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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!

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1 Answer 1

up vote 3 down vote accepted

Let $z_n:=x_n-y_n$. The problem is to find an operator $S\colon X\to X$ such that $Sx_n=z_n$ for each $n$ and $S-I$ is invertible.

Define $S(x):=\sum_{n=1}^{+\infty}f_n(x)z_n$: it's a linear operator, such that $S(x_n)=z_n$. Moreover, $$\lVert Sx\rVert\leqslant \varepsilon\lVert x\rVert\sup_n\lVert f_n\rVert,$$ hence when $\varepsilon\sup_n\lVert f_n\rVert<1$, this $S$ does the job.

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