On a theorem of W.A. Markov

On the occasion of generalizing AA Markov's famous inequality, WA Markov gave [2], as a helpful lemma, the following theorem:

If the roots of two polynomials of degree$n$, with all real roots, separate, then the roots of their derivatives also separate.

In the second part of the paper we will give a proof of this theorem, slightly different from that of WA Markov himself as well as from that of P. Monte1 [3], given about 30 years ago.

The proof we give is based on the continuity and monotonicity of the roots of the derivative of a polynomial with all real roots, with respect to the roots of the polynomial. In the first part of this paper we will analyze this monotonicity property a little.

Finally, in the third part of the paper we will give a new theorem on polynomials with all real roots, analogous to the one cited by$W$A. Markov.

In what follows we will consider only polynomials of one variable with all real roots, and by the degree of a polynomial we will understand its effective degree, even if these things are not explicitly specified. Since we are only interested in the roots of polynomials, two polynomials that differ only by a multiplicative constant can be considered equal. The emphasis on polynomials means derivation.

1. 1.

   If a polynomial has all its roots real, then its derivative also has all its roots real. There is an important, well-known property of the separation of the roots of the derivative by the roots of the polynomial. We will take this property into account below.

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The roots of a polynomial with the highest coefficient equal to 1 (so a polynomial of the form$x^{n}+$polynomial of degree 1$<n$) are continuous functions with respect to the coefficients of the polynomial, and the coefficients are continuous functions (polynomials) with respect to the roots of the polynomial. If we take into account the relations between the roots and coefficients of a polynomial, we deduce the continuity of the roots of the derivative with respect to the roots of the polynomial.
2. The monotonicity property of the roots of the derivative with respect to the roots of the polynomial can be stated in the following form:

The roots of the derivative are non-decreasing functions of the roots of the polynomial.

This property is well known and was widely used by Laguerre in his research on polynomials with all real roots.

To clarify the monotonicity property above, we introduce the relation$P_{\rightarrow}^{\mathrm{c}}Q$between two polynomials which occurs if and only if:
$1^{0}$Polynomials$P,Q$are of the same grade$n\geqq 1$.
$2^{0}$The respective roots

of these polynomials verify the inequalities

This relation is (reflexive and) transitive.
It is unnecessary to consider the case$n=0$when the previous relation makes no sense (because there are no roots). If$n=1$, in the definition, it is sufficient to maintain only inequality (2) and it is easy to see that, in this case, at least one of the relations$P\xrightarrow{\mathrm{c}}Q,Q\xrightarrow{\mathrm{c}}P$always takes place. For any$n>1$polynomials can be constructed$P,Q$for which none of the relationships$P\xrightarrow{\mathrm{c}}Q,Q\xrightarrow{\mathrm{c}}P$It is not true.

The monotonicity property of the roots of the derivative with respect to those of the polynomial is then expressed by

Theorem 1. If$P,Q$are two polynomials of degree$n>1$, from$P\xrightarrow{\mathrm{c}}Q$result$P^{\prime}\xrightarrow{\mathrm{c}}Q^{\prime}$.

We also introduce the relation$P\xrightarrow{\mathrm{cc}}Q$between two polynomials, which occurs if and only if:
$1^{0}$Polynomials$P,Q$have the same degree$n\geqq 1$and both have all simple roots.
$2^{0}$The respective roots

of these polynomials verify the inequalities

This relation, which is transitive, is a particular case of the previous relation, namely when everywhere in (1) and (2) the sign$\leqq$is replaced with$<$The two relations are also related by a mixed transitivity property, analogous to the mixed transitivity of inequality relations.$<\leq$i$\leqq$If$P\xrightarrow{\mathrm{cc}}R,R\xrightarrow{\mathrm{c}}Q$and if$Q$has all simple roots, we have$P\xrightarrow{\mathrm{cc}}Q$From$P\xrightarrow{\mathrm{cc}}R,R\xrightarrow{\mathrm{c}}S,S\xrightarrow{\mathrm{cc}}Q$result$P\xrightarrow{\mathrm{cc}}Q$.

We have the following:
If the roots of$P,Q$are continuous functions of one parameter$\lambda$on an interval containing$\lambda_{0}$and if we have$P\xrightarrow{\mathrm{cc}}Q$for$\lambda\neq\lambda_{0}$, we will have$P\xrightarrow{\mathrm{c}}Q$, but not in general$P\xrightarrow{\mathrm{cc}}Q$, for$\lambda=\lambda_{0}$This observation is also valid for pairs of relationships.$\xrightarrow{s},\xrightarrow{ss};\xrightarrow{m},\xrightarrow{mm};\xrightarrow{mc},\xrightarrow{mcmc}$, which will be considered below.

We have the following:
Theorem 2. If$P,Q$are two polynomials of degree$n>1,\operatorname{from}P\xrightarrow{\mathrm{cc}}Q$result$P^{\prime}\xrightarrow{\mathrm{cc}}Q^{\prime}$.
3. We will show that Theorem 1 follows from Theorem 2.

Indeed, either$P\xrightarrow{\mathrm{c}}Q$, (1) the roots of polynomials$P,Q,n>1$i

respectively the roots of polynomials$P^{\prime},Q^{\prime}$Let 's
consider the polynomials$P_{8},Q_{8}$of the degree$n$, having the roots respectively$x_{i}+i\varepsilon,\quad i=1,2,\ldots,n,y_{i}+(i+1)\varepsilon,i=1,2,\ldots,n$, where$\varepsilon$is a positive number. Polynomials$P_{8},Q_{8}$have all simple roots and we have$P_{\varepsilon}\xrightarrow{\mathrm{cc}}Q_{\varepsilon}$If

are respectively the roots of the polynomials$P_{\mathrm{s}}^{\prime},Q_{\mathrm{s}}^{\prime}$, we have

If we assume that Theorem 2 is true, it follows that$P_{\varepsilon}^{\prime}\xrightarrow{cc}Q_{\varepsilon}^{\prime}$and from (4) we deduce, making$\varepsilon\rightarrow 0$, that we have$P^{\prime}\xrightarrow{\mathrm{c}}Q^{\prime}$.

This proves that Theorem 1 follows from Theorem 2.
4. It remains to prove Theorem 2.

Whether$P,Q$two polynomials of degree 1l$n>1$so that we have$P\xrightarrow{\mathrm{cc}}Q$and let (1') be the roots of these polynomials respectively. Let$P_{i}$a polynomial of degree
$n$having as its roots$x_{1},x_{2},\ldots,x_{n-i},y_{n-i+1},y_{n-i+2},\ldots,y_{n}$, for$i=1,2,\ldots,n$The polynomial$P_{0}$is equal to$P$i$P_{n}$is equal to$Q$Polynomials$P_{i}$have all simple roots, but we have, in general, only$P_{i}\xrightarrow{\mathrm{c}}P_{i+1}$,$i=0,1,2,\ldots,n-1$But, if we can prove that we have

then, based on transitivity, it follows that$P^{\prime}\xrightarrow{\text{ cc }}Q^{\prime}$and Theorem 2 is proven.

It remains to prove relations (5). These relations follow from
Lemma 1. The roots of the derivative of a polynomial with all real and simple roots are increasing functions with respect to each of the roots of the polynomial.

