Show that $x_n \notin \overline { \text{span} (x_1,\dots, x_{n-1},x_{n+1},x_{n+2}, ...)}$ Let X be a banach spaces and  $(x_n)_{n\geq1}$ be a Schauder basis in $X$. 
$1)$ $\forall x\in X,$ $x=\sum_{n=1}^\infty \lambda_n x_n$ and for all $n\geq 1$, we define $x_n^*=\lambda_n$, show that $x_n^* \in X^*$ . 
$2)$ And show that $x_n \notin \overline { \text{span} (x_1,\dots, x_{n-1},x_{n+1},x_{n+2}, ...)}$ 
I can manage $(1)$ but help me for $(2)$ please thank you.
 A: The linear functionals $x^*_n$ satisfy $x^*_n(x_m)=\delta_{n,m}$. Hence, if $X_{n-}:=\operatorname{span}\big(x_k:k\in\mathbb{N}\setminus\{n\}\big)$, then $x^*_n(X_{n-})=\{0\}$ and by continuity (this is item (1) that the OP says can manage) $x^*_n(\overline{X_{n-}})=\{0\}$.  This implies that $x_n\in X\setminus \overline{X_{n-}}$.

For completeness, I present a proof of (1) which I learned from Thomas Schlumpretch.
Assume $(X,\|\;\|)$ is a Banach space and that $(\boldsymbol{e}_n:n\in\mathbb{N})$ is a Schauder basis for $X$. Thus, for any $x\in X$ there is a unique sequence $(a_n(x):n\in\mathbb{N})$ such that
$$ \lim_n\|x-\sum^n_{k=1}a_k(x) \boldsymbol{e}_n\|=0$$

*

*It is obvious that $(\boldsymbol{e}_n:n\in\mathbb{N})$ is a linearly independent sequence.

*For each $n\in\mathbb{N}$, $a_n:X\rightarrow\mathbb{C}$ is a linear map. We call the maps $a_n$'s the coordinate functionals.

For each $n\in\mathbb{N}$, define $P_n:x\mapsto \sum^n_{k=1}a_n(x) \boldsymbol{e}_n$. Each map $P_n$ is linear and has finite range. In fact

*

*$\operatorname{dim}(P_n(X))=n$

*$P_n\circ P_m=P_m\circ P_n =P_{\min(n,m)}$

*$\lim_{n\rightarrow\infty}\|x-P_nx\|=0$.

We before to the maps $P_n$'s as the canonical projections.
The following technical results will be the hammers used to hit the nail

Lemma 1: Let  $(X,\|\;\|)$ be a normed space (not necessarily a Banach space). Suppose there is a sequence  $(P_n:n\in\mathbb{N})$  of bounded linear maps satisfying 1,2 and 3 above. If $\boldsymbol{e}_1\in P_1(X)\setminus\{0\}$, and $\boldsymbol{e}_n\in \big(P_n(X)\cap\operatorname{ker}(P_{n-1})\big)\setminus\{0\}$ for $n>1$, then for each $x\in X$ there exists a unique sequence $(a_n:n\in\mathbb{N})$ in $\mathbb{C}$ such that
$$ \lim_n\big\|x-\sum^n_{k=1}a_k \boldsymbol{e}_k\big\|=0$$

