Monomials are NOT a (NOT necessarily Schauder) basis?

Question

A basis for an infinite dimensional Banach space $X$ is a set of elements $\{e_k\}=B\subseteq X$, such that

1. Each finite subset of $B$ is linearly independent.
2. The closure of the span, i.e. $\{\lim_{N\rightarrow \infty} \sum_k^N a^{(N)}_{k}e_k\}=X$

Now I note that I have purposefully written the sequences of coefficients dependent on $N$. So it is allowed to change every member of the sequence while taking $N\rightarrow \infty$.

This is exactly where the crux of my confusion lies: Is this allowed? Clearly, as long as the series converges against something in the limit under the respective norm of my space, it is part of the span. So assuming this is ok, let's move to the title of the question.

Consider the sets of all monomials. I claim it is not a basis to only $C[0,1]$ with uniform norm. Now consider the sequence $a_k^{(N)}=\delta_{kN}$. The functions $f_N(x)=\sum_k^N \delta_{kN}x^k$ converge in the uniform norm to the function 0 everywhere except 1, where it is 1. This is not a continuous function.

Where is my mistake in the line of reasoning?

Answer 1

A sequence can't converge uniformly at a point. It can either converge uniformly, full stop, or it can converge pointwise at a point. Your sequence does not converge uniformly.

Compare to the usual example of $f_n(x)=x^n$ defined on $[0,1]$, which is exactly your sequence, just written in a simpler way. It does not converge uniformly to $g(x)=\delta_{x1}$ because $\sup_{x\in[0,1]}\vert f_n(x)-g(x)\vert=1$ for all $n$.

Again for emphasis: uniform convergence is not a pointwise property. It is a global property and does not follow from pointwise convergence.

That aside, yes, it is allowed to change the coefficients at every member of the sequence. The span of all monomials is the space of polynomials, and the closure of that space contains the limit of any sequence of polynomials.