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T. Popoviciu, Sur quelques inegalités entre les fonctions convexes (II), Comptes Rendus de l’Instit. de Sci. de Roum., 2 (1938), pp. 454-458 (in French)
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113. On SOME INEQUALITIES BETWEEN CONVEX FUNCTIONS
(SECOND NOTE)
By tiberiu popoviciu, Mc. ASR
(Session of March 4, 1938)
I. We have demonstrated in the first Note that the maximum ofA_(varphi)(f)\mathrm{A}_{\varphi}(f)In(E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}, whenA(f)\mathrm{A}(f)is given, can only be reached by an elementary function of degreennat most i vertex. More precisely, there always exists an elementary functionh(x)h(x)of(E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}, having at most I vertex, such thatA(h)=A(f),A_(varphi)(h) > A_(varphi)(f)\mathrm{A}(h)=\mathrm{A}(f), \mathrm{A}_{\varphi}(h)>\mathrm{A}_{\varphi}(f), equality not being possible ifffis not an elementary function at o or I vertex.
By tiberiu popoviciu, Mc. ASR
(Session of March 4, 1938)
I. We have demonstrated in the first Note that the maximum ofA_(varphi)(f)\mathrm{A}_{\varphi}(f)In(E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}, whenA(f)\mathrm{A}(f)is given, can only be reached by an elementary function of degreennat most i vertex. More precisely, there always exists an elementary functionh(x)h(x)of(E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}, having at most I vertex, such thatA(h)=A(f),A_(varphi)(h) > A_(varphi)(f)\mathrm{A}(h)=\mathrm{A}(f), \mathrm{A}_{\varphi}(h)>\mathrm{A}_{\varphi}(f), equality not being possible ifffis not an elementary function at o or I vertex.
In particular, forn=1n=1, the maximum problem is completely solved since the value of the integralA(f)\mathrm{A}(f)completely determines the functionhh.
We now propose to examine the casen > 1n>1. It is then necessary to choose among the elementary functions at o or I vertex, the one (or ones) which gives the maximum ofA_(varphi)(f)A_{\varphi}(f).
We will use an auxiliary property, which we have already used in the previous note, and which we now state in the following form:
Eithervarphi\varphia function of the form indicated in the first note. Let thenf,f^(**)f, f^{*}two continuous, differentiable and increasing functions in the closed interval (o,I\mathrm{o}, \mathrm{I}). In addition, the functionsf,f^(**)f, f^{*}check the properties:
I^(0).f(o)=f^(**)(o)=a >= o,f(I)=f^(**)(I)=b <= I,a >= b\mathrm{I}^{0} . f(\mathrm{o})=f^{*}(\mathrm{o})=a \geq \mathrm{o}, f(\mathrm{I})=f^{*}(\mathrm{I})=b \leqq \mathrm{I}, a \geq b.
2^(0)2^{0}. We haveA(f^(**))=A(f)\mathrm{A}\left(f^{*}\right)=\mathrm{A}(f).
3. There is a numberx_(0),0 < x_(0) < 1x_{0}, 0<x_{0}<1, such that we havef^(**)(x)⋚f(x)f^{*}(x) \lesseqgtr f(x), depending on whetherx⋚x_(0),0 < x < 1x \lesseqgtr x_{0}, 0<x<1.
I^(0).f(o)=f^(**)(o)=a >= o,f(I)=f^(**)(I)=b <= I,a >= b\mathrm{I}^{0} . f(\mathrm{o})=f^{*}(\mathrm{o})=a \geq \mathrm{o}, f(\mathrm{I})=f^{*}(\mathrm{I})=b \leqq \mathrm{I}, a \geq b.
2^(0)2^{0}. We haveA(f^(**))=A(f)\mathrm{A}\left(f^{*}\right)=\mathrm{A}(f).
3. There is a numberx_(0),0 < x_(0) < 1x_{0}, 0<x_{0}<1, such that we havef^(**)(x)⋚f(x)f^{*}(x) \lesseqgtr f(x), depending on whetherx⋚x_(0),0 < x < 1x \lesseqgtr x_{0}, 0<x<1.
