Notes on Higher-Order Convex Functions (VI)

1.

We will consider functions $f=f(x)$ , uniform and defined on == bounded and closed linear set E . This set can be finite or infinite, but when it is a function of order $n$ he has at least $n+2$ points. We will designate by $\mathrm{a}=\min\mathrm{E},\mathrm{a}<\mathrm{b}=\max\mathrm{E}$ the ends of E . We will say that a function is continuous resp. semi-continuous on E if it is so on the derivative E ' of E . We will say that a subset of E is a section of E if it is formed by all the points of E belonging to a closed interval ( $\mathrm{c},\mathrm{d}$ ), $\mathrm{c}\leqslant\mathrm{d}$ . We will designate such a subset by (cEd). We will say that the sections (cFd), ( $\mathrm{c}_{1}\mathrm{Ea}_{1}$ ) are separated by E if $\mathrm{d}<\mathrm{c}_{1}$ and if the open interval $\left(\mathrm{d},\mathrm{c}_{1}\right)$ contains at least one point of E . Several sections of E are separated by E if they are two by two separated by F. We will also say that these are separate sections of E. For all other notations and properties of functions of order n the reader is asked to refer to our previous work.

A non-concave function of odd order on E is upper semicontinuous and therefore always reaches its maximum. But such a function may not reach its minimum. An even-order function $>0$ may not reach its maximum or its minimum. This is, for example, the function

\mathrm{f}(0)=\mathrm{f}(1)=0,\quad\mathrm{f}(\mathrm{x})=1-2\mathrm{x},\quad 0<\mathrm{x}<1

If the maximum or minimum of an order function $n$ is reached in $n+2$ points it is reached at all points of the smallest section which contains these points.
2. To simplify let us set $k=\frac{n}{2}$ Or $\frac{n+1}{2}$ following that $n$ is even or odd. We then have the following property

If the continuous function f is non-concave of order in on E , the set $\mathrm{E}(\mathrm{M})$ Or $\mathrm{M}=\max\mathrm{f}$ is reached is formed by at most $\mathrm{k}+1$ separate sections of E .

For simplicity we write max, min instead of max, min.
( $\varepsilon$ ) ( $\equiv$ )
The demonstration is simple. Indeed, if $\mathrm{E}(\mathrm{M})$ had another structure we could find $2\mathrm{k}+2$ points $\mathrm{x}_{1}<\mathrm{x}_{1}^{\prime}<\ldots<\mathrm{x}_{\mathrm{k}+1}<\mathrm{x}_{\mathrm{k}+1}^{\prime}$ of E so that

		$\displaystyle\mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right)=\ldots=\mathrm{f}\left(\mathrm{x}_{\mathrm{k}+\mathrm{i}}\right)=\mathrm{M}$
		$\displaystyle\mathrm{f}\left(\mathrm{x}_{\mathrm{i}}\right)<\mathrm{M},\quad\mathrm{i}=2,\ldots,\mathrm{k}+1$

and we would then have

\begin{array}[]{ll}{\left[x_{1},x_{1}^{\prime},x_{2},x_{2}^{\prime},\ldots,x_{k+1},x_{k+1}^{\prime};f\right]<0}&\text{ si }n\text{ est pair, }\\ {\left[x_{1}^{\prime},x_{2},x_{2}^{\prime},\ldots,x_{k+1},x_{k+1}^{\prime};f\right]<0}&\text{ si }n\text{ est impair, }\end{array}

which is in contradiction with the hypothesis of non-concavity of order $n$ of f.
We can specify the structure of the whole $\mathrm{E}(\mathrm{M})$ .
If n is odd and if $\mathrm{E}(\mathrm{M})$ contains at least n points it contains at least one of the ends a, b. If $\mathrm{E}(\mathrm{M})$ contains at least $\mathrm{n}+1$ points they contain both ends a, b. So for a non-concave function of odd order $\mathrm{E}(\mathrm{M})$ cannot contain more than $n+1$ points without the function reducing to a constant on E.

If $n$ is even and $E(M)$ contains more than $n$ points it contains the end b.
In general, if $\mathrm{E}(\mathrm{M})$ is not formed by a single section of E it contains at most $\mathrm{n}+1$ points.

If $n$ is odd and if $\mathrm{E}(\mathrm{M})$ is formed by k separate sections of E it contains a or b and if it is formed by $\mathrm{k}+1$ separate sections of E it contains a and b.

If n is even and if $\mathrm{E}(\mathrm{M})$ is formed by $\mathrm{k}+1$ separate sections of E it contains the end b.

Let us retain, in particular, the following property:
If the continuous function f is non-concave of order n on E, the set $\mathrm{E}(\mathrm{M})$ verifies the following property:
A. $\mathrm{E}(\mathrm{M})$ cannot be formed by at least $\mathrm{n}-\mathrm{k}+1$ separate sections of E , without containing one of the ends $\mathrm{a},\mathrm{b}$ for odd n and the end b for even n.

If E reduces to an interval $(a,b)$ and if n is odd, $\mathrm{E}(\mathrm{M})$ contains at most $k+1$ points unless $f$ is not a constant in $(a,b)$ . If $n$ is even $\mathrm{E}(\mathrm{M})$ contains or at most $\mathrm{k}+\mathrm{l}$ points or is formed by an interval $(\mathrm{c},\mathrm{b}),\mathrm{a}\leq\mathrm{c}$ .

There is no need to specify the structure of $\mathrm{E}(\mathrm{M})$ .
3. We can find analogous results for the minimum.

If the continuous function f is non-concave of order n on E, the set $\mathrm{E}(\mathrm{m})$ Or $\mathrm{m}=\min\mathrm{f}$ is reached is formed by at most $\mathrm{n}-\mathrm{k}+1$ separate sections of E .

