# 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 $$||\cdot||_Y$$.

• Maybe you can show $Y$ is isomorphic to $X$, via $\alpha \mapsto \sum_{n=1}^\infty \alpha_n e_n$. Commented Mar 10, 2022 at 11:48
• 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 $Y$. Commented Mar 10, 2022 at 13:56

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:

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.

• 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 $X$). Commented Mar 11, 2022 at 6:44
• Thanks. Can you add a reference? Commented Mar 21, 2022 at 21:54
• @LordOfNumbers I just edit it and add a reference :) Commented Mar 22, 2022 at 5:50