More generally we can assume that f* is first constantly equal to a and then is increasing.
We then have the inequalityA_(varphi)(f^(**)) > A_(varphi)(f)^(1)\mathrm{A}_{\varphi}\left(f^{*}\right)>\mathrm{A}_{\varphi}(f){ }^{1}).
The demonstration is simple. According to the representation ofvarphi\varphiby a Stieltjes integral, it suffices to demonstrate the property for the functionsvarphi_(Lambda).^(2)\varphi_{\Lambda} .{ }^{2}). This amounts to demonstrating that
The demonstration is simple. According to the representation ofvarphi\varphiby a Stieltjes integral, it suffices to demonstrate the property for the functionsvarphi_(Lambda).^(2)\varphi_{\Lambda} .{ }^{2}). This amounts to demonstrating that
G(lambda)=int_(i^(**))^(1)f^(**)dx-int_(t)^(1)fdx+lambda(t^(**)-t) > 0,quad a < lambda < b\mathrm{G}(\lambda)=\int_{i^{*}}^{1} f^{*} d x-\int_{t}^{1} f d x+\lambda\left(t^{*}-t\right)>0, \quad a<\lambda<b
Orlambda=f(l)=f^(**)(t^(**))\lambda=f(l)=f^{*}\left(t^{*}\right). Gold,t,t^(**)t, t^{*}and the first member of this inequality are functions derivable fromlambda\lambda. We have
(dG)/(d lambda)=-f^(**)(t^(**))(dt^(**))/(d lambda)+f(t)(dt)/(d lambda)+lambda((dt^(**))/(d lambda)-(dt)/(d lambda))+(t^(**)-t)=t^(**)-t\frac{d \mathrm{G}}{d \lambda}=-f^{*}\left(t^{*}\right) \frac{d t^{*}}{d \lambda}+f(t) \frac{d t}{d \lambda}+\lambda\left(\frac{d t^{*}}{d \lambda}-\frac{d t}{d \lambda}\right)+\left(t^{*}-t\right)=t^{*}-t
Ownership follows immediately from the fact thatt^(**)⋛tt^{*} \gtreqless tfollowing thatlambda⋚f(x_(0))\lambda \lesseqgtr f\left(x_{0}\right).
2. Now consider the function of(E_(n)^(b))_(n)\left(\mathrm{E}_{n}^{b}\right)_{n},
2. Now consider the function of(E_(n)^(b))_(n)\left(\mathrm{E}_{n}^{b}\right)_{n},
h(x)=a+a_(1)x+a_(2)x^(2)+dotsa_(k)x^(k)+(b-a-sum_(i=1)^(k)a_(i))[(x-mu+|x-mu|)/(2(I-mu))]^(n)h(x)=a+a_{1} x+a_{2} x^{2}+\ldots a_{k} x^{k}+\left(b-a-\sum_{i=1}^{k} a_{i}\right)\left[\frac{x-\mu+|x-\mu|}{2(\mathrm{I}-\mu)}\right]^{n}
Or,0 < mu < I,a_(i) >= 0,quad i=1,2,dots,k-I,quada_(k) > 0,k <= n-I,b-a-sum_(i=1)^(k)a_(i) > 00<\mu<\mathrm{I}, a_{i} \geqq 0, \quad i=1,2, \ldots, k-\mathrm{I}, \quad a_{k}>0, k \leqq n-\mathrm{I}, b-a-\sum_{i=1}^{k} a_{i}>0.
- This is only a special case of an inequality which, for the finite case, has been given by Messrs. GH 'Hardy, JE Littlewood, G. Pólya, yoir: ,,Some simple inequalities satisfied by convex /unctions" Messenger of. math. 58 (1929) p. 145-152. See also J. Karamata ,On an inequality relating to convex functions', Publ. Math. Univ. Belgrade t. i (1932) p. 145-148. We will return to this inequality in another work.
