Proof of Weighted Cauchy-Schwarz Inequality on a Measure Space and the Symmetrised Score Functional: Bounds and Averaging

Preliminaries. Integrals of nonnegative measurable functions are handled with claim 1 of Linearity and Monotonicity of the Lebesgue Integral (additivity, positive homogeneity, monotonicity), integrable functions with claim 2 there and with Integrable Function and the Lebesgue Integral. Sums, scalar multiples, products, absolute values and pointwise limits of measurable real-valued functions are measurable by claims 2, 3, 4 and 5 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. Two elementary facts on a measure space $(\mathsf{Z},\mathcal{Z},\nu)$, both from Lebesgue Integral of a Nonnegative Measurable Function and Simple Function and Its Integral: (F1) if $\beta:\mathsf{Z}\to[0,\infty)$ is measurable with $\int\beta\,d\nu=0$, then $\nu(\{\beta>0\})=0$, because for every natural number $k$ the simple function $\tfrac1k\mathbf{1}_{\{\beta>1/k\}}$ is at most $\beta$, so $\tfrac1k\,\nu(\{\beta>1/k\})\le\int\beta\,d\nu=0$, and $\{\beta>0\}$ is the union of the sets $\{\beta>1/k\}$, whose measure is then $0$ by countable subadditivity (claim 4 of Basic Properties of a Measure); (F2) if a measurable $u:\mathsf{Z}\to[0,\infty]$ vanishes off a set $N\in\mathcal{Z}$ with $\nu(N)=0$, then $\int u\,d\nu=0$, because every simple function $s$ with $0\le s\le u$ vanishes off $N$, so $s\le(\max s)\mathbf{1}_N$ and $\int s\,d\nu\le(\max s)\,\nu(N)=0$; the integral of $u$ is the least upper bound of these.

Step 1: claim 1. Measurability of $q$. For every natural number $k$ the map $h_k:[0,\infty)\to\mathbb{R}$, $h_k(v)=v/(v^{2}+1/k)$, is sequentially continuous on $[0,\infty)$ (a quotient of sequentially continuous maps with positive denominator, by claims 1, 2 and 4 of Arithmetic of Limits of Real Sequences), so $h_k\circ\beta$ is measurable by Sequentially Continuous Functions of Measurable Euclidean Maps are Measurable. For $v>0$, $h_k(v)\to1/v$ as $k\to\infty$ (again by the limit laws, since $1/k\to0$), while $h_k(0)=0$ for every $k$. Hence $h\circ\beta$, where $h(v)=1/v$ for $v>0$ and $h(0)=0$, is the pointwise limit of the measurable maps $h_k\circ\beta$ and is measurable. Now $q=\alpha^{2}\cdot(h\circ\beta)$ pointwise: on $\{\beta>0\}$ both sides equal $\alpha^{2}/\beta$, and on $\{\beta=0\}$ both sides vanish, as $\alpha=0$ there by hypothesis. So $q$ is measurable, being a product of measurable functions.

Case $I=0$, where $I=\int\beta\,d\nu$. By (F1), $N=\{\beta>0\}$ has $\nu(N)=0$, and $|\alpha|$ vanishes off $N$ by hypothesis. By (F2), $\int|\alpha|\,d\nu=0$, so $\alpha$ is integrable (Integrable Function and the Lebesgue Integral) with $\int\alpha^{\pm}\,d\nu\le\int|\alpha|\,d\nu=0$ by monotonicity, hence $\int\alpha\,d\nu=0$, and the asserted inequality reads $0\le0$.

