On an inequality of N. Levinson

by
TIBERIU POPOVICIU
in Cluj

1. 1.

   We will assume knowledge of the definition and main properties of divided differences. We will denote by [$x_{1},x_{2},\ldots,x_{m+1};f$the difference divided by order$m$of the function$f$on the points, or the nodes,$x_{1},x_{2},\ldots,x_{m+1}$The nodes may or may not be distinct. In cases where the nodes are not all distinct (or simple), the divided differences also involve successive derivatives of the function (over multiple nodes). In what follows, however, we will only consider divided differences over distinct nodes.

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A function is said to be non-concave respectively$(\geqq-1)$on the interval$I$, if its difference divided by an angle of order$m$non-negative respectively positive on all remainder distinct from$I$.
e$m+2$points continue if$m\geq 1$and has a rumble$m$on an open interval is con-non-concave respectively conjunction$f$is concave respectively conce d' is non-
2. The classical inequality of Jéproquement. (usual) vexes is as follows ² with respect to conc- functions

and is verified regardless of the natural number$n$, the points$x_{\alpha}\in I$,$\alpha=1,2,\ldots,n$and positive numbers$p_{\alpha},\alpha=1,2,\ldots,n$, for any function$f$non-concave of order 1 on the interval$I$.

If$n>1$and if the function$f$is convex of order 1 on$I$The equality in (1) holds if and only if the points$x_{\alpha},\alpha=1,2,\ldots,n$are all lumped together.

Several proofs of these properties are known. Note that it suffices to prove inequality (1) in the case where the points$x_{\alpha},\alpha==1,2,\ldots,n$are distinct, the function$f$being non-concave of order 1 on$I$and to show that if$n>1$And$f$is convex of order 1 on$I$, then equality in (1) is not possible.

Either$n>1$and let's choose the notations so that we have
(2)

If$p_{\alpha}>0,\alpha=1,2,\ldots,n$and if we ask

we then

Legality

then demonstrates Jensen's inequality (1).
We can state the properties expressed by this inequality also in the form of

THEOREM 1. For inequality (1) to be satisfied for all natural numbers$n$, for any group of$n$points$x_{\alpha}\in I,\alpha=1,2,\ldots,n$, and for any group of$n$positive numbers$p_{\alpha},\alpha=1,2,\ldots,n$it is necessary and sufficient that the function$f$either non-concave of order 1 on the interval$I$.

For the strict inequality (with the sign <) (1) to be satisfied for all natural numbers$n>1$any group of$n$points$x_{\alpha}\in I\alpha=1,2,\ldots,n$, not all of them together and any group of$n$positive numbers$p_{\alpha},\alpha=1,2,\ldots,n$it is necessary and sufficient that the function$f$either convex of order 1 on the interval$I$.

The necessity of the conditions in the statement follows from formula (5), noting that if$x_{1}<x_{2}<x_{3}$are any three points of$I$(subject to the inequalities mentioned), it is always possible to find two positive numbers$p,q$so that$x_{2}=\frac{px_{1}+qx_{3}}{p+q}$(For example,$p=x_{3}-x_{2},q=\left.=x_{2}-x_{1}\right)$
3. Consider$n$positive numbers$p_{\alpha},\alpha=1,2,\ldots,n$and two groups, each of$n$points,$x_{\alpha},x_{\alpha}^{\prime},\alpha=1,2,\ldots,n$Let (3) and

Levinson's inequality is written

where the points$x_{\alpha},x_{\alpha}^{\prime}$and the function$f$verify the following conditions. Suppose that$n>1$that we have (2) and that

We deduce from this$x_{n}^{\prime}<x_{n-1}^{\prime}<\ldots<x_{1}^{\prime}$And$\xi_{\alpha}<\xi_{\alpha+1}<x_{\alpha+1}\leqq$
$\leqq x_{\alpha+1}^{\prime}<\xi_{\alpha+1}^{\prime}<\xi_{\alpha}^{\prime},\quad\alpha=1,2,\ldots,n-1$, and
(9)

The equalities (9) are equivalent to$n-1$equalities (8).
Note also that we have

and it follows that, under the assumptions$n>1$, (2) and (8), we have$\left[\xi_{\alpha}^{\prime},\xi_{\alpha+1}^{\prime},x_{\alpha+1}^{\prime};f\right]\geqq$respectively$>\left[\xi_{\alpha},\xi_{\alpha+1},x_{\alpha+1};f\right],\alpha=1,2,\ldots,n-1$depending on the function$f$is non-concave respectively convex of order 2 on an interval containing the points$x_{\alpha},x_{\alpha}^{\prime},\alpha=1,2,\ldots,n$.

Taking into account (5) and (10), we have

1. 4.

   From the preceding analysis, it follows that Levinson's inequality (7) is verified for all natural numbers$n$, the points$x_{\alpha},x_{\alpha}^{\prime}\in I$,$\alpha=1,2,\ldots,n$such as one might have

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and positive numbers$p_{\alpha},\alpha=1,2,\ldots,n$, for any function$f$nonconcave of order 2 on the interval$I$.

If$n>1$and if the function$f$is convex of order 2 on$I$The equality in (7) holds if and only if the points$x_{\alpha},\alpha=1,2,\ldots,n$(so also the points$x_{\alpha}^{\prime},\alpha=1,2,\ldots,n$) are all confused.
N. LEVINSON obtained [1] the previous results, assuming that the function$f$has a third derivative$f^{\prime\prime}$We have therefore obtained a generalization of Levinson's inequality.

