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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).
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Bulletin Mathematique de la Societe des Sciences Mathematiques de Roumanie
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Romanian Society of Mathematical Sciences
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TWO REMARKS ON CONVEX FUNCTIONS
BY
I
I. Let us consider a functionf(x)f(x)non-concave on a linear setEE. IfEEis bounded we will designate bya=min Ea=\min E,a < b = max Ea<b=\max Eits extremities. The functionf(x)f(x)is bounded on every bounded subsetE_(1)E_{1}ofEEwhich contains its extremitiesCDc, dFurthermore, the maximum off(x)f(x)onE_(1)E_{1}is always reached and is only reached at least at one of the extremesCDc, dunless the function does not reduce to a constant onE_(1)E_{1}This applies to all functions.f(x) + alpha x + beta, alpha, betaf(x)+\alpha x+\beta, \alpha, \betabeing two constants. According to a remark by Mr.SSSaks^(1){ }^{1}) the converse is true and we can state the following property:
For the functionf(x)f(x)either non-concave onEEit is necessary and sufficient that, whatever the constant a and the bounded subsetE_(1)E_{1}ofEEcontaining its extremities, the functionf(x) + alpha xf(x) + α xreaches its maximum at least at one of its extremitiesCDc, d.
Indeed, we verify that the property is equivalent to the inequality of definition
|[I,x_(1),f(x_(1))],[I,x_(2),f(x_(2))],[I,x_(3),f(x_(3))]| >= 0,x_(1), < x_(2) < x_(3),x_(1),x_(2),x_(3)sub E\left|\begin{array}{lll} \mathrm{I} & x_{1} & f\left(x_{1}\right) \\ \mathrm{I} & x_{2} & f\left(x_{2}\right) \\ \mathrm{I} & x_{3} & f\left(x_{3}\right) \end{array}\right| \geq 0, x_{1},<x_{2}<x_{3}, x_{1}, x_{2}, x_{3} \subset E
of the non-concave functionf(x)f(x)
Convex functions can be characterized in the same way, but then it must be stated in the problem statement thatf(x) + alpha xf(x) + α x1. 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 pointsx_(1),x_(2)x_{1}, x_{2}ofEE, it reaches him at every point ofEEbetweenx_(1),x_(2)x_{1}, x_{2}. We
Convex functions can be characterized in the same way, but then it must be stated in the problem statement thatf(x) + alpha xf(x) + α x1. 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 pointsx_(1),x_(2)x_{1}, x_{2}ofEE, it reaches him at every point ofEEbetweenx_(1),x_(2)x_{1}, x_{2}. We
We will now seek to state a converse. We will assumeEEbounded and closed, and we will say that a function is continuous onEEif it is continuous at every point of the derivativeE^(')E^{\prime}ofEEWe therefore have the following property:
The functionf(x)f(x)finished, uniform and continuous over the closed and bounded setEE' is non-concave if when for aalpha,f(x)+alpha x\alpha, f(x)+\alpha xreached its minimum in two pointsx_(1),x_(2)x_{1}, x_{2}ofEE, it reaches him at every point ofEEbetweenx_(1)x_{1}Andx_(2)x_{2}.
First, let's note that the functionf(x) + alpha x + betaf(x) + α x + βOrbeta\betais 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, whateverx_(0)sub Ex_{0} \subset Eand different from the extremities ofEE, one can find aalpha\alphasuch that the minimum, equal to zero for a suitable choice ofbeta\beta, off(x) + alpha x + betaf(x) + α x + βbe reached forx=x_(0)x=x_{0}This is equivalent to saying that at every pointA_(0)(x_(0),f(x_(0)))A_{0}\left(x_{0}, f\left(x_{0}\right)\right), abscissax_(0)sub Ex_{0} \subset Epasses a line of support^(2){ }^{2}) of the curvey=f(x)y=f(x)We then know that the function is necessarily non-concave.
Regardless oflambda\lambdathere exists a line of support for the direction curvelambda\lambda(of angular coefficienthat(lambda)\hat{\lambda}). The curve is, moreover, not below this line. It is therefore necessary to demonstrate that ifx_(0)sub Ex_{0} \subset Ethere exists a line of support passing throughA_(0)A_{0}.
