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Let $X$ be a Banach space.

Suppose $X$ has a normalized Schauder basis $\{x_n\}_{n \in \Bbb N}$.

Let $\{y_n\}_{n \in \Bbb N}$ be a sequence in $X$ converging to $0_X$.

For each $n \in \Bbb N$, let $\{a_k^n\}_{k \in \Bbb N} \subset \Bbb C$ be the unique sequence satisfying $$\lim_{m \to \infty} \left \| y_n - \sum_{k=0}^m a_k^n x_k \right\|_X = 0.$$

I want to show that for each $k \in \Bbb N$, we have $a_k^n \rightarrow 0$ as $n \rightarrow \infty$ with respect to the norm in $\Bbb C$.

I am hoping that this result follows from the uniqueness of the Schauder basis representation.

I would like to use this result to complete some other proofs, but I'm not sure how to establish it.

Thank you for your help!

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Since $x$ is a Shauder basis then for all $y\in X$ we have unique representaion $$ y=\sum_{n=1}^\infty a_k(y) x_k $$ It is a standard fact proved in each Banach geometry course, that $a_k:Y\to\mathbb{C}:y\mapsto a_k(y)$ is a bounded linear functional (see e.g. theorem 1.1.3 in Topics in Banach space theory. F. Albiac, N. Kalton). Then for $a_k^n:=a_k(y_n)$ we have $$ \lim\limits_{n\to\infty}a_k^n :=\lim\limits_{n\to\infty}a_k(y_n) =a_k(\lim\limits_{n\to\infty}y_n) =a_k(0_X)=0 $$

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  • $\begingroup$ Actually, this boundedness you refer to is one of the results I was interested in proving. Thank you for the reference! $\endgroup$ – Open Season Apr 16 '14 at 22:07

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