- I ask the reader to refer to the first note for the assumptions made on the functions and for the notations.
We have
A(h)=a+(a_(1))/(2)+(a_(2))/(3)+dots+(a_(k))/(k+1)+((b-a-sum_(i=1)^(k)a_(l))(1-mu))/(n+1).\mathrm{A}(h)=a+\frac{a_{1}}{2}+\frac{a_{2}}{3}+\ldots+\frac{a_{k}}{k+1}+\frac{\left(b-a-\sum_{i=1}^{k} a_{l}\right)(1-\mu)}{n+1} .
Now let the function
{:[h^(**)(x)=a+a_(1)x+a_(2)x^(2)+dots+a_(k-1)x^(k-1)+(a_(k)-epsi)x^(k)+],[+(b-a-sum_(i=1)^(k)a_(i)+epsi)[(x-mu^(**)+|x-mu^(**)|)/(2(1-mu^(**)))]^(n)]:}\begin{aligned}
h^{*}(x) & =a+a_{1} x+a_{2} x^{2}+\ldots+a_{k-1} x^{k-1}+\left(a_{k}-\varepsilon\right) x^{k}+ \\
& +\left(b-a-\sum_{i=1}^{k} a_{i}+\varepsilon\right)\left[\frac{x-\mu^{*}+\left|x-\mu^{*}\right|}{2\left(1-\mu^{*}\right)}\right]^{n}
\end{aligned}
Ormu^(**)=mu-epsi(n-k+(k+1)mu)/((k+1)(b-a-sum_(i=1)^(k)a_(i)+epsi))\mu^{*}=\mu-\varepsilon \frac{n-k+(k+1) \mu}{(k+1)\left(b-a-\sum_{i=1}^{k} a_{i}+\varepsilon\right)}Andepsi\varepsilona positive number such thata_(k)-epsi > 0,b-a-Sigma^(k)a_(i)+epsi > 0,0 < mu^(**) < 1a_{k}-\varepsilon>0, b-a-\Sigma^{k} a_{i}+\varepsilon>0,0<\mu^{*}<1, which is quite feasible.
The functionh^(**^(**))h^{\stackrel{*}{*}}is still an elementary function with at most r vertices belonging(E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}. We havemu^(**) < mu_(0)\mu^{*}<\mu_{0}AndA(h^(**))=A(h)\mathrm{A}\left(h^{*}\right)=\mathrm{A}(h). We also check that the differenceh^(**)-hh^{*}-his zero forx=0,Ix=0, \mathrm{I}, is negative in the neighborhood of o and can only be zero at a single point outside o and i. I, the property of the previous No. applies and we have
A_(varphi)(h^(**)) > A_(varphi)(h).\mathrm{A}_{\varphi}\left(h^{*}\right)>\mathrm{A}_{\varphi}(h) .
It immediately follows thatA_(varphi)(h^(**))\mathrm{A}_{\varphi}\left(h^{*}\right)grows whenepsi\varepsilongrows. Now,epsi\varepsiloncan grow until one of the equalitiesepsi=a_(k),mu^(**)=0\varepsilon=a_{k}, \mu^{*}=0takes place, therefore
When the functionvarphi\varphiAndA(f)\mathrm{A}(f)are given, the maximum ofA_(varphi)(f)A_{\varphi}(f)In(E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}can only be achieved ifffis either a functionh_(lambda)(x)h_{\lambda}(x), or a polynomial of degree at most equal tonn.
Regardless offfIn(E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}we havea <= A(f) <= (a+b)/(2)a \leqq \mathrm{~A}(f) \leqq \frac{a+b}{2}. For a functionh_(lambda)(x)h_{\lambda}(x)we have
A(h_(lambda))=a+((b-a)(r-lambda))/(n+I)\mathrm{A}\left(h_{\lambda}\right)=a+\frac{(b-a)(\mathrm{r}-\lambda)}{n+\mathrm{I}}
SOa <= A(h_(lambda)) <= (na+b)/(n+I)a \leqq \mathrm{~A}\left(h_{\lambda}\right) \leqq \frac{n a+b}{n+\mathrm{I}}. It follows that a functionh_(lambda)h_{\lambda}can only be maximizing ifa <= A(f) <= (na+b)/(n+I)a \leqq \mathrm{~A}(f) \leqq \frac{n a+b}{n+\mathrm{I}}. It is clear that ifA(f) > (na+b)/(n+I)\mathrm{A}(f)>\frac{n a+b}{n+\mathrm{I}}The maximizing function (or functions) can only be a polynomial.
