If \(f(x)\) belongs to \(K_\varepsilon^*\) and \(f(1)=1\), then the equation

\[ f^{(k)}(x)-\widetilde P_n^{(k)}(x)=0,\qquad n=1,2,\ldots, \]

has one and only one root \(\xi_k\in(0,1)\) for \(k=1,2,\ldots,n\).

Lemma 4. If \(f(x)\) and \(g(x)\) belong to \(K_\varepsilon^*\), and the equation

\[ f^{(k)}(x)-g^{(k)}(x)=0,\qquad k=1,2,\ldots, \]

has a root \(\xi_{k_0}\in(0,1)\) for some nonnegative integer \(k_0\), then this equation has a root \(\xi_k\in(0,1)\) also for every \(k>k_0\).

Let \(n_1<n_2<\cdots<n_k<\cdots\) be the sequence of indices for which \(\varepsilon_{n_1},\varepsilon_{n_2},\ldots,\varepsilon_{n_k},\ldots\) are the first terms of the combination (4). Then \(\varepsilon_{n_k+m}=1\) for \(m=1,2,\ldots,p-2\); \(\varepsilon_{n_k}=\varepsilon_{n_k+p-1}=1\); \(x_{n_k+m}=0\) for \(m=1,2,\ldots,p-3\); \(x_{n_k}=x_{n_k+p-2}=1\). We prove the inequality

\[ \frac{p-2}{(p-1)!}\,P_{n_k}(1)\le P_{n_k+p-2}(1). \tag{6} \]

Indeed:

\[ P_{n_k+p-2}(x)=\frac{1}{(p-3)!}\int_{x_0}^{x}\tau_0\,dt_1\int_{x_1}^{t_1}\tau_1\,dt_2\cdots \int_{x_{n_k}-1}^{t_{n_k}-1}\tau_{n_k-1}\,dt_{n_k} \int_{t_{n_k}}^{1}t^{p-3}\,dt= \]

\[ =\frac{1}{(p-2)!}P_{n_k}(x)- \frac{1}{(p-1)!}\int_{x_0}^{x}\tau_0\,dt_1\int_{x_1}^{t_1}\tau_1\,dt_2\cdots \]

\[ \cdots\int_{x_{n_k}-1}^{t_{n_k}^{\,p-1}}\tau_{n_k-1}\,du \ge \frac{1}{(p-2)!}P_{n_k}(x)-\frac{1}{(p-1)!}P_{n_k}(x). \]

Putting \(x=1\), we obtain (6).

* The possibility of such a classical proof was pointed out to me by Prof. Ya. A. Tagamlitskii.

Lemma 5. If \(f(x)\in K_\varepsilon^*\), then:

I. \(\left|f^{(n)}(1-x_n)\right|\leqslant \left|f^{(n+1)}(1-x_{n+1})\right|\), \(n=0,1,2,\ldots\)

II. \(-f^{(n_k)}(0)\geqslant \dfrac{f(1)}{P_{n_k}(1)}\).

III. \(f^{(n_k+p-3)}(1)\leqslant \dfrac{f(1)}{P_{n_k+p-2}(1)}\).

Proof.

I.
\[ \left|f^{(n)}(1-x_n)\right| = \left|f^{(n)}(1-x_n)-f^{(n)}(x_n)\right| = |1-2x_n|\,\left|f^{(n+1)}(\xi)\right| \leqslant \left|f^{(n+1)}(1-x_{n+1})\right|. \]

II.
\[ f(1)= \int_{x_0}^{1}\tau_0dt_1 \int_{x_1}^{t_1}\tau_1dt_2\cdots \int_{x_{n_k-1}}^{t_{n_k}-1}\tau_{n_k-1}\varepsilon_{n_k}f^{(n_k)}(t)\,dt \leqslant \]
\[ \leqslant \varepsilon_{n_k}f^{(n_k)}(0)P_{n_k}(1) = -f^{(n_k)}(0)P_{n_k}(1). \]

III. If \(f(x)\) is a nonvanishing function, then \(f(1)\neq 0\). From Lemma 3 it follows that the equations
\[ \varphi(x)=\frac{1}{f(1)}f^{(n_k+p-3)}(x)-\widetilde P_{n_k+p-2}^{(n_k+p-3)}(x)=0 \]
and
\[ \varphi'(x)=0 \tag{1} \]
have, respectively, roots \(\xi'\) and \(\xi''\in(0,1)\). It is not hard to see that \(0<\xi''<\xi'\), and
\[ \widetilde P_{n_k+p-2}^{(n_k+p-3)}(x)=\frac{x}{P_{n_k+p-2}(1)}. \]
We assert that \(\varphi(1)\leqslant 0\). Indeed, suppose that \(\varphi(1)>0\). Then we would have
\[ \varphi(1)=\varphi(1)-\varphi(\xi') =(1-\xi')(\eta'-\xi'')\varphi''(\eta')>0, \]
where \(1>\eta'>\xi'>\xi''\); consequently,
\[ \varphi''(\eta')>0. \]
But
\[ \varphi_x=\frac{1}{f(1)}f^{(n_k+p-1)}(x)\leqslant 0,\qquad x\in[0,1], \]
i.e. \(\varphi(1)\leqslant 0\). On the other hand,
\[ \varphi(1)=\frac{1}{f(1)}f^{(n_k+p-3)}(1)-\frac{1}{P_{n_k+p-2}(1)}\leqslant 0, \]
which proves III.

