Question about Schauder bases in C([0,1]). 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:


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*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.)

*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.)
 A: 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 $L_1(\mathbb T)$. 
A: I would say, monomials are not a Schauder 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.
