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