If the parallel to the axisOxO xled byA_(0)A_{0}If the line of support is a supporting line, the property is proven. Otherwise, let's draw the supporting line.Delta_(0)\Delta_{0}parallel toOx^(3)\mathrm{O} x^{3}This line contains at least one point of the curvey=f(x)y=f(x); eitherHASHASone of these points and, to clarify, let's assume its abscissax_(1) < x_(0)x_{1}<x_{0}Let us now consider all the lines of support.Delta_(lambda)\Delta_{\lambda}(of direction)lambda\lambda) of the curveCCrepresented byy=f(x)y=f(x)whenx sub Ex \subset Evaries within the closed interval(a,x_(0))\left(a, x_{0}\right)and eitherC^(')C^{\prime}the curve represented byy=f(x)y=f(x)whenx sub Ex \subset Evaries within the closed interval (x_(0),bx_{\mathbf{0}}, b). All the points ofC^(')C^{\prime}are above any rightDelta_(lambda)\Delta_{\lambda}withlambda <\lambda<o. Whenlambda\lambdaStarting from 0, two cases are possible:
I. We arrive (by virtue of continuity) at a valuelambda^(') >= 0\lambda^{\prime} \geq 0such asDelta_(lambda^('))\Delta_{\lambda^{\prime}}is a line of support ofCCIn this case, this lineDelta_(lambda^('))\Delta_{\lambda^{\prime}}contains at least one point ofCCand at least one point ofC^(')C^{\prime}Therefore, it is necessary thatA_(0)A_{0}also belongs to this line, and the property is proven.
II. The curveC^(')C^{\prime}is aboveDelta_(lambda)\Delta_{\lambda}regardless oflambda\lambdaI say that this is impossible. Indeed,Delta_(lambda)\Delta_{\lambda}is not below the rightdelta_(lambda)\delta_{\lambda}, of directionlambda\lambda, through the projection ofA_(0)A_{0}onDelta_(0)\Delta_{0}The curveC^(')C^{\prime}must therefore be abovedelta_(lambda)\delta_{\lambda}regardless oflambda\lambdawhich is clearly impossible.
I. We arrive (by virtue of continuity) at a valuelambda^(') >= 0\lambda^{\prime} \geq 0such asDelta_(lambda^('))\Delta_{\lambda^{\prime}}is a line of support ofCCIn this case, this lineDelta_(lambda^('))\Delta_{\lambda^{\prime}}contains at least one point ofCCand at least one point ofC^(')C^{\prime}Therefore, it is necessary thatA_(0)A_{0}also belongs to this line, and the property is proven.
II. The curveC^(')C^{\prime}is aboveDelta_(lambda)\Delta_{\lambda}regardless oflambda\lambdaI say that this is impossible. Indeed,Delta_(lambda)\Delta_{\lambda}is not below the rightdelta_(lambda)\delta_{\lambda}, of directionlambda\lambda, through the projection ofA_(0)A_{0}onDelta_(0)\Delta_{0}The curveC^(')C^{\prime}must therefore be abovedelta_(lambda)\delta_{\lambda}regardless oflambda\lambdawhich is clearly impossible.
The stated property is therefore proven.
3. - Suppose, in particular, thatEElet be a closed and finite interval (a,ba, b), we can state the following property:
3. - Suppose, in particular, thatEElet be a closed and finite interval (a,ba, b), we can state the following property:
The functionf(x)f(x)finite and continuous in the finite and closed interval (a,ba, b) is convex if, for any numberalpha\alpha, the functionf(x) + alpha xf(x) + α xreaches 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)={[I-sqrt(I-(xI)^(2)),","quad0 <= x < I],[(2-x),I <= x <= 2]:}g(x)= \begin{cases}\mathrm{I}-\sqrt{\mathrm{I}-(x-\mathrm{I})^{2}} & , \quad 0 \leqq x<\mathrm{I} \\ (2-x) & \mathrm{I} \leqq x \leqq 2\end{cases}
It is easy to verify thatg(x) + alpha xg(x) + α xalways reaches its minimum and at a single point, the functiong(x)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. - EitherEEa linear set. I say thatEEis decomposed into two consecutive subsetsE_(1)E_{1}AndE_(2)E_{2}if:^(@)E_(1),E_(2){ }^{\circ} E_{1}, E_{2}contain only points ofE,2^(@)E, 2^{\circ}every point ofEEbelongs to one and only one of the setsE_(1),E_(2),3^(@)E_{1}, E_{2}, 3^{\circ}every point ofE_(1)E_{1}is to the left of every point ofE_(2)E_{2}Any point betweenmaxE_(1)\max E_{1}AndminE_(2)\min E_{2}is a decomposition point. Such a point may or may not belong to one of the setsE_(1),E_(2)E_{1}, E_{2}It can also happen that one of the setsE_(1),E_(2)E_{1}, E_{2}either empty, the other then coincides withEEIn this case, there is no decomposition point.
We know that:
Iff(x)f(x)is non-concave onEEwe can break down the wholeEEin two consecutive subsetsE_(1),E_(2)E_{1}, E_{2}such that on each the function is monotonic.
Iff(x)f(x)is non-concave onEEwe can break down the wholeEEin two consecutive subsetsE_(1),E_(2)E_{1}, E_{2}such that on each the function is monotonic.