The structure of $\mathrm{E}(\mathrm{m})$ can be specified easily. For example, as soon as $\mathrm{E}(\mathrm{m})$ is not reduced to a section it contains at most $\mathrm{n}+1$ points. For even-order functions the structure of $\mathrm{E}(\mathrm{m})$ is also deduced from the structure of $\mathrm{E}(\mathrm{M})$ , noting that the function $-\mathrm{f}(-\mathrm{x})$ is also non-concave of order n .

In particular therefore
B. If n is even, the set $\mathrm{E}(\mathrm{m})$ cannot be formed by at least $\mathrm{k}+1$ separate sections of E without containing the end a.

For non-concave functions of odd order, we retain the following property
C. If n is odd, the set $\mathrm{E}(\mathrm{m})$ is formed by at most k separate sections of E .

If E reduces to an interval $(\mathrm{a},\mathrm{b}),\mathrm{F}(\mathrm{m})$ contains or at most $\mathrm{n}-\mathrm{k}+1$ points or is reduced to an interval which contains the endpoint a for even n.
4. If the function $f$ is non-concave of order $n$ the same is true of the function $\mathrm{f}-\mathrm{P},\mathrm{P}$ being a polynomial of degree $\mathrm{n}\left({}^{1}\right)$ . If f is not a polynomial of order n we can always determine the polynomial $P$ so that the whole $E(M)$ , $\mathrm{E}(\mathrm{m})$ corresponding to $\mathrm{f}-\mathrm{P}$ are formed by the maximum number of sections
(1) A polynomial of degree n for us is a polynomial of effective degree $\leq\mathbf{n}$ .

Lies of E, so that $\mathrm{E}(\mathrm{M})+\mathrm{E}(\mathrm{m})$ be formed by $\mathrm{n}+2$ separate sections of È. in fact, $T_{i}$ the best approximation polynomial of Tchebycheff of the deme of the function $f$ on $E$ . $T_{i}$ is therefore the (unique) polynomial for which

\min\max|\mathrm{f}-\mathrm{P}|\text{, }

reached when P is a polynomial of degree i. We then have the -irant property ( ¹ )

If $\mathrm{T}_{\mathrm{n}}$ is the Chebyshefi polynomial of degree nd $\geq$ the function $f$ cone-shaped and non-concave of order n, not reducing to a polynomial of degree n, == can find $\mathrm{n}+2$ and only $\mathrm{n}+2$ consecutive points where the difference $\mathrm{T}_{\mathrm{n}}$ reached the maximum value $\left|\mathrm{f}-\mathrm{T}_{\mathrm{n}}\right|$ with alternating signs.

This property is equivalent to the following
If f is a continuous function of order n, not reducing to a polynomial of degree n, the polynomials $\mathrm{T}_{\mathrm{n}},\mathrm{T}_{\mathrm{n}+1}$ are distinct, therefore $\mathrm{T}_{\mathrm{n}+1}$ is actually of degree $\mathrm{n}+1$ .

If f is non-concave of order n , the function $\mathrm{f}-\mathrm{T}_{\mathrm{n}}$ therefore enjoys the property that $\mathrm{E}(\mathrm{M})$ is formed by $\mathrm{k}+1$ And $\mathrm{E}(\mathrm{m})$ by $\mathrm{n}-\mathrm{k}+1$ separate sections of E , unless, of course, f is not polynomial.

Let us also note the following property
If $\mathrm{T}_{\mathrm{n}}$ is the Tchebycheff polynomial of degree n of the continuous function f and of order 11, the max value $\left|\mathrm{f}-\mathrm{T}_{\mathrm{n}}\right|$ is necessarily reached at the ends a and b. We have moreover $\left[\mathrm{f}(\mathrm{a})-\mathrm{T}_{\mathrm{n}}(\mathrm{a})\right]\left[\mathrm{f}(\mathrm{b})-\mathrm{T}_{\mathrm{n}}(\mathrm{b})\right]\geq$ Or $\leq 0$ depending on whether n is odd or even.
5. We will now establish, and this is the main aim of this work, the reciprocals of the preceding properties and first those of the properties $\mathrm{A},\mathrm{B},\mathrm{C}$ . Before arriving at these properties we will deal with an auxiliary problem.

Let $\mathrm{x}_{1}>\mathrm{x}_{2}>\ldots>\mathrm{x}_{\mathrm{r}},\mathrm{r}$ points of E such that each open interval $\left(\mathrm{x}_{\mathrm{i}},\mathrm{x}_{\mathrm{i}+1}\right),\mathrm{i}=1,2,\ldots,r-1$ contains at least one point of E . The points $\mathrm{x}_{\mathrm{r}},\mathrm{x}_{1}$ may or may not coincide with the ends $\mathrm{a},\mathrm{b}$ . Let us pose

\text{ et }\begin{array}[]{ll}F_{0}=\left(x_{1}Eb\right)&\text{ si }x_{1}<b\\ F_{r}=\left(aEx_{r}\right)&\text{ si }a<x_{r}\end{array}

We therefore have, depending on the case, $r-1,r$ Or $r+1$ sets $F_{i}$ . We will also assume that if $r=1,a<x_{1}<b$ and if $r=2$ we don't have at the same time $\mathrm{a}=\mathrm{x}_{2},\mathrm{x}_{1}=\mathrm{b}$ . So the number $\mathrm{s}+2$ sets $\mathrm{F}_{\mathrm{i}}$ is still $\geq 2$ , SO $s\geq 0$ .

