Let $E,F$ be Banach spaces and let $U\subseteq F$ be a subspace and let $(T_n)$ be a sequence of bounded operators between $E$ and $F$ such that $T_n(E)\subseteq U$ for each $n\in\mathbb N$.
If there exists a bounded operator $T:E\to F$ such that $T_n\to T$ in $\mathcal{L}(E,F)$ and $TE\subseteq U$, then is the convergence true in $\mathcal{L}(E,U)$ as well?
My thoughts: I think the result should directly follow from closed graph theorem, although I do not know how to formally show this. The source of the question above is the following observation:
Let $A$ be a closed operator on a Banach space $E$ and let $0\in\sigma(A)$ be pole of order one. Then $\lambda R(\lambda,A)$ converges to the spectral projection $P$ as $\lambda\to 0$ in $\mathcal{L}(E,D(A))$.
My teacher said that it is sufficient to show that $\lambda A R(\lambda,A)$ converges to $AP$ in $\mathcal{L}(E)$. However, I wondered if one can alternatively use closed graph theorem?
Edit: As mentioned in the comments, in order for the first question to be a generalization of the second, one should have that $U$ is is equipped with a stronger norm (compared to $F$) under which $U$ is complete.