Proof of Weighted Square-Summable Sequence Spaces are Real Hilbert Spaces: the Standard Orthonormal Basis and the Embedding of a Hilbert Space under Bounded Weights

Each result cited is universally quantified over the data in its own statement. Throughout, series of real numbers, their convergence and their sums are those of Series of Real Numbers §convergent. Two elementary facts are used repeatedly. (P) If $(t_{m})_{m\in\mathbb{N}}$ is a sequence of real numbers and $k\in\mathbb{N}$ with $t_{m}=0$ for every $m\ne k$, then the partial sums $\sum_{m=1}^{n}t_{m}$ equal $0$ for $n<k$ and $t_{k}$ for $n\ge k$, so the series $\sum_{m=1}^{\infty}t_{m}$ converges with sum $t_{k}$. (Q) If $v\in X_{a}$ and $K\in\mathbb{N}$, then the terms $a_{k}v_{k}^{2}$ are nonnegative, since $a_{k}>0$, so by Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §dominates

$$0\le a_{K}v_{K}^{2}\le\sum_{k=1}^{K}a_{k}v_{k}^{2}\le\sum_{k=1}^{\infty}a_{k}v_{k}^{2}=\langle v,v\rangle_{a},$$

the second inequality because every term of the finite sum is nonnegative.

Claim 1. Vector space. By Weighted Square-Summable Sequence Spaces §space, termwise sums and real multiples of elements of $X_{a}$ lie in $X_{a}$. The zero sequence $0$ lies in $X_{a}$, all terms of $\sum_{k}a_{k}0^{2}$ being $0$, and $v+0=v$ termwise. For $v\in X_{a}$ the sequence $(-1)v=(-v_{k})_{k}$ lies in $X_{a}$ and $v+(-1)v=0$. Conditions 1, 2 and 5 to 8 of Vector Space over a Field hold because they hold in each term by the field axioms of $\mathbb{R}$. Hence $X_{a}$ is a real vector space with zero vector the zero sequence, which is its only zero vector by claim 1 of Elementary Identities in a Vector Space; by claim 2 there the additive inverse of $w$ is $(-w_{k})_{k}$, so $v-w=(v_{k}-w_{k})_{k}$. Inner product. By Weighted Square-Summable Sequence Spaces §inner-product the series defining $\langle v,w\rangle_{a}$ converges for $v,w\in X_{a}$. We check conditions (a) to (d) of Real Inner Product Space §inner-product; let $v,w,z\in X_{a}$ and $\lambda\in\mathbb{R}$. (a) The series for $\langle v,w\rangle_{a}$ and $\langle w,v\rangle_{a}$ have the same terms $a_{k}v_{k}w_{k}=a_{k}w_{k}v_{k}$. (b) The $k$-th term of the series for $\langle v+w,z\rangle_{a}$ is $a_{k}v_{k}z_{k}+a_{k}w_{k}z_{k}$; since $\sum_{k}a_{k}v_{k}z_{k}$ and $\sum_{k}a_{k}w_{k}z_{k}$ converge, Elementary Properties of Series of Real Numbers §linearity gives $\langle v+w,z\rangle_{a}=\langle v,z\rangle_{a}+\langle w,z\rangle_{a}$. (c) The $k$-th term of the series for $\langle\lambda v,w\rangle_{a}$ is $\lambda(a_{k}v_{k}w_{k})$, so $\langle\lambda v,w\rangle_{a}=\lambda\langle v,w\rangle_{a}$ by Elementary Properties of Series of Real Numbers §linearity. (d) $0\le\langle v,v\rangle_{a}$ by Weighted Square-Summable Sequence Spaces §inner-product. If $\langle v,v\rangle_{a}=0$, then for each $K\in\mathbb{N}$ fact (Q) gives $0\le a_{K}v_{K}^{2}\le0$, so $a_{K}v_{K}^{2}=0$, hence $v_{K}^{2}=0$ because $a_{K}>0$, and $v_{K}=0$; thus $v$ is the zero sequence. So $\langle\cdot,\cdot\rangle_{a}$ is an inner product on $X_{a}$. Since $|v|_{a}=\sqrt{\langle v,v\rangle_{a}}$ is the nonnegative real number whose square is $\langle v,v\rangle_{a}$, it is the norm of Real Inner Product Space §norm, and the distance of Real Inner Product Space §distance is $d_{a}(v,w)=|v-w|_{a}$. Completeness. By Real Hilbert Space §hilbert and Complete Metric Space it remains to show that every Cauchy sequence $(v^{j})_{j\in\mathbb{N}}$ in $(X_{a},d_{a})$ converges to a point of $X_{a}$. For $i,j,k\in\mathbb{N}$ the $k$-th term of $v^{j}-v^{i}$ is $v^{j}_{k}-v^{i}_{k}$, so by (Q)

