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.

  • $\begingroup$ $T$ is uniquely determined by the image of a countable basis. $\endgroup$ Nov 24, 2020 at 17:55

1 Answer 1


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!

  • $\begingroup$ 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 $X(v) = x v, \forall v \in \mathcal{H}$. $\endgroup$
    – C_M
    Nov 24, 2020 at 18:06
  • $\begingroup$ @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! $\endgroup$ Nov 24, 2020 at 18:18
  • $\begingroup$ @C_M oh yes sorry, thanks for catching the typo: $X_x(\alpha_n)= x_n\alpha_n$ $\endgroup$ Nov 24, 2020 at 18:21
  • $\begingroup$ I just edited your answer without prior asking -- bear with me. If you do not agree, then consider to rollback the changes. $\endgroup$
    – Hanno
    Dec 11, 2020 at 14:29

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