We can also state these properties in the form of
Theorem 2. For inequality (7) to be satisfied, for all natural numbers$n$, for any group of$2n$points$x_{\alpha}x_{\alpha}^{\prime}\in I\alpha=1,2,\ldots,n$, such that we have (12) and for any group of$n$positive numbers$p_{\alpha},\alpha=1,2,\ldots,n$it is sufficient that the function$f$and the function must continue$f$either non-concave of order 2 on the interval$I$.

For the strict inequality (with the sign <) (7) to be satisfied for all natural numbers$n>1$, for any group of$2n$points$x_{\alpha},x_{\alpha}^{\prime}\in I\alpha=1,2,\ldots,n$, such that one has (12), the$x_{\alpha}$not all lumped together and for every group of$n$positive numbers$p_{\alpha},\alpha=1,2,\ldots,n$it is sufficient that the function$f$and the function must continue$f$either convex of order 2 on the interval$I$.

In the demonstration, the restriction that the points$x_{\alpha},\alpha=1,2,\ldots,n$being distinct does not restrict generality since by virtue of (12), if$x_{\alpha}=x_{\beta}$We have$x_{\alpha}^{\prime}=x_{\beta}^{\prime}$and vice versa.

The necessity of the conditions in the statement follows from formula (7). If we posit$n=2,x_{1}=x,x_{2}=x_{2}^{\prime}=x+\frac{3}{2}h,x_{1}^{\prime}=x+3h,p_{1}=1,p_{2}==2$, We have$\xi^{2}=x+h,\xi_{2}^{\prime}=x+2h$, the first member of formula (11) becomes$\Delta_{h}^{3}f(x)=f(x+3h)-3f(x+2h)+3f(x+h)-f(x)==6h^{3}[x,x+h,x+2h,x+3h;f]$But, we know [2], that if$\Delta_{h}^{3}f(x)\geqq 0$respectively$>0$, for all$x,x+3h\in I,h>0$, the function$f$assumed to be continuous, is non-concave, respectively convex of order 2 on$I$.

Given the necessity of the theorem's condition, the hypothesis of the function's continuity can generally be replaced by a less restrictive hypothesis (measurability, being bounded on$I$, etc.).
5. The proof we gave above of Levinson's inequality is based on the fact that the successive means (3) and (6) of the points$x_{\alpha}$and points$x_{\alpha}^{\prime}$are formed with the same weights and
these points are linked by the equalities (9). In reality, to be able to use a formula such as (10), it is only sufficient that one has

and not necessarily (9). By imposing on the points$x_{\alpha},x_{\alpha}^{\prime}$The restriction (13), without all the equalities (9) being verified, allows us to obtain other theorems, analogous to Theorem 2.6.
As an example, suppose that we have

assuming that the averages$\xi_{\alpha},\xi_{\alpha}^{\prime}$are given by (3) and (6). We deduce that

If inequalities (2) are satisfied, we also have$x_{1}^{\prime}<x_{2}^{\prime}<\ldots<x_{n}^{\prime},\xi_{\alpha}^{\prime}--\xi_{\alpha}=y,\alpha=1,2,\ldots,n$and we always$\xi_{\alpha}<\xi_{\alpha+1}<x_{\alpha+1},\alpha=1,2,\ldots$,$\ldots,n-1$We have the formula.

and it follows that, under the assumptions (2), (15) (or (14)) and$y>0$, 1a difference (10) (or (16)) still verifies the non-concavity property of order 2 mentioned above.

As above, we deduce
Theorem 3. For inequality (7) to be verified for all natural numbers$n$, for any group of$2n$points$x_{\alpha},x_{\alpha}^{\prime}\in I,\alpha=1,2,\ldots,n$, such that one has (15) with$y>0$and for any group of$n$positive numbers$p_{\alpha},\alpha==1,2,\ldots,n$it is sufficient that the function$f$and it is necessary that the function be continuous (or measurable, or bounded etc.)$f$either non-concave of order 2 on the interval$I$.

For the strict inequality (with the sign <) (7) to be satisfied for all natural numbers$n>1$, for any group of$2n$points$x_{\alpha},x_{\alpha}^{\prime}\in I,\alpha=1,2$,$\ldots,n$, THE$x_{\alpha}$not all together, such as one has (15) with$y>0$and for any group of$n$positive numbers$p_{\alpha},\alpha=1,2,\ldots,n$it is sufficient that the function$f$and the continuous function (measurable, bounded, etc.) must be convex of order 2 on the interval$I$.

To establish the necessity of the conditions in the statement, it suffices to take$n=3$And$x_{1}=x,x_{2}=x_{1}^{\prime}=x+h,x_{3}=x_{2}^{\prime}=x+2h,x_{3}^{\prime}=x+3h$and inequality (7) becomes even$\Delta_{n}^{3}f(x)\geqq 0$with$h>0$7.
Inequality (7), under the hypotheses of Theorem 3 ($I$(being an open interval) can be demonstrated for a function$f$non-concave (respectively convex) of order 2, noting that this function has a
non-concave (respectively convex) derivative of order 1. By Jensen's inequality, the function of$y$,