Proof of The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability

Each result cited is universally quantified over the data in its own statement.

Throughout, $d_{a}(x,y)=|x-y|_{a}$ for $x,y\in X^{a}$. Since $0<a_{1}\le\bar{a}$, the weights $a_{k}$ being positive by Weight Sequences and the Weighted Fisher Information Relative to a Diagonal Gaussian Measure on a Hilbert Space §weights, the number $\bar{a}$ is positive, and so are $\bar{a}^{-1}$ and the positive square root $\bar{a}^{1/2}$; we write $\bar{a}^{-1/2}$ for the multiplicative inverse of $\bar{a}^{1/2}$, which is positive as well. For $k\in\mathbb{N}$ put $\lambda_{k}=\bar{a}\,a_{k}^{-1}$; multiplying $a_{k}\le\bar{a}$ by the positive number $a_{k}^{-1}$ gives $1\le\lambda_{k}$. Also $a_{k}^{1/2}a_{k}^{-1}=a_{k}^{-1/2}$, since $(a_{k}^{1/2})^{2}=a_{k}$.

Step 0 (comparison with the weighted coefficient subspace). The sequence $(e_{k})_{k\in\mathbb{N}}$ is an orthonormal basis of the real Hilbert space $X$ by Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §space, and $1\le\lambda_{k}$ for every $k$, so The Weighted Coefficient Subspace Determined by an Orthonormal Basis and a Sequence of Weights applies with $H=X$, this basis and the weights $(\lambda_{k})_{k\in\mathbb{N}}$; let $V$, $\langle\cdot,\cdot\rangle_{V}$, $|\cdot|_{V}$ and $d_{V}$ be the subspace, inner product, norm and distance named there. For $x\in X$ one has $\lambda_{k}x_{k}^{2}=\bar{a}\,(a_{k}^{-1}x_{k}^{2})$ and $a_{k}^{-1}x_{k}^{2}=\bar{a}^{-1}(\lambda_{k}x_{k}^{2})$ for every $k$, so by the multiples part of Elementary Properties of Series of Real Numbers §linearity the series $\sum_{k}\lambda_{k}x_{k}^{2}$ converges if and only if $\sum_{k}a_{k}^{-1}x_{k}^{2}$ converges; that is, $V=X^{a}$ by The Noise Space of a Weight Sequence on a Hilbert Space with an Orthonormal Basis §space. For $x,y\in X^{a}$ the series $\sum_{k}a_{k}^{-1}x_{k}y_{k}$ converges with sum $\langle x,y\rangle_{a}$ by The Noise Space of a Weight Sequence on a Hilbert Space with an Orthonormal Basis §inner-product, so Elementary Properties of Series of Real Numbers §linearity gives

$$\langle x,y\rangle_{V}=\sum_{k=1}^{\infty}\bar{a}\,a_{k}^{-1}x_{k}y_{k}=\bar{a}\,\langle x,y\rangle_{a},\qquad |x|_{V}=\bar{a}^{1/2}|x|_{a},\qquad d_{V}(x,y)=\bar{a}^{1/2}d_{a}(x,y),$$

the second identity because both sides are nonnegative with square $\bar{a}\langle x,x\rangle_{a}$, and the third by applying the second to $x-y\in X^{a}$.