$$a_{k}\bigl(v^{j}_{k}-v^{i}_{k}\bigr)^{2}\le\sum_{l=1}^{K}a_{l}\bigl(v^{j}_{l}-v^{i}_{l}\bigr)^{2}\le|v^{j}-v^{i}|_{a}^{2}\qquad(k\le K).\tag{1}$$

Coordinates. Fix $k\in\mathbb{N}$ and $\varepsilon>0$; then choose $J$ with $d_{a}(v^{j},v^{i})<\varepsilon a_{k}^{1/2}$ for all $i,j\ge J$, where $a_{k}^{1/2}>0$ is the power of Real Power of a Nonnegative Real Number §power, with $(a_{k}^{1/2})^{2}=a_{k}$ by Properties of Real Powers of Nonnegative Real Numbers §agreement and $0<a_{k}^{1/2}$ by Properties of Real Powers of Nonnegative Real Numbers §values. By (1), $a_{k}(v^{j}_{k}-v^{i}_{k})^{2}<a_{k}\varepsilon^{2}$, so $(v^{j}_{k}-v^{i}_{k})^{2}<\varepsilon^{2}$ and $|v^{j}_{k}-v^{i}_{k}|<\varepsilon$ for $i,j\ge J$. Thus $(v^{j}_{k})_{j}$ is a Cauchy sequence of real numbers and converges to a real number $v_{k}$. Let $v=(v_{k})_{k\in\mathbb{N}}$. Uniform estimate. Let $\varepsilon>0$; choose $J$ with $d_{a}(v^{j},v^{i})<\varepsilon$ for all $i,j\ge J$; then fix $j\ge J$ and $K\in\mathbb{N}$. By (1), $\sum_{l=1}^{K}a_{l}(v^{j}_{l}-v^{i}_{l})^{2}<\varepsilon^{2}$ for every $i\ge J$. As $i\to\infty$ the left side converges to $\sum_{l=1}^{K}a_{l}(v^{j}_{l}-v_{l})^{2}$ by the limit laws for sums and products applied to its finitely many terms, so this finite sum is at most $\varepsilon^{2}$ by the order properties of limits. Hence the partial sums of the series $\sum_{l}a_{l}(v^{j}_{l}-v_{l})^{2}$, whose terms are nonnegative, are bounded above by $\varepsilon^{2}$, and by Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §criterion the series converges with sum their supremum, which is at most $\varepsilon^{2}$. So $w^{j}=(v^{j}_{l}-v_{l})_{l}$ belongs to $X_{a}$ and $|w^{j}|_{a}\le\varepsilon$ for every $j\ge J$. Taking $\varepsilon=1$ and $j=J$, the sequence $v=v^{J}+(-1)w^{J}$ (termwise) belongs to $X_{a}$; then $v^{j}-v=w^{j}$ termwise for every $j$. Given $\varepsilon>0$, apply the uniform estimate with $\varepsilon/2$ to get $J$ with $d_{a}(v^{j},v)=|w^{j}|_{a}\le\varepsilon/2<\varepsilon$ for $j\ge J$. Hence $v^{j}\to v$ in $(X_{a},d_{a})$, and $X_{a}$ is a real Hilbert space.