Lemma 6. If \(f(x)\) and \(g(x)\) belong to \(K_\varepsilon^*\) and \(f(1)=1\),
\[ g(1)<\frac{p-2}{(p-1)!}, \]
then the function \(h(x)=f(x)-g(x)\) also belongs to \(K_\varepsilon^*\).

Proof. From Lemma 5 and (6) we find
\[ \left|g^{(n_k+1)}(1)\right| \leqslant \left|g^{(n_k+p-3)}(1)\right| \leqslant \frac{g(1)}{P_{n_k+p-2}(1)} < \frac{p-2}{(p-1)!}\frac{(p-1)!}{p-2}\frac{1}{P_{n_k}(1)} \leqslant \left|f^{(n_k)}(0)\right|, \]
i.e.
\[ \left|g^{(n_k+1)}(1)\right|<\left|f^{(n_k)}(0)\right|. \tag{7} \]

It is not hard to see that
\[ -f^{(n_k)}(x)\geqslant (1-x)\left|f^{(n_k)}(0)\right|, \qquad -g^{(n_k)}(x)\leqslant (1-x)\left|g^{(n_k+1)}(1)\right|, \tag{8} \]
since \(\varepsilon_{n_k}=-1\), \(\varepsilon_{n_k+1}=\varepsilon_{n_k+2}=1\). From (7) and (8) it follows that the equation
\[ f^{(n_k)}(x)-g^{(n_k)}(x)=0 \]
has no solution in the interval \((0,1)\). Hence, by Lemma 4, it follows that the function \(h(x)=f(x)-g(x)\) is regularly monotone on the segment \([0,1]\), since the sequence \(\{n_k\}\) is unbounded. On the other hand,
\[ h^{(n)}(x_n)=0,\qquad n=0,1,2,\ldots, \]
and \(h(x)\geqslant 0\) for \(x\in[0,1]\). From this it follows that \(h(x)\in K_\varepsilon^*\).

As we have already noted, there exists a nonvanishing function \(R_\varepsilon(x)\in K_\varepsilon^*\) for which
\[ R_\varepsilon(1)=1. \]

Lemma 7. If \(f(x)\in K_{\varepsilon}^{*}\), then \(f(x)=AR_{\varepsilon}(x)\), where \(A=f(1)\geq 0\).

Proof. The function

\[ h(x)=R_{\varepsilon}(x)-cf(x)\in K_{\varepsilon}^{*}, \tag{9} \]

if \(0\leq c<\dfrac{p-2}{f(1)(p-1)}\). Denote by \(c_{0}\) the greatest constant \(c\) satisfying (9). Such a constant exists by virtue of the closedness of \(K_{\varepsilon}^{*}\). We shall prove that \(R_{\varepsilon}(x)-c_{0}f(x)\equiv 0\). If this is not so, then \(R_{\varepsilon}(1)-c_{0}f(1)>0\), and, by Lemma 6, there exists a constant \(\rho>0\) such that \(h_{1}(x)=h(x)-\rho f(x)\in K_{\varepsilon}^{*}\). But \(h_{1}(x)=R_{\varepsilon}(x)-(c_{0}+\rho)f(x)\), which contradicts the choice of \(c_{0}\).

Using Lemma 2, one can show that the limit exists

\[ \lim_{n\to\infty}\frac{P_{n}(x)}{P_{n}(1)}=R_{\varepsilon}(x),\qquad x\in[0,1]. \]

It is not difficult to see that Lemma 7 is merely another formulation of Theorem 1.

Let us note that condition (A) is satisfied if and only if the sequence of type numbers \({}^{(5)}\) \(\lambda_{1},\lambda_{2},\ldots,\lambda_{n},\ldots\) contains a bounded partial subsequence.

If the sequence \(\varepsilon\) is periodic and its primitive period \(q\geq 3\), then the sequence \(\varepsilon\) satisfies condition (A).

In \({}^{(1)}\) it is proved that if \(\varepsilon\) is periodic and \(q\) is its period, then \(R_{\varepsilon}(x)\) satisfies the differential equation

\[ f^{(q)}(x)=\alpha^{q}f(x),\qquad \alpha>0. \]

This assertion follows from Lemma 7 owing to the fact that, if \(f(x)\in K_{\varepsilon}^{*}\), then also \(f^{(q)}(x)\in K_{\varepsilon}^{*}\).

Mathematical Institute
of Sofia University
Sofia, Bulgaria

Received
3 VI 1957

References

1. B. Sendov, DAN, 110, No. 1, 27 (1956).
2. Ya. A. Tagamlitski, Yearbook of Sofia Univ., 48, book 1, part 1, 69 (1953—1954).
3. S. N. Bernstein, Collected Works, 1, 1952, pp. 370—425.
4. S. N. Bernstein, Collected Works, 2, 1954, pp. 493—516.
5. S. N. Bernstein, Collected Works, 1, 1952, pp. 350—360.