# Functionla Analysis separability

1. Apr 13, 2014

### Funky1981

Let T: l^2 -> l^2 be bounded linear operators. K=L(l^2,l^2) be the space of T, Prove that K=L(l^2,l^2) is not separable

I know that if a space contains an uncountable number of non intersecting open balls then it is not separable. But how can I apply this statement here ( I mean how to construct such open balls) And are there any easier way to do it ???

2. Apr 13, 2014

### dextercioby

I don't know a direct proof of the result, but I have to ask which topology you're considering on K.

3. Apr 13, 2014

### micromass

Staff Emeritus
A diagonal operator has the form

$$T(x_1,x_2,x_3,...) = (\alpha_1 x_1, \alpha_2 x_2, \alpha_3 x_3, ...)$$

where $\alpha_k\in \mathbb{C}$.

It can easily be checked that $T$ is bounded if and only if the sequence $(\alpha_k)_k$ is bounded. Furthermore, we have $\|T\|= \|(\alpha_k)_k\|_\infty$. This yields an isometric embedding $\ell^\infty\rightarrow \mathcal{B}(\ell^2)$. Thus it suffices to show $\ell^\infty$ is not separable, which is well-known.