Whether$g$a polynomial of degree$n(\geqq 1)$with all the roots$\alpha_{1}<\alpha_{2}<<\ldots:<\alpha_{n}$real and simple. By the statement of Lemma 1 we understand that each of the roots of the derivative of the polynomial$f_{\alpha}=(x-\alpha)g$is an increasing function of$\alpha$These roots$\beta_{1}<\beta_{2}<\ldots<\beta_{n}$are continuous functions of$\alpha$and remain distinct. We will first prove that they are strictly monotone functions of$\alpha$Indeed, if, for example,$\beta_{k}$would not be a strictly monotonic function of$\alpha$, we could find two different values$\alpha^{\prime},\alpha^{\prime\prime}$his/hers$\alpha$for which the polynomials

to have a common root$\beta_{k}$This is impossible, however, because any common root of the polynomials (6) would have to be a common root of the polynomials$g,g^{\prime}$, which contradicts the fact that$g$has only simple roots. With this strict monotonicity of the roots$\beta_{i}$is demonstrated. It remains only to specify the meaning of this monotony. If we take into account the fact that the roots of the derivative are separate from those of the polynomial and if we observe that

we immediately deduce that the sense of monotony is increasing for each of the roots$\beta_{i}$.

With this Lemma 1 is proven.
Other proofs of Lemma 1 can be given. Proofs can be given based on some considerations analogous to those made in Part II and Part III of this paper. We will not deal with such proofs.

Observation. If$\alpha_{1}^{\prime}<\alpha_{2}^{\prime}<\ldots<\alpha_{n-1}^{\prime}$are the roots of the polynomial$g^{\prime}$, the roots$\beta_{i},i=1,2,\ldots,n$varies respectively in the intervals$\left(-\infty,\alpha_{1}\right]$,$\left[\alpha_{i-1},\alpha_{i}^{\prime}\right],i=2,3,\ldots,n-1,\left[\alpha_{n-1}^{\prime},\infty\right)$, if$n>2$If$n=1$, the root$\beta_{1}$varies from$-\infty$TO$\infty$, and if$n=2$ROOTS$\beta_{1},\beta_{2}$varies respectively in the intervals$\left(-\infty,\frac{\alpha_{1}+\alpha_{2}}{2}\right],\left[\frac{\alpha_{1}+\alpha_{2}}{2},\infty\right)$.

1. 5.

   We will now deal with the reported proof of WA Markov's theorem.

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We introduce the relationship$P\xrightarrow{s}Q$between two polynomials, which occurs if and only if:
$1^{\circ}$Polynomials$P,Q$have the same degree$n\geqq 1$.
$2^{\circ}$The respective roots (1) of these polynomials verify the inequalities

If$P\xrightarrow{\mathrm{\penalty 10000\ s}}Q$or$Q\xrightarrow{\mathrm{\penalty 10000\ s}}P$, we can say that the roots of polynomials$P,Q$separate.

Generally from$P\xrightarrow{\mathrm{\penalty 10000\ s}}Q$result$P\xrightarrow{\mathrm{c}}Q$, and for$n=1$, relationships$P\xrightarrow{\mathrm{c}}Q$,$P\xrightarrow{\mathrm{\penalty 10000\ s}}Q$are equivalent.

Based on a previous observation, for any$n>1$we can find the polynomials$P,Q$of the degree$n$so that none of the relationships$P\xrightarrow{\mathrm{\penalty 10000\ s}}Q,Q\xrightarrow{\mathrm{\penalty 10000\ s}}P$not to be checked.

WA Markov's theorem can be stated in the following form:
Theorem 3. If$P,Q$are two polynomials of degree$n>1$, from$P\xrightarrow{\mathrm{\penalty 10000\ s}}Q$result$P^{\prime}\xrightarrow{\mathrm{s}}Q^{\prime}$.

If$n=2$, Theorem 3 follows from Theorem 1. Indeed, from$P\xrightarrow{s}Q$result$P\xrightarrow{\mathrm{c}}Q$from which, based on Theorem 1, it follows$P^{\prime}\xrightarrow{\mathrm{c}}Q^{\prime}$This relationship is however (for$n=2$) equivalent to$P^{\prime}\xrightarrow{\mathrm{s}}Q^{\prime}$and the property is proven.