Now consider a continuous function $f$ which cancels out at the points $x_{i}$ and such that each $F_{i}$ contains at least one point $x_{i}$ where it takes a positive value

Let's ask

R=\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{r}\right)

and either

\mu=\min\max_{(\mathrm{E})}(\mathrm{f}-\mathrm{RQ}),

when $Q$ runs through the set of polynomials of degree s.
We have $\mu>0$ .
We obviously have $\mu\geqslant 0$ . Either

\delta=\left[x_{1},x_{2},\ldots,x_{r},x_{o}^{\prime},x_{1}^{\prime},\ldots,x_{r}^{\prime};f\right],

with the condition of deleting the point $\mathrm{x}^{\prime}{}_{,}$ if $\mathrm{x}_{1}=\mathrm{b}$ and the point $\mathrm{x}^{\prime}{}_{\mathrm{r}}$ if $\mathrm{x}_{\mathrm{r}}=\mathrm{a}.\delta$ is therefore a divided difference of order $\mathrm{r}+\mathrm{s}+1$ . On the other hand
(1)

\left[x_{1},x_{2},\ldots,x_{r},x_{o}^{\prime},x_{1}^{\prime},\ldots x_{r}^{\prime};f-RQ\right]=\delta,

whatever the polynomial $Q$ of degree $s$ .
If we had $\mu=0$ we could find, whatever $\frac{|\delta|}{2}>\varepsilon>0$ , a polynomial $Q$ such as

\left[\mathrm{x}_{1},\mathrm{x}_{2},\ldots,\mathrm{x}_{\mathrm{r}},\mathrm{x}_{0}^{\prime},\mathrm{x}_{1}^{\prime},\ldots,\mathrm{x}_{\mathrm{r}}^{\prime};\mathrm{f}-\mathrm{R}Q\right]>-\varepsilon>\frac{\delta}{2}\text{ ou }<\varepsilon<\frac{\delta}{2}

following that $x_{1}=b$ Or $x_{1}<b$ , which is in contradiction with (1).
6. Let us now demonstrate the

Lemma I. If the polynomial $Q=c_{0}\mathrm{x}^{s}+c_{1}\mathrm{x}^{s-1}+\ldots+c_{s}$ , Or $-Q$ checks for inequality

\mathrm{R}(\mathrm{x})Q(\mathrm{x})<\mathrm{A},\quad\mathrm{x}\subset\mathrm{E},\quad(\mathrm{\penalty 10000\ A}>0),

(2)

we can find a positive number B , depending on A but not on the polynomial Q, such that we have

\left|c_{i}\right|<B,i=0,1,\ldots,s

This lemma follows from the following
Lemma II. If the polynomial $Q=c_{0}x^{s}+c_{1}x^{s-1}+\ldots+c_{s}$ , Or $-Q$ checks for inequality
(3)

(-1)^{\mathrm{i}}\mathrm{Q}\left(\mathrm{y}_{\mathrm{i}}\right)<\mathrm{A}^{*},\quad\mathrm{i}=1,2,\ldots,\mathrm{\penalty 10000\ s}+2,\quad\left(\mathrm{\penalty 10000\ A}^{*}>0\right),

we can find a positive number $\mathrm{B}^{*}$ , depending on $\mathrm{A}^{*}$ but not of the polynomial Q, such that we have

\left|c_{i}\right|<B^{*},\quad i=0,1,\ldots,s.

Let us first demonstrate the property for the first coefficient $\mathrm{c}_{0}$ . If we pose $\mathrm{W}=\left(\mathrm{x}-\mathrm{y}_{1}\right)\left(\mathrm{x}-\mathrm{y}_{2}\right)\ldots\left(\mathrm{x}-\mathrm{y}_{\mathrm{s}+2}\right)$ , the Lagrange interpolation formula gives us

c_{0}=\sum_{i=1}^{s+1}\frac{\left(y_{i}-y_{s+2}\right)Q\left(y_{i}\right)}{W^{\prime}\left(y_{i}\right)}=\sum_{i=2}^{s+2}\frac{\left(y_{i}-y_{1}\right)Q\left(y_{i}\right)}{W^{\prime}\left(y_{i}\right)}

But

(-1)^{\mathrm{i}}\mathrm{\penalty 10000\ W}^{\prime}\left(\mathrm{y}_{\mathrm{i}}\right)>0,\quad\mathrm{i}=1,2,\ldots,\mathrm{\penalty 10000\ s}+2

and we deduce from it

-A^{*}\cdot\sum_{i=2}^{s+2}\frac{y_{1}-y_{i}}{(-1)^{i}W^{\prime}\left(y_{i}\right)}<c_{0}<A^{*}\cdot\sum_{i=1}^{s+1}\frac{y_{i}-y_{s+2}}{(-1)^{i}W^{\prime}\left(y_{i}\right)}.

()

Now suppose that the property is true for the coefficients $\mathrm{c}_{0}$ , $\mathrm{c}_{1},\ldots,\mathrm{c}_{\mathrm{j}-1}$ and demonstrate it for the coefficient $\mathrm{c}_{\mathrm{j}}$ . We have

\left|c_{i}\right|<B_{1}^{*},\quad i=0,1,\ldots,j-1,

(1) We deduce an analogous limitation if - ? verifies inequality (3).

Bit being a constant depending on $A^{*}$ but no $Q$ . If we pose $Q_{1}=c_{j}x^{s}i+-c_{j+},x^{s-j-1}+\ldots+c_{s}$ , We have
$(-1)^{i}Q_{1}\left(y_{i}\right)=(-1)^{i}Q\left(y_{i}\right)-(-1)^{i}\left(c_{0}x^{s}+c_{1}x^{s-1}+\ldots+c_{j-1}x^{sj+1}\right)$ , Bonc

\begin{gathered}(-1)^{i}Q\left(y_{i}\right)<A_{1}^{*}=A^{*}+B_{1}^{*}\max_{t=1,2,\ldots,s+2}\left(y_{t}{}^{s}+\left|y_{t}\right|^{s-1}+\ldots+y_{t}{}^{s-j+1}\right)\\ i=1,2,\ldots,s+2-j\end{gathered}

And $\mathrm{A}_{1}^{*}$ depends on $\mathrm{A}^{*}$ only. The limitation of $\mathrm{c}_{\mathrm{j}}$ thus comes back to the limitation of a first coefficient.