Proof: For $x\in X$,
\begin{align}
P_{n-1}(P_n(x)-P_{n-1}(x))&=P_{n-1}(x)-P_{n-1}(x)=0\\
P_n(x)-P_{n-1}(x)&=P_n(P_n(x)-P_{n-1}(x))\in P_n(X)
\end{align}
Hence $P_n(x)-P_{n-1}(x)\in P_n(X)\cap\operatorname{ker}(P_{n-1})$. Notice that $P_{n-1}(X)$ is a proper subspace of $P_n(X)$ of codimension $1$, and that the choice of $e_n$ yields
$$P_n(X)=P_{n-1}(X)\oplus\operatorname{span}(\boldsymbol{e}_n)$$
Consequently $P_n(x)-P_{n-1}(x)=a_n\boldsymbol{e}_n$ for some $a_n\in\mathbb{C}$. Setting $P_0=0$ we obtain by that
$$\|x-P_n(x)\|=\big\|x-\sum^n_{k=1}(P_j(x)-P_{j-1}(x))\big\|=\|x-\sum^n_{k=1}a_k\boldsymbol{e}_k\|\xrightarrow{n\rightarrow\infty}0$$
To prove that $(a_n:n\in\mathbb{N})$ is unique, suppose $x=\sum^\infty_{n=0}b_n\boldsymbol{e}_n$ (convergence in $\|\;\|$). Notice that  for $k\leq m$, $P_m(\boldsymbol{e}_k)=\boldsymbol{e}_k$; and for $m< k$, $P_m(\boldsymbol{e}_k)=P_m(P_{k-1}(\boldsymbol{e}_k))=P_m(0)=0$. Then, by continuity of the maps $P_n$, for each $m\in\mathbb{N}$
$$ a_m\boldsymbol{e}_m=(P_m-P_{m-1})(x)=\lim_n(P_m-P_{m-1})\big(\sum^n_{k=1}b_k\boldsymbol{e}_k\big)=b_m\boldsymbol{e}_m$$
Hence, $a_m=b_m$. $\Box$

Lemma 2: Suppose $(X,\|\;\|)$ is a normed space and assume the $(\boldsymbol{e}_n:n\in\mathbb{N})\subset X$ is such that for each $x\in X$, there is a unique sequence $(a_n:n\in\mathbb{N})\subset\mathbb{C}$ such that
$$\lim_n\big\|x-\sum^n_{k=1}a_k \boldsymbol{e}_k\big\|=0$$
If the canonical projections $P_n$ defined above are uniformly bounded, i.e.  $C:=\sup_{n\in\mathbb{N}}\|P_n\|<\infty$, then $(\boldsymbol{e}_n:n\in\mathbb{N})$ is a Schauder basis for the Banach-Hausdorff completion $(\overline{X},\|\;\|_{\overline{X}})$ of $(X,\|\;\|)$.

Remark: Recall that for any normed space $(X,\|\;\|)$ there is a Banach space $(\overline{X},\|\;\|_{\overline{X}})$ and a norm isometry $\iota:X\rightarrow\overline{X}$ such that $\iota(X)$ is dense in $\overline{X}$. The space $\overline{X}$ is unique up-to isometries. We call such space $\overline{X}$ the Banach-Hausdorff completion of $X$.
Proof: Since each $P_n$ has finite range, the extension $\overline{P_n}:\overline{X}\rightarrow\overline{X}$ of $P_n$ satisfies $\overline{P_n}(\overline{X})=\operatorname{span}(\boldsymbol{e}_k:1\leq k\leq n)$ which is closed in $\overline{X}$ (for it is a finite dimensional subspace). Then, $\overline{P_n}$ satisfies properties (1) and (2) in Lemma 1. As for property (3), if $x\in\overline{X}$ and $(x_k:k\in\mathbb{N})\subset X$ is such that $\|x_k-x\|_{\overline{X}}$, then
$$\begin{align}
\|x-\overline{P_n}(x)\|_{\overline{X}}&\leq \|x-x_k\|_{\overline{X}}+\|P_n(x-x_k)\|_{\overline{X}}+\|P_n(x_k)-x_n\|\\
&\leq \|x-x_k\|_{\overline{X}}+C\|x-x_k)\|_{\overline{X}}+\|P_n(x_k)-x_n\|\xrightarrow{k\rightarrow\infty}0
\end{align}
$$
The conclusion follows by applying Lemma 1 to $\overline{X}$. $\Box$
We now state and prove the main result:

Theorem: Let $(X,\|\;\|)$ be a Banach space with a Schauder basis $(\boldsymbol{e}_n:n\in\mathbb{N})$ with coordinate functionals $(a_n:n\in\mathbb{N})$ and canonical projections $(P_n:n\in\mathbb{N})$. Then, $P_n$ is bounded for every $n$; furthermore,
$$C:=\sup_{n\in\mathbb{N}}\|P_n\|<\infty,$$
$a_n\in X^*$ (the dual of $X$) for each $n$, and
$$\|a_n\|_{X^*}=\frac{\|P_n-P_{n-1}\|}{\|\boldsymbol{e}_n\|_{X}}\leq \frac{2C}{\|\boldsymbol{e}_n\|_{X}}$$

