# Is the set of compact operators closed with the strong topology?

It is well-known that the space of compact operators over Banach spaces is closed within the norm topology. My question:

Let $$X$$ be a Banach space.

Considering the strong topology (defined by seminorms) on $$\mathcal{L}(X)$$ i.e., $$\{p_x(T):=Tx \text{ for all } T\in \mathcal{L}(X),\, x\in X \}$$. Do we still have the same result over the strong topology.

$$\textbf{Question}$$:

$$(T_n)_{n\in\mathbb{N}}$$ a sequence of compact operators which converges strongly to $$T\in \mathcal{L}(X)$$, is $$T$$ compact ?

• Have you thought about the case where the $T_n$ are 'approximation' operators? (e.g. $T_nx = (x_1,\dots,x_n,0,0,\dots)$ on $l^p$)
– DCM
Jul 10, 2021 at 17:41
• On Hilbert space every bounded operator is a strong limit of a sequence of compact operators. Jul 10, 2021 at 20:08
• The in the two previously mentioned examples one has that every operator is the limit of a sequence of operators with finite-dimensional range (called finite-rank operators), which are a fortiori compact operators. In fact the set of finite-rank operators is dense $\mathcal{L}(X)$ in the strong topology, but there cannot be a sequence of finite-rank operators approximating the identity in the strong topology unless the space $E$ has the bounded approximation property, which not all separable Banach spaces have, but is implied by the existence of a Schauder basis. Jul 12, 2021 at 15:44