Harmonic analogue of the Weierstrass approximation theorem

Question

The Weierstrass approximation theorem says that, given any continuous function $f(x)$ on a closed interval, there is a polynomial which approximates it arbitrarily closely. I'm looking for a theorem of the form

Given any nice enough function $f(x)$ on a closed interval, there is a finite Fourier series $$a_{0}+\sum_{n=1}^k a_n\cos(nx)+b_n\sin(nx)$$ which approximates it arbitrarily closely (with the possible exception of a finite number of points or even an infinite number of points with measure 0).

A Google search has turned up Carleson's theorem which seems relevant but as I'm not familiar with Fourier analysis I'm not confident making that call.

Answer 1

I am not sure what you meant by Fourier series here. Fourier series is very specific in that the coefficients are fixed. Weierstrass Theorem for periodic function states that:

If $f$ is periodic and continuous in $[-\pi, \pi]$, then given ε $>0$ , there is a trigonometric polynomial $T{_n}(x)$ (what looks like a Fourier series)

$T{_n}(x) = {a_0} + \sum\limits_{k = 1}^n {\left( {{a_k}\cos (kx) + {b_k}\sin (kx)} \right)} $

such that $|f(x) - {T_n}(x)| < \varepsilon $ for all $x$ in $[-\pi, \pi]$.

Here the coefficients may be different for different ε .

If you are looking for Fourier series, then you have the celebrated Carleson Hunt Theorem which states that for any periodic function $f$ in the L^p spaces, ${L^p}[ - \pi ,\pi ]$, $p$ ≥ $2$, its Fourier series converges almost everywhere to $f(x)$.

I.e., there is a set $E$ of measure zero such that for all $x$ in $[-\pi ,\pi ] - E$,

$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{A_0}}}{2} + \sum\limits_{k = 1}^n {\left( {{A_k}\cos (kx) + {B_k}\sin (kx)} \right)} } \right) = f(x)$,

where $\frac{{{A_0}}}{2} + \sum\limits_{k = 1}^\infty {\left( {{A_k}\cos (kx) + {B_k}\sin (kx)} \right)} $

is the Fourier series of $f$.

These two possibilities are quite different.

The Carleson Hunt Theorem does not give uniform convergence so you may not be able to approximate the function. For different $x$ you may need different number of terms of the Fourier series to give the same approximation.