Isometric homomorphism between bounded operators of Hilbert pace and bounded sequences

Question

Let $H$ be a real separable Hilbert space and $\ell^\infty$ the space of bounded sequences in $\mathbb{R}$. Let $$\phi: \ell^\infty \to B(H),\quad a\mapsto (x\mapsto \sum_na_n\langle x,e_n\rangle e_n)$$ I want to (show) $\ell^\infty\cong B(H)$, $B_{\text{fin}}(H)\cong c_{00}$ and $c_0\cong K(H)$. These denote the bounded, eventually zero and zero/compact sequences/operators. Now we have for $\phi$: $$\|\phi(a)\|^2=\| \sum_na_n\langle x,e_n\rangle e_n)\|^2=\sum_n|a_n|^2|\langle x,e_n\rangle |^2\leq\|a\|_{\ell^2}^2\|x\|^2$$ hence $\phi(a)$ is bounded and hence in $B(H)$. And $$\sum_n a_n\| \langle e_m,e_n\rangle e_n\|=|a_m|$$ so $\|\phi(a_n)\|=\|a_n\|_{\ell^\infty}$

I am not sure how to show $\phi$ is an isomorphism (if it is). Any hins or references very appreciated.

Answer 1

You wrote

$\|\phi(a)\|^2=\| \sum_na_n\langle x,e_n\rangle e_n)\|^2=\sum_n|a_n|^2|\langle x,e_n\rangle |^2\leq\|a\|_{\ell^2}^2\|x\|^2.$

Two things are not correct: if $a=(a_n) \in l^{\infty}$, then, in general $a=(a_n) \notin l^{2}$, hence $\|a\|_{\ell^2}$ makes no sense.

But we have

$\|\phi(a)(x)\|^2=\| \sum_na_n\langle x,e_n\rangle e_n)\|^2=\sum_n|a_n|^2|\langle x,e_n\rangle |^2\leq\|a\|_{l^{\infty}}^2\|x\|^2.$

Hence $\|\phi(a)\| \le \|a\|_{l^{\infty}}.$

Answer 2

It is not. For example, operator $A$ that switches first and second coordinates: $A(\langle x_1, x_2, x_3 \ldots\rangle) = \langle x_2, x_1, x_3, \ldots\rangle$ isn't in image of $\phi$, as $\phi(a)(e_1) = a_1 e_1 \perp e_2$, but $A e_1 = e_2 \not\perp e_2$.

Image of $\phi$ is exactly all bounded operators for which $e_1, e_2, \ldots$ is eigenbasis - and there are many operators (even in $B_{fin}$) for which it is not true.