To prove it is a Schauder basis? In Classical Banach Spaces I and II by Lindenstrauss and Tzafriri
proposition 1.a.3
Let $\{x_n\}_{n=1}^\infty$ be a sequence of vectors in $X.$ Then $\{x_n\}_{n=1}^\infty$ is a Schauder basis of $X$ if and only if the following three conditions hold.

*

*$x_n \neq 0$ for all $n.$

*There is a constant $K$ so that, for every choice of scalars $\{a_i\}_{i=1}^\infty$ and integers $n <m,$ we have
$$\|\sum_{i=1}^n a_i x_i\| \leq K \|\sum_{i=1}^m a_i x_i\|.$$

*The closed linear span of $\{x_n\}_{n=1}^\infty$ is all of $X.$
Then in the rest of the book to show that say $\{x_i\}_{i=1}^\infty$ is a Schauder basis of a Banach space X, he always say by proposition 1.a.3 we have to show that the operators $\{P_n\}_{n=1}^\infty,$ defined by $P_n x = \sum_{i=1}^n x_i^*(x) x_i,$ are uniformly bounded
Q1 How proposition 1.a.3 enables us to do that?
Any help will be appreciated
 A: Probably the authors mean that when $P_n$ are uniformly bounded then $(x_i)$ is Schauder basis on $\overline{\text{span}\{x_i:\,i\geq 1\}}$, since it cannot be on $X$ necessarily.
For a counterexample, consider $X=\ell_{\infty}$ with the sup norm. Then, $\ell_\infty$ does not admit a Schauder basis since it is not separable. On the other hand, if $x_n$ is the sequence with $x_n(k)=0$ when $k\neq n$ and $x_n(k)=1$ when $n=k$, then the corresponding $P_n$'s are uniformly bounded and $(x_n)$ is a Schauder basis on $c_{0}$ not on $\ell_{\infty}$.
Now, suppose that the $P_n$'s are uniformly bounded, then $(x_n)$ is a Schauder basis on $\overline{\text{span}\{x_n:\,n\geq 1\}}$. To show this, observe now that we are only need to prove the second condition of the proposition. Let $n<m$ and put $K=\sup_{n}||P_n||<\infty$. Observe that
$$
\begin{align}
\sum_{k=1}^{n}x_k^*(x)x_k=P_n\biggl(\sum_{k=1}^{m}x_k^*(x)x_k\biggr).
\end{align}
$$
Therfore,
$$\begin{align}
\biggl|\biggl|\sum_{k=1}^{n}x^*(x)x_k\biggr|\biggr|&=\biggl|\biggl| P_n\biggl(\sum_{k=1}^{m}x_k^*(x)x_k\biggr)\biggr|\biggr|\\
&\leq ||P_n||\cdot \biggl|\biggl|\sum_{k=1}^{m}x_k^*(x)x_k\biggr|\biggr|\\
&\leq K\cdot \biggl|\biggl|\sum_{k=1}^{m}x_k^*(x)x_k\biggr|\biggr|\\
\end{align}
$$
And condition 2 is satisfied with basis constant $K$.