Claim (hilbert). By Step 0 and The Weighted Coefficient Subspace Determined by an Orthonormal Basis and a Sequence of Weights §subspace, $X^{a}=V$ is a linear subspace of $X$, so it contains $0_{X}$ and is closed under the sums and real multiples of $X$; with these restricted operations it is a real vector space with zero vector $0_{X}$, and by the same clause $\langle\cdot,\cdot\rangle_{V}$ is an inner product on it in the sense of Real Inner Product Space §inner-product. Since $\langle\cdot,\cdot\rangle_{a}=\bar{a}^{-1}\langle\cdot,\cdot\rangle_{V}$ with $\bar{a}^{-1}$ positive, symmetry, additivity and homogeneity in the first argument pass from $\langle\cdot,\cdot\rangle_{V}$ to $\langle\cdot,\cdot\rangle_{a}$; moreover $0\le\bar{a}^{-1}\langle x,x\rangle_{V}=\langle x,x\rangle_{a}$, and $\langle x,x\rangle_{a}=0$ forces $\langle x,x\rangle_{V}=0$, hence $x=0_{X}$. So $\langle\cdot,\cdot\rangle_{a}$ is an inner product on $X^{a}$, and its norm in the sense of Real Inner Product Space §norm, the nonnegative square root of $\langle x,x\rangle_{a}$, is $|x|_{a}$ by The Noise Space of a Weight Sequence on a Hilbert Space with an Orthonormal Basis §inner-product. For completeness, let $(x^{m})_{m\in\mathbb{N}}$ be a Cauchy sequence in $(X^{a},d_{a})$. Given $\varepsilon>0$, choose $M$ with $d_{a}(x^{m},x^{n})<\bar{a}^{-1/2}\varepsilon$ for $m,n\ge M$; then $d_{V}(x^{m},x^{n})<\varepsilon$ by Step 0. So $(x^{m})$ is Cauchy in $(V,d_{V})$, and by The Weighted Coefficient Subspace Determined by an Orthonormal Basis and a Sequence of Weights §complete it converges in $(V,d_{V})$ to some $x\in V=X^{a}$. Then $d_{a}(x^{m},x)=\bar{a}^{-1/2}d_{V}(x^{m},x)\to0$, so $(x^{m})$ converges to $x$ in $(X^{a},d_{a})$. Hence $X^{a}$ is a real Hilbert space in the sense of Real Hilbert Space §hilbert.

Claim (basis). By The Weighted Coefficient Subspace Determined by an Orthonormal Basis and a Sequence of Weights §subspace each $e_{j}$ lies in $V=X^{a}$, and by The Weighted Coefficient Subspace Determined by an Orthonormal Basis and a Sequence of Weights §coefficients and Step 0, $\langle x,e_{j}\rangle_{a}=\bar{a}^{-1}\langle x,e_{j}\rangle_{V}=\bar{a}^{-1}\lambda_{j}x_{j}=a_{j}^{-1}x_{j}$ for $x\in X^{a}$ and $j\in\mathbb{N}$. Hence $f_{k}=a_{k}^{1/2}e_{k}\in X^{a}$, the latter being a linear subspace, and by homogeneity and symmetry of $\langle\cdot,\cdot\rangle_{a}$,

$$\langle x,f_{k}\rangle_{a}=a_{k}^{1/2}\langle x,e_{k}\rangle_{a}=a_{k}^{1/2}a_{k}^{-1}x_{k}=a_{k}^{-1/2}x_{k}\qquad(x\in X^{a},\ k\in\mathbb{N}).$$

The coordinates of $f_{j}$ are $\langle f_{j},e_{k}\rangle=a_{j}^{1/2}\langle e_{j},e_{k}\rangle$, which is $a_{j}^{1/2}$ for $k=j$ and $0$ for $k\ne j$ by the orthonormality of $(e_{k})$ (Orthogonality, Orthogonal Complement and Orthonormal Families in a Real Inner Product Space §orthonormal). So $\langle f_{j},f_{k}\rangle_{a}=a_{k}^{-1/2}\langle f_{j},e_{k}\rangle$ equals $1$ for $j=k$ and $0$ for $j\ne k$; in particular $|f_{k}|_{a}=1$, and $(f_{k})_{k\in\mathbb{N}}$ is an orthonormal sequence in $X^{a}$ in the sense of Orthogonality, Orthogonal Complement and Orthonormal Families in a Real Inner Product Space §orthonormal. If $x\in X^{a}$ satisfies $\langle x,f_{k}\rangle_{a}=0$ for every $k$, then $a_{k}^{-1/2}x_{k}=0$, so $x_{k}=\langle x,e_{k}\rangle=0$ for every $k$, and $x=0_{X}$ because $(e_{k})$ is an orthonormal basis of $X$ (Orthonormal Basis of a Real Hilbert Space §basis). Since $0_{X}$ is the zero vector of $X^{a}$ by Claim (hilbert), $(f_{k})_{k\in\mathbb{N}}$ is an orthonormal basis of $X^{a}$ by Orthonormal Basis of a Real Hilbert Space §basis.

