Talk:Hermite–Hadamard inequality

Copying here from other talk pages:

Likely violation of SPIP: Everything Retkes

Everything involving Retkes' name here looks like pollution to me (source: PhD in pure maths). I smell self-promotion: Someone with the time and expertise should clean up, and possibly mark several of these pages for deletion according to WP:SPIP — Preceding unsigned comment added by 172.254.63.183 (talk) 06:08, 4 May 2013‎

Dear X. I would like to turn your attention to this article:

Some applications of Retkes' identity P Kórus - Archiv der Mathematik, 2015 - Springer

Abstract. We present some formulas for certain numeric sums related to the Riemann zeta function. The main tool used in our investigation is Retkes’ identity. We get a formula for ζ(3) with the Euler beta function in it. Mathematics Subject Classification. Primary 11Y60; Secondary 11M06, 33B15. Keywords. Retkes’ identity, Zeta(3), Binomial coefficients, Alternating sums. White Tiger (talk) 11:49, 29 July 2018 (UTC)

Section "A corollary on Vandermonde-type integrals"

I've removed much of the low-density crud from this section. According to User:Tudor987, whe whole section was created by User Vezér who self-promoted in many ways his non-notable results in Wikipedia. This section is likely from his work, which I can't access—it's behind a paywall, and I won't be on my University's campus to access the physical copy for another month. Without a simple (i.e. 1-paragraph) proof, it's probably not on-topic for the article. Bernanke's Crossbow (talk) 02:26, 26 June 2019 (UTC)

Article is incomplete

The inequality is an "if and only if":

First part

Let $I$ be an interval of $\mathbb {R}$. If $f:I\to \mathbb {R}$ is a continuous function that satisfies

$f\left({\frac {x+y}{2}}\right)\leq {\frac {1}{y-x}}\int _{x}^{y}f(t)\,dt,\quad \forall x,y\in I,\quad x\leq y,$

then $f$ is convex. (Note that the continuity of $f$ is essential, otherwise we could lower the value $f$ at a point without breaking the inequality, but $f$ would no longer be convex).

Proof. Fix $a,b\in I$, $a<b$ and set $g(x)=f(x)-{\dfrac {b-x}{b-a}}f(a)-{\dfrac {x-a}{b-a}}f(b)$, then $g(x)$ satisfies the same inequality as for $f$ because they differ only by a linear function. We have $g(a)=g(b)=0$. If $g(x)>0$ for some $x\in (a,b)$, let $\sup _{x\in [a,b]}g(x)=g(c)$. WLOG suppose that $c\leq {\dfrac {a+b}{2}}$ (the case $c\geq {\dfrac {a+b}{2}}$ is similar), then

$g(c)\leq {\frac {1}{2(c-a)}}\int _{a}^{2c-a}g(t)\,dt,$

which implies $g(x)=g(c)>0$ for all $x\in [a,2c-a]$, contradicting $g(a)=0$.

Second part

Let $I$ be an interval of $\mathbb {R}$. If $f:I\to \mathbb {R}$ is a continuous function that satisfies

${\frac {1}{y-x}}\int _{x}^{y}f(t)\,dt\leq {\frac {f(x)+f(y)}{2}},\quad \forall x,y\in I,\quad x\leq y,$

then $f$ is convex. (Note that the continuity of $f$ is essential, otherwise we could raise the value $f$ at a point without breaking the inequality, but $f$ would no longer be convex).

Proof. Fix $a,b\in I$, $a<b$ and set $g(x)=f(x)-{\dfrac {b-x}{b-a}}f(a)-{\dfrac {x-a}{b-a}}f(b)$, then $g(x)$ satisfies the same inequality as for $f$ because they differ only by a linear function. We have $g(a)=g(b)=0$. If there is $c\in (a,b)$ such that $f(c)>{\dfrac {b-c}{b-a}}f(a)+{\dfrac {c-a}{b-a}}f(b)$, which is to say $g(c)>0$, define

$a':=\sup\{x\in [a,c]:g(x)\leq 0\},\quad b':=\inf\{x\in [c,b]:g(x)\leq 0\},$

then $a'<c<b'$, $g(a')=g(b')=0$ by continuity, and $g(x)>0$ for all $x\in (a',b')$, and now

${\frac {1}{b'-a'}}\int _{a'}^{b'}g(t)\,dt\leq {\frac {g(a')+g(b')}{2}}$

is violated. 129.104.241.17 (talk) 22:38, 6 February 2025 (UTC)