Show that $l^2$ is a Hilbert space

Question

Let $l^2$ be the space of square summable sequences with the inner product $\langle x,y\rangle=\sum_\limits{i=1}^\infty x_iy_i$.
(a) show that $l^2$ is H Hilbert space.

To show that it's a Hilbert space I need to show that the space is complete. For that I need to construct a Cauchy sequence and show it converges with respect to the norm. However, I find it confusing to construct a Cauchy sequence of sequences?

Answer 1

In this answer, I will use $x_n$ as a sequence in $l^2$ and write $x_n(k)$ as the $k$-th member of that sequence.

The norm in the Hilbert space is given by $\|x\| = \sqrt{\langle x, x \rangle}$. We wish to show that if a sequence $\{ x_n \} \subset l^2$ is Cauchy, then it converges in $l^2$.

Suppose that $\{x_n\}$ is such a Cauchy sequence. Let $\{ e_k \}$ be the collection of sequences for which $e_k(i) = 1$ if $i=k$ and zero if $i\neq k$.

Then $\langle x_n, e_k \rangle = x_n(k)$. Notice that $$|x_n(k) - x_m(k)| = |\langle x_n - x_m, e_k \rangle| \le \|x_n-x_m\| \| e_k\| = \|x_n-x_m\|$$ for all $k$ (also note that this convergence is uniform over $k$). Therefore the sequence of real numbers given by $\{x_n(k)\}_{n\in \mathbb{N}}$ is Cauchy for each $k$, and thus converges. Call the limit of this sequence $\tilde x(k)$.

Let $\tilde x = (\tilde x(k))_{k\in\mathbb{N}}$. We wish to show that $\tilde x \in l^2$.

Consider $$\sum_{k=1}^\infty |\tilde x(k)|^2=\sum_{k=1}^\infty |\lim_{n\to\infty} x_n(k)|^2=\lim_{n\to\infty} \sum_{k=1}^\infty |x_n(k)|^2=\lim_{n\to\infty}\|x_n\|^2.$$

The exchange of limits is justified, since the convergence of $\lim_{n\to\infty} x_n(k)$ is uniform over $k$. Finally, since $\{ x_n \}$ is Cauchy, the inequality, $$| \|x_m\| - \|x_n\| | < \| x_m - x_n\|$$ implies that $\|x_n\|$ is a Cauchy sequence of real numbers, and so $\|x_n\|$ converges. Thus $\tilde x$ is in $l^2$.

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Edit: Completing the proof as per the comments.

We have thus shown that $\tilde x$ is in $l^2$. $\tilde x$ is the most likely candidate for the Cauchy sequence to converge to, and it has been demonstrated to be in our space. What remains is to show that $$\| x_n - \tilde x\| \to 0$$ as $n \to \infty$.

We will utilize a generalized form of the dominated convergence theorem for series. This states that if $a_{n,k} \to b_k$ for all $k$, $a_{n,k} < d_{n,k}$ and $\sum_{k} d_{n,k} \to \sum_{k} D_k < \infty$, then $\lim_{n \to \infty} \sum_{k=0}^\infty a_{n,k} = \sum_{k=0}^\infty b_k$. (here $a_{n,k}, b_k, d_{n,k}, D_{k}$ are all real numbers)

Writing $$\| x_n - \tilde x\|^2 = \sum_{k=0}^\infty |x_n(k) - \tilde x(k)|^2.$$

We see that in this case $a_{n,k} = |x_{n}(k) - \tilde x(k)|^2$, $b_k = 0$, and we must find a $d_{n,k}$ that "dominates" $a_{n,k}$ to finish the proof.

Now note that $|x_n(k) - \tilde x(k)|^2 \le 2 |x_n(k)|^2 + 2 |\tilde x(k)|^2$ and $$\lim_{n \to \infty} \sum_{n=0}^\infty ( 2 |x_n(k)|^2 + 2 |\tilde x(k)|^2) = \sum_{k=0}^\infty (2 |\tilde x(k)|^2 + 2 | \tilde x(k)|^2).$$ Recall that we demonstrated $\lim_{n \to \infty} \sum_{n=0}^\infty |x_n(k)|^2 = \sum_{n=0}^\infty |\tilde x(k)|^2$ in the first half. Thus $D_k$ is played by $4|\tilde x(k)|^2$ in this case.

Thus by the dominated convergence theorem we may conclude that $$\sum_{k=0}^\infty |x_n(k)-\tilde x(k)|^2 \to 0.$$

Answer 2

Let $(\mathbf{x_n})$ be a Cauchy sequence in $l^2$, where $\mathbf{x_n} = (x_1^{(n)},x_2^{(n)},\ldots)$, i.e., given $\epsilon >0$ there exists a natural number $N$ such that for all $m,n\geq N$ \begin{equation} \|\mathbf{x_n}-\mathbf{x_m}\| = \left(\sum_\limits{j=1}^{\infty}|x_j^{(n)}-x_j^{(m})|^2\right)^{\frac{1}{2}} <\epsilon \end{equation} In particular, it follows that for every $j=1,2,\ldots$ we have \begin{align} |x_j^{(n)}-x_j^{(m)}| < \epsilon && (m,n\geq N). \end{align} That is for each fixed $j$ the sequence $(x_j^{(n)},x_j^{(n)},\ldots)$ is a Cauchy sequence in the scalar field $\mathbb{R}$ or $\mathbb{C}$ and hence it converges. Let $x_j^{(n)} \to x_j$ as $n \to \infty$. Using these limits now define $\mathbf{x} = (x_1,x_2,\ldots)$. Now, with this basic setting try to show that $(\mathbf{x_n})$ converges to $\mathbf{x}$.

Answer 3

Hint : Let $(x^n)$ be a Cauchy sequence. We can write $x^n=\sum _k x_k^ne_k$ where $(e_k)$ is the canonical basis. Show $|x_k^n-x_k^m|\leq \|x^n-x^m\|$ so $(x_k^n)_n$ is a Cauchy sequence in $\mathbb{C}$, so there exist $x_k$ such that $\lim x_k^n=x_k$ for every $k$. Define $x=\sum x_ke_k$ and prove that $\|x^n-x\|\to 0$.

Answer 4

Pervin = Pervin, W. J.: Foundations of General Topology, New York: Academic press, 1964.
See the proof of Pervin, p.119, Theorem 7.1.2.
All you need to do is replace $H$ in the proof with $l^2$.
The proof is divided into two steps:
Step 1: ($\{x_n\}$ is Cauchy) $\Rightarrow$ ($x_n\to x$) [Pervin, p.119, l.$-$13--l.$-$1].
Step 2: $x\in l^2$ [Pervin, p.120, l.1--l.8].