Proof of Elementary Properties of Series in a Real Inner Product Space

Throughout, $(s_{n})$ and $(t_{n})$ denote the partial sums of $(x_{k})$ and $(y_{k})$. We write $S$ for the successor map on $\mathbb{N}$ and $n+1$ for $S(n)$, and use the recursion of Finite Sum Notation in a Vector Space: $s_{1}=x_{1}$ and $s_{n+1}=s_{n}+x_{n+1}$.

An observation on indices. Every $k\in\mathbb{N}$ is either $1$ or of the form $S(j)$ for some $j\in\mathbb{N}$: the set of $k$ with this property contains $1$ and contains $S(m)$ whenever it contains $m$, hence is all of $\mathbb{N}$ by induction as in Natural Numbers. Moreover, if $N\in\mathbb{N}$ and $k\in\mathbb{N}$ satisfies $S(N)\le k$, then $k=S(j)$ for some $j$ with $N\le j$. Indeed $k\ne1$: if $k=1$ then $S(N)\le1$, while $1\le N$ by claim 4 of Properties of the Order on the Natural Numbers and $N\le S(N)$ by claims 5 and 1 of that lemma, so transitivity gives $N\le1$ and $S(N)\le N$, hence $N=S(N)$ by claim 2 and therefore $N<N$, which claim 2 excludes. So $k=S(j)$ for some $j\in\mathbb{N}$; and if $N\le j$ failed then $j<N$ by claim 3, hence $j\le N$ and $S(j)\le S(N)$ by claims 1 and 6, while $S(j)\ne S(N)$ because $S$ is injective by Natural Numbers, so $S(N)\le S(j)$ and $S(j)\le S(N)$ would give $S(j)=S(N)$ by claim 2, a contradiction.

Claim 1. By claims 2 and 3 of Properties of Finite Sums of Vectors the $n$-th partial sums of $(x_{k}+y_{k})$ and of $(\lambda x_{k})$ are $s_{n}+t_{n}$ and $\lambda s_{n}$. If $(s_{n})$ converges to $x$ and $(t_{n})$ to $y$, then $(s_{n}+t_{n})$ converges to $x+y$ and $(\lambda s_{n})$ converges to $\lambda x$ by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §linear-limits. This is the assertion.

Claim 2. Suppose $(s_{n})$ converges to $x$ and let $\varepsilon$ be positive, so that $\tfrac{\varepsilon}{2}$ is positive with $\tfrac{\varepsilon}{2}+\tfrac{\varepsilon}{2}=\varepsilon$ by claim 8 of Elementary Order Arithmetic in an Ordered Field. Choose $N$ with $|s_{n}-x|<\tfrac{\varepsilon}{2}$ for $N\le n$ and put $N'=S(N)$. For $k$ with $N'\le k$, the observation gives $k=S(j)$ with $N\le j$, and the recursion gives $x_{k}=s_{k}-s_{j}$, whence by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §triangle

using also $|x-s_{j}|=|s_{j}-x|$, which holds because $d$ is symmetric. Hence $(x_{k})$ converges to $0_{E}$.

Claim 3. We show $s_{n}=z_{n+1}-z_{1}$ by induction: $s_{1}=x_{1}=z_{2}-z_{1}$, and if $s_{n}=z_{n+1}-z_{1}$ then $s_{n+1}=s_{n}+x_{n+1}=(z_{n+1}-z_{1})+(z_{n+2}-z_{n+1})=z_{n+2}-z_{1}$. If $(z_{k})$ converges to $z$, then $(z_{n+1})_{n\in\mathbb{N}}$ converges to $z$, since $N\le n$ implies $N\le n+1$ by claims 5 and 1 of Properties of the Order on the Natural Numbers, and hence $(s_{n})$ converges to $z-z_{1}$ by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §linear-limits. Conversely, if $(s_{n})$ converges to $p$, then $(z_{n+1})_{n\in\mathbb{N}}=(s_{n}+z_{1})$ converges to $p+z_{1}=:z$ by the same reference; given a positive $\varepsilon$, choose $N$ with $|z_{n+1}-z|<\varepsilon$ for $N\le n$ and put $N'=S(N)$, so that for $N'\le k$ the observation gives $k=S(j)=j+1$ with $N\le j$ and hence $|z_{k}-z|<\varepsilon$. Thus $(z_{k})$ converges to $z$, and the sum of the series is $z-z_{1}$.

Claim 4. Since $d(s_{n},s_{m})=|s_{n}-s_{m}|$, the stated condition says exactly that $(s_{n})$ is a Cauchy sequence in $(H,d)$. If the series converges, say $(s_{n})$ converges to $x$, then for positive $\varepsilon$ and $N$ with $|s_{n}-x|<\tfrac{\varepsilon}{2}$ for $N\le n$ we get $|s_{n}-s_{m}|\le|s_{n}-x|+|x-s_{m}|<\varepsilon$ for $N\le m$ and $N\le n$, by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §triangle. Conversely, a Cauchy sequence in $(H,d)$ converges, because $H$ is a real Hilbert space and $(H,d)$ is therefore complete.

