Separating space of a linear map ($S(T)=\{y:\exists x_n\to 0,T{x}_n\to y\}$)

For an everywhere-defined linear map $T:E\to F$ between Banach spaces, its separating space measures the possible limiting defect in continuity at zero. It is a closed linear subspace: linear combinations combine witnessing sequences, and diagonal choices with $\Vert x_j\Vert <1/j$ and $\Vert T{x}_j-y_j\Vert <1/j$ handle a convergent sequence $y_j$ of defects. The closed graph theorem gives $T$ continuous exactly when $S(T)=\{0\}$, since a limit $(x_n,T{x}_n)\to (x,y)$ has $y-Tx\in S(T)$.

Let $S:E\to Y$ be linear between Banach spaces, let $T_n:E\to E$ be bounded, and let $U_n:Y\to Z_n$ be bounded. If $U_{n}S{P}_m$ is continuous whenever $m>n$, then $U_{n}S{P}_n$ is continuous for all sufficiently large $n$. After rescaling the bounded operators, assume their norms are at most one. If the conclusion failed, discontinuity would allow successively tiny vectors $x_j$ whose diagonal images dominate all previous contributions. Choose their norms also to control the continuous maps $U_{n_i}S{P}_{n_i+1}$ on subsequent tails. The convergent sum $x={\sum }_j{P}_{n_j}{x}_j$ would then satisfy $\Vert U_{n_j}Sx\Vert >j$ despite $\Vert U_{n_j}Sx\Vert \le \Vert Sx\Vert$, a contradiction. The tail is factored through $P_{n_i+1}$ before applying $S$, so no continuity of $S$ is assumed.