Lemma II is therefore proven. Note that we can take for B* a number of the form $\mathrm{K}.\mathrm{A}^{*},\mathrm{\penalty 10000\ K}$ being independent of $\mathrm{A}^{*}$ and the polynomial Q .

The lemma $I$ results easily. Just take for $y_{i}$ the points $x_{i}$ and to note that from (2) results an inequality of the form (3) for $Q$ Or $-Q$ , with the value

\mathrm{A}^{*}=\frac{\mathrm{A}}{\min_{(\mathrm{i})}\left|\mathrm{R}\left(\mathrm{x}_{\mathrm{i}}^{\prime}\right)\right|}

We will say, to simplify the language, that a polynomial $Q$ of degree s for which the minimum is reached, is a polynomial of best upper bound of f on the set E. From Lemma I, we deduce, by classical reasoning, that

There exists at least one polynomial with the best upper bound.
The proof consists of first noting that max ( $\mathrm{f}-\mathrm{RQ}$ ) is a continuous function of the coefficients of $Q$ . We then see that it is sufficient to consider the polynomials $Q$ for which

f\left(x^{\prime},-R(x)Q(x)\leqslant\max f,\quad x\in E\right.

so the polynomials $Q$ for which

-R(x)Q(x)\leqslant\max f-\min f,\quad x\subset E.

7.

We will now demonstrate that

If $Q$ is a polynomial with best upper bound, the function $\mathrm{f}-\mathrm{RQ}$ reached the value $\mu$ on each of the sets $\mathrm{F}_{\mathrm{i}}$ .

Let us assume the opposite and let.

\text{ (t) }\quad\mathrm{F}_{\mathrm{i}_{1}},\mathrm{\penalty 10000\ F}_{\mathrm{i}_{2}},\ldots,\mathrm{\penalty 10000\ F}_{\mathrm{i}_{\mathrm{t}}},\mathrm{i}_{1}<\mathrm{i}_{2}<\ldots<\mathrm{i}_{\mathrm{t}},\mathrm{t}\leqslant\mathrm{\penalty 10000\ s}+1

those of the sets $F_{i}$ on which the value $\mu$ is reached. We therefore have

		$\displaystyle\max_{\left(F_{ij}\right)}(f-RQ)=\mu,\quad j=2,\ldots,t$
		$\displaystyle\max_{\left(\tau_{i}\right)}(f-RQ)<\mu\quad i\neq i_{j},j=2,\ldots,t$

We can then find a $\mu_{1}<\mu$ such as

f-RQ\leq\mu_{1},x\subset F_{i},i\neq i,\quad j=1,2,\ldots,t

Let's construct the polynomial $\mathrm{U}=\Pi\left(\mathrm{x}-\mathrm{x}_{\mathrm{j}}\right)$ where j takes only the values is +1 for which $\mathrm{i}_{\mathrm{j}+\mathrm{i}}-\mathrm{i}_{\mathrm{i}}$ is odd, $1\leqslant\mathrm{j}\leqslant\mathrm{t}-1$ . If $\mathrm{t}=1,\mathrm{U}=1$ . It is still so if all the differences $i_{j+1}-i_{j}$ are even.

We see that U is of degree s and that RU has the same sign at all points of the sets (1). Let $\lambda$ a constant of this same sign; $\lambda$ RU is therefore positive on sets (4) except on their ends where this polynomial vanishes.

This being said, we can easily see that we can take $|\lambda|$ small enough that we have

		$\displaystyle f-R(Q+\lambda U)<\mu,x\subset F_{i_{j}},j=2,\ldots,t$
		$\displaystyle f-R(Q+\lambda U)\leq\frac{\mu+\mu_{1}}{2}<\mu,x\subset F_{i},i\neq i_{j},j=2,\ldots,t$

And $Q$ would not be a best-bound polynomial. The property is therefore demonstrated.

We can also note that
if Q is a polynomial of best upper bound, the set on which $\mathrm{f}-\mathrm{RQ}$ reached the value $\mu$ cannot contain any of the points $\mathrm{x}_{\mathrm{i}}$ and cannot be formed by less than $\mathrm{s}+2$ separate sections of E .

In the following, the existence of a polynomial with the best upper bound will be sufficient, but we can demonstrate the following property

There exists a single polynomial with the best upper bound.
Suppose the opposite and let $Q,Q_{1}$ two distinct polynomials with best upper bound. If $Q_{2}=\frac{Q+Q_{1}}{2}$ , We have

	$\displaystyle\mathrm{f}-\mathrm{RQ}_{2}=\frac{1}{2}$	$\displaystyle{\left[(\mathrm{f}-\mathrm{RQ})+\left(\mathrm{f}-\mathrm{R}Q_{1}\right)\right]\leqslant\mu,\quad\mathrm{x}\subset\mathrm{E}}$		(5)
		$\displaystyle\max\left(\mathrm{f}-\mathrm{RQ}_{2}\right)\geqslant\mu$

SO $Q_{2}$ is still a polynomial with the best upper bound. We immediately deduce that there exists at least $s+2$ points, different from points $\mathrm{x}_{\mathrm{i}}$ , Or $\mathrm{f}-\mathrm{RQ}=\mu$ and, according to ( $\overline{\mathbf{5}}$ ), at least $\mathrm{s}+2$ such points where

