# (Fake proof) Bounded linear operators $\mathcal{L}(X,Y)$ between separable Banach spaces $X$, $Y$ is itself a separable Banach space?

Fake proof: Let $$X$$ and $$Y$$ be separable Banach spaces. We know that the space $$\mathcal{L}(X,Y)$$ of bounded operators $$X \to Y$$ endowed with the operator norm is itself a Banach space. (Cf. [1][2][3].) So we need to show that $$\mathcal{L}(X,Y)$$ is separable.

It is known (cf. p.14 of Alexander S. Kechris, Classical Descriptive Set Theory) that the closed unit ball of $$\mathcal{L}(X,Y)$$ with the subspace topology inherited from the strong operator topology of $$\mathcal{L}(X,Y)$$ is separable. (Cf. [4][5].)

However, if the unit ball of a normed vector space is separable, then the entire vector space must be separable. (Cf. [6][7].) Therefore $$\mathcal{L}(X,Y)$$ is separable (in the strong operator topology), so "$$\mathcal{L}(X,Y)$$ is a separable Banach space".

Why proof is fake (?): The main issue is that the operator norm topology on $$\mathcal{L}(X,Y)$$ is being conflated with the strong operator topology on $$\mathcal{L}(X,Y)$$. Is that correct?

The strong operator topology on $$\mathcal{L}(X,Y)$$ is separable but need not be (complete) metrizable. In contrast, the operator norm topology on $$\mathcal{L}(X,Y)$$ is (complete) metrizable but need not be separable. (Cf. [8][9].) Is that correct?

E.g. in general the strong operator topology on $$\mathcal{L}(X,Y)$$ is generated by semi-norms [10] in some sense related to the operator norm, but in general is not even metrizable [11] and thus clearly not induced by the operator norm and not a Banach space. Cf. also this question [12].

You can see this clearly when $$X=Y=\ell^2(\mathbb N)$$. You can embed $$A=\ell^\infty(\mathbb N)$$ into $$L(X)$$ as multiplication operators. That is $$M_xy=(x_ny_n).$$ This embedding is isometric when you consider the operator norm. As $$A$$ is not separable, neither is $$L(X)$$. On the other hand, $$L(X)$$ is separable in the strong/weak operator topology since $$X$$ is separable and the finite-rank operators are dense in $$L(X)$$ in the weak/strong operator topology.
• To clarify, what is the relationship between compactness and separability of the unit ball being used here? Or is it just that the proof that (unit ball separable) $\implies$ (whole space is separable) requires the assumption that the unit ball is compact? Is this referring to a special case of Banach-Alaoglu? en.wikipedia.org/wiki/… Sorry, my functional analysis is really rusty and was never very good to begin with, that's why I was confused about this at first before realizing the argument is wrong. Mar 3, 2022 at 21:09