Claim 5. Let $\sigma_{n}=\sum_{k=1}^{n}|x_{k}|$ denote the partial sums of the convergent series $\sum_{k=1}^{\infty}|x_{k}|$.

We first show that $|s_{n}-s_{m}|\le\sigma_{n}-\sigma_{m}$ whenever $m\le n$, by induction on $n$. If $n=1$ then $m\le1$ and $1\le m$ by claim 4 of Properties of the Order on the Natural Numbers, so $m=1$ by claim 2 and both sides are $0$. Suppose the assertion holds for $n$ and let $m\le n+1$. If $m=n+1$ both sides are $0$. Otherwise $m\le n$ by claim 5 of Properties of the Order on the Natural Numbers, and then, by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §triangle and the inductive hypothesis,

The same induction, started from $|s_{1}|=|x_{1}|=\sigma_{1}$, gives $|s_{n}|\le\sigma_{n}$ for every $n$.

Now let $\varepsilon$ be positive. By claim 3 of Elementary Properties of Series of Real Numbers there is $N$ with $|\sigma_{n}-\sigma_{m}|<\varepsilon$ for $N\le m$ and $N\le n$. Given such $m,n$, the order on $\mathbb{N}$ is total, so one of $m\le n$ and $n\le m$ holds; in the first case $|s_{n}-s_{m}|\le\sigma_{n}-\sigma_{m}\le|\sigma_{n}-\sigma_{m}|<\varepsilon$ by claim 3 of Properties of the Absolute Value in an Ordered Field, and in the second the same bound holds after exchanging $m$ and $n$, since $|s_{n}-s_{m}|=|s_{m}-s_{n}|$. By claim 4 the series $\sum_{k=1}^{\infty}x_{k}$ converges; write $x$ for its sum.

Finally, $\bigl||s_{n}|-|x|\bigr|\le|s_{n}-x|$ by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §reverse-triangle, so $(|s_{n}|)$ converges to $|x|$, while $(\sigma_{n})$ converges to $\sum_{k=1}^{\infty}|x_{k}|$. Since $|s_{n}|\le\sigma_{n}$ for every $n$, claim 1 of Order Properties of Limits of Real Sequences gives $|x|\le\sum_{k=1}^{\infty}|x_{k}|$.

Claim 6. By claim 4 of Properties of Finite Sums of Vectors, $Ts_{n}=\sum_{k=1}^{n}Tx_{k}$, so $(Ts_{n})$ is the sequence of partial sums of $(Tx_{k})$. The map $T$ is continuous by Elementary Properties of Bounded Linear Maps and Functionals on Real Inner Product Spaces §lipschitz-continuous, so if $(s_{n})$ converges to $x$ then $(Ts_{n})$ converges to $Tx$ by Continuity Between Metric Spaces is Equivalent to Sequential Continuity. This is the assertion.

Claim 7. Let $\Lambda:E\to\mathbb{R}$ be given by $\Lambda u=\langle u,w\rangle$. By Elementary Properties of Bounded Linear Maps and Functionals on Real Inner Product Spaces §functionals the map $\Lambda$ is a bounded linear functional on $E$, and by Elementary Properties of Bounded Linear Maps and Functionals on Real Inner Product Spaces §real-line it is the same thing as an element of $\mathcal{L}(E,\mathbb{R})$, where $\mathbb{R}$ denotes the real inner product space fixed there, with norm $|\cdot|$ and distance $d_{\mathbb{R}}$. Claim 6 applied to $\Lambda$ shows that the series $\sum_{k=1}^{\infty}\langle x_{k},w\rangle$ converges in that space with sum $\bigl\langle\sum_{k=1}^{\infty}x_{k},w\bigr\rangle$, and claim 8 identifies this with convergence as a series of real numbers, with the same sum.

Claim 8. Let $(a_{k})$ be a sequence of real numbers, let $\sigma_{n}$ be its $n$-th partial sum in the sense of Series in a Real Inner Product Space and $\tau_{n}$ its $n$-th partial sum in the sense of Series of Real Numbers. Both satisfy the same recursion: $\sigma_{1}=a_{1}$ and $\sigma_{n+1}=\sigma_{n}+a_{n+1}$ by Finite Sum Notation in a Vector Space, since the vector addition of the real inner product space $\mathbb{R}$ is the addition of the field $\mathbb{R}$; and $\tau_{1}=a_{1}$ and $\tau_{n+1}=\tau_{n}+a_{n+1}$ by claim 1 of Properties of Finite Sums. Hence $\sigma_{n}=\tau_{n}$ for every $n$ by induction. The distance of the real inner product space $\mathbb{R}$ is $d_{\mathbb{R}}$ by Elementary Properties of Bounded Linear Maps and Functionals on Real Inner Product Spaces §real-line, and convergence of a sequence of real numbers agrees with convergence in $(\mathbb{R},d_{\mathbb{R}})$ by claim 1 of Convergence and the Cauchy Condition for Real Sequences Agree with Those in the Real Line as a Metric Space. Therefore the two notions of convergence of the series agree, and the sums coincide.