Two remarks on convex functions

1 month ago

Tiberiu Popoviciu

(original), convexity, Mathematical Analysis, not-at-ICTP, paper

Abstract

Authors

T. Popoviciu

Keywords

?

Paper coordinates

T. Popoviciu, Deux remarques sur les fonctions convexes, Bull. de la Sect. sci. de l’Acad. Roum., 20 (1938) no. ?, pp. 45-49 (in French).

PDF

https://ictp.acad.ro/wp-content/uploads/2025/11/1938-c-142-Popoviciu-Deux-remarques-sur-les-fonctions-convexes_opt.pdf

About this paper

Journal

Bulletin Mathematique de la Societe des Sciences Mathematiques de Roumanie

Publisher Name

Romanian Society of Mathematical Sciences

DOI

Print ISSN

Online ISSN

[Zbl 0021.11605, JFM 64.1024.03]

google scholar link

??

Paper (preprint) in HTML form

1938-c -142- Popoviciu- Two remarks on convex functions

Original text

Rate this translation

Your feedback will be used to help improve Google Translate

TWO REMARKS ON CONVEX FUNCTIONS

BY

TIBERIU POPOVICIUNote presented by MG Tzitzéica, MAR

I

I. Let us consider a function

f (x)

non-concave on a linear set

E

. If

E

is bounded we will designate by

has = min E

,

has < b = max E

its extremities. The function

f (x)

is bounded on every bounded subset

E_{1}

of

E

which contains its extremities

c, d

Furthermore, the maximum of

f (x)

on

E_{1}

is always reached and is only reached at least at one of the extremes

c, d

unless the function does not reduce to a constant on

E_{1}

This applies to all functions.

f (x) + α x + β, α, β

being two constants. According to a remark by Mr.

S

Saks

^{1}

) the converse is true and we can state the following property:

For the function

f (x)

either non-concave on

E

it is necessary and sufficient that, whatever the constant a and the bounded subset

E_{1}

of

E

containing its extremities, the function

f (x) + α x

reaches its maximum at least at one of its extremities

c, d

.

Indeed, we verify that the property is equivalent to the inequality of definition

| \begin{array}{lll} I & x_{1} & f (x_{1}) \\ I & x_{2} & f (x_{2}) \\ I & x_{3} & f (x_{3}) \end{array} | \geq 0, x_{1}, < x_{2} < x_{3}, x_{1}, x_{2}, x_{3} \subset E

of the non-concave function

f (x)

Convex functions can be characterized in the same way, but then it must be stated in the problem statement that

f (x) + α x

1. If a non-concave function reaches its maximum at only one of its endpoints.
2. If a non-concave function reaches its minimum at two points

x_{1}, x_{2}

of

E

, it reaches him at every point of

E

between

x_{1}, x_{2}

. We

We will now seek to state a converse. We will assume

E

bounded and closed, and we will say that a function is continuous on

E

if it is continuous at every point of the derivative

E^{'}

of

E

We therefore have the following property:

The function

f (x)

finished, uniform and continuous over the closed and bounded set

E

' is non-concave if when for a

α, f (x) + α x

reached its minimum in two points

x_{1}, x_{2}

of

E

, it reaches him at every point of

E

between

x_{1}

And

x_{2}

.

First, let's note that the function

f (x) + α x + β

Or

β

is a constant and enjoys the same property. We will take into account the fact that a continuous function on a closed set always reaches its minimum.

The whole difficulty lies in showing that, whatever

x_{0} \subset E

and different from the extremities of

E

, one can find a

α

such that the minimum, equal to zero for a suitable choice of

β

, of

f (x) + α x + β

be reached for

x = x_{0}

This is equivalent to saying that at every point

{HAS}_{0} (x_{0}, f (x_{0}))

, abscissa

x_{0} \subset E

passes a line of support

^{2}

) of the curve

y = f (x)

We then know that the function is necessarily non-concave.

Regardless of

λ

there exists a line of support for the direction curve

λ

(of angular coefficient

\hat{λ}

). The curve is, moreover, not below this line. It is therefore necessary to demonstrate that if

x_{0} \subset E

there exists a line of support passing through

{HAS}_{0}

.

