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 '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!

  • $\begingroup$ 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 '20 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 '20 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 '20 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 '20 at 14:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.