Let $(X,||\cdot||_X)$ be a Banach space and $(e_n)_{n\in\mathbb{N}}$ a Schauder basis of X. How can I prove that $Y:=\{\alpha:\mathbb{N}\to\mathbb{R} \space| \lim_{N\to\infty}\sum_{n=0}^N\alpha_n e_n \;\text{exists}\}$ is a Banach space with the norm $||\alpha||_Y:=\sup_{N}||\sum_{n=0}^N\alpha_n e_n||_X$.

Especially how can I show that $||\cdot||_Y$ is complete. So far I could only show that if we have a Cauchy sequence $(\alpha^{(l)})_{l\in\mathbb{N}}\subset Y$ then we have pointwise $\alpha^{(l)}_j\to\alpha_j$ as $l\to\infty$ for some $a_j\in\mathbb{R}$. This way one can define $\alpha:\mathbb{N}\to\mathbb{R}$. But I don't know how to show $\alpha^{(l)}\to\alpha$ as $l\to\infty$ in $||\cdot||_Y$.

  • $\begingroup$ Maybe you can show $Y$ is isomorphic to $X$, via $\alpha \mapsto \sum_{n=1}^\infty \alpha_n e_n$. $\endgroup$
    – Landau
    Commented Mar 10, 2022 at 11:48
  • 1
    $\begingroup$ Outline: Take your Cauchy sequence, and choose $r$ so that $\Vert \sum_{i=1}^n (\alpha_i^{(p)}-\alpha_i^{(r)})x_i\Vert$ is small for $p>r$. You may also make $\Vert \sum_{i=n}^m \alpha_i^{(r)} x_i\Vert$ small for $n,m$ sufficiently large. This implies $\Vert\sum_{i=n}^m \alpha_ix_i\Vert$ can be made small. It then follows your sequence converges to $(\alpha_i)$ in $Y$. $\endgroup$ Commented Mar 10, 2022 at 13:56

1 Answer 1


Define $$\Lambda: Y \to X: \alpha\mapsto \lim_{N\to \infty}\sum_{n=1}^N \alpha_ne_n.$$ We show $\Lambda$ is an isomorphism.

  • By the definition of $Y$-norm, $\Lambda$ is contractive, hence continuous.
  • $\Lambda$ is onto, since {$e_n$} is a Schauder basis.
  • $\Lambda$ is injective. Assume $\Lambda \alpha =0$, that is $\sum_{n=1}^\infty \alpha_ne_n=0$. Again, since {$e_n$} is a Schauder basis , the $\omega$-linear independence of $\{e_n\}$ gives $\alpha_n=0$ for all $n$.
  • It remains to show $\Lambda^{-1}$ is bounded. This is the key step. The following proposition is an equivalent definition of Schauder basis.

Proposition: $\{e_n\}$ is Schauder basis if and only if it is $\omega$-linear independence basis, and there exists a constant $C>0$, such that $\|P_N\|\leq C$. Here $P_N$ is the projection $$P_N:X\to X: x=\sum_{n=1}^\infty \alpha_n e_n \mapsto\sum_{n=1}^N\alpha_ne_n.$$

With this proposition we see $\|\Lambda^{-1}\|\leq C$.


  1. above proposition is not trivial. It can be found in text books and I will add references later.
  2. You can't use Banach inverse operator theorem to replace the fourth step, since we don't know $Y$ is Banach yet.

One reference is Topics in Banach spaces theory, Proposition 1.1.9. As David Mitra said, it this proposition may equivalent with that $X$ is isometric to $Y$. I'm sorry for not checking it when I post this answer. Anyway, this book will provide a self-contained proof to your question.

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    $\begingroup$ Showing $Y$ is complete can be done without the result that the $P_n$ are uniformly bounded (in the treatments I have seen, the boundedness result is proved using the result that $Y$ is isomorphic to $X$). $\endgroup$ Commented Mar 11, 2022 at 6:44
  • $\begingroup$ Thanks. Can you add a reference? $\endgroup$ Commented Mar 21, 2022 at 21:54
  • $\begingroup$ @LordOfNumbers I just edit it and add a reference :) $\endgroup$
    – Landau
    Commented Mar 22, 2022 at 5:50

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