IfE_(1)E_{1}is emptyf(x)f(x)is non-decreasing onEEand ifE_(2)E_{2}is emptyf(x)f(x)is non-increasing onEEIn general, the function is non-increasing onE_(1)E_{1}and non-decreasing onE_(2)^(4)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 iff(x)f(x)is non-concave we have
(I)quad f(x_(2)) <= max[f(x_(1)),f(x_(3))],x_(1) < x_(2) < x_(3),x_(1),x_(2),x_(3)sub E\quad f\left(x_{2}\right) \leqq \max \left[f\left(x_{1}\right), f\left(x_{3}\right)\right], x_{1}<x_{2}<x_{3}, x_{1}, x_{2}, x_{3} \subset E.
2. - From the inequality of definition, it immediately follows that iff(x)f(x)is non-concave we have
(I)quad f(x_(2)) <= max[f(x_(1)),f(x_(3))],x_(1) < x_(2) < x_(3),x_(1),x_(2),x_(3)sub E\quad f\left(x_{2}\right) \leqq \max \left[f\left(x_{1}\right), f\left(x_{3}\right)\right], 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 whoseppnth power, withp >= 1p \geq 1, is non-concave. Functions that satisfy
Inequality (I) can be viewed precisely as the limiting case forp rarr+oop \rightarrow+\infty.
We propose to demonstrate that:
Eitherf(x)f(x)a finite, uniform function defined on the set E. So that we can decomposeEEin two consecutive subsetsE_(1),E_(2)E_{1}, E_{2}on property (I).
Eitherf(x)f(x)a finite, uniform function defined on the set E. So that we can decomposeEEin two consecutive subsetsE_(1),E_(2)E_{1}, E_{2}on property (I).
We assume that ifE_(1),E_(2)E_{1}, E_{2}are not empty the function is non-increasing onE_(1)E_{1}and non-decreasing onE_(2)E_{2}Otherwise, the function -f(x)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 functionf(x)f(x), finite, uniform and defined on the setEEverifying property (1), we can decompose the setEEin two consecutive subsets such that on each the function is monotonic.
3. - Let us therefore demonstrate this last property. Letm = min f(x)m = min f(x)
which can be a finite number or-oo-\infty. Either
3. - Let us therefore demonstrate this last property. Letm = min f(x)m = min f(x)
which can be a finite number or-oo-\infty. Either
{:(2)xi_(1)","xi_(2)dots","xi_(n)","dots:}\begin{equation*} \xi_{1}, \xi_{2} \ldots, \xi_{n}, \ldots \tag{2} \end{equation*}
a sequence of points, distinct or not, such thatf(xi_(n))rarr mf\left(\xi_{n}\right) \rightarrow mForn rarr oon \rightarrow \inftyWe can always assume that the sequence (2) is either convergent or tends towards+oo+\inftyOr-oo-\inftyTo simplify, we can say that the sequence tends towards a pointxi\xifinite or infinite. It's easy to see that ifm=-oom=-\inftythere is only one pointxi\xiThree cases may arise:
I.xi\xiis finished. Consider two pointsx_(1) < x_(2)x_{1}<x_{2}to the left ofxi\xiWe cannot havef(x_(1)) < f(x_(2))f\left(x_{1}\right)<f\left(x_{2}\right)since then we could find axi_(n) > x_(2)xi_{n}>x_{2}such as one hasf(xi_(n)) < f(x_(2))f\left(\xi_{n}\right)<f\left(x_{2}\right)By takingx_(3)=xi_(n)x_{3}=\xi_{n}This contradicts inequality (I). Therefore, it is necessary thatf(x_(1)) >= f(x_(2))f\left(x_{1}\right) \geq f\left(x_{2}\right)We see in the same way that ifx_(1) < x_(2)x_{1}<x_{2}are to the right ofxi\xiwe havef(x_(1)) <= f(x_(2))f\left(x_{1}\right) \leq f\left(x_{2}\right)So, if the pointxi\xidoes not belong toEEThe property has been demonstrated. It remains to be shown that this is still the case ifxi sub E\xi \subset ETwo scenarios are possible:I^(@)f(xi)=m\mathrm{I}^{\circ} f(\xi)=mwhich is only possible ifmmis finished. In this case, we obviously havef(x_(1)) >= f(xi)f\left(x_{1}\right) \geq f(\xi)Forxi > x_(1)sub E\xi>x_{1} \subset EAndf(xi) <= f(x_(2))f(\xi) \leq f\left(x_{2}\right)Forxi < x_(2)sub E\xi<x_{2} \subset EThe property therefore results in this case as well, andxi\xibelongs indifferently toE_(1)E_{1}OrE_(2),2^(@)f(xi) > mE_{2}, 2^{\circ} f(\xi)>m. In this casexi\xican only be a limit on one side. If it is a left (right) limit, then we havef(xi) <= f(x_(2))f(\xi) \leqq f\left(x_{2}\right)for everythingxi < x_(2)sub E[f(x_(2)) >= f(xi):}\xi<x_{2} \subset E\left[f\left(x_{2}\right) \geq f(\xi)\right.for everything{: xi > x_(1)sub E]\left.\xi>x_{1} \subset E\right]and the property still results, the point of decompositionxi\xibelonging toE_(2)E_{2}(resp. toE_(1)E_{1}II
.xi=-oo\xi=-\inftyThis can only happen ifEEis unbounded below. We then see, as above, that ifx_(1) < x_(2)x_{1}<x_{2}are two points ofEEwe havef(x_(1)) <= f(x_(2))f\left(x_{1}\right) \leqq f\left(x_{2}\right)The function is therefore non-decreasing onEEIII
.xi=+oo\xi=+\inftySimilarly, we can see that the function is non-increasing onEE.
