Complete space, Hilbert space and inner product

Question

I have proven that $(x,y) = \sum_{n=0}^{\infty} (1+n^2) x_n \overline{y_n}$ is an inner product.

Now, I have to prove that $H = \{x=(x_n)_{n\in\mathbb{N}} \in \mathbb{C}^{\mathbb{N}}, \sum_{n=0}^{\infty} (1+n^2) |x_n|^2 < +\infty \} $ is a Hilbert space, i.e. I have to prove that $H$ is complete.

To do it, I would like to prove that all Cauchy sequence is convergent. But how to do it? I really need help for this question...

Answer 1

Hint:

Define $U : \ell^2 \to H$ as

$$U(x_1, x_2, x_3, \ldots) = \left(\frac{x_1}{\sqrt{1+1^2}}, \frac{x_2}{\sqrt{1+2^2}}, \frac{x_3}{\sqrt{1+3^2}}, \ldots\right)$$

Clearly $U$ is linear and surjective. $U $ is also an isometry:

$$\|Ux\|^2 = \sum_{n=1}^\infty (1+n^2)\left|\frac{x_n}{1+n^2}\right| = \sum_{n=1}^\infty |x_n|^2 = \|x\|_2^2 $$

Therefore, $U$ is an isometric isomorphism so $H$ is complete because $\ell^2$ is complete.

Answer 2

Correspond each $x\in H$ to $y_{x}$ that $y_{x}\in l^{2}$, where $(y_{x})_{n}=\sqrt{1+n^{2}}x_{n}$. If $(x^{m})_{m}\subseteq H$ is a Cauchy sequence, then so is $(y_{x^{m}})_{m}$ in $l^{2}$. We know that $l^{2}$ is complete, so $y_{x^{m}}\rightarrow y$ for some $y\in l^{2}$. Now it is routine to check that $x^{m}\rightarrow z$, where $z_{n}=\dfrac{1}{\sqrt{1+n^{2}}}y_{n}$.