# Prove that $\ell_2$ is a Hilbert space

Can someone check my proof? Thank you.

The space $\ell_2:=\ell_2(\Bbb Z)$ is the set of convergent sequences in $\Bbb C^{\Bbb Z}$ under the inner product defined by

$$(\mathbf x|\mathbf y)_2:=\sum_{k\in\Bbb Z}x_k\bar y_k$$

It is clear that $\ell_2$ is an (inner product) vector space, sum and scalar product are closed operations in $\ell_2$:

1) Scalar product: $\|k\mathbf x\|_2=|k|\cdot\|\mathbf x\|_2<\infty$

2) Sum of vectors: $$\|\mathbf {x+y}\|_2^2=\sum_{k\in\Bbb Z}|x_k+y_k|^2\le\sum_{k\in\Bbb Z}|x_k|^2+\sum_{k\in\Bbb Z}|y_k|^2+2\sum_{k\in\Bbb Z}|x_ky_k|<\infty$$

where $\sum_{k\in\Bbb Z}|x_ky_k|<\infty$ comes from the Hölder inequality:

$$\sum_{k\in\Bbb Z}|x_ky_k|\le\left(\sum_{k\in\Bbb Z}|x_k|^2\right)^{1/2}\left(\sum_{k\in\Bbb Z}|y_k|^2\right)^{1/2}$$

Then it remains to show that $\ell_2$ is complete. Suppose that $(\mathbf x_n)$ is a Cauchy sequence in $\ell_2$, then exists some $N\in\Bbb N$ such that

$$\|\mathbf x_n-\mathbf x_m\|_2<\epsilon,\quad n,m\ge N$$

In other words

$$\|\mathbf x_n-\mathbf x_m\|_2^2=\sum_{k\in\Bbb Z}|x_{n,k}-x_{m,k}|^2<\epsilon^2$$

In particular we have that $(x_{n,k})_{n\in\Bbb N}$ is a Cauchy sequence for each $k\in\Bbb Z$, and because $\Bbb C$ is a complete space under the standard metric then $(x_{n,k})_{n\in\Bbb N}\to \tilde x_k$.

Let define $\mathbf{\tilde x}:=(\tilde x_k)_{k\in\Bbb Z}$. We must show now that $(\mathbf x_n)\to\mathbf{\tilde x}$, observe that

$$\sum_{|k|\le p}|x_{n,k}-x_{m,k}|^2\le\sum_{k\in\Bbb Z}|x_{n,k}-x_{m,k}|^2<\epsilon^2,\quad\forall p\in\Bbb N,\forall n,m\ge N$$

hence

$$\lim_{m\to \infty}\sum_{|k|\le p}|x_{n,k}-x_{m,k}|^2=\sum_{|k|\le p}|x_{n,k}-\tilde x_k|^2\le\epsilon^2,\quad\forall p\in\Bbb N$$

and taking again limits we finally conclude

$$\lim_{p\to\infty}\sum_{|k|\le p}|x_{n,k}-\tilde x_k|^2=\sum_{k\in\Bbb Z}|x_{n,k}-\tilde x_k|^2\le \epsilon^2$$

It remains to show that $\|\mathbf{\tilde x}\|_2<\infty$. Now observe that

$$\big|\|\mathbf x_n\|_2-\|\mathbf{\tilde x}\|_2\big|\le\|\mathbf x_n-\mathbf{\tilde x}\|_2<\epsilon$$

Hence $\|\mathbf{\tilde x}\|_2<\infty$ provided that $\|\mathbf x_n\|_2<\infty$.$\Box$

• When you say "ℓ2(ℤ) is closed" you mean "ℓ2(ℤ)is complete" – VictorZurkowski Apr 4 '17 at 4:22
• The inequality $‖xn−x̃ ‖^2≤‖xn−xm‖^2$ is not true, nor it is necessary for the comclusion. But you are on the right track. Note first that for every finite N, $∑_{|k|<=N}|xn,k−xk|^2<=ϵ^2$, then conclude $‖xn−x̃ ‖2≤ϵ$. – VictorZurkowski Apr 4 '17 at 4:27
• Ah, thank you @Victor, I had some doubt about this inequality. But what you propose is not very clear to me, how we can fix $\epsilon^2$ for any $N\in\Bbb N$ to conclude what you propose? – Masacroso Apr 4 '17 at 4:39
• For large n,m: $∑_{|k|<=N}|xn,k−xmk|2 <= ∑_{k∈ℤ}|xn,k−xm,k|2 < ϵ^2$, in particular $∑_{|k|<=N}|xn,k−xmk|2 <= ϵ^2$. Pass to the limit in m (which you can do on FINITE sum without further assumption). – VictorZurkowski Apr 4 '17 at 4:44
• You have all the ingredients, but the end is muddy, because you are (implicitly) using $‖x̃ ‖2<∞$ when you write $∣∣‖xn‖2−‖x̃ ‖2∣∣≤‖xn−x̃ ‖2<ϵ$. To remove that wrinkle, you can use the triangle inequality for finitely many components (taking only finitely many terms), to show the partial sums $∑_{|k|≤p}|x̃ k|2$ are bounded uniformly in p. – VictorZurkowski Apr 4 '17 at 5:11