# Stone-Weierstrass, uniform convergence, and sums

Assume $\{F_n\}_{n\in \mathbb{N}} \subset C(X)$ is some series of continuous complex functions on the compact Hausdorff space $X$. Assume also that the $F_n$ separate the points of $X$, and that the $F_n$ have the property that every product can be written as a finite sum, i.e. $F_{n_1} F_{n_2} = \sum_{k=1}^K a_k F_k$.

Let $\mathcal{A} = \mathsf{span} \{F_n | n\in\mathbb{N}\}$ be the complex linear span of $\{F_n\}_n$. Because products can be written as sums, the span, $\mathcal{A}$, is the subalgebra of $C(X)$ generated by the $F_n$.

By Stone-Weierstrass, $\mathcal{A}$ is a dense subset of $C(X)$ under the sup-norm, i.e. any function $f\in C(X)$ can be written as the uniformly convergent limit of a sequence $\{f_j\}_j$ in $\mathcal{A}$. This is all well and good, but what I would really like is to write each $f\in C(X)$ as an infinite linear combination of the $F_n$.

That is, each of the $f_j$ in the sequence that uniformly converges to $f$ is a finite linear combination $f_j = \sum_{k=1}^{K_n} a_{j,k}\, F_k$, but does that imply that I can write the limit, $f$, as an infinite sum $f = \sum_{k=1}^{\infty} a_k F_k$ - and how? If not, are there some extra conditions that I can impose to make this possible?

• Let's assume the $F_n$ are linearly independent. Then you have projections $p_j:\mathcal A\to {\mathbb C} F_j$ and what you need is that these projections are continuous, i.e., that they extend to continuous linear maps $C(X)\to {\mathbb C} F_j$. – user1688 Feb 16 '16 at 9:41
• I think the classical example of polynomials (with the $x^k$ as your basis) speak for itself: every continuous functions is a limit of polynomial, but only analytic functions can be written locally as $\sum a_k x^k$ – Simon Henry Feb 16 '16 at 10:10
• Simon Henry's comment answers the first question. There remains to find conditions on the $F_n$ such that every continuous $f$ is an infinite linear combination. Even the Fourier system $F_n(x)=e^{inx}$ ($n\in\mathbb Z$) doesn't satisfy this... – Jean Duchon Feb 16 '16 at 13:01
• The condition given by Anton isn't quite enough. I think that you are looking for the concept of a Schauder basis, which means that your condition is fulfilled, together with uniquenss of the representation. Grunblum's criterion (for which google) may be the kind of condition you are looking for. As pointed out by Simon Henry, the monomials do not form a Schader basis, but the so-called Schauder functions do (see any respectable text on Banach spaces). The product condition is irrelevant here. – dalry Feb 16 '16 at 13:05

As I said in the comment, this is clearly not true in general: the algebra of polynomial function on $[0,1]$ is dense among all continuous functions and generated by the $x^i$ but only functions that are analytic on the unit disk can be sum of a series of the form $\sum_i a_i x^i$.

Now there is indeed a few case where it works:

If you have a scalar product on your space of functions, that the $F_i$ are orthogonal for this scalar product and that the space they generates is also dense for the norm induced by the scalar product (so for example if uniform convergence implies convergence for the scalar product) then your function can be approximated by a series $\sum a_i F_i$ which converge normally in the sense of the scalar product (but not necessarily uniformly).

This is very easy to prove:any element will be the limit of its projection on the succesive space generated by the first $n$ generators and the orthogonality relation between the generator ensure that the successive projection are the partial sum of a series.

This apply for example to the case of the algebra of trigonometric functions which gives Fourier series. In this case of course there is some well known trick involving convolution by well chosen kernel to recover uniform convergence under some additional conditions.

There is of course other kind of example: what you are looking for in general is called a Schauder basis.

• Thank you very much for the answer. Is there a citable theorem for the property that a sequence of mutually orthogonal functions, whose span is dense, yields a Schauder basis? And: is it correctly understood that this approach only works for finding a Schauder basis for a Hilbert space (I.e., would not work for $C(X)$)? I think it would be useful to add to your answer what the general conditions are for a total sequence to be a Schauder basis. Another commenter cites Grunblum's theorem, perhaps that would be good to include in the answer? – James Feb 19 '16 at 10:02