Case $I>0$. By claim 3 of Image Measures, Measures with Densities, and Change of Variables, $\nu'(A)=\int\mathbf{1}_A\,(\beta/I)\,d\nu$ ($A\in\mathcal{Z}$) is a measure on $(\mathsf{Z},\mathcal{Z})$ with $\nu'(\mathsf{Z})=\int\beta\,d\nu/I=1$, so $(\mathsf{Z},\mathcal{Z},\nu')$ is a probability space, and for every measurable $F:\mathsf{Z}\to[0,\infty]$ one has $\int F\,d\nu'=\int F\,(\beta/I)\,d\nu$, while a measurable $F:\mathsf{Z}\to\mathbb{R}$ is $\nu'$-integrable if and only if $F\beta/I$ is $\nu$-integrable, with equal integrals. Put $X=\alpha\cdot(h\circ\beta)$, a measurable real-valued function, equal to $\alpha/\beta$ on $\{\beta>0\}$ and to $0$ on $\{\beta=0\}$. Then $X^{2}\beta=q$ pointwise (on $\{\beta>0\}$ both are $\alpha^{2}/\beta$; on $\{\beta=0\}$ both vanish), so

and $X$ is a square-integrable random variable on $(\mathsf{Z},\mathcal{Z},\nu')$, hence integrable there. Also $X\beta=\alpha$ pointwise (on $\{\beta=0\}$ both vanish), so $|X|\beta/I=|\alpha|/I$; the $\nu'$-integrability of $|X|$ therefore gives $\int|\alpha|\,d\nu=I\int|X|\,d\nu'<\infty$, so $\alpha$ is $\nu$-integrable, and by the change-of-density identity for integrable functions, $\int\alpha\,d\nu=I\int X\,d\nu'=I\,\mathbb{E}'[X]$, where $\mathbb{E}'$ is the expectation under $\nu'$. The constant $1$ is square-integrable with mean-square norm $1$, so claim 1 of Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm gives $|\mathbb{E}'[X]|=|\mathbb{E}'[X\cdot1]|\le\lVert X\rVert_{2}$, that is, $\mathbb{E}'[X]^{2}\le\mathbb{E}'[X^{2}]$ (squaring is monotone on $[0,\infty)$ by Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field). Therefore

which is the claim.

Step 2: claim 2. First, since every term of $\mathsf{D}(a,b)$ is nonnegative with a positive coefficient, $\mathsf{D}(a,b)=0$ holds exactly when $a=b_1=\dots=b_n=0$; this is used throughout. (b). Let $(a,b)\in[0,\infty)^{1+n}$ and put $\Sigma=a+\sum_jb_j\ge0$. Since $|w_j|\le\lVert w\rVert_1$ for every $j$ and $a,b_j\ge0$, the absolute-value bound for finite sums (Comparison and Absolute Value Bounds for Finite Sums of Real Numbers) gives

Moreover $\mathsf{D}(a,b)\ge\frac{1}{2n}\Sigma$, because $\frac12\ge\frac1{2n}$ and $a\ge0$. If $\mathsf{D}(a,b)>0$ then $\Sigma>0$ (otherwise $a=b_1=\dots=b_n=0$ and $\mathsf{D}(a,b)=0$), and

if $\mathsf{D}(a,b)=0$ then $\Phi_w(a,b)=0\le2n\lVert w\rVert_1^{2}\Sigma$. Nonnegativity of $\Phi_w$ is clear, as a quotient of a square by a positive number or $0$.

(a). Each coordinate map $(a,b)\mapsto a$, $(a,b)\mapsto b_j$ is sequentially continuous on $\mathbb{R}^{1+n}$, since the distance between two points bounds the difference of any fixed coordinate (Euclidean Distance on $\mathbb{R}^n$); hence $\mathsf{N}_w$ and $\mathsf{D}$, finite linear combinations of coordinates, are sequentially continuous by claims 1 and 3 of Arithmetic of Limits of Real Sequences. Let $(a^k,b^k)_{k\in\mathbb{N}}$ be a sequence in $[0,\infty)^{1+n}$ converging to $(a,b)\in[0,\infty)^{1+n}$. If $\mathsf{D}(a,b)>0$, then $\mathsf{D}(a^k,b^k)\to\mathsf{D}(a,b)>0$, so $\mathsf{D}(a^k,b^k)>0$ for all large $k$, and $\Phi_w(a^k,b^k)=\mathsf{N}_w(a^k,b^k)^{2}/\mathsf{D}(a^k,b^k)\to\mathsf{N}_w(a,b)^{2}/\mathsf{D}(a,b)=\Phi_w(a,b)$ by claims 2 and 4 of Arithmetic of Limits of Real Sequences. If $\mathsf{D}(a,b)=0$, then $(a,b)=0$ and $\Phi_w(a,b)=0$, while by (b), $0\le\Phi_w(a^k,b^k)\le2n\lVert w\rVert_1^{2}\,(a^k+\sum_jb^k_j)\to0$, so $\Phi_w(a^k,b^k)\to0=\Phi_w(a,b)$ by the comparison of limits (claim 1 of Order Properties of Limits of Real Sequences). Thus $\Phi_w$ is sequentially continuous on $[0,\infty)^{1+n}$.

