I recently encountered the following somewhat random-looking problem in my research. At first I thought that should not be too hard, but now, the more I think about it, the more interesting it seems.
So I would like to show (or see a counterexample to) the following statement:
Let $\mathscr{S} \subseteq \mathrm{B}(E)$ be a set of injective bounded operators on a Banach space $E$. Suppose that for each $x \in E$, the set $\{Tx \mid T \in \mathscr{S}\} \subseteq E$ is contained in a finite-dimensional subspace. Then $\mathscr{S}$ is contained in a finite-dimensional subspace of $\mathrm{B}(E)$.
This smells like a uniform boundedness principle, as the assumption is that the set $\{Tx \mid T \in \mathscr{S}\}$ is bounded with respect to the fine bornology on $E$.
Edit2: There was an example by Matthew Daws where $\mathscr{S} = \{T_f \mid | f \in E^*, \|f\| \leq 1\}$ with $T_fx = f(x)x_0$ for some fixed nonzero $x_0 \in E$. This has infinite-dimensional span but $\{T x \mid T \in \mathscr{S}\} = \mathbb{C} x_0$ is one-dimensional.
Edit3: After some further thoughts, I now believe that the correct assumption to make is that all the operators are injective (although this can be weakened in obvious ways). In particular, this excludes the example of Matthew Daws.
A similar (though somewhat different and not obviously equivalent) version of the question is then the following: Suppose that $T_1, \dots, T_n$ are injective operators from a vector space $E$ to a vector space $F$, which are linearly independent in the space of all such operators. Then there exists $x \in E$ such that $T_1 x, \dots, T_n x$ are linearly independent.
Let $\{T_0, T_1, T_2, \dots\}$ be some infinite sequence inside $\mathscr{S}$.
As remarked by Narutaka Ozawa, we can define the sets $$A_n = \left\{x \in X \mid \forall k \in \mathbb{N} \,\exists \alpha_1, \dots, \alpha_n : T_k x = \sum_{i=0}^n \alpha_i T_ix \right\}.$$ Then our hypothesis implies that each $x$ is contained in some $A_n$, hence their union is all of $E$. I am not quite sure how to show that the $A_n$ are closed yet, but assuming this, one of the Baire theorems gives that there exists $n_0 \in \mathbb{N}$ such that $A_{n_0}$ has non-empty interior. Hence we have $$ \forall k \in \mathbb{N} \, \forall x \in A_{n_0}\, \exists \alpha_1, \dots, \alpha_{n_0} : T_k x = \sum_{i=0}^{n_0} T_i x.$$ The remaining task is now to swap the quantifiers so that the $\alpha_i$ do not depend on $x$ (once we achieved this, we are done, as because $A_{n_0}$ has non-empty interior, operators are determined by their values on $A_{n_0}$).
So far we didn't use any assumption on $\mathscr{S}$ (e.g., that the $T_n$ are injective or that the sequence converges) and a variation of the example of Matthew shows that we can't get away without this (or some other) additional hypothesis. As Narutaka wrote, one can use the convergence hypothesis to obtain the additional restriction $\sum_{i=0}^n |\alpha_i| \leq 1+\varepsilon$ on the coefficients (assuming also that $T_0 = \mathrm{id}$). But it is not clear to me yet how this helps with the problem.
Edit: In a previous version of this post was concerned with the following statement:
Let $T_1, T_2, \dots$ be a sequence of operators on a Banach space $E$ converging strongly to the identity (they are surjective isometries, not sure if that is relevant). Suppose that for each $x \in E$, the sequence $T_1x, T_2x, \dots$ is contained in a finite-dimensional subspace. Then there exists a sequence $\lambda_1, \lambda_2, \dots$ in $\mathbb{C}$ (necessarily converging to one), a subspace $V \subset E$ of finite codimension and $n_0 \in \mathbb{N}$ such that for all $n \geq n_0$ and all $x \in V$, we have $T_n x = \lambda_n x$.
As hinted at by Narutaka Ozawa, this statement is (rather obviously) false, a counterexample is whenever the sequence is contained in a finite-dimensional subspace of $\mathrm{B}(E)$. Now it seems to me that this is corrected by the above version.