We also introduce the relationship$P\xrightarrow{\text{ ss }}Q$between two polynomials, which occurs if and only if:
$1^{\circ}$Polynomials$P,Q$have the same degree$\geqslant 1$and both have all their roots simple.
$2^{\circ}$The respective roots ($1^{\prime}$) of these polynomials verify the inequalities

from$P\xrightarrow{\mathrm{ss}}Q$result$P\xrightarrow[\rightarrow]{\mathrm{cc}}Q$, and for$n=1$these relations are equivalent.
We have the following particular case of WA Markov's theorem:
Theorem 4. If$P,Q$are two polynomials of degree$n>1$, from$P\xrightarrow{\mathrm{ss}}Q$, it results$P^{\prime}\xrightarrow{\mathrm{ss}}Q^{\prime}$.

As above it is demonstrated that for$n=2$, Theorem 4 follows from Theorem 2.
6. It is sufficient to prove Theorem 4, because then Theorem 3 follows. To show this we proceed in the same way as in No. 3, where we showed that Theorem 1 follows from Theorem 2.

If we have$P^{\mathrm{s}}Q$and if we now consider the polynomials$P_{\varepsilon},Q_{\varepsilon}$having respectively as roots$x_{i}+(2i-1)\varepsilon,i=1,2,\ldots,n,y_{i}+2i\varepsilon,i=1,2,\ldots,n$, where$\varepsilon$is a positive number, we have$P_{\varepsilon}\xrightarrow{ss}Q_{\varepsilon}$If we assume that Theorem 4 is true, it follows that$P_{\varepsilon}^{\prime}\mathrm{ss}Q_{\varepsilon}^{\prime}$If we do$\varepsilon\rightarrow 0$, his roots$P_{\varepsilon}^{\prime},Q_{\varepsilon}^{\prime}$tend respectively towards its roots$P^{\prime},Q^{\prime}$and we deduce$P^{\prime}\xrightarrow{\mathrm{s}}Q^{\prime}$. Theorem 3 is proved.
7. Theorem 4 is proved based on Theorem 2, Lemma 1 and the continuity of the roots of the derivative.

If$P\xrightarrow{\text{ ss }}Q$, it results$P\xrightarrow{\mathrm{cc}}Q$, so$P^{\prime\text{ cc }}Q^{\prime}$. To show that we even have$P^{\prime}\xrightarrow{\text{ ss }}Q^{\prime}$, it is sufficient to prove that the derivatives of the polynomials$P,Q$(which all have simple roots) cannot have any common root. Indeed, it is easy to see that, in this case, the relation$P^{\prime}\xrightarrow{\text{ ss }}Q^{\prime}$it is maintained when its roots$P$grow towards their respective roots$Q$.

But, if$\left(1^{\prime}\right)$are the roots of the polynomials$P,Q$of the degree$n$, the relationship$P\xrightarrow{ss}Q$is equivalent to equality

where$a$is a non-zero constant,$\operatorname{iar}a_{1},a_{2},\ldots,a_{n}$ARE$n$constants different from zero and of the same sign. Moreover, the product$aa_{i}$is of opposite sign to the highest coefficient of$P$, so with the sign of$P$for$x$very big.

By derivation from (8) we deduce

From here it is seen that if$P^{\prime},Q^{\prime}$would have a common root, this should also cancel the polynomial$\frac{D}{P}$, which is impossible, because, by hypothesis,$P$has all its roots simple.
8. Relationships$P\xrightarrow{\mathrm{\penalty 10000\ s}}Q,P\xrightarrow{\mathrm{ss}}Q$can also be extended to the case when the polynomial$P$is of the degree$n$and the polynomial$Q$of the degree$n-1$If$n>1$i

are respectively the roots of$P$and his$Q$, the relationship$P\xrightarrow{\mathrm{\penalty 10000\ s}}Q$occurs if and only if

It can still be said that then the roots of$P$i$Q$they separate.
The relationship$P\xrightarrow{ss}Q$occurs if and only if, in addition, the roots of$P$i$Q$are all simple and instead of inequalities (11) we have

We then have:
Corollary 1. If$P$is a polynomial of degree$n$i$Q$a polynomial of degree$n-1,n>1$, from$P\xrightarrow{\mathrm{\penalty 10000\ s}}Q$result$P^{\prime}\xrightarrow{\mathrm{s}}Q^{\prime}$.