Claim (embedding). For $x\in X^{a}=V$, The Weighted Coefficient Subspace Determined by an Orthonormal Basis and a Sequence of Weights §subspace and Step 0 give $|x|\le|x|_{V}=\bar{a}^{1/2}|x|_{a}$; both sides being nonnegative, squaring gives $|x|^{2}\le\bar{a}\,|x|_{a}^{2}$.

Claim (partial-sums). Fix $k$. If $x^{m}\to x$ in $(X,d)$, then $x^{m}_{k}=\langle x^{m},e_{k}\rangle\to\langle x,e_{k}\rangle=x_{k}$ by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §continuity; so the coordinate function $x\mapsto x_{k}$ is continuous on $X$ by Continuity Between Metric Spaces is Equivalent to Sequential Continuity §on-subset. By Continuity of Sums and Products of Real-Valued Functions on a Metric Space §on-set (with $A=X$), the products $x\mapsto a_{k}^{-1}x_{k}x_{k}$ are continuous, and by the same clause and induction on $N$ so is their finite sum $S_{N}$. Next, $S_{N+1}(x)-S_{N}(x)=a_{N+1}^{-1}x_{N+1}^{2}\ge0$. Finally, $(S_{N}(x))_{N\in\mathbb{N}}$ is the sequence of partial sums of the series $\sum_{k}a_{k}^{-1}x_{k}^{2}$, whose terms are nonnegative; by Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §criterion this series converges, that is $x\in X^{a}$, if and only if $\{S_{N}(x):N\in\mathbb{N}\}$ is bounded above, and then its sum, which is $\langle x,x\rangle_{a}=|x|_{a}^{2}$ by The Noise Space of a Weight Sequence on a Hilbert Space with an Orthonormal Basis §inner-product, equals $\sup_{N}S_{N}(x)$.

Claim (closed-balls). Let $B_{R}=\{x\in X^{a}:|x|_{a}\le R\}$ and let $x$ lie in the closure of $B_{R}$ in $(X,d)$. By Sequential Characterization of the Closure in a Metric Space there is a sequence $(x^{m})$ in $B_{R}$ converging to $x$ in $(X,d)$. For each $N$, Claim (partial-sums) gives $S_{N}(x^{m})\le\sup_{N'}S_{N'}(x^{m})=|x^{m}|_{a}^{2}\le R^{2}$, and $S_{N}(x^{m})\to S_{N}(x)$ by the continuity of $S_{N}$ and Continuity Between Metric Spaces is Equivalent to Sequential Continuity §sequential; hence $S_{N}(x)\le R^{2}$. So $(S_{N}(x))_{N}$ is bounded above by $R^{2}$, and by Claim (partial-sums) $x\in X^{a}$ with $|x|_{a}^{2}=\sup_{N}S_{N}(x)\le R^{2}$, whence $|x|_{a}\le R$ as $0\le R$. Thus the closure of $B_{R}$ is contained in $B_{R}$; it contains $B_{R}$ by claim 1 of The Closure is the Smallest Closed Superset, so $B_{R}$ equals its closure and is closed by claim 4 of that theorem.

Claim (borel). Every $x\in X^{a}$ satisfies $|x|_{a}\le m$ for some $m\in\mathbb{N}$ (Archimedean property), so $X^{a}=\bigcup_{m\in\mathbb{N}}B_{m}$. Each $B_{m}$ is closed by Claim (closed-balls), hence in $\mathcal{B}(X)$ by claim 1 of Borel Measurability and Bounded Integration on a Metric Space, and a countable union of members of the $\sigma$-algebra $\mathcal{B}(X)$ belongs to it; so $X^{a}\in\mathcal{B}(X)$. For $N\in\mathbb{N}$ let $h_{N}=S_{N}\cdot\mathbf{1}_{X^{a}}$. The function $S_{N}$ is continuous, hence Borel by claims 2 and 3 of Borel Measurability and Bounded Integration on a Metric Space; $\mathbf{1}_{X^{a}}$ is Borel by claim 1 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions; so $h_{N}$ is Borel by claim 3 of that lemma. For $x\in X^{a}$ the sequence $(S_{N}(x))_{N}$ is nondecreasing and bounded above with supremum $|x|_{a}^{2}$ (Claim (partial-sums)), so $h_{N}(x)=S_{N}(x)\to|x|_{a}^{2}=n_{a}(x)$ by A Bounded Monotone Sequence of Real Numbers Converges §nondecreasing; for $x\notin X^{a}$, $h_{N}(x)=0=n_{a}(x)$ for every $N$. Hence $n_{a}$ is the pointwise limit of the Borel functions $h_{N}$ and is Borel by claim 5 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions.

