A Surprising Proof of an Inequality

Problem : Prove that for real numbers $x_1,\dots ,x_n$,

$$\sum _{i=1}^n\sum _{j=1}^n{|x_i-x_j|}^r\le \sum _{i=1}^n\sum _{j=1}^n{|x_i+x_j|}^r$$

for any $r\in [0,2]$.

Solution : The cases $r=0,2$ can be checked by hand. So, we assume $r\in (0,2)$ and begin by noting that the integral

$$\mathcal{I}(\alpha )={\int }_0^{\mathrm{\infty }}\frac{1-\cos \left(\alpha x\right)}{x^{r+1}}\text{d}x$$

is positive and finite. Also, it is easy to prove using the substitution $y=|\alpha |x$ that $\mathcal{I}(\alpha )=\sqrt{|\alpha |}\cdot \mathcal{I}(1)$.

Now, using this, we have

$$\begin{array}{rl}|a+b{|}^r-|a-b{|}^r& =\frac{1}{\mathcal{I}(1)}{\int }_0^{\mathrm{\infty }}\frac{\cos ((a-b)x)-\cos ((a+b)x)}{x^{r+1}}\text{d}x\\ & =\frac{1}{\mathcal{I}(1)}{\int }_0^{\mathrm{\infty }}\frac{2\sin (ax)\sin (bx)}{x^{r+1}}\text{d}x\end{array}$$

and hence

$$\sum _{i=1}^n\sum _{j=1}^n{|x_i-x_j|}^r-\sum _{i=1}^n\sum _{j=1}^n{|x_i+x_j|}^r=\frac{2}{\mathcal{I}(1)}{\int }_0^{\mathrm{\infty }}\frac{{(\sum _{i=1}^n\sin (x_{i}x))}^2}{x^{r+1}}\text{d}x\ge 0$$

hence completing the proof.

Prokhorov’s Theorem

The aim is to give a proof of Prokhorov’s Theorem.

Theorem : Let $\{{\mu }_n:n\ge 1\}$ be a sequence of probability measures on $({\mathbb{R}}^d,\mathcal{B}\left({\mathbb{R}}^d\right))$. If $\{{\mu }_n:n\ge 1\}$ is tight, then every sequence has a further subsequence that converges weakly to some probability measure.

We will use the following (well known) fact in our proof.

Fact : Let $\{{\mu }_n:n\ge 1\}$ be a sequence of probability measures on $({\mathbb{R}}^d,\mathcal{B}\left({\mathbb{R}}^d\right))$. Then, weak convergence ${\mu }_n\stackrel{d}{\to }{\mu }_0$ is equivalent to $\int g\text{d}{\mu }_n\to \int g\text{d}{\mu }_0$ for all compactly supported continuous functions.

With this at hand, we give the proof of the theorem.

Proof of Theorem : Let $K^N=[-N,N{]}^d$. Then, ${\mu }_n$ induces a subprobability ${\nu }_n^N:={\mu }_n{|}_{K^N}$ on $K^N$. By weak${}^{\ast }$ compactness, we can find a subsequence ${\nu }_{n_k}^N$ such that ${\nu }_{n_k}^N\stackrel{w^{\ast }}{\to }{\nu }^N$ for a subprobability measure ${\nu }^N$ on $K^N$. Repeating the argument starting with this subsequence on $K^{N+1}$ gives a new limit ${\nu }^{N+1}$. Repeating this procedure countably many times (and noticing that ${\nu }^N={\nu }^{N+1}$ on $K^N$), we get

$$\mu (A):=\underset{N\to \mathrm{\infty }}{lim}{\nu }^N(A\cap K^N)$$

for $A\in \mathcal{B}\left({\mathbb{R}}^d\right)$.

Now, we will show that $\mu$ is a probability measure. From here, it follows using the mentioned fact that our subsequence converges weakly to $\mu$.

Clearly $\mu \left(\mathrm{\Omega }\right)\le 1$ as each ${\nu }^N$ is a subprobability. Given $\epsilon >0$, tightness yields $N_{\epsilon }$ such that ${\mu }_n\left(K^{N_{\epsilon }}\right)\ge 1-\epsilon$ for all $n\ge 1$. It follows that $\mu \left(\mathrm{\Omega }\right)\ge \mu \left(K^{N_{\epsilon }}\right)\ge 1-\epsilon$. This shows that $\mu \left(\mathrm{\Omega }\right)=1$. It remains to prove that $\mu$ is $\sigma$-additive. Clearly $\mu$ is monotone. Let first

$$A=\bigcup _{i=1}^I{A}_i$$

be a disjoint union of finitely many measurable sets. Then

$$|\mu \left(\bigcup _{i=1}^I{A}_i\right)-\sum _{i=1}^I\mu \left(A_i\right)|\le |\mu (\bigcup _{i=1}^I{A}_i\cap K^{N_{\epsilon }})-\sum _{i=1}^I\mu (A_i\cap K^{N_{\epsilon }})|+(I+1)\epsilon =(I+1)\epsilon$$

since the restriction of $\mu$ to $K^{N_{\epsilon }}$ (which is ${\nu }^{N_{\epsilon }}$) is $\sigma$-additive. Thus $\mu$ is finitely additive. It follows that $\mu$ is $\sigma$-superadditive as

$$\mu (A)\ge \mu \left(\bigcup _{i=1}^I{A}_i\right)=\sum _{i=1}^I\mu \left(A_i\right)\to \sum _{i=1}^{\mathrm{\infty }}\mu \left(A_i\right)$$

by letting $I\to \mathrm{\infty }$.

Conversely, using again that the restriction of $\mu$ to $K^{N_{\epsilon }}$ is $\sigma$-additive, we have

$$\begin{array}{rl}\mu (A)& =\mu (A\setminus K^{N_{\epsilon }})+\mu (\bigcup _{i=1}^{\mathrm{\infty }}{A}_i\cap K^{N_{\epsilon }})\\ & =\mu (A\setminus K^{N_{\epsilon }})+\sum _{i=1}^{\mathrm{\infty }}\mu (A_i\cap K^{N_{\epsilon }})\\ & \le \epsilon +\sum _{i=1}^{\mathrm{\infty }}\mu \left(A_i\right)\end{array}$$

thus completing the proof.