3. It remains to find the maximizing polynomials of(E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}. Either
3. It remains to find the maximizing polynomials of(E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}. Either
P(x)=a+a_(1)x+a_(2)x^(2)+dotsa_(n)x^(n),P(x)=a+a_{1} x+a_{2} x^{2}+\ldots \mathrm{a}_{n} x^{n},
a_(i) >= 0,i=I,2dots,n,sum_(i=1)^(n)a_(i)=b-aa_{i} \geq 0, i=\mathrm{I}, 2 \ldots, n, \sum_{i=1}^{n} a_{i}=b-a, a polynomial of(E_(a)^(b))^(n)\left(\mathrm{E}_{a}^{b}\right)^{n}.
Suppose that the given value ofA=A(f)\mathrm{A}=\mathrm{A}(f)checks the inequalities
and thatA(P)=A\mathrm{A}(\mathrm{P})=\mathrm{A}. The polynomial of(E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n},
and thatA(P)=A\mathrm{A}(\mathrm{P})=\mathrm{A}. The polynomial of(E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n},
{:(2)P^(**)(x)=a+alphax^(i)^(1)+betax^(j):}\begin{equation*}
\mathrm{P}^{*}(x)=a+\alpha x^{i}{ }^{1}+\beta x^{j} \tag{2}
\end{equation*}
is then completely determined by the conditionA(P^(**))=A(P)\mathrm{A}\left(\mathrm{P}^{*}\right)=\mathrm{A}(\mathrm{P}). We have
alpha=j(j+r)(A-(ja+b)/(j+I)),quad beta=j(j+r)[((j-I)a+b)/(j)-A].\alpha=j(j+\mathrm{r})\left(\mathrm{A}-\frac{j a+b}{j+\mathrm{I}}\right), \quad \beta=j(j+\mathrm{r})\left[\frac{(j-\mathrm{I}) a+b}{j}-\mathrm{A}\right] .
But, the polynomialP^(**)-P\mathrm{P}^{*}-\mathrm{P}can only be cancelled at a single point different fromooAndIIand, moreover, this polynomial is negative in the neighborhood of 0. It follows thatA_(varphi)(P^(**)) > A_(varphi)(P)\mathrm{A}_{\varphi}\left(\mathrm{P}^{*}\right)>\mathrm{A}_{\varphi}(\mathrm{P}).
So finally if we have inequality (1), the (unique) maximizing function is the polynomial (2).
