Notes on Higher-Order Convex Functions (II)

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Tiberiu Popoviciu

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T. Popoviciu, Notes sur les fonctions convexes d’ordre supérieur (II), Mathematica, 12 (1936), pp. 227-233 (in French).

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1936 d -Popoviciu- Mathematica – Notes sur les fonctions convexes d’ordre superieur (II)

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1936 d -Popoviciu- Mathematica - Notes on convex functions of higher order (II)

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NOTES ON HIGHER ORDER CONVEX FUNCTIONS (II)

byTiberiu Popoviciuin Cernăuți.

Received June 8, 1936.

On univalent functions and maltivalent functions of a real variable.

Either $f (x)$ a real, definite and uniform function in the finite and open interval ( $has, b$ ), $has < b$ . We therefore assume that $f (x)$ has a finite and well-determined value at any point of ( $has, b$ ) except at the ends where it is not defined. The monotonicity of such a function will be understood in the strict sense. It will therefore be a question of increasing or decreasing functions. Similarly, in the following, convexity and concavity should not be confused with non-concavity and non-convexity respectively.

The function

f (x)

has a maximum (or upper bound) at each point

M (f; x)

and a minimum (or lower bound)

m (f; x)

. The number

M (f; x)

is the limit of the maximum of

f (x)

in an interval containing the point

x

when this interval tends towards the point

x

. The number

m (f; x)

has a similar definition. The number

M (f; x)

can become equal to

+ \infty

and the number

m (f; x)

can become equal to

- \infty

and these two circumstances can also occur at the same time. We always have

m (f; x) \leq f (x) \leq\leq M (f; x)

.

We will attach to the function

f (x)

another function

F (x)

which takes to the point

x

all values between

m (f; x)

And

M (f; x)

, these ends included if they are finished. If

M (f; x) = + \infty

the function

f (x)

takes, to the point

x

, all values greater than some fixed number and there is an analogous property if

m (f; x) = - \infty

. When

M (f; x) = + \infty

And

m (f; x) = - \infty

at the same time, the function

F (x)

takes all possible values at point

x

. The function

F (x)

is generally multifaceted. So that

f (x)

is continuous it is necessary and sufficient that the attached function

F (x)

be uniform, the functions

f (x)

And

F (x)

then coincide everywhere.
2. The function

f (x)

is univalent if it takes each of its values only once, and this notion of univalence can obviously also be extended to multifaceted functions. A continuous and univalent function is necessarily monotone but univalence alone is not sufficient to assert that the function is monotone (

^{1}

).

We propose to demonstrate the following property:
I. For the function

f (x)

is monotonous it is necessary and sufficient that the function

F (x)

, attached to

f (x)

, or univalent.

The condition is obviously necessary and it remains to be shown that it is also sufficient. We will demonstrate it in the following Nos.
3. Let us therefore suppose that

F (x)

is univalent. We can easily see that then at every point

x

the numbers

M (f; x), m (f; x)

are finished. Let's score the points

has, b

on the real axis and take a point

x

of

(has, b)

. Let us represent the points representing the values

m (f; x)

And

M (f; x)

. Let AA' be the vertical which contains the values of

F (x)

to the point

x^{(2)}

. Let us also represent the lines BB ', CC ' parallel to the real axis and passing through the points representing the values

m (f; x), M (f; x)

The plan is thus broken down into several regions. First there is the band between BB', CC' including the edges. Then four CM quadrants

(f; x) {HAS}^{'}, {HAS}^{'} M (f; x) C

',

B^{'} m (t; x) HAS, HAS m (f; x) B

numbered respectively by (I), (II), (III), (IV). The only values taken by the function

F (x)

in the band are on the segment

m (f; x) M (f; x)

. The strip BB', CC' can also be reduced to a single straight line parallel to the real axis. Property I will be demonstrated if we establish how the values of the function

F (x)

, taken outside the point

x

, are distributed in the quadrants (I)-(IV).
4. I. The values taken to the left of

x

are either all in quadrant (II) or all in quadrant (III).