The property follows from Theorem 3 by a passage to the limit. To show this, let (10) be the roots of$P$i$Q$, which verifies the relationship$P\xrightarrow{s}Q$Let us consider the polynomial$R=\left(x-y_{n}\right)Q$of the degree$n$If$x_{n}\leqq y_{n}$, we have$P\xrightarrow{\mathrm{\penalty 10000\ s}}R$, from which, based on Theorem 3 , we deduce$P^{\prime}\xrightarrow{\mathrm{s}}R^{\prime}$If we do$y_{n}\rightarrow\infty$, one of its roots$R^{\prime}$(the largest) tends to$\infty$and the others to the respective roots of$Q^{\prime}$Taking into account the continuity of the roots of the derivative with respect to the roots of the polynomial, it is seen that by doing$y_{n}\rightarrow\infty$, from$P^{\prime}\xrightarrow{\mathrm{s}}R^{\prime}$result$P^{\prime}\xrightarrow{\mathrm{s}}Q^{\prime}$.

We also have:
Consecration 2. If$P$is a polynomial of degree$n$i$Q$a polynomial of degree$n-1,n>1$, from$P\xrightarrow{\mathrm{ss}}Q$result$P^{\prime}\xrightarrow{\mathrm{ss}}Q^{\prime}$:

This property follows from Theorem 4 as well as Corollary 1 from Theorem 3. We form the polynomial as above$R$If$P\xrightarrow{\text{ ss }}Q$i$x_{n}<y_{n}$, we have$P^{\text{ss }}R$and therefore, based on the theorem$4,P^{\prime\text{ ss }}R^{\prime}$. Hence, if we do$y_{n}\rightarrow\infty$, it results$P^{\prime}\xrightarrow{s}Q^{\prime}$, and to show that we even have$P^{\prime}\xrightarrow{ss}Q^{\prime}$it is enough to prove that if$P^{\mathrm{ss}}Q$, polynomials$P^{\prime},Q^{\prime}$they cannot have any common root.

If$P\xrightarrow{\mathrm{ss}}Q$we have formula (8), where$a=0$i$a_{i},i=1,2,\ldots,n$. are$n$constants different from zero and of the same sign. Formula (9) shows us that$P^{\prime},Q^{\prime}$they cannot have any common root.

Corollary 2 is proved.
9. As an application, consider a sequence of orthogonal polynomials

It is known that its roots$\Pi_{n}$are all real and simple and that its roots$\Pi_{n-1}$are separated, in a strict sense, by the roots of$\Pi_{n}$In other words for$n>1$, we have$\Pi_{n}\xrightarrow{ss}\Pi_{n-1}$From consequence 2 it therefore also follows:

Consequence 3. If$\Pi_{n-1},\Pi_{n}$are two consecutive terms of a sequence of orthogonal polynomials$(n>1)$, we have$\Pi_{n}^{\prime}\xrightarrow{\mathrm{ss}}\Pi_{n-1}^{\prime}$.

1. 10.

   We will now deal with a theorem analogous to WA Markov's.

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We introduce the relationship$P\xrightarrow{m}Q$between two polynomials, which occurs if and only if:
$1^{\circ}$Polynomials$P,Q$have the same degree$n\geqq 1$.
$2^{\circ}$The roots (1) of these polynomials verify the inequalities

as well as equality

If$n=1$we keep only the equality (13) from the definition.
If$n=1$the relationship$P^{\mathrm{m}}\rightarrow Q$means that$P,Q$have the same root, so that - according to the meaning adopted at the beginning of this work - they are equal. It is clear that for any$n\geqq 1$we can find two polynomials$P,Q$of the degree$n$, so that none of the relationships$P\xrightarrow{\mathrm{\penalty 10000\ m}}Q,Q\xrightarrow{\mathrm{\penalty 10000\ m}}P$not to be checked.