\mathrm{f}-\mathrm{RQ}=\mathrm{f}-\mathrm{RQ}_{1}\text{, ou }\mathrm{Q}=\mathrm{Q}_{1}

SO $Q\equiv Q_{1}$ . Uniqueness results from the sole fact that the value $\mu$ is reached in at least $\mathrm{s}+1$ points.
8. Let us now return to functions of order n. Let us demonstrate that

Theorem I. If f is a continuous function defined on È and if, whatever the polynomial P of degree n and the section $\mathrm{E}_{1}$ of E, the set $\mathrm{E}_{1}(\mathrm{M})$ corresponding to $\mathrm{f}-\mathrm{P}$ verifies property A, the function f is non-concave of order n on $\mathrm{E}\left({}^{1}\right)$ .

It suffices to demonstrate that if f is not non-concave of order no one can find a $E_{1}$ and a polynomial $P$ such as $E_{1}(M)$ does not verify ownership $A$ .

If f is not aon-concave of order no can find n +2 points $x_{1}>x_{2}>\ldots>x_{n+2}$ such that we have
(v)

\left[\mathrm{x}_{1},\mathrm{x}_{2},\ldots;\mathrm{x}_{\mathrm{n}+2};\mathrm{f}\right]<0.

We can then find a polynomial $S$ of degree n such that the function $f_{1}=f-S$ checks for equalities

\begin{gathered}f_{1}\left(x_{1}\right)=f_{1}\left(x_{3}\right)=\ldots=f_{1}\left(x_{2k+1}\right)=0\\ f_{1}\left(x_{2}\right)=f_{1}\left(x_{1}\right)=\ldots=f_{1}\left(x_{2n-2k+2}\right)=2\rho>\rho\end{gathered}

Just take $S=G-\rho$ , Or $G$ is the best approximation polynomial of degree $n$ of $f$ on the points $x_{1}$ And $\rho>0$ this best approximation.

So then $Q$ the polynomial of best upper bound of $f_{1}$ on the whole

$E=\left(x_{n};{}_{2}Ex_{1}\right)$ , taking as points $x_{i}$ the points $x_{1},x_{3},\ldots,x_{2k+1}$ . We immediately see that for $\mathrm{P}=\mathrm{G}-\rho+\mathrm{RQ}$ property A is not verified for EP on $\mathrm{E}_{1}$ .

We immediately deduce
Theorem II. If f is a continuous function defined on E and if, whatever the polynomial P of even degree n and the section $\mathrm{E}_{1}$ of F, the set $\mathrm{E}_{2}(\mathrm{\penalty 10000\ m})$ corresponding to $\mathrm{f}-\mathrm{P}$ verifies property B , the function f is non-concarve of even order n on E .

For non-concave functions of odd order we have the following property
Theorem III. If i is a continuous function defined on È and if, whatever the polynomial P of odd degree n, the set $\mathrm{E}(\mathrm{m})$ corresponding to $\mathrm{f}-\mathrm{P}$ verifies property C , the function is non-concave of odd order n on E .

The proof is done as for Theorem I. If f is not non-concave of order n we can find $\mathrm{n}+2$ points $\mathrm{x}_{1}>\mathrm{x}_{2}>\ldots,>\mathrm{x}_{\mathrm{n}+2}$ such that we have the inequality ('). We can then find, as above, a polynomial $S$ of degree in such that if $f_{1}=f-S$ we have

		$\displaystyle f_{1}\left(x_{2}\right)=f_{1}\left(x_{4}\right)=\ldots=f_{1}\left(x_{2k}\right)=0$
		$\displaystyle f_{1}\left(x_{1}\right)=f_{1}\left(x_{3}\right)=\ldots=f_{1}\left(x_{n+2}\right)<0$

So then $Q$ the polynomial of best upper bound of $-f_{1}$ on the whole $E$ , taking as points $x_{i}$ the points $x_{2},x_{4},\ldots,x_{2k}$ . We immediately see that for $\mathrm{P}=\mathrm{S}-\mathrm{RQ}$ property C is not verified by $\mathrm{f}-\mathrm{P}$ on E.
9. Let us make some remarks on the previous theorems. We can always consider only the polynomials $P$ which all cancel out at the same point, for example we can only consider polynomials divisible by $\mathrm{x}[\mathrm{P}()=0]$ .

In the demonstration of theorems I, II the hypothesis $n>1$ intervenes implicitly. But these theorems remain true for $\mathrm{n}=0,1$ . For $\mathrm{n}=0$ we have the following property, which is almost obvious

If f is defined on E and if, whatever the section $\mathrm{E}_{1}=(\mathrm{cEd})$ of $\mathrm{E},\mathrm{c},\mathrm{d}<\mathrm{E}$ , we $a\mathrm{f}(\mathrm{d})=\max\mathrm{f}$ , the function f is non-decreasing on E .
( is ₁ )
For $\mathrm{n}=1$ we have the following theorem, due to MS Saks $\left({}^{1}\right)$ .
If f is a function defined on E and if, whatever the section E, of E and the constant $\alpha$ , the function $\mathrm{f}+\alpha\mathrm{x}$ reaches its maximum at at least one of the ends of $\mathrm{E}_{1}$ the function f is non-concave of order 1 on E .