The demonstration is very simple. Let us suppose the opposite. It is clear that at any point to the left of

x

the values of

F (x)

are all in one of the quadrants (II), (III). Now let

x_{1}

And

x_{2}

two points to the left of

x

such as values

F (x_{1})

are all in (II) and the values

F (x_{2})

all in (III). Let us designate by J

I_{1}

the interval (

x_{1}, x_{2}

) and take its middle

x_{4}^{'}

. Let us take as an interval

I_{2} (x_{1}, x_{4}^{'})

Or (

x_{1}^{'}, x_{2}

) such that at the ends the values are in different quadrants. Repeating the process we construct a sequence of intervals

I_{1}, I_{2}, \dots, I_{m}, \dots

such that each is half of the previous one and that at the ends the values of

F (x)

are always in different dials. The intervals

I_{m}

tend, for

m \to \infty

, towards a limit point

ξ

. It immediately follows that

M (f; ξ \geq M (f; x)

And

m (f; ξ) \leq m (f; x)

, which is clearly in contradiction with the univalence of

F (x)

.

We demonstrate in the same way that:
III. The values taken to the right of

x

are either all in quadrant (I) or all in quadrant (IV).

Finally, to specify the distribution of all values of

F (x)

, we will demonstrate the following property:

IIII. Values taken outside the point

x

are either all in quadrants (I) and (III) or all in quadrants (II) and (IV).

It will be enough to demonstrate, for example, that the values cannot all be in quadrants (I) and (II). Let us suppose the opposite and let

F (x_{0})

a value taken at a point

x_{0}

to the left of

x_{0}

. Under the definition of

M (f; x)

and properties already demonstrated, all values taken by

F (x)

to the right of

x

must be smaller than

F (x_{0})

. On the other hand, still by virtue of the definition of

M (f; x)

, the function

F (x)

takes left of

x

values as close as one wants to

M (f; x)

. It would therefore follow that the value

M (f; x)

is taken at any point to the right of

x

which is impossible.

From what we have demonstrated it is easy to deduce that

f (x)

is increasing or decreasing. Property I is therefore completely established.
5. Although the function

f (x)

, is not defined at points

has

And

b

we can extend the function

F (x)

, attached to

f (x)

, on these points the numbers

M (f; has), m (f; has)

And

M (f; b), m (f; b)

being defined. We can then easily see that if the function

F (x)

thus extended is univalent, it is necessarily bounded. It is easy to deduce the following property:
I. For the function

f (x)

is monotone is bounded it is necessary and sufficient that the attached function

F (x)

, extended on the ends a and

b

, or univalent.
6. Let us now move on to the study of convex or concave functions of order

n (n \geq 1)

We assume that we know the main properties and the notations that we have used in this theory.

^{(3)}

. Let us first say what we mean by a multivalent (uniform or multifaceted) function of order

n + 1

. Such a function takes at most

n + 1

times the same value as a polynomial of degree

n

(more exactly of degree at most equal to

n

) and this regardless of the polynomial considered. Multivalence therefore has a different meaning here as usual.

Let us suppose that the function

f (x)

be continuous and multivalent of order

n + 1

. Let

x_{1} < x_{2} < \dots < x_{n + 1} n + 1

points in (

has, b

) And

P (x_{1}, x_{2}, \dots, x_{n + 1}; f ∣ x)

the Lagrange polynomial taking the values

f (x_{i})

to the point

x_{i}

. The difference

f (x_{i}) - P (x_{1}, x_{2}, \dots, x_{n + 1}; f ∣ x)

is different from zero and keeps the same sign in each of the open intervals

(a, x_{1}), (x_{1}, x_{2}), \dots, (x_{n + 1}, b)

. Now, I say that in two consecutive intervals this difference takes values of opposite signs. Let us suppose in fact that the values in (

x_{i}, x_{i + 1}

), (

x_{i + 1}, x_{i + 2}

)

(x_{0} = a, x_{n + 2} = b)

are of the same sign. Let us then take a point

x^{'}

in the meantime (

x_{l}, x_{i + 1}

) and consider the Lagrange polynomial

P (x_{1}, x_{2}, \dots, x_{i}, x^{'}, x_{i + 2}, \dots, x_{n + 1}; f ∣ x)

. The equation

P (x_{1}, x_{2}, \dots, x_{n + 1}; f ∣ x) = P (x_{1}, x_{2}, \dots, x_{i}, x^{'}, x_{l + 2}, \dots, x_{n + 1}; f ∣ x)

is not identically verified and is of degree

n

She has the roots

x_{1}, x_{2}, \dots, x_{i}, x_{i + 2}, \dots, x_{n + 1}

, so no other root. Taking into account the continuity we easily deduce that, if

x

' is close enough to

x_{i + 1}

, there is at least one point

x^{''}

In (

x_{i + 1}, x_{i + 2}

) such as

f (x^{''}) = P {(x_{1}, x_{2}, \dots, x_{1}, x^{'}, x_{1 + 2}, \dots, x_{n + 1}; f ∣ x^{''})}^{'}

