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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

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3 Answers 3

up vote 2 down vote accepted

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).

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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 $[1/2,1]$.

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Very nice answer. – Ramiro Nov 13 at 12:29


"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.

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