How to show that the monomials are not a Schauder basis for $C[0,1]$ why the monomials are not a Schauder basis for $C[0,1]$?
$p_n(x)=x^n$ such that $(p_n)$ does not form a Schauder basis for $C[0,1]$
span$\lbrace p_n : n\ge 0\rbrace$ is dense in $C[0,1]$ by Weierstrass approxmation theorem, but I cannot figure out why they are not a Schauder basis.
Could you please explain
 A: Let
$$
f(x)=\begin{cases}
0 & \text{if $0\le x\le1/2$,}\\
x-1/2 & \text{if $0<1/2< x\le1$.}
\end{cases}$$
Suppose that $\sum_{n=0}^\infty a_nx^n$ converges uniformly to $f$ on $[0,1]$. Then $\sum_{n=0}^\infty a_nx^n$ converges uniformly to $0$ on $[0,1/2]$. This implies that $a_n=0$ for $n=0,1,2,\dots$, which is impossible since $\sum_{n=0}^\infty a_nx^n$ converges to $x-1/2$ on $[1/2,1]$.
A: Having a dense span is just one of the conditions a family $\{e_n\}$ must satisfy to be a Schauder basis. The much more restrictive condition is that there are (continuous) coordinate linear forms $\xi_n$ such that for every $x\in X$ we have
$$\lim_{n\to\infty} \left\lVert x - \sum_{k=0}^n \xi_k(x)\cdot e_k\right\rVert = 0.$$
For the space $C([0,1])$, the convergence is uniform convergence, and for the sequence of monomials - $e_k(t) = t^k$ - that would mean that every continuous function had a representation
$$f(t) = \sum_{k=0}^\infty \xi_k(f)\cdot t^k$$
as a power series, which converges uniformly on $[0,1]$. But the limit function of a convergent power series is (real-) analytic, and only a tiny minority of continuous functions is real-analytic (and of those that are, a lot have singularities in the unit disk).
A: See 
"Shukurov, Aydin. The power system is never a basis in the space of continuous functions. Amer. Math. Monthly  122  (2015), no. 2, 137"
for the proof of the most general fact. 
