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 $k$.


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?

  • $\begingroup$ Do you want to prove it with the Banach-Steinhaus theorem or do you want to see some alternative proof? $\endgroup$
    – Demophilus
    Commented Nov 1, 2017 at 10:03
  • $\begingroup$ @Demophilus Primarily with the Banach-Steinhaus, but alternative ways are also welcome. $\endgroup$
    – Konstantin
    Commented Nov 1, 2017 at 10:04

1 Answer 1


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 $x =\sum_{k=1}^\infty c_k x_k$ .

  • 1
    $\begingroup$ 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). $\endgroup$ Commented Nov 3, 2017 at 0:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .