Proof of The Weighted Coefficient Subspace Determined by an Orthonormal Basis and a Sequence of Weights

Each result cited is universally quantified over the data in its own statement. Throughout, $x_{k}=\langle x,e_{k}\rangle_{H}$ and likewise for other vectors of $H$; by Orthonormal Expansions in a Real Hilbert Space §parseval the series $\sum_{k=1}^{\infty}x_{k}^{2}$ converges with sum $|x|_{H}^{2}$ for every $x\in H$. By claim 1 of Elementary Arithmetic in an Ordered Field and transitivity, $0\le\lambda_{k}$ for every $k$, and $\lambda_{k}$ is positive, since $\lambda_{k}=0$ would give $1\le0$. By orthonormality, $\langle e_{j},e_{k}\rangle_{H}$ is $1$ if $k=j$ and $0$ otherwise.

We use repeatedly the following remark. If a sequence $(s_{n})_{n\in\mathbb{N}}$ of real numbers, or of vectors of $H$, satisfies $s_{n}=s$ for every $n$ with $j\le n$, then it converges to $s$: given a positive $\varepsilon$, every such $n$ satisfies $|s_{n}-s|=0<\varepsilon$.

Claim 1. The set $V$ is a linear subspace containing each $e_{j}$. By Elementary Identities in a Real Inner Product Space §zero the vector $0_{H}$ has $(0_{H})_{k}=0$ for every $k$, so every partial sum of $\sum_{k=1}^{\infty}\lambda_{k}(0_{H})_{k}^{2}$ is $0$ by claim 7 of Properties of Finite Sums, and the series converges by the remark; thus $0_{H}\in V$. Let $x,y\in V$ and $c\in\mathbb{R}$. By Elementary Identities in a Real Inner Product Space §bilinear we have $(x+y)_{k}=x_{k}+y_{k}$ and $(cx)_{k}=c\,x_{k}$ for every $k$. Applying Products and Sums of Weighted Square-Summable Sequences of Real Numbers §sums with the nonnegative weights $\lambda_{k}$ and the sequences $(x_{k})$, $(y_{k})$ shows that $\sum_{k=1}^{\infty}\lambda_{k}(x_{k}+y_{k})^{2}$ and $\sum_{k=1}^{\infty}\lambda_{k}(c\,x_{k})^{2}$ converge, so $x+y\in V$ and $cx\in V$. Hence $V$ is a linear subspace of $H$.

Fix $j\in\mathbb{N}$. The terms of $\sum_{k=1}^{\infty}\lambda_{k}\langle e_{j},e_{k}\rangle_{H}^{2}$ vanish for $k\ne j$ and equal $\lambda_{j}$ for $k=j$, so by claim 7 of Properties of Finite Sums every partial sum with $j\le n$ equals $\lambda_{j}$; by the remark the series converges with sum $\lambda_{j}$, so $e_{j}\in V$.

The pairing. For $x,y\in V$, Products and Sums of Weighted Square-Summable Sequences of Real Numbers §products with the weights $\lambda_{k}$ shows that $\sum_{k=1}^{\infty}\lambda_{k}|x_{k}y_{k}|$ and $\sum_{k=1}^{\infty}\lambda_{k}x_{k}y_{k}$ converge, so $\langle x,y\rangle_{V}$ is defined.

The inner product axioms of Real Inner Product Space §inner-product. Symmetry holds because $\lambda_{k}x_{k}y_{k}=\lambda_{k}y_{k}x_{k}$ for every $k$. For additivity and homogeneity in the first argument, let $x,y,w\in V$ and $c\in\mathbb{R}$; then $\lambda_{k}(x+y)_{k}w_{k}=\lambda_{k}x_{k}w_{k}+\lambda_{k}y_{k}w_{k}$ and $\lambda_{k}(cx)_{k}w_{k}=c\,\lambda_{k}x_{k}w_{k}$ for every $k$, so Elementary Properties of Series of Real Numbers §linearity gives $\langle x+y,w\rangle_{V}=\langle x,w\rangle_{V}+\langle y,w\rangle_{V}$ and $\langle cx,w\rangle_{V}=c\,\langle x,w\rangle_{V}$.