We have already given Theorem III for $\mathrm{n}=1$ in a previous work. $\left({}^{2}\right)$ .
The assumption of continuity of f cannot be removed in general, but it can be replaced by less restrictive assumptions. For example, by upper semi-continuity for the maximum and lower semi-continuity for the minimum. As the example of functions of order 0 or 1 shows, we can completely do without such assumptions under certain conditions. We can also impose on $\mathrm{f}-\mathrm{P}$ less restrictive conditions and thus characterize functions of order 11. Thus, for example, we have the property, almost obvious.
(1) S. Saks "O funkjach wypuklych i podharmonicznych" Mathesis Polska, 6, 43-55 (1931).
(2) Tiberiu Popoviciu "Two remarks on convex functions" Bulletin de l'Acad. Rounaine, 20, 45-19 (1939).

If f is defined on E and if, whatever the section $\mathrm{E}_{1}=(\mathrm{cE}\mathrm{d})$ of E, we $\boldsymbol{a}\mathrm{f}(\mathrm{c})=\min_{\left(1_{1}\right)}\mathrm{f}$ Or $\mathrm{f}(\mathrm{d})=\max_{\left(\equiv_{1}\right)}\mathrm{f}$ , the function f is non-decreasing on E .
10. Properties of sets $\mathrm{E}(\mathrm{M},\mathrm{E}(\mathrm{m})$ polynomials are very precise $\mathrm{T}_{\mathrm{n}}^{\prime},\mathrm{T}_{\mathrm{n}+1}$ best approximation of a continuous function of order n.

We will now assume that E is a closed interval $(\mathrm{a},\mathrm{b})$ and the function $f$ continues in $(a,b)$ .

Let us pose ( ¹ )

	$\displaystyle\tau=\tau\left(\mathrm{x}_{1},\mathrm{x}_{2},\ldots,\mathrm{x}_{\mathrm{n}+2};\mathrm{f}\right)=$	$\displaystyle\frac{\mathrm{U}\left(\mathrm{x}_{1},\mathrm{x}_{2},\ldots,\mathrm{x}_{\mathrm{n}+2};\mathrm{f}\right)}{\sum_{\mathrm{i}=1}^{\mathrm{n}+2}\left\|\mathrm{\penalty 10000\ V}\left(\mathrm{x}_{1},\mathrm{x}_{2},\ldots,\mathrm{x}_{\mathrm{i}-1},\mathrm{x}_{\mathrm{i}+1},\ldots,\mathrm{x}_{\mathrm{n}+2}\right)\right\|}$
	$\displaystyle\mathrm{x}_{1}$	$\displaystyle>\mathrm{x}_{2}>\ldots>\mathrm{x}_{\mathrm{n}+2}.$

We assume that the reader is familiar with the properties of best approximation polynomials established by Messrs. E. Borel ( ² ) and $\mathbf{Ch}$ . of the Poussin Valley ( ³ ). In particular, recall that the best approximation of order n of the function $f$ in an interval is equal to the maximum of $|\tau|$ when the $x_{i}$ remain in this interval. If $x_{1},>x_{2},>\ldots>x_{n}\vdash_{2}$ are points for which this maximum is reached, $f-T_{n}$ takes the values alternately $\pm|\tau|$ on these points. The polynomial $T_{n}$ is characterized completely by the fact that $f-T_{n}$ reached the maximum value $\left|\mathrm{f}-\mathrm{T}_{\mathrm{n}}\right|$ in at least $\mathrm{n}+2$ consecutive points with alternating signs.
Mr. Ch. de la Vallée Poussin notes that $|\tau|$ is a continuous function of the $\mathrm{x}_{\mathrm{i}}$ . But if we take into account that there is a continuous correspondence between continuous functions and their polynomials $\mathrm{T}_{\mathrm{n}}$ we can easily see that
$\tau$ is a continuous function when the $\mathrm{x}_{\mathrm{i}}$ remain in an interval.
We can also demonstrate this property directly.
It should also be noted that the maximum or minimum, assumed to be non-zero, of $\tau$ can only be achieved for distinct values of the $x_{i}$ .
11. Let us now prove the following lemma:

Lemma III. If $\tau$ does not remain constantly non-positive and if $\tau^{\prime}==\tau\left(\mathrm{x}_{1}^{\prime},\mathrm{x}_{2}^{\prime},\ldots,\mathrm{x}_{n+2}^{\prime}\right.$ ; f) $\mathrm{x}_{1}^{\prime}>\mathrm{x}_{2}^{\prime}>\ldots>\mathrm{x}_{n+2}^{\prime}$ is the maximum $(>0)$ of $\tau$ in the meantime ( $\mathrm{a},\mathrm{b}$ ), $\tau^{\prime}$ is the best approximation of order n of f in the interval ( $\mathrm{x}_{\mathrm{n}+2}^{\prime},\mathrm{x}^{\prime}{}_{1}$ ).

Let P be the best approximation polynomial of degree n on the points $\mathrm{x}^{\prime}{}_{1},\mathrm{x}^{\prime}{}_{2},\ldots,\mathrm{x}^{\prime}{}_{\mathrm{n}+{}_{2}}$ . We will show that P is the polynomial of best approximation of degree $n$ of $f$ in the meantime $\left(x_{n+_{2}}^{\prime},x_{1}^{\prime}\right)$ . To do this it is necessary and sufficient to demonstrate that the function $f_{1}=f-P$ remains between $-\tau^{\prime}$ And $\tau^{\prime}$ In $\left(x_{n+2}^{\prime},x_{1}^{\prime}\right)$ . Let's take the point $\mathrm{x}_{\mathrm{j}-1}^{\prime}>\mathrm{x}>\mathrm{x}_{\mathrm{j}+1}^{\prime},\mathrm{j}=1,2,\ldots,\mathrm{n}+2,\mathrm{x}_{0}^{\prime}=\mathrm{x}_{1}^{\prime},\mathrm{x}_{\mathrm{n}+3}^{\prime}=\mathrm{x}_{\mathrm{n}+2}^{\prime}$ . We have

