If we have
$l^2= \{ a \in \mathbb{R}^{\mathbb{N}} | \sum_{k=0}^{\infty} |a_k|^{2} < \infty \}$ and $||a||_2 = (\sum_{k=0}^{\infty} |a_k|^2 )^{1/2}$
1
Proposition: $l^2 $ is a vector space.
We need to show that $\lambda a \in l^2$ and $(a+b)\in l^2$ for $\lambda\in \mathbb{R}$ and $a,b\in l^2$ .
for the scalar multiplication: $\lambda a = \sum _{k=0}^{\infty}|\lambda a_k|^2 = \sum_{k=0}^{\infty}|\lambda|^2|a_k|^2=|\lambda|^2\sum_{k=0}^{\infty}|a_k|^2$
the closure under addition: $a+b = \sum_{k=0}^{\infty}|a_{k}+b_{k}|^2\le \sum_{k=0}^{\infty}|a_k|^2 +2\sum_{k=0}^{\infty}|a_kb_k|+\sum_{k=0}^{\infty}|b_k|^2 $
The first and last term are finite because $a$ and $b$ are taken from $l^2$, but how do I deal with $2\sum_{k=0}^{\infty}|a_kb_k|$ ?
2
Proposition: $||.||_{2}$ is a norm on $l^2$.
We need to show that from $||x||_2 = 0$ it follows that $x=0$, that $||\lambda x||_2 = |\lambda| ||x||_2$ and $||x+y||_2 \le ||x||_2 + ||y||_2$ for $\lambda \in \mathbb{R}$ and $x,y \in l^2$.
If $||x||_2 = (\sum_{k=0}^{\infty}|x_k|^2)^{1/2} = 0 $, then $x_k = 0$ for all $k\in \mathbb{N}$.
For the scalar multiplication: $||\lambda x ||_2 = (\sum_{k=0}^{\infty}|\lambda x_k|^2 )^{1/2} = (|\lambda|^2\sum_{k=0}^{\infty}| x_k|^2 )^{1/2} = |\lambda| (\sum_{k=0}^{\infty}| x_k|^2 )^{1/2} = |\lambda| ||x||_2 $.
For the triangular inequality property: $||x+y||_2^2= \sum_{k=0}^{\infty}|x_k+y_k|^2 \le \sum_{k=0}^{\infty}|x_{k}|^2 +2\sum_{k=0}^{\infty}|x_k y_k|+\sum_{k=0}^{\infty}|y_k|^2$.
It is the same as in 1 when showing for closure under addition.... is $\sum_{k=0}^{\infty}|x_{k}y_{k}| \le (\sum_{k=0}^{\infty} | x_{k}|^2)^{1/2}(\sum_{k=0}^{\infty}|y_{k}|^2)^{1/2}$, how can one show that this inequality is true? If this was proven then:
$||x+y||_2^2= \sum_{k=0}^{\infty}|x_k+y_k|^2 \le \sum_{k=0}^{\infty}|x_{k}|^2 +2\sum_{k=0}^{\infty}|x_k y_k|+\sum_{k=0}^{\infty}|y_k|^2 \le ||x||_2^2+2||x||_2||y||_2 + ||y||_2^2 = (||x||_2+||y||_2)^2$
3
Proposition: $(l^2,||.||_2)$ is a Banach space.
We need to show completeness by showing that every Cauchy sequence converges.
If we suppose that we have elements $(a_k)_{n\in \mathbb{N}} \in l^2$ that form a Cauchy sequence and an $\epsilon > 0$ then there must be an $N$ such that : $||a_k - a_m ||_2^2 = \sum_{n=0}^{\infty} |(a_{k})_n-(a_{m})_n|^2 < \epsilon ^2 $ with $k,m\ge N $ from this it follows that: $|(a_{k})_n-(a_{m})_n|< \epsilon,$ so every subsequence is a Cauchy sequence and has a limit.
Are 1,2, and 3 correct now?