For definiteness, let $x\in V$. The terms $\lambda_{k}x_{k}^{2}$ are nonnegative, so $0\le\langle x,x\rangle_{V}$ by Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §dominates. Multiplying $1\le\lambda_{k}$ by the nonnegative number $x_{k}^{2}$, using claim 5 of Elementary Arithmetic in an Ordered Field, gives $0\le x_{k}^{2}\le\lambda_{k}x_{k}^{2}$ for every $k$, so Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §comparison yields

If $\langle x,x\rangle_{V}=0$ then $|x|_{H}^{2}\le0$, while $0\le|x|_{H}^{2}$, so $\langle x,x\rangle_{H}=|x|_{H}^{2}=0$ and $x=0_{H}$ by condition (d) of Real Inner Product Space §inner-product for $H$. Thus $V$ with $\langle\cdot,\cdot\rangle_{V}$ is a real inner product space, and $|x|_{V}^{2}=\langle x,x\rangle_{V}$.

Finally $|x|_{H}\le|x|_{V}$: both numbers are nonnegative, and if $|x|_{V}<|x|_{H}$ then claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field would give $|x|_{V}^{2}<|x|_{H}^{2}$, contradicting the last display; trichotomy gives the inequality.

Claim 2. Let $x\in V$ and $j\in\mathbb{N}$. The terms of $\sum_{k=1}^{\infty}\lambda_{k}x_{k}\langle e_{j},e_{k}\rangle_{H}$ vanish for $k\ne j$ and equal $\lambda_{j}x_{j}$ for $k=j$, so as above every partial sum with $j\le n$ equals $\lambda_{j}x_{j}$ and $\langle x,e_{j}\rangle_{V}=\lambda_{j}x_{j}$. Taking $x=e_{j}$ gives $|e_{j}|_{V}^{2}=\langle e_{j},e_{j}\rangle_{V}=\lambda_{j}\langle e_{j},e_{j}\rangle_{H}=\lambda_{j}$.

Claim 3. Let $(x^{(m)})_{m\in\mathbb{N}}$ be a Cauchy sequence in $(V,d_{V})$. Since $|v|_{H}\le|v|_{V}$ for $v\in V$ by claim 1, it is a Cauchy sequence in $(H,d_{H})$, and $H$ is complete, so it converges in $H$ to some $x\in H$. Note that for every $k$ and every $v\in H$, The Cauchy-Schwarz Inequality in a Real Inner Product Space and $|e_{k}|_{H}=1$ give $|\langle v,e_{k}\rangle_{H}|\le|v|_{H}$; applying this to $v=x^{(l)}-x$ and using Elementary Identities in a Real Inner Product Space §bilinear gives $|x^{(l)}_{k}-x_{k}|\le|x^{(l)}-x|_{H}$ for every $l$. Since $(x^{(l)})$ converges to $x$ in $H$, the sequence $l\mapsto|x^{(l)}-x|_{H}$ converges to $0$, so claim 3 of Order Properties of Limits of Real Sequences shows that $x^{(l)}_{k}$ converges to $x_{k}$ as $l$ grows, for each fixed $k$.

Let $\varepsilon$ be positive and choose $M\in\mathbb{N}$ with $|x^{(m)}-x^{(l)}|_{V}<\varepsilon$ for all $m,l$ with $M\le m$ and $M\le l$; then $|x^{(m)}-x^{(l)}|_{V}^{2}<\varepsilon^{2}$ by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. Fix such an $m$ and fix $n\in\mathbb{N}$. For every $l$ with $M\le l$ the vector $x^{(m)}-x^{(l)}$ lies in $V$, so Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §dominates applied to its defining series gives

For each $k$ the sequence $l\mapsto\lambda_{k}(x^{(m)}_{k}-x^{(l)}_{k})^{2}$ converges to $\lambda_{k}(x^{(m)}_{k}-x_{k})^{2}$ by claims 2 and 3 of Arithmetic of Limits of Real Sequences, and hence, by claim 1 of that theorem applied $n-1$ times, that is by induction on the number of summands, the left-hand side converges to $\sum_{k=1}^{n}\lambda_{k}(x^{(m)}_{k}-x_{k})^{2}$ as $l$ grows. By claim 1 of Order Properties of Limits of Real Sequences,

This holds for every $n\in\mathbb{N}$, and the terms are nonnegative, so by Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §criterion the series $\sum_{k=1}^{\infty}\lambda_{k}(x^{(m)}_{k}-x_{k})^{2}$ converges and its sum, the supremum of its partial sums, is at most $\varepsilon^{2}$. Since $(x^{(m)}-x)_{k}=x^{(m)}_{k}-x_{k}$, this says that $x^{(m)}-x\in V$ with $|x^{(m)}-x|_{V}^{2}\le\varepsilon^{2}$, whence $|x^{(m)}-x|_{V}\le\varepsilon$ by the argument used at the end of claim 1.

