Integrals Inequality

Let's rewrite $A,B,C$ using Riemann sums:

$\large \begin{aligned} A&=\int_0^1f(x)\ dx\\ &=\lim_{N\to\infty}\frac{1}{N}\sum_{i=0}^{N}f\left(\frac{i}{N}\right) \end{aligned}$

$\large \begin{aligned} B&=\frac{1}{\int_0^1\frac{dx}{f(x)}}\\ &=\frac{1}{\displaystyle\lim_{N\to\infty}\frac{1}{N}\displaystyle\sum_{i=0}^{N}\frac{1}{f\left(\frac{i}{N}\right)}}\\ &=\lim_{N\to\infty}\frac{N}{\displaystyle\sum_{i=0}^{N}\frac{1}{f\left(\frac{i}{N}\right)}} \end{aligned}$

$\large \begin{aligned} C&=e^{\int_0^1\ln\left(f(x)\right)}\\ &=e^{\displaystyle\lim_{N\to\infty}\frac{1}{N}\displaystyle\sum_{i=0}^{N}\ln\left(f\left(\frac{i}{N}\right)\right)}\\ &=e^{\displaystyle\lim_{N\to\infty}\ln\left(\sqrt[N]{\displaystyle\prod_{i=0}^{N}f\left(a+\frac{i}{N}\right)}\right)}\\ &=\lim_{N\to\infty}\sqrt[N]{\prod_{i=0}^{N}f\left(\frac{i}{N}\right)} \end{aligned}$

Now, it should be obvious that $A$ is the arithmetic mean of $f\left(\frac{i}{N}\right)$ for integers $i=0$ to $N$ , $B$ is the harmonic mean, and $C$ is the geometric mean. Since $f(x)>0$ for all $x\in [0,1]$ , by the AM-GM-HM inequality, we know that $B\le C\le A$ , with equality occurring only and only when $f(x)$ is a constant function almost everywhere for $x\in [0,1]$ .

Definition. Let $U$ be a convex subset of a vector space over $\mathbb{R}$ . A function $f:U\longrightarrow\mathbb{R}$ is called convex just in case $f(px+qy)\leq pf(x)+qf(y)$ for all convex linear combinations $px+qy$ , $x,y\in U$ and $p,q\in[0,1]$ , such that $p+q=1$ .

Definition. Let $U$ be a convex subset of a vector space over $\mathbb{R}$ . A function $f:U\longrightarrow\mathbb{R}$ is called concave just in case $f(px+qy)\geq pf(x)+qf(y)$ for all convex linear combinations, $px+qy$ .

Lemma. Let $U\subseteq\mathbb{R}$ be convex and $f:U\longrightarrow\mathbb{R}$ be a $C^{2}$ function. Then a sufficient condition for convexivity (respectively concavity) of $f$ is that $f''\geq 0$ (respectively $\leq 0$ .

Proof. Clearly for trivial convex linear combinations, $px+qy$ , that is $x=y$ or $p=0$ or $q=0$ , it holds that $f(px+qy)=pf(x)+qf(y)$ . Thus it suffices to consider convex linear combinations, $px+qy$ , with $x\neq y$ and $p,q\neq 0$ and to show that $f''\geq 0$ be a sufficient condition for $f(px+qy)\leq pf(x)+qf(y)$ . Wlog consider the case $x<y$ . So in particular $x<px+qy<y$ .

• By the mean valued theorem and $C^{1}$ differentiability, there exist $x^*\in(x,px+qy)$ as well as $y^*\in(px+qy,y)$ , such that $f(px+qy)-f(x)=(px+qy-x)f'(x^*)=-q(y-x)f'(x^*)$ and $f(px+qy)-f(y)=(px+qy-y)f'(y^*)=p(y-x)f'(y^*)$ .
• Since $x^*<px+qy<y^*$ and by $C^{2}$ differentiability, there further exists $c\in(x^*,y^*)$ , such that $f'(y^*)-f(x^*)=(y^*-x^*)f''(c)$ .

It follows that

$\begin{array}{c}[mc]{rcl} f(px+qy)-(pf(x)+qf(y)) &= &p(f(px+qy)-f(x))+q(f(px+qy)-f(y))\\ &= &-pq(y-x)f'(x^*)+qp(y-x)f'(x^*)\\ &= &pq(y-x)(f'(y^*)-f'(x^*))\\ &= &pq(y-x)(y^*-x^*)f''(c).\\ \end{array}$

The final expression is $\geq 0$ , since $p,q>0$ , $x<y$ , $x^*<y^*$ and by the condition that $f''\geq 0$ . Hence $f(px+qy)-(pf(x)+qf(y))\geq 0$ . Thus $f$ is convex. A symmetric argument can be performed for convexity. QED

Examples. $f=\log(\cdot)$ defined on $(0,\infty)$ is twice differentiable with $f''=-1/x^{2}<0$ , hence concave. The function $f=\cdot^{-1}$ defined on $(0,\infty)$ is twice differentiable with $f''=2/x^{3}>0$ , hence convex.

Lemma (Jensen's Inequality). Let $U\subseteq\mathbb{R}$ be convex and $X$ be a $U$ -valued random variable. If $f$ is a convex (respectively concave) function on $U$ and both $X,f(X)\in L^{1}$ , then

$f(\mathbb{E}[X])\leq\mathbb{E}[f(X)]$

(respectively $\geq$ in the case of a concave function).

Proof (Sketch). Note that integrals are approximated by convex linear sums. Appealing to this and the property of convexivity (resp. concavity) yields the result. QED.

Corollary. Let $X$ be a r. v. which is a.e. positive. Suppose $X,1/X,\log(X)$ are in $L^{1}$ . Then $1/\mathbb{E}[1/X]\leq\exp(\mathbb{E}[\log(X)])\leq\mathbb{E}[X]$ . Proof. Since $\log$ is concave it holds by the above lemma that $\log(\mathbb{E}[X])\geq\mathbb{E}[\log(X)]$ . By strict monotonicity of $f=\log(\cdot)$ this implies $\mathbb{E}[X]\geq \exp(\mathbb{E}[\log(X)])$ . Since $1/X$ and $\log(1/X)=-\log(X)$ are in $L^{1}$ , we get for free that

$\mathbb{E}[1/X]\geq \exp(\mathbb{E}[\log(1/X)])=\exp(-\mathbb{E}[\log(X)])=1/\exp(\mathbb{E}[\log(X)]).$

From this it follows that $1/\mathbb{E}[1/X]\leq \exp(\mathbb{E}[\log(X)])$ . Hence the claim holds. QED.

Remark. In general one has that a sufficient condition for $f^{-1}\mathbb{E}[f(X)]\leq\mathbb{E}[X]$ is that $f$ be strictly monoton and concave, or that $f$ be strictly antitone and convex. (See the above lemma and the above concrete example in the corollary.)