In the book 'Classical Banach Spaces I and II' by Lindenstrauss, page $1,$ Chapter $1$, he stated the following:
Let $(X,\|\cdot\|)$ be a Banach space with a (Schauder) basis $(x_n)_{n=1}^{\infty}.$ For every $x = \sum_{n=1}^{\infty}a_nx_n$ in $X$ the expression
$$|||x||| = \sup_{n}\left\Vert \sum_{i=1}^n a_ix_I \right\Vert$$ is finite. Evidently, $|||\cdot|||$ is a norm on $X$ and $\|x\| \leq |||x|||$ for every $x \in X.$ A simple argument shows that $X$ is complete also with respect to $|||\cdot|||.$
Question: How to use a simple argument to show that $(X, |||\cdot|||)$ is complete? I manage to prove all statements before it. Below is my partial attempt (actually is not even partial).
Suppose that $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence in $(X, |||\cdot|||).$ Since $\|\cdot\| \leq |||\cdot|||$ and $(X,\|\cdot\|)$ is a Banach space, there exists an $x \in X$ such that $\lim_{n}\|x_n - x\| = 0.$
I think I should I show that $x_n$ converges to $x$ in the norm $|||\cdot|||.$ But I have no idea how to show at all.
Any hint would be appreciated.
$\newcommand{\nv}[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1\right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}\newcommand{\veps}{\varepsilon}$Let $\{y_n\}_{n=1}^{\infty}$ be a Cauchy sequence in $(X,\nv{\cdot})$, and for every $n\in\mathbf{N}$ write in the Schauder basis, $$y_n=\sum_{i=1}^{\infty}a^{(n)}_ix_i.$$ For any fixed $i\in\mathbf{N}$ that $$|a_i^{(n)}-a_i^{(m)}|\|x_i\|=\|(a_i^{(n)}-a_i^{(m)})x_i\|\leq\left\|\sum_{j=1}^{i}(a_j^{(n)}-a_j^{(m)})x_j\right\|+\left\|\sum_{j=1}^{i-1}(a_j^{(n)}-a_j^{(m)})x_j\right\|\leq 2\nv{y_n-y_m}.$$ Since $x_i\neq 0$, we deduce that $\{a_i^{(n)}\}_{n\in\mathbf{N}}$ is a Cauchy sequence in $\mathbf{R}$, and so $a_i^{(n)}\to a_i$ for some sequence of scalars $a_i$.
Now fix $\veps>0$ . Since $\{y_n\}$ is Cauchy with respect to $\nv{\cdot}$ there exists some $N_0\in\mathbf{N}$ such that for all $n,m\geq N_0$ $\nv{y_n-y_m}<\veps/3$. Now fix any $m\geq N_0$, $M\in\mathbf{N}$, and for all $n\in\mathbf{N}$ define $$z_n^{m,M}=\sum_{i=1}^{M}(a_i^{(n)}-a_i^{(m)})x_i.$$ We see that $\|z_n^{m,M}\|\leq\nv{y_n-y_m}<\veps/3$ for all $n\geq N_0$. Since $z_n^{m,M}\to z^{m,M}:=\sum_{i=1}^{M}(a_i-a_i^{(m)})x_i$ as $n\to\infty$ we see that $\|z^{m,M}\|\leq\veps/3$. Since $M$ was arbitrary we see that for all $m\geq N_0$, \begin{align} \sup_{M}\left\|\sum_{i=1}^{M}(a_i - a_i^{(m)})x_i\right\|\leq\frac{\veps}{3}.\tag{1} \end{align} Now we show that $\sum_{i=1}^{\infty}a_ix_i$ converges in $(X,\|\cdot\|)$. Fix any $m\geq N_0$ and since $y_m\in X$ we see that $\sum_{i=1}^{\infty}a_i^{(m)}x_i$ is convergent in $X$, and so the tail of the sequence can be made arbitrarily small. Rigorously, there exists some $N_1\in\mathbf{N}$ such that $M_2>M_2>N_1$ implies $$\left\|\sum_{i=M_1}^{M_2}a_i^{(m)}x_i\right\|<\frac{\veps}{3}.$$ The triangle inequality gives us $$ \left\|\sum_{i=M_1}^{M_2}a_ix_i\right\| = \left\|\sum_{i=M_1}^{M_2}a_i^{(m)}x_i + \sum_{i=1}^{M_2}(a_i-a_i^{(m)})x_i - \sum_{i=1}^{M_1-1}(a_i-a_i^{(m)})x_i\right\|\leq\veps.$$ Hence $\{\sum_{i=1}^{n}a_ix_i\}_{n\in\mathbf{N}}$ is a Cauchy sequence in $(X,\|\cdot\|)$; since $X$ is complete, $y:=\sum_{i=1}^{\infty}a_ix_i$ converges in $X$.
In light of the above convergence and (1) we see that $\nv{y-y_m}\leq\veps/3$ for all $m\geq N_0$. Since $\veps$ was arbitrary $y_m\to y$ as $m\to\infty$ with respect to $\nv{\cdot}$, which establishes the completeness of $(X,\nv{\cdot})$.