\tau\left(\mathrm{x}_{1}^{\prime},\mathrm{x}_{2}^{\prime},\ldots,\mathrm{x}_{\mathrm{j}-1}^{\prime},\mathrm{x},\mathrm{x}_{\mathrm{j}+1}^{\prime},\ldots,\mathrm{x}_{\mathrm{n}+2}^{\prime};\mathrm{f}_{1}\right)\leqslant\tau^{\prime}

But

=\tau^{\prime}+\frac{(-1)^{n\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{j-1}^{\prime},x,x_{j+1}^{\prime},\ldots,x_{n+1}^{\prime},f_{1}\right)=}}{\sum_{i=1}^{n+2}\mid V\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{j-1}^{\prime},x_{j+1}^{\prime},\ldots,x_{n+2}^{\prime}\right)\left[f_{1}(x)-(-1)^{j-1}\tau^{\prime}\right]}

it in the sum of the denominator replaces $\mathrm{x}_{\mathrm{j}}^{\prime}$ by x .
So we have

(-1)^{j-1}\left[f_{1}(x)-(-1)^{j-1}\tau^{\prime}\right]\leqslant 0,\quad x\subset\left(x_{j+1}^{\prime},x_{j-1}^{\prime}\right),

=which proves the property.
We prove it in exactly the same way as
Lemma IV. If $\tau$ does not remain constantly non-negative and if $\tau^{\prime\prime}==\tau\left(\mathrm{x}^{\prime\prime}{}_{1},\mathrm{x}^{\prime\prime}{}_{2},\ldots,\mathrm{x}^{\prime\prime}{}_{\mathrm{n}+2};\mathrm{f}\right),\mathrm{x}^{\prime\prime}{}_{1}>\mathrm{x}^{\prime\prime}{}_{2}>\ldots>\mathrm{x}^{\prime\prime}{}_{\mathrm{n}+2}$ is the minimum ( $\langle 0$ ) of $\tau$ in the interval $(\mathrm{a},\mathrm{b}),-\tau^{\prime\prime}$ is the best approximation of order n of f in the interralle $\left(\mathrm{x}^{\prime\prime}{}_{\mathrm{n}+2},\mathrm{x}^{\prime\prime}{}_{1}\right)$ .

Note that in both lemmas the points $\mathrm{x}^{\prime}{}_{1},\mathrm{x}^{\prime}{}_{2},\ldots,\mathrm{x}^{\prime}{}_{\mathrm{n}+2}$ resp- $\mathrm{x}^{-}\mathrm{x}^{\prime\prime}{}_{2},\ldots,\mathrm{x}^{\prime\prime}{}_{\mathrm{n}+2}$ are points where $\mathrm{f}-\mathrm{T}_{\mathrm{n}}$ reached with alternating signs, the maximum meleur $\left|\mathrm{f}-\mathrm{T}_{\mathrm{n}}\right|,\mathrm{T}_{\mathrm{n}}$ and the maximum being taken in the interval ( $\mathrm{x}_{\mathrm{n}+{}_{2}},\mathrm{x}_{1}^{\prime}$ ) esp. $\left(x^{\prime\prime}{}_{n+2},x^{\prime\prime}{}_{1}\right)$ .

Lemma V. If the continuous function f is not of order n in the interwall $(\mathrm{a},\mathrm{b})$ , we can find a subinterval $(\mathrm{c},\mathrm{d})$ of $(\mathrm{a},\mathrm{b})$ such that if $\tau_{1}(>0)$ , $\tau_{2}(<0)$ are the maximum and minimum of $\tau$ In $(\mathrm{c},\mathrm{d})$ we have $\tau_{1}=-\tau_{2}$ .

Always be

0<\tau^{\prime}=\max_{(\mathrm{a},\mathrm{\penalty 10000\ b})}\tau,0>\tau^{\prime\prime}=\min_{(\mathrm{a},\mathrm{\penalty 10000\ b})}\tau

If we have $\tau^{\prime}=-\tau^{\prime\prime}$ the property is demonstrated $(c,d)=(a,b)$ . Let us suppose that $\tau^{\prime}>-\tau^{\prime\prime}$ . We can find, by Lemma IV, an interval $\left(a^{\prime},b^{\prime}\right)$ Or

\min_{\left(a^{a^{\prime}},b^{\prime}\right)}\tau=\tau^{\prime\prime},\max_{\left(a^{\prime},b^{\prime}\right)}=\tau_{1}^{\prime}\leqslant-\tau^{\prime\prime}

If $\tau_{1}^{\prime}=-\tau^{\prime\prime}$ the property is demonstrated, $(\mathrm{c},\mathrm{d})=\left(\mathrm{a}^{\prime},\mathrm{b}^{\prime}\right)$ . Let us suppose that $\tau_{1}^{\prime}<\cdots\tau^{\prime\prime}$ and either

\begin{gathered}\max_{\left(a_{1},b_{1}\right)}\tau=\tau_{1}^{*},\min_{\left(a_{1}b_{1}\right)}\tau=\tau_{2}^{*}\\ a_{1}=\frac{a+\lambda a^{\prime}}{1+\lambda},b_{1}=\frac{b+\lambda b^{\prime}}{1+\lambda},\quad\lambda\subset(0,+\infty)\end{gathered}

But, $\boldsymbol{\tau}_{1}^{*}+\boldsymbol{\tau}_{2}^{*}$ is obviously a continuous function of $\lambda$ And

\begin{array}[]{ll}\boldsymbol{\tau}_{1}^{*}+\boldsymbol{\tau}_{2}^{*}=\boldsymbol{\tau}^{\prime}+\boldsymbol{\tau}^{\prime\prime}>0,&\text{ pour }\lambda=0,\\ \boldsymbol{\tau}_{1}^{*}+\boldsymbol{\tau}_{2}^{*}\rightarrow\boldsymbol{\tau}_{1}^{\prime}+\boldsymbol{\tau}^{\prime\prime}<0,&\text{ pour }\lambda\rightarrow+\infty.\end{array}

So there is a positive value of $\lambda$ for which $\tau_{1}^{*}=-\tau^{*}{}_{2}$ , which demonstrates the property.