Claim (pairs). Let $\delta:X\times X\to X$, $\delta(z)=\pi_{2}(z)-\pi_{1}(z)$. For $k\in\mathbb{N}$, additivity and homogeneity of the inner product give $\langle\delta(z),e_{k}\rangle=\langle\pi_{2}(z),e_{k}\rangle-\langle\pi_{1}(z),e_{k}\rangle$. The maps $\pi_{1},\pi_{2}$ are Borel by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §product-sigma and the coordinate function $x\mapsto x_{k}$ is Borel by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §continuity, so $z\mapsto\langle\pi_{i}(z),e_{k}\rangle$ is Borel by claim 4 of Borel Measurability and Bounded Integration on a Metric Space, and their difference is Borel by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. By Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §measurable, applied with $(\Omega,\mathcal{F})=(X\times X,\mathcal{B}(X\times X))$ and $Y=\delta$, the map $\delta$ is Borel. Therefore $D_{a}=\delta^{-1}(X^{a})\in\mathcal{B}(X\times X)$, as $X^{a}\in\mathcal{B}(X)$ by Claim (borel) (Measurable Function and Real-Valued Measurable Function), and $c_{a}=n_{a}\circ\delta$ is Borel by claim 4 of Borel Measurability and Bounded Integration on a Metric Space. It is nonnegative because $n_{a}$ takes the values $|x|_{a}^{2}\ge0$ and $0$.

Claim (measurable). By Claims (hilbert) and (basis), Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality applies with $E=X^{a}$, the orthonormal basis $(f_{k})_{k\in\mathbb{N}}$ and the given $(S,\mathcal{S})$; its $\mathcal{B}(E)$ is the Borel $\sigma$-algebra of $X^{a}$ for $d_{a}$. For $v:S\to X^{a}$ its coordinates there are, by Claim (basis), $v_{k}(s)=\langle v(s),f_{k}\rangle_{a}=a_{k}^{-1/2}\langle v(s),e_{k}\rangle$.

Suppose $v$ is measurable. Then each $v_{k}$ is measurable by Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality §operations, and so is $s\mapsto\langle v(s),e_{k}\rangle=a_{k}^{1/2}v_{k}(s)$ by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions.

Conversely, suppose each $s\mapsto\langle v(s),e_{k}\rangle$ is measurable, and put $g_{k}=v_{k}$, measurable by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. For each $s$, $g_{k}(s)^{2}=a_{k}^{-1}\langle v(s),e_{k}\rangle^{2}$, and $\sum_{k}g_{k}(s)^{2}$ converges because $v(s)\in X^{a}$ (The Noise Space of a Weight Sequence on a Hilbert Space with an Orthonormal Basis §space). By Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality §synthesis the map $w:S\to X^{a}$, $w(s)=\sum_{k=1}^{\infty}g_{k}(s)f_{k}$, is measurable with $w_{k}=g_{k}=v_{k}$ for every $k$. For each $s$, additivity of $\langle\cdot,\cdot\rangle_{a}$ gives $\langle w(s)-v(s),f_{k}\rangle_{a}=w_{k}(s)-v_{k}(s)=0$ for every $k$, so $w(s)-v(s)=0_{X}$ because $(f_{k})$ is an orthonormal basis of $X^{a}$ (Orthonormal Basis of a Real Hilbert Space §basis). Thus $v=w$ is measurable.

Finally, if $v:S\to X$ is measurable with respect to $\mathcal{S}$ and $\mathcal{B}(X)$ and takes values in $X^{a}$, then each $s\mapsto\langle v(s),e_{k}\rangle$ is measurable by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §measurable, applied with $(\Omega,\mathcal{F})=(S,\mathcal{S})$ and $Y=v$, and the criterion just proved shows that $v$ is measurable as a map into $X^{a}$.