Let $X$ be a Banach space and $(x_n)$, $(y_n)$, $(f_n)$ be bounded sequences in $X$, $X$, $X^*$ respectively such that $f_m(x_n)=\delta_{mn}$ $\forall m,n$ and $\epsilon=\Sigma\|x_n-y_n\|<\infty$. Why must there exist an invertible operator $T:X\rightarrow X$ such that $Tx_n=y_n$ $\forall n$, provided $\epsilon$ is sufficiently small?
I found this on a set problem sheet, so the mention of the $f_n$ must be a hint. Perhaps $T$ is given by composing some linear combination of the $f_n$ with the inverse of some other such combination?
We know that the sequence $(f_n)$ is bounded, say by a positive constant $M$. Then $|1-\Sigma_nf_i(y_n)|<M\epsilon$ for all $i$ - does this help?
By taking linear combinations of the $f_n$, we can map the $(x_n)$ to an arbitrary sequence in $\mathbb{R}$. Maybe the numbers $f_m(y_n)$ are relevant?
Many thanks for any help with this!