If $B$ is a closed subspace of a Banach space $E$ and $L_n,L$ are bounded linear operators on $E$ with $L_nB⊆B$ and $L_n\to L$, then $LB⊆B$

Question

Let

• $E$ be a $\mathbb R$-Banach space
• $B$ be a closed subspace of $E$
• $(L_n)_{n\in\mathbb N}\subseteq\mathfrak L(E)$ with $$L_nB\subseteq B\;\;\;\text{for all }n\in\mathbb N\tag1$$ and $L\in\mathfrak L(E)$ with $$\left\|L_n-L\right\|_{\mathfrak L(E)}\xrightarrow{n\to\infty}0\tag2.$$

Are we able to conclude $LB\subseteq B$?

This seems to be easy, but I don't find a starting point. Maybe we need to show that $\mathfrak L(B)$ is a closed subspace of $\mathfrak L(E)$ and conclude from that?

Answer 1

For each $x\in B$, we have $$Lx=\lim_{n\to\infty}L_nx.$$ Since $L_nx\in B$ for each $n$ and $B$ is closed, the result follows.