(c). For $c\ge0$, $\mathsf{N}_w(ca,cb)=c\,\mathsf{N}_w(a,b)$ and $\mathsf{D}(ca,cb)=c\,\mathsf{D}(a,b)$ by homogeneity of finite sums. If $c>0$ and $\mathsf{D}(a,b)>0$, then $\mathsf{D}(ca,cb)>0$ and $\Phi_w(ca,cb)=c^{2}\mathsf{N}_w(a,b)^{2}/(c\,\mathsf{D}(a,b))=c\,\Phi_w(a,b)$; if $c>0$ and $\mathsf{D}(a,b)=0$ then $\mathsf{D}(ca,cb)=0$ and both sides vanish; if $c=0$ then $(ca,cb)=0$ and both sides vanish.

(d). Under the hypotheses, $\mathsf{D}(a,b)\ge\frac12a+\frac{1}{2n}\cdot n(1-\delta)a=(1-\delta/2)\,a>0$, so $\Phi_w(a,b)=\mathsf{N}_w(a,b)^{2}/\mathsf{D}(a,b)\le\mathsf{N}_w(a,b)^{2}/((1-\delta/2)a)$. Finally $(1+\delta)(1-\delta/2)=1+\tfrac{\delta}{2}(1-\delta)\ge1$ for $0\le\delta\le1$, so $1/(1-\delta/2)\le1+\delta$, which gives (d).

Step 3: claim 3. The map $z\mapsto(\ell_0(z),\ell_1(z),\dots,\ell_n(z))$ has measurable components and takes values in the nonempty set $E=[0,\infty)^{1+n}$, on which $\Phi_w$ is sequentially continuous by claim 2(a); so $z\mapsto\Phi_w(\ell_0(z),\dots,\ell_n(z))$ is measurable by Sequentially Continuous Functions of Measurable Euclidean Maps are Measurable. Define the measurable functions

so that $\bar\beta(z)=\mathsf{D}(\ell_0(z),\dots,\ell_n(z))$ and $\alpha(z)=\mathsf{N}_w(\ell_0(z),\dots,\ell_n(z))$ for every $z$. By linearity of the integral for nonnegative functions, $\int\bar\beta\,d\nu=\mathsf{D}(A,B_1,\dots,B_n)<\infty$; each $\ell_i$ is integrable (nonnegative with finite integral), so $\alpha$ is integrable with $\int\alpha\,d\nu=\sum_jw_j(A-B_j)=\mathsf{N}_w(A,B_1,\dots,B_n)$ by linearity for integrable functions. Where $\bar\beta(z)=0$, every $\ell_i(z)$ vanishes, hence $\alpha(z)=0$. The function $q$ of claim 1 built from $(\alpha,\bar\beta)$ equals $\Phi_w(\ell_0(z),\dots,\ell_n(z))$ pointwise, by the definition of $\Phi_w$ (where $\bar\beta(z)>0$ it is $\alpha(z)^{2}/\bar\beta(z)$, and where $\bar\beta(z)=0$ both are $0$). If $\int q\,d\nu=\infty$ there is nothing to prove. Otherwise claim 1 applies and gives $\mathsf{N}_w(A,B)^{2}\le\mathsf{D}(A,B)\int q\,d\nu$, writing $B=(B_1,\dots,B_n)$. If $\mathsf{D}(A,B)>0$, dividing yields $\Phi_w(A,B)\le\int q\,d\nu$; if $\mathsf{D}(A,B)=0$, then $\Phi_w(A,B)=0\le\int q\,d\nu$. This proves claim 3. $\blacksquare$