# Banach space

• Jan 26th 2010, 06:58 AM
bigdoggy
Banach space
How would I show that the vector space $\displaystyle \ell^1$ with the 2 norm $\displaystyle ||\cdot||_2$is not a Banach space?

$\displaystyle \ell ^1= ( (x_i)^\infty_{i=1} : \Sigma^\infty_{i=1}|x_i|< \infty)$

I would perhaps have to show that it doesn't converge in the appropriate norm..? I am not sure how..!
• Jan 26th 2010, 12:18 PM
Opalg
Quote:

Originally Posted by bigdoggy
How would I show that the vector space $\displaystyle \ell^1$ with the 2 norm $\displaystyle ||\cdot||_2$is not a Banach space?

$\displaystyle \ell ^1= ( (x_i)^\infty_{i=1} : \Sigma^\infty_{i=1}|x_i|< \infty)$

I would perhaps have to show that it doesn't converge in the appropriate norm..? I am not sure how..!

You need to show that the space is not complete in that norm. To see that, start by taking a sequence that is in $\displaystyle \ell^2$ but not in $\displaystyle \ell^1$. For example, you could take the element $\displaystyle (x_i)$ with $\displaystyle x_i = 1/i$. Then consider the sequence of elements $\displaystyle x^{(n)}$ in $\displaystyle \ell^1$ defined by $\displaystyle \textstyle x^{(n)}(i) = \begin{cases}1/i&\text{if }i\leqslant n,\\ 0&\text{if }i>n.\end{cases}$

Then $\displaystyle (x^{(n)})$ is Cauchy for the 2-norm, but it does not have a 2-norm limit in the space $\displaystyle \ell^1$ (because it is trying to converge to an element that isn't in $\displaystyle \ell^1$).