Unclear limit in showing that $\ell^2$ spaces are complete

Question

I'm working my way through the book Introduction to Hilbert Spaces with Applications and trying to follow the early example (1.4.6) showing that $\ell ^2$ spaces are complete (Cauchy sequences $(a_n) \in \ell ^2$ have their converge in $\ell ^2$). At this point only basic results of normed vector spaces are shown, neither the Dominated nor Monotone convergence theorems are presented.

They start by declaring the Cauchy sequence $(a_n) \in \ell ^2$ with elements $a_n = (\alpha_{n1}, \alpha_{n2}, ...)$. By the definition of Cauchy sequences and the $\ell ^2$ norm they get that there exists $N$ such that

\begin{equation} \sum_{k=1}^{\infty}|\alpha_{mk} - \alpha_{nk}|^2 < \epsilon^2 \tag{1} \end{equation}

for $m, n > N$. They conclude that this means that for each $k$, $|\alpha_{mk} - \alpha_{nk}| < \epsilon$ so $(\alpha_{nk})$ is a Cauchy sequence in $\mathbb{C}$ and therefore converges to a limit $\alpha_k \in \mathbb{C}$. Now to the part that is unclear to me. They state that from (1), by letting $m \rightarrow \infty$, we get

\begin{equation} \sum_{k=1}^{\infty}|\alpha_{k} - \alpha_{nk}|^2 \le \epsilon^2 \tag{2} \end{equation}

for $n > N$, without mentioning how. It seems like this limit is supposed to be obvious, but the only methods I can figure out or find rely on being able to state that ($\stackrel{!}{=}$)

\begin{equation} \lim_{m \rightarrow \infty} \sum_{k=1}^{\infty}|\alpha_{k} - \alpha_{nk}|^2 = \lim_{m \rightarrow \infty} \lim_{K \rightarrow \infty} \sum_{k=1}^{K}|\alpha_{k} - \alpha_{nk}|^2 \stackrel{!}{=} \lim_{K \rightarrow \infty} \sum_{k=1}^{K}\lim_{m \rightarrow \infty}|\alpha_{mk} - \alpha_{nk}|^2 = \sum_{k=1}^{\infty}|\alpha_{k} - \alpha_{nk}|^2 \end{equation}

using later results like the aforementioned convergence theorems. Is there some better method I'm completely missing here?

Answer 1

Do it in a few steps. Take any constant $n>N$ and $K\in\mathbb{N}$. For all $m>N$ we have:

$$ \sum_{k=1}^K |\alpha_{mk}-\alpha_{nk}|^2\leq\epsilon^2 $$

Now, this is a finite sum so there is no problem to use arithmetic of limits. By taking $m\to\infty$ we get the inequality $\sum_{k=1}^K |\alpha_k-\alpha_{nk}|^2\leq\epsilon^2$. And since this is true for all $K\in\mathbb{N}$ (since $K$ was an arbitrary natural number), by taking $K\to\infty$ we get $\sum_{k=1}^\infty |\alpha_k-\alpha_{nk}|^2\leq\epsilon^2$.

By the way, it is important to write non strict inequalities. After all, strict inequalities are not always preserved when taking a limit.

Answer 2

It may be convenient to consider elements of $\ell_2$ as functions $X:\mathbb{N}\rightarrow\mathbb{C}$ such that $\|X\|^2_2:=\sum_{k\geq1}|X(k)|^2<\infty$.

Suppose $\{X_n:n\in\mathbb{N}\}$ is a Cauchy sequence. To see that there is $X\in \ell_2$ such that $\|X-X_n\|_2\xrightarrow{n\rightarrow\infty}0$, one may proceed as follows:

1. Show first that $X_n$ convergent point wise, that is, for any $k\in \mathbb{N}$, $\lim_{n\rightarrow\infty}X_n(k)=: X(k)$ exists. Then $X$ will be our candidate for a limit in $\ll_2$. Pointwise convergence will follow from the simple observation that $$|X_n(k)-X_m(k)|\leq \|X_n-X_m\|_2$$ for each $k$, and $n,m$. For $\{X_n\}$ being a Cauchy sequence in $\ell_2$ implies that $\{X_n(k):n\in\mathbb{N}\}$ is a Cauchy sequence in $\mathbb{R}$ for each $k$.

2. Now that the point wise limit is established, let show that $X\in \ell_2$. There exists $N\in\mathbb{N}$ such that for all $n,m\geq N$, $$\|X_n-X_m\|_2\leq1$$ Then for any $K$ $$ \sum^K_{k=1}|X_n(k)|^2\leq \|X_n\|^2_2\leq (\|X_n-X_N\|_2+\|X_N\|_2)^2\leq 2(1+\|X_N\|^2_2)$$

Letting $n\rightarrow\infty$ shows that $$ \sum^K_{k=1}|X(k)|^2\leq 2(1+\|X_N\|^2_2) $$ for all $K$. Therefore $\|X\|^2_2=\lim_{K\rightarrow\infty}\sum^{K}_{k=1}|X(k)|^2<\infty$.

3. Show that $X_n$ converges to $X$ in $\ell_2$. Let $\varepsilon>0$. Since $\{X_n\}$ is Cauchy, there exists $N_\varepsilon$ such that $$ \|X_n-X_m\|_2<\varepsilon\quad\text{for all}\quad n,m\geq N_\varepsilon $$ Since $X\in\ell_2$, there exits $K_\varepsilon$ such that $$ \begin{align} \sum_{k>K_\varepsilon}|X(k)|^2&<\varepsilon^2\\ \sum_{k>K_\varepsilon}|X_N(k)|^2&<\varepsilon^2 \end{align} $$

$$ \begin{align} \|X-X_n\|_2&\leq \|X-X_N\|_2+\|X_N-X_n\|_2\\ &< \Big(\sum^{K_\varepsilon}_{k=1}|X(k)-X_N(k)|^2\Big)^{1/2} + \Big(\sum_{k>K_\varepsilon}|X(k)|^2+|X_N(k)|^2\Big)^{1/2} + \varepsilon\\ &< \Big(\sum^{K_\varepsilon}_{k=1}|X(k)-X_N(k)|^2\Big)^{1/2} + 3\varepsilon\\ &=\lim_{n\rightarrow\infty}\Big(\sum^{K_\varepsilon}_{k=1}|X_n(k)-X_N(k)|^2\Big)^{1/2} +3\varepsilon\leq 4\varepsilon \end{align} $$ since the limit is taken over a finite sum and $$\Big(\sum^{K_\varepsilon}_{k=1}|X_n(k)-X_N(k)|^2\Big)^{1/2}\leq\|X_n-X_N\|_2<\varepsilon$$

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