Question about Schauder basis The question is : Let $B$ be a Banach space and suppose $\{x_n\}$ the Schauder basis and $M$ be the
space of sequence of scalars $\{a_n\}$ such that the sup norm of power series of $a_n x_n$ converges. 
I am able to prove that B is separable. But I have two more questions.


*

*How to construct a bijection from $B$ to $M$. 

*For any basis $(x_n)$ of a Banach space B, why the coefficient $f$ always continuous so that $f \in B^*$?
 A: I assume that by "sup norm of power series" you mean the space $M = \{(\alpha_n)_{n=1}^\infty\;|\; \sum_{n=1}^\infty a_nx_n \text{ converges}\}$ with the norm $||(\alpha_n)_{n=1}^\infty||_M := \sup_{m} ||\sum_{i=1}^m \alpha_ix_i ||$.
Then the answers are:


*

*The bijection is given: $T:x = \sum_{i=1}^\infty \alpha_i x_i \mapsto (\alpha_n)_{n=1}^\infty$. This function is continuous because of $||T^{-1}(\alpha_n)_{n=1}^\infty|| = ||\sum_{i=1}^\infty \alpha_ix_i|| \leq ||(\alpha_n)_{n=1}^\infty||_M$ and the open mapping theorem (M is a Banach space which can be verified through computation).
Note that this is well defined, since $\{x_i\}_{i=1}^\infty$ is a Schauder basis.

*The coefficient function $f_i:x \mapsto \alpha_i$ is continuous since the spaces $B$ and $M$ are isomorphic (by $T$) and the Projection operators $P_m:x \mapsto \sum_{i=1}^m \alpha_ix_i$ are continuous (since $P_m$ corresponds to the map $(\alpha_n)_{n=1}^\infty \mapsto (\beta_n)_{n=1}^\infty$ with $\beta_n = \alpha_n$ for $n \leq m$ and $\beta_n = 0$ for $n > m$, and this map is continuous by a trivial computation) and $f_i = P_i-P_{i-1}$ for $i > 1$ and $f_1 = P_1$.

