Show that $(Y,||\cdot||_Y)$ is a Banach space 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 $||\cdot||_Y$.
 A: 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 $\|\Lambda^{-1}\|\leq C$.
Remark:

*

*above proposition is not trivial. It can be found in text books and I will add references later.

*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.