Taking $\varepsilon=1$ in the last paragraph produces one $m$ with $x^{(m)}-x\in V$; as $x^{(m)}\in V$ and $V$ is a linear subspace, $x=x^{(m)}-(x^{(m)}-x)\in V$. Now let $\varepsilon$ be positive and apply the last paragraph to the positive number $\tfrac{\varepsilon}{2}$, obtaining $M$ with $|x^{(m)}-x|_{V}\le\tfrac{\varepsilon}{2}<\varepsilon$ for every $m$ with $M\le m$. Hence $(x^{(m)})$ converges to $x$ in $(V,d_{V})$. So every Cauchy sequence in $(V,d_{V})$ converges in $V$, and $V$ is a real Hilbert space by Real Hilbert Space §hilbert.

Claim 4. Let $x\in H$ and for $n\in\mathbb{N}$ let $s_{n}=\sum_{k=1}^{n}x_{k}e_{k}$. Each $e_{k}$ lies in $V$ by claim 1, and $V$ is closed under sums and scalar multiples, so induction on $n$, using the recursion of claim 1 of Properties of Finite Sums of Vectors, gives $s_{n}\in V$ for every $n$. By Orthonormal Expansions in a Real Hilbert Space §expansion the sequence $(s_{n})$ converges to $x$ in $H$, so $x$ lies in the closure of $V$ in $(H,d_{H})$ by Sequential Characterization of the Closure in a Metric Space. As $x\in H$ was arbitrary, that closure is $H$, which is to say that $V$ is dense in $H$.

Claim 5. Fix $k\in\mathbb{N}$. The number $\mu_{k}$ is positive: it is nonnegative, and $\mu_{k}=0$ would give $\tfrac{1}{\lambda_{k}}=\mu_{k}^{2}=0$ by claim 4 of Properties of Natural Number Powers in a Field and hence $1=\lambda_{k}\tfrac{1}{\lambda_{k}}=0$, which is false by claim 6 of Elementary Order Arithmetic in an Ordered Field. Put $f_{k}=\mu_{k}e_{k}$, which lies in $V$ by claim 1.

Let $j,k\in\mathbb{N}$. By the additivity and homogeneity established in claim 1 and by claim 2, $\langle f_{j},f_{k}\rangle_{V}=\mu_{j}\mu_{k}\langle e_{j},e_{k}\rangle_{V}=\mu_{j}\mu_{k}\lambda_{k}\langle e_{j},e_{k}\rangle_{H}$, which is $0$ for $j\ne k$ and $\mu_{j}^{2}\lambda_{j}=1$ for $j=k$. In particular $|f_{j}|_{V}^{2}=1$, and $|f_{j}|_{V}=1$ because both numbers are nonnegative and squaring is strictly monotone on the nonnegative numbers by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. So $(f_{k})_{k\in\mathbb{N}}$ is an orthonormal sequence in $V$.

Suppose $x\in V$ satisfies $\langle x,f_{k}\rangle_{V}=0$ for every $k$. Then $\mu_{k}\lambda_{k}x_{k}=\mu_{k}\langle x,e_{k}\rangle_{V}=0$ by claim 2 and homogeneity, and $\mu_{k}\lambda_{k}$ is positive, so $x_{k}=0$ for every $k$. Since $(e_{k})$ is an orthonormal basis of $H$ and $x\in H$, this gives $x=0_{H}$, which is also the zero vector of $V$ by claim 1 of A Linear Subspace is a Vector Space and Inherits an Inner Product. Hence $(f_{k})_{k\in\mathbb{N}}$ is an orthonormal basis of the real Hilbert space $V$ of claim 3, and A Real Hilbert Space with an Orthonormal Basis is Separable §separable shows that $(V,d_{V})$ is separable.