which is in contradiction with the property of multivalence.
Consider

n + 3

points

x_{1} < x_{2} < \dots < x_{n + 3}

. The differences

f (x) - P (x_{1}, x_{2}, \dots, x_{n + 1}; f ∣ x), f (x) - P (x_{2}, x_{3}, \dots, x_{n + 2}; f ∣ x)

then present values of the same signs at the point

x_{n + 3}

. We can easily see by noticing, for example, that the second Lagrange polynomial is deduced from the first by a continuous deformation. If we now consider

m \geq n + 2

points
(1)

x_{1} < x_{2} < \dots < x_{m}

we see that the differences in order

n + 1

[x_{i} < x_{i + 1}, \dots, x_{i + n + 1}; f], i = 1, 2, \dots, m - n - 1

^{(3)}

See Tiberiu Popoviciu, "On some properties of functions of one and two real variables". Mathematica t. V11l, p. 1-86.
are all different from zero and of the same sign. Any divided difference taken on

n + 2

any point of the sequence (1) will be different from zero and will have the same sign. Finally, any two divided differences

[x_{1}, x_{2}, \dots, x_{n + 2}; f], [x_{1}^{'}, x_{2}^{'}, \dots, x_{n + 2}^{'}]

are different from zero and have the same sign. It is sufficient to arrange all the points

x_{i}, x_{i}

into a sequence (1). We can therefore state the following property:
II. If the function

f (x)

is continuous and multivalent of order

n + 1

it is necessarily convex or concave of order n.
7. Multivalence of order

n + 1

is not sufficient to assert the convexity or concavity of order

n (^{4})

. We will demonstrate the following property:
III. For the function

f (x)

either convex or concave of order

n

it is necessary and sufficient that the function

F (x)

, attached to

f (x)

be multivalent in order

n + 1

.

The condition is obviously necessary. It remains to be shown that it is also sufficient.
8. I say first that:

If

F (x)

is multivalent in order

n + 1

it is bounded at every point of (a, b).

Suppose, for example, that at a point

x, M (f; x) = + \infty

. There then necessarily exists a sequence of points

x_{1}, x_{2}, \dots, x_{m}, \dots

of

(a, b)

tending towards

x

and a sequence of values

F (x_{1}), F (x_{2}), \dots, F (x_{m}), \dots

, taken at these points, and tending towards

+ \infty; F (x_{m}) \to + \infty

For

m \to \infty

. Either

P (x)

the Lagrange polynomial taking the values

F (x_{1}), F (x_{2}), \dots \dots, F (x_{n}), F (x_{m}) (m > n)

to the points

x_{1}, x_{2}, \dots, x_{n}, x_{m}

. We can easily see that, for

m

large enough, there is a value

F (x)

to the point

x

such as

F (x) = P (x)

This results from the continuity of a

^{(4)}

The function

f (x) = {\begin{array}{cc} \frac{x^{2}}{2} & 0 < x \leq\leq \frac{1}{2} \\ - \frac{4 x^{2} - 12 x + 1}{8} & \frac{1}{2} < x < 1 \end{array}

is indeed bivalent but is not convere or concave of order 1. This function is even discontinuous at the point

\frac{1}{2}

. It is easy to construct a ple example for

n

any
polynomial and is in contradiction with the property of multivalence of order

n + 1

.
9. Let us move on to the demonstration of property III and therefore suppose that

F (x)

be multivalent in order

n + 1

. Let us consider again

n + 1

points

x_{1} < x_{2} \dots < x_{n + 1}

In

(a, b)

and certain determined values

F (x_{1}), F (x_{2}), \dots, F (x_{n + 1})

of the function

F (x)

at these points. Either

P (x)

the Lagrange polynomial taking the values

F (x_{i})

to the points

x_{t}

. We have the following property:

{III}_{I}

. The difference

F (x) - P (x)

is different from zero and takes values of the same sign in each of the open intervals (

a, x_{1}

),

(x_{1}, x_{2}), \dots, (x_{n + 1}, b)

.

In other words, the values of the function

F (x)

remain, in each interval

(x_{i}, x_{i + 1}) (x_{0} = a, x_{n + 2} = b)

, on the same side of the polynomial

P (x)

The demonstration is done exactly as for property II.

