I'm currently reading a non harmonic Fourier series book by Robert M Young, and came across the following, Theroem 3 on page 23. \ \textbf{Theorem 3}: If $\{x_n\}$ is a basis for a Banach Space $X$ and if $\{f_n\}$ is the associated sequence of coefficient functionals, then each $f_n \in X^{*}$, the space of bounded linear functionals on $X$. There exists a constant $M$ such that: \begin{equation} 1 \leqq \Vert x_n \Vert \cdot \Vert f_n \Vert \leqq M \quad (n=1,2,3,..) \end{equation}
\textbf{Proof}: Introduce the vector space $Y$ consisting of those sequences of scalars $\{c_n\}$ for which the series $\sum_{n=1}^{\infty}c_n x_n$ is convergent in $X$. If $\{c_n\} \in Y$ then the number \begin{equation} \Vert \{c_n\} \Vert = \sup_{n}\Vert \sum_{i=1}^{n} c_i x_i \Vert \end{equation} satisfies all properties of the norm. First we must show that $Y$ is a Banach Space with this norm. We realise that since $\Vert \{ c_n \} \Vert = \sup_{n} \Vert \sum_{i=1}^{n} c_i x_i \Vert $, then $\Vert \{ c_n \} \Vert \geq 0$.After
Next we will show that $X$ and $Y$ are isomorphic.The map $T: Y \rightarrow X$ defined by \begin{equation} \{c_n\} \rightarrow \sum_{n=1}^{\infty} c_n x_n \end{equation} is a linear mapping since $\{x_n\}$ is a basis for $X$, and it is also one to one and onto (Bijective).
Since \begin{equation} \Vert \sum_{n=1}^{\infty} c_n x_n \Vert \leqq \sup_{n} \Vert \sum_{i=1}^{n} c_i x_i \Vert \end{equation} it follows that the mapping $T$ must be continuous, and the open mapping theorem guaranteed that $T^{-1}$ is also continuous.This proves that $X$ and $Y$ are isomorphic. Suppose now that $x=\sum_{n=1}^{\infty}c_n x_n$ is a fixed, arbitrary element of $X$. Then for every $n$, \begin{equation} \vert f_n(x) \vert =\vert c_n \vert =\frac{\Vert c_n x_n \Vert}{\Vert x_n \Vert} \leqq \frac {\Vert \sum_{i=1}^{n} c_i x_i \Vert + \Vert \sum_{i=1}^{n-1} c_i x_i \Vert }{\Vert x_n \Vert} \end{equation} \begin{equation} \leqq \frac{2 \sup_{n} \Vert \sum_{i=1}^{n} c_i x_i \Vert}{\Vert x_n \Vert}= \frac{2\Vert T^{-1}x\Vert}{\Vert x_n \Vert} \leqq \frac{2\Vert T^{-1}\Vert \Vert x \Vert}{\Vert x_n \Vert} \end{equation}
This proves that each $f_n$ is bounded, and since it is linear, therefore continuous, and that $\Vert f_n \Vert \leqq \frac{2 \Vert T^{-1} \Vert}{\Vert x_n \Vert}$
Choosing $M=2\Vert T^{-1} \Vert$, we have \begin{equation} \Vert x_n \Vert \cdot \Vert f_n \Vert \leqq M \end{equation} for every $n$. Finally, since $1=f_n(x_n)$, it is true that \begin{equation} f_n(x_n) \leqq \Vert f_n \Vert \cdot \Vert x_n \Vert \end{equation}
This completes the proof. $\blacksquare$ \bigskip \newline
What I'm having trouble undertanding is how I can show that Y is a Banach space equipeed with this norm? He leaves it as an excercise for the reader, and the proof I uploaded here has a few of my own touches to it (drawn out to help me understand) but I can't seem ti figure this bit out. Any advice?
Let ${(c_n^{1})},{(c_n^{2})},...$ be a Cauchy sequence. Let $n_0$ be such that $ \|\sum\limits_{k=1}^{n} (c_k^{1}x_k-c_k^{m}x_k)\|<\epsilon$ for all $n$ for $l,m \geq n_0$. $\cdots$ (1)
Then, for each $n$, $\|c_n^{1}x_n-c_n^{m}x_n\| <2\epsilon$ for $l,m \geq n_0$. Since $x_n \neq 0$ we see that $(c_n^{l})_{l\geq 1}$ is Cauchy for each fixed $n$. Let $c_n$ be the limit of this sequence as $l \to \infty$. Now we get $ \|\sum\limits_{k=1}^{n} (c_n^{1}x_k-c_n^{m}x_k)\|\leq \epsilon$ for $l \geq n_0$ by letting $m \to \infty$ in (1). Is it clear now that ${c_n^{1}},{c_n^{2}},...$ converges to $(c_n)$ in the norm of $Y$?