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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?

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    If the main purpose of your question is to ask for checking and criticism of your proof (as opposed to asking for any proof of the mentioned fact), you should use ([tag:proof-verification]) tag to make this clear.2017-01-08

1 Answers 1

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Suppose that $a$ and $b$ are in $l^2(\mathbb R)$. You showed that to show that $a+ b$ is also in $l^2(\mathbb R)$, you needed to show that $\sum_{n=0}^\infty |a_nb_n|$ is finite. This is a corollary of the famous result, the Cauchy-Schwarz Inequality.

Proposition (Cauchy-Schwarz) For real numbers $a_1, a_2, \ldots, a_n$ and $b_1, b_2,\ldots, b_n$, $ \left(\sum_{k=1}^n a_kb_k \right)^2 \leq \left(\sum_{k=1}^n a_k^2\right) \left(\sum_{k=1}^n b_k^2\right). $ The proof is given in most linear algebra texts. A hint to the proof can also be found on MathWorld: http://mathworld.wolfram.com/CauchysInequality.html.