Hilbert space $\mathcal{H}$ is separable if and only if $\mathcal{B}(\mathcal{H})$ is separable

Let $$\mathcal{H}$$ be a Hilbert space. Then $$\mathcal{H}$$ is separable if and only if $$\mathcal{B}(\mathcal{H})$$, the space of all bounded linear operators on $$\mathcal{H}$$, is separable.

I am looking to determine whether or not the above statement is true, and if it is, how to prove it. Honestly, I do not know how to even approach the question.

If $$\mathcal{H}$$ is separable, then it has an orthonormal basis $$(e_n)_{n \in \mathbb{N}}$$ such that $$x = \sum_{n=1}^\infty \langle x, e_n \rangle e_n.$$

Then, for every $$T \in \mathcal{B}(\mathcal{H})$$, we have that $$Tx = \sum_{n=1}^\infty \langle x, e_n \rangle Te_n.$$

How can the above help us in defining a countably dense subset in $$\mathcal{B}(\mathcal{H})$$? For the converse implication, I do not know how to start.

• $T$ is uniquely determined by the image of a countable basis. Nov 24 '20 at 17:55

The statement can't be true: let $$\mathcal{H}$$ be a separable Hilbert space. Since it is separable, consider a countable orthonormal basis $$\{\alpha_1,\alpha_2,\dots\}$$ for $$\mathcal{H}$$. For each bounded sequence $$x=(x_n)$$ of real numbers, that is an element of $$l_{\infty}$$, consider the diagonal operator on $$\mathcal{H}$$, defined by $$X_{x}(\alpha_n)= x_n\alpha_n$$ for $$\alpha_n$$ an orthonormal basis vector.
Then $$\|X_x\|_\text{op} = \|x\|_{l_\infty}$$ (check!), and this induces an isometric inclusion of $$l_{\infty}$$ into the space of all linear bounded operators on $$\mathcal{H}$$, and $$l_{\infty}$$ is not separable!
• What is $x$ in your answer? A real number? What has this got to do with $\ell^\infty$? If $X(\alpha_n) = x \alpha_n$, then $X(v) = x v, \forall v \in \mathcal{H}$.
• @C_M $x$ is an element of $l_{\infty}$, and for each $x\in l_{\infty}$, we consider all such linear bounded operators $X_x$. The point is that $\vert \vert X_x\vert \vert =\vert \vert x\vert \vert _{l_{\infty}}$. Maybe I should also edit and inlcude this in the body of the answer. What it has to do with $l_{\infty}$ is that doing so will isometrically embed $l_{\infty}$ in the space of all bounded linear operator! Nov 24 '20 at 18:18
• @C_M oh yes sorry, thanks for catching the typo: $X_x(\alpha_n)= x_n\alpha_n$ Nov 24 '20 at 18:21