$\mathcal{L}(\ell_2,\ell_2)$ is not separable and a isomerty $T\colon\ell_\infty\to\mathcal{L}(\ell_2,\ell_2)$

Question

I have to prove that the operator $T\colon\ell_\infty\to\mathcal{L}(\ell_2,\ell_2)$ such $T((a_j)_{j=1}^{\infty})((b_j)_{j=1}^{\infty})=(a_jb_j)_{j=1}^{\infty}$ is an linear isometry.

How can I show that $T$ is an isometry, $( ||T(b_n||)=||b_n|| )$

can I use this to show that $\mathcal{L}({\ell_2,\ell_2})$ is not separable? I know that every normed separable space is isometrically isomorphic to a subspace of $\ell_\infty$ so I have to prove that $T$ is not a bijection. Is this true?

Answer 1

Hints:

1. Bound $\|T(a)(b)\|_2$ in terms of $\|a\|_\infty$ and $\|b\|_2$.
2. What does $T(a)$ do to the "standard unit vector" $e_j$?
3. $\ell_\infty$ contains an uncountable set $S$ such that $\|s-t\|\ge 1$ for all $s,t \in S$ distinct.

Answer 2

Your $T $ is definitely not a bijection, but you cannot use it as you want. The way to use it is that you simply get that $\mathcal L (\ell_2,\ell_2) $ contains a non-separable subspace. A separable metric space cannot contain a non-separable subspace, so you get the non-separability.

For the first part, \begin{align} \|Ta\|^2&=\sup\{\|Ta (b)\|_2^2:\ \|b\|_2=1\} =\sup\{\,\sum |a_jb_j|^2:\ \sum|b_j|^2=1\,\}=\|a\|_\infty^2. \end{align}