If the parallel to the axis

O x

led by

{HAS}_{0}

If the line of support is a supporting line, the property is proven. Otherwise, let's draw the supporting line.

Δ_{0}

parallel to

O x^{3}

This line contains at least one point of the curve

y = f (x)

; either

HAS

one of these points and, to clarify, let's assume its abscissa

x_{1} < x_{0}

Let us now consider all the lines of support.

Δ_{λ}

(of direction)

λ

) of the curve

C

represented by

y = f (x)

when

x \subset E

varies within the closed interval

(has, x_{0})

and either

C^{'}

the curve represented by

y = f (x)

when

x \subset E

varies within the closed interval (

x_{0}, b

). All the points of

C^{'}

are above any right

Δ_{λ}

with

λ <

o. When

λ

Starting from 0, two cases are possible:
I. We arrive (by virtue of continuity) at a value

λ^{'} \geq 0

such as

Δ_{λ^{'}}

is a line of support of

C

In this case, this line

Δ_{λ^{'}}

contains at least one point of

C

and at least one point of

C^{'}

Therefore, it is necessary that

{HAS}_{0}

also belongs to this line, and the property is proven.
II. The curve

C^{'}

is above

Δ_{λ}

regardless of

λ

I say that this is impossible. Indeed,

Δ_{λ}

is not below the right

δ_{λ}

, of direction

λ

, through the projection of

{HAS}_{0}

on

Δ_{0}

The curve

C^{'}

must therefore be above

δ_{λ}

regardless of

λ

which is clearly impossible.

The stated property is therefore proven.
3. - Suppose, in particular, that

E

let be a closed and finite interval (

has, b

), we can state the following property:

The function

f (x)

finite and continuous in the finite and closed interval (

has, b

) is convex if, for any number

α

, the function

f (x) + α x

reaches its minimum at a single point.

The assumption of continuity cannot be eliminated in these statements to draw the preceding conclusions. Let us consider, for this purpose, the function

g (x) = {\begin{cases} I - \sqrt{I - (x - I)^{2}} & , 0 ≦ x < I \\ (2 - x) & I ≦ x ≦ 2 \end{cases}

It is easy to verify that

g (x) + α x

always reaches its minimum and at a single point, the function

g (x)

However, it is not convex. On the contrary, one can substitute for continuity a less restrictive hypothesis, for example lower semi-continuity.

II

I. - Either

E

a linear set. I say that

E

is decomposed into two consecutive subsets

E_{1}

And

E_{2}

if:

^{\circ} E_{1}, E_{2}

contain only points of

E, 2^{\circ}

every point of

E

belongs to one and only one of the sets

E_{1}, E_{2}, 3^{\circ}

every point of

E_{1}

is to the left of every point of

E_{2}

Any point between

max E_{1}

And

min E_{2}

is a decomposition point. Such a point may or may not belong to one of the sets

E_{1}, E_{2}

It can also happen that one of the sets

E_{1}, E_{2}

either empty, the other then coincides with

E

In this case, there is no decomposition point.

We know that:
If

f (x)

is non-concave on

E

we can break down the whole

E

in two consecutive subsets

E_{1}, E_{2}

such that on each the function is monotonic.

If

E_{1}

is empty

f (x)

is non-decreasing on

E

and if

E_{2}

is empty

f (x)

is non-increasing on

E

In general, the function is non-increasing on

E_{1}

and non-decreasing on

E_{2}^{4}

).

But there are more general functions that possess the same property. We propose to characterize these functions by inequalities between three of their values.
2. - From the inequality of definition, it immediately follows that if

f (x)

is non-concave we have
(I)

f (x_{2}) ≦ max [f (x_{1}), f (x_{3})], x_{1} < x_{2} < x_{3}, x_{1}, x_{2}, x_{3} \subset E

.

However, this inequality does not characterize non-concave functions. The inequality is satisfied, for example, by any non-negative function whose

p

nth power, with

p \geq 1

, is non-concave. Functions that satisfy

Inequality (I) can be viewed precisely as the limiting case for

p \to + \infty

.

