A geometric mean form of the Hermite-Hadamard inequality, for negative powers

Question

The following inequality appeared in the analysis of a random approximation algorithm: $$ \int_u^{u+1} x^p\ \mathrm{dx} \leq \sqrt{u^p(u+1)^p}\text{, for } -1\leq p\leq 0, u\geq 1. $$

This resembles the well-known Hermite-Hadamard inequality for convex functions $$ \int_a^b f(x)\ \mathrm{dx} \leq (b-a)\frac{f(a)+f(b)}{2} $$ but with the right-hand side being the geometric mean, rather than the arithmetic mean $(u^p+(u+1)^p)/2$.

Computationally, this inequality appears to hold, but I have been unable to find a proof. Is this a known result? If so, is there a broader class of convex functions for which it holds?

Answer 1

The inequality in question is a particular case (with $v=u+1$) of the inequality $$\int_u^v x^p\, dx \le(v-u)u^{p/2}v^{p/2}\tag{1}$$ for $v\ge u>0$, where without loss of generality (wlog) $p\in(-1,0]$. By the homogeneity in $(u,v)$, wlog $u=1$, and then (1) can be rewritten as $$g(v):=g_p(v):=v^{p+1}-1-(p+1)(v-1) v^{p/2}\le0\tag{2}$$ for $v\ge1$.

Note that
$$g'(v)=\tfrac12\, (p+1) v^{p/2-1}h_v(p),\tag{3}$$ where $$h_v(p):=2 v (v^{p/2}-1)+p-p v,$$ so that $h_v(p)$ is convex in $p$. Also, $h_v(0)=0$ and $h_v(-1)=-(\sqrt v-1)^2\le0$. So, $h_v\le0$ and hence $g'\le0$, which implies that $g$ is decreasing, from $g(1)=0$. Thus, (2) follows.

---

One may also note that $h_v(-2)=0$, so that $h_v(p)\le0$ for all $p\in[-2,0]$, whence, by (3), $g'=g'_p\ge0$ for $p\in[-2,-1)$, which implies that $g$ is increasing, from $g(1)=0$. So, for $p\in[-2,-1)$, inequality (2) switches the direction, and (1) continues to hold -- because (2) was obtained by multiplying both sides of (1) by $p+1$. (The extension to $p\in[-2,-1)$ was previously suggested by Alapan Das.)

Answer 2

This inequality fails to hold e.g. when $p=-3$ and $u=1$. (Then the left-hand side of the inequality is $3/8$, whereas the right-hand side is $2^{-3/2}$, so that the ratio of the left-hand side to the right-hand side is $\sqrt{9/8}>1$.)

---

With the additional condition that $p\ge-1$, the inequality holds, in fact for all real $u>0$. Indeed, without loss of generality $-1<p\le0$. Let $R(u)$ denote the ratio of the left-hand side to the right-hand side. Then $$R_1(u):=R'(u)S(u)=\frac{u^{-p} (u+1)^p \left((2-p) u-p+2 u^2\right)}{(p+2) u+2 u^2}-1,$$ where $$S(u):=\frac{2 (p+1) u^{1-p/2} (u+1)^{1+p/2}}{(p+2) u+2 u^2}>0,$$ so that $R'(u)$ has the same sign as $R_1(u)$. Moreover, $$R'_1(u)=\frac{p (p+1) (p+2) u^{-p-2} (u+1)^p}{(p+2 u+2)^2}\le0.$$ Also, $R_1(\infty-)=0$. So, $R_1\ge0$ and hence $R(u)$ is increasing in $u>0$, to $R(\infty-)=1$.

So, $R(u)\le1$ for $u>0$, as desired.

Answer 3

Substituting $x\rightarrow \frac{x}{u}=z$ and taking $1/u=y\leq 1$ we define a function $f(y)=\frac{y}{(1+y)^{r/2}}-\int_{1}^{1+y} z^{-r} dz$ where $r=-p, t=t/2$.

Now, $f(0)=0$ and $f'(y)=\frac{(1+y)^t-ty(1+y)^{t-1}-1}{(1+y)^{t}}$

Again, as $(1+y)>1$ we define $f_1(y)=(1+y)^{t+1}-ty(1+y)^t-(1+y), f_1(0)=0$

Hence, $f'_1(y)=(1+y)^t-t^2y(1+y)^{t-1}-1$. So, we carry on doing this and get $f_n(y)=T_n(1+y)^{t+1}-t^{n}y(1+y)^{t}-(1+y) ;f_n(0)>0$

Here, $T_{n+1}=(1+t)T_{n}-t^{n}$.

$T_3=(1+t-t^2)>1$ as $t<0.5 \Rightarrow T_4=(1+t)T_3-t^4=T_3+T_3t-t^4>1$ and hence, $T_n >1$ $\forall n \in \mathbb N$

For, $n>>0, t^n \rightarrow 0$ as $0\leq t \leq 1/2$ $f_n(y)=(1+y)(T_n(1+y)^t-1)>0 \rightarrow f_{n-1}(y)>0 \rightarrow ... \rightarrow f(y)>0$

[For, large $n>>0$ as $f_n(0)=\epsilon_n>0$ we expect to find some $N_0$ such that $t^ny(1+y)^t<\epsilon_n$ for all $n>N_0$ and $0\leq y \leq 1$]

Hence, proved. So, it must hold for $-2\leq p \leq 0$

Answer 4

Here is a "more conceptual" and more general proof:

For real $u>0$ and $p<0$, let
$$m(p):=\Big(\int_u^{u+1}x^p\,dx\Big)^{1/p},$$ so that $$\ln m(p)=\frac{l(p)}p,$$ where $l(p):=\ln\int_u^{u+1}x^p\,dx$.

For each real $x>0$, $x^p$ is log convex in $p$ and hence $l$ is convex -- see e.g. this. Also, $l(0-)=0$. So, the "average slope" $\dfrac{l(p)}p$ of $l$ over the interval $(p,0)$ is increasing in $p<0$; here one can use e.g. the special-case l'Hospital-type rule for monotonicity, Proposition 4.1. So, $m$ is increasing on $(-\infty,0)$. So, $$m(p)\le m(-2)\quad\text{for}\quad p\in(-\infty,-2]$$ and $$m(p)\ge m(-2)\quad\text{for}\quad p\in[-2,0).$$

The latter inequality is just another way to write the inequality in question, which thus holds for all $p\in[-2,0]$ and all real $u>0$.

For $p\in(-\infty,-2]$, the direction of the inequality in question changes to the opposite.