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I would like to check a statement about Schauder bases in $C([0,1])$ to be sure that I don't lie to my students on Monday. The statement(s) that I would like to check are:

  1. The family of monomials $\{1,t,t^2,t^3,\dots\}$ is a topological basis but not a Schauder basis in $C([0,1])$ because there's not a unique choice of coefficients converging to a given continuous function. (An example I thought of: there's a polynomial approximation of $|t|$ on $[-1,1]$ using only even polynomials, but then that's a polynomial approximation of $t$ on $[0,1]$ with zero $t$ coefficient.)

  2. Ditto for trigonometric polynomials.

The reason I ask is because when searching for this on the internet, I came across a statement claiming that trigonometric polynomials weren't a Schauder basis because in general (there's that phrase again!) Fourier series don't converge uniformly for continuous functions. That seems to me like a load of dingo's kidneys (not the convergence statement, but the deduction from it) but - and here's the clincher - the statement was made by someone whose answers on MO I've found to be generally reliable. (To be clear, the statement wasn't made on MO and was somewhere fairly obscure and I'm not going to "name and shame" because I don't want to embarrass that person - if I'm right - or myself - if I'm wrong.)

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While you are busy trying to not embarrass yourself, maybe you also want to change your example so that the second clause reads "...that's a polynomial approximation of t on [0,1] with zero t coefficient." – Willie Wong Oct 30 '10 at 19:11
Also, assuming the phrase "Fourier series don't converge uniformly for continuous functions", you actually do have a heuristic that the trigonometric functions form not a Schauder basis. The point is that for it to be a Schauder basis, the representation of an arbitrary function must converge with respect to the norm topology of $C[0,1]$, which (I assume you use the sup norm) is uniform convergence, no? So the existence of continuous functions whose Fourier series converges pointwise, but not uniformly, puts a dent in the definition. – Willie Wong Oct 30 '10 at 19:19
Willie, thanks for the first (now corrected). As for the second, the Fourier series stuff is evidence towards the result, and probably could be made into a proof (albeit a convoluted one), the thing that has to be ruled out is that there may be some other way of choosing the coefficients that has absolutely nothing to do with Fourier Series, but which made the trigonometric functions into a Schauder basis. I'm pretty sure that this is true, and without the statement I came across I wouldn't have worried, but given that statement, I want to check. – Loop Space Oct 30 '10 at 20:08
Actually if a trigonometric series converges uniformly to a function $f$, then the series must be the Fourier series of $f$. This is statement 1.41 on page 6 of Zygmund's {\it Trigonometric Series}. – Hany Oct 30 '10 at 20:27
@Hany: Indeed: if $\sum_k a_k e^{ikt}$ converges uniformly to $f$ then it's pretty elementary that $\int f(t) e^{-ikt} \ dt = \sum_l a_l \int e^{ilt-ikt} \ dt = a_k$. – Matthew Daws Oct 30 '10 at 20:34
up vote 7 down vote accepted

I would say, monomials are not a Shauder basis for $C[0,1]$ because functions that admits a representation are analytic functions on the unit disc. Actually, $f(t)=\sum_k a_k t^k$ immediately implies differentiability at $t=0,$ which is enough to conclude. So, it's more a matter of non-existence than non-unicity.

With trigonometric polynomials it is a bit more delicate. But the reason is again non-existence of the uniform convergent expansion for some continuous function. Of course, if $f$ admits a representation as uniform limit of a series $\sum_k c_k e^{ikt}$ then the series is its Fourier series, so the claim that $\{ e^{ikt} \}_{k\in\mathbb{Z} }$ is not a Schauder basis for the continuous $2\pi$-periodic functions is equivalent to the statement that a Fourier series of a continuous $2\pi$-periodic function may fail to converge uniformly. (I do not have an example handy; one can also show it indirectly as a consequence of Uniform Boundedness principle if I remember well).

[edit] For instance, Wheeden & Zygmund's book Measure and Integral (p 227) has a nice example of a continuous function $f$ whose Fourier series is unbounded at $0$. The idea is defining $f$ as a uniform limit of a normally convergent series $\sum_{j=1}^{\infty} Q_j(t)$ which is formally made out of an unbounded trigonometric series after a rearrangement and a parenthesization. In other words, each $Q_j$ is a linear combination of $e^{ikt}$ with $k\in I_j,$ and the $I_j$'s are pairwise disjoint finite subsets of $\mathbb{Z}$). So computing the Fourier series of $f,$ one finds again the bad series.

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Hmmm. Maybe the point where I'm confused is earlier than I thought. If I take a sequence of polynomials converging uniformly to some continuous functions, then from what you say here then I have to conclude that there is no way (in general) to make that sequence into a series in which the coefficients of each monomial converge. – Loop Space Oct 30 '10 at 20:30
Exact; e.g. if $P_n$ is a sequence of polynomials uniformly converging to $|t|$ on $[-1,1]$, the coefficients would vary like crazy! – Pietro Majer Oct 30 '10 at 20:58
The usual example in $C([0,1])$ is $\vert t-\frac 12\vert$. Its Fourier series does not converge uniformly. – Hany Oct 30 '10 at 22:29
Yes; and talking about the distinction "topological vs Schauder basis", it is worth observing that for orthonormal families, the two notions coincide (that's the magic of the Hilbert structure). As to the name, the term "total" is sometimes used for a set whose linear span is dense. – Pietro Majer Oct 31 '10 at 22:53
Okay, I'm accepting this answer for the statement: "if f admits a representation as uniform limit of a series $\sum_k c_k e^{i k t}$ then the series is its Fourier series". That's the bit I hadn't grokked. More generally, I was so used to being able to pass back and forth between sequences and series that the difference between a topological basis (sequences) and Schauder basis (series) wasn't evident. Thanks for helping me see the light! I've tried to summarise what I've learnt at: no doubt I've made yet more elementary errors. – Loop Space Nov 1 '10 at 10:40

I'd like to expand a bit on Pietro Majer's remark concerning the relation with the principle of uniform boundedness.

Indeed, suppose that the trigonometric system is a Schauder basis of $C(\mathbb T)$, i.e. the sequence of partial Fourier sums $f_n=S_n f$ converges uniformly to $f$ for every $f\in C(\mathbb T)$. By the principle of uniform boundedness the norms of the operators $S_n:f\mapsto f_n$ should be bounded with some constant for all $n\in\mathbb N$. But Lebesgue showed that $$\|S_n\|_{C(\mathbb T)\to C(\mathbb T)} = L_n:=\frac{1}{2\pi}\int_{0}^{2\pi}\left|\frac{\sin\frac{(2n+1)t}{2}}{\sin\frac{t}{2}}\right|dt\to\infty.$$ Hence the trigonometric system is not a basis of $C(\mathbb T)$.

By the way, the same argument shows that the trigonometric system cannot be a basis of $L_1(\mathbb T)$.

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