We propose to demonstrate that:
Either

f (x)

a finite, uniform function defined on the set E. So that we can decompose

E

in two consecutive subsets

E_{1}, E_{2}

on property (I).

We assume that if

E_{1}, E_{2}

are not empty the function is non-increasing on

E_{1}

and non-decreasing on

E_{2}

Otherwise, the function -

f (x)

must verify inequality (I).

It is easy to see that the condition is necessary. It remains to show that it is also sufficient, therefore:

If the function

f (x)

, finite, uniform and defined on the set

E

verifying property (1), we can decompose the set

E

in two consecutive subsets such that on each the function is monotonic.
3. - Let us therefore demonstrate this last property. Let

m = min f (x)

which can be a finite number or

- \infty

. Either

\begin{matrix} (2) & ξ_{1}, ξ_{2} \dots, ξ_{n}, \dots \end{matrix}

a sequence of points, distinct or not, such that

f (ξ_{n}) \to m

For

n \to \infty

We can always assume that the sequence (2) is either convergent or tends towards

+ \infty

Or

- \infty

To simplify, we can say that the sequence tends towards a point

ξ

finite or infinite. It's easy to see that if

m = - \infty

there is only one point

ξ

Three cases may arise:
I.

ξ

is finished. Consider two points

x_{1} < x_{2}

to the left of

ξ

We cannot have

f (x_{1}) < f (x_{2})

since then we could find a

ξ_{n} > x_{2}

such as one has

f (ξ_{n}) < f (x_{2})

By taking

x_{3} = ξ_{n}

This contradicts inequality (I). Therefore, it is necessary that

f (x_{1}) \geq f (x_{2})

We see in the same way that if

x_{1} < x_{2}

are to the right of

ξ

we have

f (x_{1}) \leq f (x_{2})

So, if the point

ξ

does not belong to

E

The property has been demonstrated. It remains to be shown that this is still the case if

ξ \subset E

Two scenarios are possible:

I^{\circ} f (ξ) = m

which is only possible if

m

is finished. In this case, we obviously have

f (x_{1}) \geq f (ξ)

For

ξ > x_{1} \subset E

And

f (ξ) \leq f (x_{2})

For

ξ < x_{2} \subset E

The property therefore results in this case as well, and

ξ

belongs indifferently to

E_{1}

Or

E_{2}, 2^{\circ} f (ξ) > m

. In this case

ξ

can only be a limit on one side. If it is a left (right) limit, then we have

f (ξ) ≦ f (x_{2})

for everything

ξ < x_{2} \subset E [f (x_{2}) \geq f (ξ)

for everything

ξ > x_{1} \subset E]

and the property still results, the point of decomposition

ξ

belonging to

E_{2}

(resp. to

E_{1}

II
.

ξ = - \infty

This can only happen if

E

is unbounded below. We then see, as above, that if

x_{1} < x_{2}

are two points of

E

we have

f (x_{1}) ≦ f (x_{2})

The function is therefore non-decreasing on

E

III
.

ξ = + \infty

Similarly, we can see that the function is non-increasing on

E

.

The property is therefore completely proven.
We can also note that while equality is never possible in (I), the function is strictly monotonic (decreasing and increasing) on the decomposition subsets

E_{1}

And

E_{2}

and vice versa.

Seminarul Matematic, Universitatea Cernăuți

$^{1}$ ) S. Saks, O funkcjach wypuklych i podharmonicznych, “Mathesis Polska”, t. VI (1931), pp. 43-65.
$^{2}$ ) A support line is a line that passes through a point on the curve which is entirely located on one side of that line.
$^{3}$ The reader is asked to draw the figure.
$^{4}$ ) The example $E =$ interval $(0, 1), f (0) = 1, f (x) = x$ For $0 < x ≦ 1$ shows us that $E_{1}$ can well be formed by a single point (similarly $E_{2}$ ). We agree, of course, that any function defined at a single point is monotonic and indifferently increasing or decreasing.

1938