$\{x^k\}^\infty_{k=0}$ is not basis in $C[0;1]$ [duplicate]

$\{x^k\}^\infty_{k=0}$ is not basis in $C[0;1]$.

I know how to show that a system of functions is basis if it is orthogonal (Parseval's identity), but I do not understand how to show it for arbitrary system.

Possibly, we should find a continuous function that has some problems with approximation with an infinite sum of monomials, but I have no idea what contradiction we need to obtain.

marked as duplicate by ziggurism, Community♦Dec 19 '17 at 15:53

• Like, a Hamel basis? Anything that's not a polynomial will not be in the span of those vectors. For example, anything with infinitely many roots. – ziggurism Dec 18 '17 at 15:26
• The answers here might be of interest: math.stackexchange.com/questions/695421/… – Lionel Ricci Dec 18 '17 at 15:28
• Assume a basis should be finitely linear combinated to form every continuous function on [0, 1], then exp(x) is not the case, since it is an infinite power series. – xbh Dec 18 '17 at 15:40
• You should mention the definition of the basis. – Yurii Savchuk Dec 18 '17 at 16:00

Let $F$ be normed space. The system of elements $\{f_k\}_{k=0}^{\infty}$ is a basis for $F$, if for every $f\in F$ $$f=\sum_{k=0}^{\infty} \alpha_k f_k:=\lim_{N\to\infty}\sum_{k=0}^{N} \alpha_k f_k,$$ and this representation is unique, where by convergence we mean convergence in norm.

The system of functions you gave is not a basis in $C[0;1]$. Indeed, if one wrote that $$f=\sum_{i=0}^{\infty}a_k x^k,$$ for given $f\in C[0;1]$, he would obtain that $f$ is infinitely differentiable. Obviously, it isn't true for any given continuous function.

• You should also mention the definition of the basis. – Yurii Savchuk Dec 18 '17 at 16:00

The subspace $P[0,1]$ generated by $\{x_k\}_{k=0}^\infty$ consists of the polynomials. So all you need to do is show that there are functions in $C[0,1]$ that are not polynomials.

You can do this by constructing a function with infinitely many zeroes. Or by consider $f(x)=\sqrt x$, which is not differentiable at zero. Any function that is not infinitely differentiable will do.

One might be tempted to write "infinite sums" (i.e., series). It is very important to understand that a series is not a sum, but a limit of sums; and talking about "limit" implies that we have a metric or at least a topology. With $C[0,1]$, the usual metric is the one given by the uniform norm.

Now, this is interesting: if we consider all uniformly convergent series $\sum_{k=0}^\infty a_k x^k$, the set of functions we obtain is the set of analytic functions $A[0,1]$. On the other hand, the closure of $P[0,1]$ is all of $C[0,1]$ via Weierstrass' Theorem.