1.7 The ell_2 space

Definition 1 We define the ${\ell }_2$ space of infinite sequences of finite energy as

On this space, define the inner product $〈x,,,y〉={\sum }_{i=1}^{\infty }{x}_i\overline{y_i}$ , and obtain the induced norm ${∥x∥}_2=\sqrt{{\sum }_{i=1}^{\infty }{\left|x_i\right|}^2}$ , which we term the ${\ell }_2$ norm.

Here ${\mathbb{R}}^{\infty }$ refers to the set of all infinite sequences of real values. Note that this implies that the sequence $|x_i|$ must converge to zero as $i\to \infty$ . Note also can then say that the ${\ell }_2$ space consists of all infinite sequences with finite ${\ell }_2$ norm.

Theorem 1 The ${\ell }_2$ space is a Hilbert space.

To prove this theorem, we need the following lemma.

Lemma 1 A Cauchy sequence in a normed space is bounded.

Proof of Lemma 1: Let $\left\{x_n\right\}$ be a Cauchy sequence and let $n_0$ be an integer such that $\parallel x_n-x_{n_0}\parallel <1$ for $n>n_0$ . For $n>N$ , we have $\parallel x_n\parallel =\parallel x_n-x_{n_0}+x_{n_0}\parallel \le \parallel x_n-x_{n_0}\parallel +\parallel x_{n_0}\parallel \le 1+\parallel x_{n_0}\parallel$ . Now set $M={max}_{1\le n\le n_0}\parallel x_n\parallel +1$ . Then, $\parallel x_n\parallel <M$ for all $n=1,2,3,...$ .

Proof of Theorem 1: We show that ${\ell }_2$ is complete by proving that all Cauchy sequences converge in ${\ell }_2$ . Assume that $\left\{x^{\left(n\right)}\right\}$ is a Cauchy sequence in ${\ell }_2$ . Then, if $x^{\left(n\right)}=\left(x_1^{\left(n\right)},,,x_2^{\left(n\right)},,,...\right)$ , we have that there exists some $n_0$ such that if $j,k>n_0$ , then for each $a=1,2,...$ ,

Therefore, each sequence $\left\{x_a^{\left(n\right)}\right\}$ for each $a=1,2,\cdots$ is a Cauchy sequence in the space $(\mathbb{C},d_0)$ , which is complete. Thus, each sequence $\left\{x_a^{\left(n\right)}\right\}$ must converge in $\mathbb{C}$ to some value $x_a^{*}$ .

Define $x^{*}=(x_1^{*},x_2^{*},...)$ . We show that $x^{*}\in {\ell }_2$ . According to Lemma 1, the sequence $\left\{x_n\right\}$ is bounded by some constant $M$ . For each pair $a,k=1,2,...$ , we have

This inequality is valid for each value of $n$ , and so we must have

Additionally, his inequality is valid for each value of $k$ , and so we must have

Thus, we have shown that $x^{*}\in {\ell }_2$ . The last point we need to show is that $x^{\left(n\right)}\to x^{*}$ . First, since the sequence $\left\{x^{\left(n\right)}\right\}$ is Cauchy, we have that there exists an $n_0$ such that if $j,k\ge n_0$ and for each $l=1,2,...$ , we have

Observe also that for each $i$ there exists $n_{0,i}$ such that if $k\ge n_{0,i}$ , then $\left|x_i^{\left(k\right)},-,x_i^{*}\right|<\frac{2^{-i}ϵ}{4}$ ; therefore, if $k\ge max\left(n_0,,,{sup}_i,n_{0,i}\right)$ , we have

where the first inequality come from the fact that ${|a+b|}^2\le {2\left(\right|a|}^2+{\left|b\right|}^2)$ , which is easy to check. Since this is true for each $l=1,2,...$ , then it follows that if $j\ge n_0^{*}:=max\left(n_0,,,{sup}_i,n_{0,i}\right)$ , then

This shows that $x^{\left(n\right)}\to x^{*}$ .