If we take to

F (x_{l})

all values between

m (f; x_{l}), M (f; x_{l i})

and for each point

x_{l}

, we obtain a family of polynomials

P (x)

. This family has an upper limit function

ϕ_{1} (x)

and a lower limit function

ϕ_{2} (x)

. Any point located in the band formed by the functions

ϕ_{1} (x), ϕ_{2} (x)

enjoys the property that there exists a polynomial of the family passing through this point. We can therefore notice that the values of

F (x)

taken outside the points

x_{l}

are all outside the band included between

ϕ_{1} (x)

And

ϕ_{2} (x)

.

Let us now demonstrate the following property:

{III}_{II}

. The difference

F (x) - P (x)

is of opposite sign in two consecutive intervals (

x_{i}, x_{i + 1}

), (

x_{i + 1}, x_{i + 2}

).

Let us suppose the opposite therefore that, for example,

F (x) > P (x)

in these intervals. For

F (x_{t + 1})

let's take the value

M (f; x_{t + 1})

. Either

x^{'}

a point in (

x_{i}, x_{i + 1}

) And

F (x^{'})

one of the values taken at this point. Let us construct the Lagrange polynomial

Q (x)

taking the values

F (x_{1})

,

F (x_{2}), \dots, F (x_{i}), F (x^{'}), F (x_{i + 2}), \dots, F (x_{n + 1})

at the corresponding points. The difference

P (x) - Q (x)

is a polynomial of degree

n

which cancels out for

x = x_{1}, x_{2}, \dots, x_{i}, x_{i + 2}, \dots, x_{n + 1}

, so at no other point of (

a, b

). The property then results exactly like property III.
10. To complete the demonstration we still have to show that:

III

_{III}

. The function

F (x)

is uniform in every respect.
Let us keep the previous notations. It suffices to demonstrate that we have

M (f; x_{1}) = m (f; x_{1})

. To demonstrate this, let us assume that

M (f; x_{1}) > m (f; x_{1})

. To fix the ideas, let us suppose that

F (x) > P (x)

to the left of

x_{1}

; we then have

F (x) < P (x)

in the meantime (

x_{1}, x_{2}

). Let's take

F (x_{1}) = m (f; x_{1})

and let the points be fixed

x_{3}, x_{4}, \dots, x_{n + 1}

as well as the values

F (x_{3}), F (x_{4}), \dots, F (x_{n + 1})

. Under the definition of

M (f; x_{1})

we can take for

x_{2}

successively the terms of a sequence of points

x_{2}^{'}, x_{2}^{''}, \dots, x_{2}^{(m)}, \dots

tending towards

x_{1}

and such that the corresponding values

F (x_{2}^{(m)})

tend towards

M (f; x_{4})

For

m \to \infty

. Let us designate by

P_{m} (x)

the polynomial deduced from

P (x)

by replacing

x_{2}

And

F (x_{2})

by

x_{2}^{(m)}

And

F (x_{2}^{(m)})

respectively. If

x_{0}

is a point to the left of

x_{1}

We have

\begin{matrix} F (x_{0}) < P_{m} (x_{0}) . \\ P_{m} (x_{0}) \to - \infty pour m \to \infty \end{matrix}

But
and so we arrive at an impossibility. Here we take into account a simple property which results from the continuity of a polynomial (5). We must therefore have

M (f; x_{1}) = m (f; x_{1})

which demonstrates property IIIII.

It therefore follows that:

{III}_{IV}

. If the function

F (x)

, attached to

f (x)

, is multivalent in order

n + 1

, the function

f (x)

is continuous.

Property III then follows from Property II.
11. Exactly as in No. 9 we can still state the following property:

III'. For the function

f (x)

either bounded and convex or concave of order

n

it is necessary and sufficient that the attached function

F (x)

, extended on the ends a and b, is multivalent of order

n + 1

.
(5). This can be demonstrated using the Lagrange interpolation formula.
polynomial and is in contradiction with the property of multivalence of order

n + 1

.
9. Let us move on to the demonstration of property III and therefore suppose that

F (x)

be multivalent in order

n + 1

. Let us consider again

n + 1

points

x_{4} < x_{2} \dots < x_{n + 1}

In

(a, b)

and certain determined values

F (x_{1}), F (x_{2}), \dots, F (x_{n + 1})

of the function

F (x)

at these points. Either

P (x)

the Lagrange polynomial taking the values

F (x_{i})

to the points

x_{i}

. We have the following property:

{III}_{I} . La

difference

F (x) - P (x)

is different from zero and takes values of the same sign in each of the open intervals (

a, x_{1}

),

(x_{1}, x_{2}), \dots, (x_{n + 1}, b)

.