I.xi\xiis finished. Consider two pointsx_(1) < x_(2)x_{1}<x_{2}to the left ofxi\xiWe cannot havef(x_(1)) < f(x_(2))f\left(x_{1}\right)<f\left(x_{2}\right)since then we could find axi_(n) > x_(2)xi_{n}>x_{2}such as one hasf(xi_(n)) < f(x_(2))f\left(\xi_{n}\right)<f\left(x_{2}\right)By takingx_(3)=xi_(n)x_{3}=\xi_{n}This contradicts inequality (I). Therefore, it is necessary thatf(x_(1)) >= f(x_(2))f\left(x_{1}\right) \geq f\left(x_{2}\right)We see in the same way that ifx_(1) < x_(2)x_{1}<x_{2}are to the right ofxi\xiwe havef(x_(1)) <= f(x_(2))f\left(x_{1}\right) \leq f\left(x_{2}\right)So, if the pointxi\xidoes not belong toEEThe property has been demonstrated. It remains to be shown that this is still the case ifxi sub E\xi \subset ETwo scenarios are possible:I^(@)f(xi)=m\mathrm{I}^{\circ} f(\xi)=mwhich is only possible ifmmis finished. In this case, we obviously havef(x_(1)) >= f(xi)f\left(x_{1}\right) \geq f(\xi)Forxi > x_(1)sub E\xi>x_{1} \subset EAndf(xi) <= f(x_(2))f(\xi) \leq f\left(x_{2}\right)Forxi < x_(2)sub E\xi<x_{2} \subset EThe property therefore results in this case as well, andxi\xibelongs indifferently toE_(1)E_{1}OrE_(2),2^(@)f(xi) > mE_{2}, 2^{\circ} f(\xi)>m. In this casexi\xican only be a limit on one side. If it is a left (right) limit, then we havef(xi) <= f(x_(2))f(\xi) \leqq f\left(x_{2}\right)for everythingxi < x_(2)sub E[f(x_(2)) >= f(xi):}\xi<x_{2} \subset E\left[f\left(x_{2}\right) \geq f(\xi)\right.for everything{: xi > x_(1)sub E]\left.\xi>x_{1} \subset E\right]and the property still results, the point of decompositionxi\xibelonging toE_(2)E_{2}(resp. toE_(1)E_{1}II
.xi=-oo\xi=-\inftyThis can only happen ifEEis unbounded below. We then see, as above, that ifx_(1) < x_(2)x_{1}<x_{2}are two points ofEEwe havef(x_(1)) <= f(x_(2))f\left(x_{1}\right) \leqq f\left(x_{2}\right)The function is therefore non-decreasing onEEIII
.xi=+oo\xi=+\inftySimilarly, we can see that the function is non-increasing onEE.
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 subsetsE_(1)E_{1}AndE_(2)E_{2}and vice versa.
We can also note that while equality is never possible in (I), the function is strictly monotonic (decreasing and increasing) on ​​the decomposition subsetsE_(1)E_{1}AndE_(2)E_{2}and vice versa.
Seminarul Matematic, Universitatea Cernăuți
- ^(1){ }^{1}) S. Saks, O funkcjach wypuklych i podharmonicznych, “Mathesis Polska”, t. VI (1931), pp. 43-65.
- ^(2){ }^{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){ }^{3}The reader is asked to draw the figure. - ^(4){ }^{4}) The exampleE=E=interval(0,1),f(0)=1,f(x)=x(0,1), f(0)=1, f(x)=xFor0 < x <= 10<x \leqq 1shows us thatE_(1)E_{1}can well be formed by a single point (similarlyE_(2)E_{2}). We agree, of course, that any function defined at a single point is monotonic and indifferently increasing or decreasing.