Proof: For each $x\in X$ define
$$
\newcommand\vertiii[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 
    \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}
 \vertiii{x}=\sup_{n\in\mathbb{N}}\|P_nx\|_{X}
$$
Since $\|x\|_X=\lim_n\|P_nx\|_X$, it follows that  $\|x\|_X\leq \vertiii{x}$; moreover, it is straight forward to check that $\vertiii{\;}$ defines a norm on $X$. The operator induced norm of each $P_n$ with respect to $\vertiii{\;}$ satisfies
\begin{align}
\vertiii{P_n}&=\sup_{\vertiii{x}=1}\vertiii{P_nx}=\sup_{\vertiii{x}=1}\sup_{m\in\mathbb{N}}\|P_mP_nx\|_X\\
&=\sup_{\vertiii{x}=1}\sup_{m\in\mathbb{N}}\|P_{\min(m,n)}x\|_X=\sup_{\vertiii{x}=1}\max_{1\leq m\leq n}\|P_nx\|_X\leq\vertiii{x}\leq 1
\end{align}
This means that the canonical projections $P_n$ are uniformly bounded operators on $(X,\vertiii{\,})$. For any $x\in X$
$$
\vertiii{x-P_nx}=\sup_m\|P_mx-P_{\min(m,n)}x\|_X=\sup_{m\geq n}\|P_mx-P_nx\|_X\xrightarrow{n\rightarrow\infty}0$$
Let $(\overline{X},\vertiii{\;}_{\overline{X}})$ denote the Banach completion of $X$, and for each $n\in\mathbb{N}$, let $\overline{P_n}$ be the unique extension of $P_n$ to $\overline{X}$. Lemma 1 and Lemma 2 applied to $(X,\vertiii{\;})$ with $(\boldsymbol{e}_n:n\in\mathbb{N})$ and $(P_n:n\in \mathbb{N})$, the sequence $(\boldsymbol{e}_n:n\in\mathbb{N})$ is a Schauder basis for the Banach completion $(\overline{X},\vertiii{\;}_{\overline{X}}$ of $(X,\vertiii{\;})$, and for any $x\in \overline{X}$,
$\overline{P}_n(x)=\sum^n_{k=1}a_n(x)\boldsymbol{e}_n$ converges to $x$ in $(\overline{X},\vertiii{\;}_{\overline{X}})$.
We now show that $(X,\vertiii{\;})$ is actually complete. Let $\bar{x}\in \overline{X}$; there is a unique sequence $(a_n:n\in\mathbb{N})$ such that
$$\overline{x}=\sum^\infty_{n=1}a_n\boldsymbol{e}_n$$
where convergence is in $(\overline{X},\vertiii{\;}_{\overline{X}})$. Since $\|\;\|_X\leq\vertiii{\;}$, it follows that $\overline{P_n}\overline{x}=\sum^n_{k=1}a_k\boldsymbol{e}_n\in X$ is a Cauchy sequence in $(X,\|\;\|_X)$. Since $(X,\|\;\|_X)$ is complete, there is $x\in X$ such that $x=\sum^\infty_{n=1}a_n\boldsymbol{e}_n$. Since the series representation is unique, it follows that $x=\overline{x}$, that is, $X=\overline{X}$.
Since $I:(X,\vertiii{\;})\rightarrow(X,\|\;\|_x)$ is a bounded bijection, $I^{-1}$ is bounded by the open map theorem; hence, $\|\;\|_X$ and $\vertiii{\;}$ are equivalent and
$$\sup_n\sup_{\|x\|_X=1}\|P_nx\|_X=\sup_{\|x\|_X=1}\sup_n\|P_nx\|_X=\sup_{\|x\|_X=1}\vertiii{x}=\sup_{\|x\|_X=1}\vertiii{I^{-1}x}<\infty
$$
Consequently, $\{P_n:n\in\mathbb{N}\}$ is uniformly bounded as operators from $X$ to itself.
The last statement follows from
$$|a_n(x)|\|\boldsymbol{e}_n\|_X=\|P_n(x)-P_{n-1}(x)\|_X\leq \|P_n-P_{n-1}\|\|x\|_X$$
where $\|P_n-P_{n-1}\|=\sup_{\|u\|_X=1}\|P_n(u)-P_{n-1}(u)\|_X$. $\Box$