We demonstrate the same way if $\tau^{\prime}<-\tau^{\prime\prime}$ based on Lemma III.
12. Let always $f$ a continuous function and suppose that it is not Zorder n in $(\mathrm{a},\mathrm{b})$ . Consider the subinterval $(\mathrm{c},\mathrm{d})$ defined by Lemma V.

Either $\mathrm{T}_{\mathrm{n}}$ the best approximation polynomial of degree n of f in ( c , d ). We can then find $\mathrm{n}+2$ points $\mathrm{x}_{1}^{\prime}>\mathrm{x}_{2}^{\prime}>\ldots>\mathrm{x}_{\mathrm{n}+2}^{\prime}$ Or

\tau\left(\mathrm{x}_{1}^{\prime},\mathrm{x}_{2}^{\prime},\ldots,\mathrm{x}_{\mathrm{n}+2}^{\prime};\mathrm{f}\right)=\tau_{1}

(7)

and where $\mathrm{f}-\mathrm{T}_{\mathrm{n}}$ takes, with alternating signs, the max value $\mathrm{f}-\mathrm{T}_{\mathrm{n}}$ . We can also find $\mathrm{n}+2$ points $\mathrm{x}^{\prime\prime}{}_{1},>\mathrm{x}^{\prime\prime}{}_{2}>\ldots>\mathrm{x}^{\prime\prime}{}_{n+2}$ Or

\tau\left(\mathrm{x}^{\prime\prime}{}_{1},\mathrm{x}^{\prime\prime}{}_{2},\ldots,\mathrm{x}^{\prime\prime}{}_{\mathrm{n}+2};\mathrm{f}\right)=\tau_{2}

(8)

and where $f-T_{n}$ takes, with alternating signs, the max values $f-T_{n}!$ .
Relations (7), (8) and $\tau_{1}=-\tau_{2}>0$ show us that we can always choose among the points $\mathrm{x}_{\mathrm{i}}^{\prime},\mathrm{x}_{\mathrm{i}}^{\prime\prime}$ a sequence of at least $\mathrm{n}+3$ consecutive points where $\mathrm{f}-\mathrm{T}_{\mathrm{n}}$ takes, with alternating signs, the maximum value $\mid\mathrm{f}-\mathrm{T}_{\mathrm{n}}$ . It follows that $\mathrm{T}_{\mathrm{n}}$ is also the polynomial of best degree approximation $\mathrm{n}+1$ of f in (c, d) therefore

Lemma VI. If the continuous function f is not of order n in the interval ( $\mathrm{a},\mathrm{b}$ ), we can find a subinterval of ( $\mathrm{a},\mathrm{b}$ ) where the best approximation polynomials $\mathrm{T}_{\mathrm{n}},\mathrm{T}_{\mathrm{n}+1}$ of degrees $\mathrm{n},\mathrm{n}+1$ coincide, so where the polynomial of best degree approximation $\mathrm{n}+1$ is of effective degree $\leq\mathrm{n}$ .

If we notice that $\mathrm{T}_{\mathrm{n}}\equiv\mathrm{T}_{\mathrm{n}+\mathrm{a}}$ also takes place in any interval where f reduces to a polynomial of degree n, we deduce the

Theorem IV. If f is a continuous function in the closed interval $(\mathrm{a},\mathrm{b})$ and if, whatever the closed subinterval $(\mathrm{c},\mathrm{d})$ of $(\mathrm{a},\mathrm{b})$ , the Chebysheff polynomial $\mathrm{T}_{\mathrm{n}+1}$ of degree $\mathrm{n}+1$ of f in ( $\mathrm{c},\mathrm{d}$ ) is actually of degree $\mathrm{n}+1$ , the function f is convex or concave of order n in $(\mathrm{a},\mathrm{b})$ .

Note that if $f$ is not in order $n$ in the meantime ( $a,b$ ), in the meantime $(\mathrm{c},\mathrm{d})$ from Lemma V the function is certainly not polynomial of order n. It is easy to deduce the following result

Theorem V. If f is a continuous function in the closed interval ( $\mathrm{a},\mathrm{b}$ ) and if, whatever the closed subinterval $(\mathrm{c},\mathrm{d})$ of $(\mathrm{a},\mathrm{b})$ , the Chebysheff polynomial $\mathrm{T}_{\mathrm{n}}$ of degree 11 of f in ( $\mathrm{c},\mathrm{d}$ ) checks for equalities $\left|\mathrm{f}(\mathrm{c})-\mathrm{T}_{\mathrm{n}}(\mathrm{c})\right|==\left|\mathrm{f}(\mathrm{d})-\mathrm{T}_{\mathrm{n}}(\mathrm{d})\right|=\max_{(\mathrm{c},\mathrm{d})}\left|\mathrm{f}-\mathrm{T}_{\mathrm{n}}\right|$ , the function f is of order n in $(\mathrm{a},\mathrm{b})$ .

Indeed, for a function which is not of order n it is sufficient to shorten the interval a little $(\mathrm{c},\mathrm{d})$ of Lemma V to find an interval where the property is not satisfied.

Cernăuti, April 11, 1939

Notes on Higher-Order Convex Functions (VI)

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