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A sequence $\{x_k\}$ in a Banach space $X$ is a Schauder basis of $X$ if every element $x\in X$ has a unique representation $$ x = \sum_{k=1}^\infty c_k x_k $$ with the series converging in the norm of $X$. A basis $\{x_k\}$ is an unconditional basis if the series $\sum_{k=1}^\infty d_k x_k$ converges for all sequences $\{d_k\}$ satisfying $|d_k|\leq |c_k|$ for each $k$.

QUESTION:

If $\{x_k\}$ is an unconditional basis, the Banach-Steinhaus theorem implies that there is a constant $M$ such that $$ \left\|\sum_{k=1}^\infty d_k x_k\right\|\leq M \left\|\sum_{k=1}^\infty c_k x_k\right\| $$ whenever $|d_k|\leq |c_k|$ for all $k$. How is it possible to show it?

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  • $\begingroup$ Do you want to prove it with the Banach-Steinhaus theorem or do you want to see some alternative proof? $\endgroup$ – Demophilus Nov 1 '17 at 10:03
  • $\begingroup$ @Demophilus Primarily with the Banach-Steinhaus, but alternative ways are also welcome. $\endgroup$ – Konstantin Nov 1 '17 at 10:04
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One attempt that doesn't use the Banach-Steinhaus theorem. I'm still wandering why the author mentioned the theorem.

1) There are $\{\lambda_k\}_{k=1}^\infty$ such that $|\lambda_k|\leq\infty$ and $$ \sum_{k=1}^\infty d_k x_k = \sum_{k=1}^\infty \lambda_k c_k x_k $$

2) We could define a linear operator $T_\lambda$ and it's bounded by the definition of the unconditional basis $$ T_\lambda x = \sum_{k=1}^\infty \lambda_k c_k x_k < \infty $$

3) By the definition of a bounded operator $\|T_\lambda x\| \leq M\|x\| $

4) The result follows by observing that $x =\sum_{k=1}^\infty c_k x_k$ .

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  • $\begingroup$ Two problems. First: $T_{\lambda}$ is not bounded by definition (at least the one you're giving), it is only well-defined. Second: you want to have one constant $M$ for all sequences $d_k$ and $c_k$ and you only proved it for $d_k = \lambda_kc_k$. You need to prove that $||T_\lambda|| \le M$ for any $\lambda \in l^{\infty}$ (and that's where Banach-Steinhaus would come in handy). $\endgroup$ – Petr Naryshkin Nov 3 '17 at 0:04

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