In other words, the values of the function

F (x)

remain, in each interval

(x_{l}, x_{i + 1}) (x_{0} = a, x_{n + 2} = b)

, on the same side of the polynomial

P (x)

. The demonstration is done exactly as for the property

I_{I}

.

If we take to

F (x_{l})

all values included against

m (f; x_{i}), M (f; x_{i l})

ef for each point

x_{i}

, we obtain a family of polynomials

P (x)

. This family has an upper limit function

ϕ_{1} (x)

and a lower limit function

ϕ_{2} (x)

. Any point located in the band formed by the functions

ϕ_{1} (x), ϕ_{2} (x)

enjoys the property that there exists a polynomial of the family passing through this point. We can therefore notice that the values of

F (x)

taken outside the points

x_{i}

are all outside the band included between

ϕ_{1} (x)

And

ϕ_{2} (x)

.

Let us now demonstrate the following property:
III II . The difference

F (x) - P (x)

is of opposite sign in two consecutive intervals

(x_{i}, x_{l + 1}), (x_{l + 1}, x_{l + 2})

.

Let us suppose the opposite therefore that, for example,

F (x) > P (x)

in these intervals. For

F (x_{i + 1})

let's take the value

M (f; x_{t + 1})

. Either

x^{'}

a point in

(x_{l}, x_{i + 1})

And

F (x^{'})

one of the values taken at this point. Let us construct the Lagrange polynomial

Q (x)

taking the values

F (x_{1})

,

F (x_{2}), \dots, F (x_{i}), F (x^{'}), F (x_{i + 2}), \dots, F (x_{n + 1})

at the corresponding points. The difference

P (x) - Q (x)

is a polynomial of degree

n

which cancels out for

x = x_{1}, x_{2}, \dots, x_{i}, x_{i + 2}, \dots, x_{n + 1}

, so at no other point of (

a, b

). The property then results exactly like property III.
10. To complete the demonstration we still have to show that:

III III. The function

F (x)

is uniform at all points.
Let us keep the previous notations. It suffices to demonstrate that we have

M (f; x_{1}) = m (f; x_{1})

. To demonstrate this, let us assume that

M (f; x_{1}) > m (f; x_{1})

. To fix the ideas, let us suppose that

F (x) > P (x)

to the left of

x_{1}

; we then have

F (x) < P (x)

in the meantime (

x_{1}, x_{2}

). Let's take

F (x_{1}) = m (f; x_{1})

and let the points be fixed

x_{3}, x_{4}, \dots, x_{n + 1}

as well as the values

F (x_{3}), F (x_{4}), \dots, F (x_{n + 1})

. Under the definition of

M (f; x_{1})

we can take for

x_{2}

successively the terms of a sequence of points

x_{2}^{'}, x_{2}^{''}, \dots, x_{2}^{(m)}, \dots

tending towards

x_{1}

and such that the corresponding values

F (x_{2}^{(m)})

tend towards

M (f; x_{4})

For

m \to \infty

. Let us designate by

P_{m} (x)

the polynomial deduced from

P (x)

by replacing

x_{2}

And

F (x_{2})

by

x_{3}^{(m)}

And

F (x_{2}^{(m)})

respectively. If

x_{0}

is a point to the left of

x_{1}

We have

F (x_{0}) < P_{m} (x_{0})

But

P_{m} (x_{0}) \to - \infty pour m \to \infty

and we therefore arrive at an impossibility. Here we take into account a simple property which results from the continuity of a polynomial (

^{5}

). So we must have

M (f; x_{1}) = m (f; x_{1})

which demonstrates property III

_{11}

.

It therefore follows that:

{III}_{IV}

. If the function

F (x)

, attached to

f (x)

, is multivalent in order

n + 1

, the function

f (x)

is continuous.

Property III then follows from Property II.
11. Exactly as in No. 9 we can still state the following property:

III'. For the function

f (x)

either bounded and convex or concave of order

n

it is necessary and sufficient that the attached function

F (x)

, extended on the ends a and b, is multivalent of order

n + 1

.
(5). This can be demonstrated using Lagrange's interpolation formula.

$^{(1)}$ For example, the function
is indeed univalent but is not monotonic.
(²) The reader is asked to draw the figure.

1936

Notes on Higher-Order Convex Functions (II)

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NOTES ON HIGHER ORDER CONVEX FUNCTIONS (II)

On univalent functions and maltivalent functions of a real variable.

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