4. We can now state our final result:
4. We can now state our final result:
Ifvarphi\varphiis a definite, non-decreasing, non-concave (or non-convex) function in the term interval(O,I)(\mathrm{O}, \mathrm{I})and ifffis a function of(E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}, we have the inequality
{:[(3)int_(o)^(1)varphi(f)dx <= (ou >= )int_(o)^(1)varphi{a+j(j+1)[(A-(ja+b)/(j+1))+:}],[+(((j-1)a+b)/(j)-A)x]x^(j)}^(1)dx],[" si "(ja+b)/(j+I) <= A <= ((j-1)a+b)/(j)","quad2 <= j <= n]:}\begin{gather*}
\int_{o}^{1} \varphi(f) d x \leqq(o u \geq) \int_{o}^{1} \varphi\left\{a+j(j+1)\left[\left(\mathrm{A}-\frac{j a+b}{j+1}\right)+\right.\right. \tag{3}\\
\left.\left.+\left(\frac{(j-1) a+b}{j}-\mathrm{A}\right) x\right] x^{j}\right\}^{1} d x \\
\text { si } \frac{j a+b}{j+\mathrm{I}} \leqq \mathrm{~A} \leqq \frac{(j-1) a+b}{j}, \quad 2 \leqq j \leqq n
\end{gather*}
and inequality
{:[(4)int_(o)^(1)varphi(f)dx <= (ou >= )(b+na-(n+r)A)/(b-a)varphi(a)+],[+((n+1)(A-a))/(n(b-a)root(n)(b-a))int_(a)^(b)(varphi(x)dx)/(root(n)((x-a)^(n-1)))]:}\begin{align*}
& \int_{o}^{1} \varphi(f) d x \leqq(o u \geqq) \frac{b+n a-(n+\mathrm{r}) \mathrm{A}}{b-a} \varphi(a)+ \tag{4}\\
& +\frac{(n+1)(\mathrm{A}-a)}{n(b-a) \sqrt[n]{b-a}} \int_{a}^{b} \frac{\varphi(x) d x}{\sqrt[n]{(x-a)^{n-1}}}
\end{align*}
ifa <= A <= (na+b)/(n+I)a \leqq \mathrm{~A} \leqq \frac{n a+b}{n+\mathrm{I}}.
If in addition the functionvarphi\varphiis convex (or concave), the equality in (3) is only possible if
f(x)=a+j(j+I)[(A-(ja+b)/(j+I))+(((j-I)a+b)/(j)-A)x]x^(j)f(x)=a+j(j+\mathrm{I})\left[\left(\mathrm{A}-\frac{j a+b}{j+\mathrm{I}}\right)+\left(\frac{(j-\mathrm{I}) a+b}{j}-\mathrm{A}\right) x\right] x^{j}
and the equality in (4) that if
f(x)=a+(b-a)[(x-lambda+∣x-lambda)/(2(I-lambda))]^(n),quad lambda=(b+na-(n+1)A)/(b-a).f(x)=a+(b-a)\left[\frac{x-\lambda+\mid x-\lambda}{2(\mathrm{I}-\lambda)}\right]^{n}, \quad \lambda=\frac{b+n a-(n+1) \mathrm{A}}{b-a} .
If we haveA=(ja+b)/(j+I)\mathrm{A}=\frac{j a+b}{j+\mathrm{I}}, formula (3) or (4) can also be written
int_(o)^(1)varphi(f)dx <= (ou >= )(1)/(jroot(j)(b-a))int_(n)^(b)(varphi(x)dx)/(root(j)((x-a)^(j-1)))\int_{o}^{1} \varphi(f) d x \leqq(o u \geqq) \frac{1}{j \sqrt[j]{b-a}} \int_{n}^{b} \frac{\varphi(x) d x}{\sqrt[j]{(x-a)^{j-1}}}
- In the casen=0,fn=0, fis a continuous and non-decreasing function. The maximum ofA_(varphi)(f)\mathrm{A}_{\varphi}(f)is not affected by any functionff. We have
maxA_(varphi)(f)=((b-A)varphi(a)+(A-a)varphi(b))/(b-a)\max \mathrm{A}_{\varphi}(f)=\frac{(b-\mathrm{A}) \varphi(a)+(\mathrm{A}-a) \varphi(b)}{b-a}
but equality cannot take place ifvarphi\varphiis convex andffcontinuous. The maximum is, in this case, reached by limit functions of(E_(a)^(b))_(0)\left(\mathrm{E}_{a}^{b}\right)_{0}. Such a maximizing function is, for example, the function
f(x)={[a",",0 <= x <= (b-A)/(b-a)],[b",",(b-A)/(b-a) <= x <= I]:}f(x)= \begin{cases}a, & 0 \leqq x \leqq \frac{b-\mathrm{A}}{b-a} \\ b, & \frac{b-\mathrm{A}}{b-a} \leqq x \leqq \mathrm{I}\end{cases}
Ifvarphi\varphiis a concave function we have the same property for the minimum of the integralA_(varphi)(f)\mathrm{A}_{\varphi}(f).