Claim 2. Fix $k$ and write $r_{k}$ for the reciprocal of $a_{k}^{1/2}$, which exists since $0<a_{k}^{1/2}$ by Properties of Real Powers of Nonnegative Real Numbers §values; by Properties of Real Powers of Nonnegative Real Numbers §agreement, $a_{k}^{1/2}$ is the nonnegative square root of $a_{k}$, so $a_{k}=(a_{k}^{1/2})^{2}$. The terms $a_{m}((f^{a}_{k})_{m})^{2}$ vanish for $m\ne k$, so by (P) the series $\sum_{m}a_{m}((f^{a}_{k})_{m})^{2}$ converges and $f^{a}_{k}\in X_{a}$. For $v\in X_{a}$, the terms $a_{m}v_{m}(f^{a}_{k})_{m}$ vanish for $m\ne k$, so by (P)

$$\langle v,f^{a}_{k}\rangle_{a}=a_{k}v_{k}r_{k}=a_{k}^{1/2}\,a_{k}^{1/2}\,r_{k}\,v_{k}=a_{k}^{1/2}v_{k}.\tag{2}$$

Applying (2) with $v=f^{a}_{j}$: for $j\ne k$, $\langle f^{a}_{j},f^{a}_{k}\rangle_{a}=a_{k}^{1/2}\cdot0=0$, and $\langle f^{a}_{k},f^{a}_{k}\rangle_{a}=a_{k}^{1/2}r_{k}=1$, so $|f^{a}_{k}|_{a}=1$, $1$ being the nonnegative real number with square $1$. Thus $(f^{a}_{k})_{k}$ is an orthonormal sequence in the real Hilbert space $X_{a}$ of claim 1. If $v\in X_{a}$ satisfies $\langle v,f^{a}_{k}\rangle_{a}=0$ for every $k$, then $a_{k}^{1/2}v_{k}=0$ by (2), hence $v_{k}=0$ because $a_{k}^{1/2}\ne0$, for every $k$; so $v$ is the zero sequence, the zero vector of $X_{a}$ by claim 1. By Orthonormal Basis of a Real Hilbert Space §basis, $(f^{a}_{k})_{k}$ is an orthonormal basis of $X_{a}$.

Claim 3. Let $x\in X$. Since $(e_{k})_{k}$ 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, Orthonormal Expansions in a Real Hilbert Space §parseval shows that $\sum_{k}x_{k}^{2}$ converges with sum $|x|^{2}$, where $x_{k}=\langle x,e_{k}\rangle$. By Elementary Properties of Series of Real Numbers §linearity, $\sum_{k}\bar{a}x_{k}^{2}$ converges with sum $\bar{a}|x|^{2}$. Since $0\le a_{k}x_{k}^{2}\le\bar{a}x_{k}^{2}$ for every $k$, Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §comparison shows that $\sum_{k}a_{k}x_{k}^{2}$ converges, so $\iota_{a}(x)\in X_{a}$, and $|\iota_{a}(x)|_{a}^{2}=\sum_{k}a_{k}x_{k}^{2}\le\bar{a}|x|^{2}$. For $x,y\in X$ and $\lambda\in\mathbb{R}$, conditions (b) and (c) of Real Inner Product Space §inner-product for the inner product of $X$ give $\langle x+y,e_{k}\rangle=x_{k}+y_{k}$ and $\langle\lambda x,e_{k}\rangle=\lambda x_{k}$ for every $k$, so $\iota_{a}(x+y)=\iota_{a}(x)+\iota_{a}(y)$ and $\iota_{a}(\lambda x)=\lambda\iota_{a}(x)$ termwise; $\iota_{a}$ is linear. If $\iota_{a}(x)=\iota_{a}(y)$, then $z=x+(-1)y$ satisfies $\iota_{a}(z)=\iota_{a}(x)+(-1)\iota_{a}(y)=0$, that is $\langle z,e_{k}\rangle=0$ for every $k$, so $z=0_{X}$ by Orthonormal Basis of a Real Hilbert Space §basis; since $(-1)y=-y$ by claim 5 of Elementary Identities in a Vector Space, $x=z+y=y$. Hence $\iota_{a}$ is injective.