Proposition 10.13.2.label Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$ be equicontinuous, then for any ideal $\sigma \subset \mathfrak{B}(E)$, $\alg$ is a bounded subset of $B_{\sigma}(E; F)$.
Proof. Let $S \in \sigma$ and $U \in \cn_{F}(0)$, then there exists $V \in \cn_{E}(0)$ such that $\bigcup_{T \in \alg}T(V) \subset U$. Since $S$ is bounded, there exists $\lambda \in K$ such that $S \subset \lambda V$. Therefore $\bigcup_{T \in \alg}T(S) \subset \lambda U$, and $\alg$ is bounded in $B_{\sigma}(E; F)$.$\square$