TITLE: Is the Fourier series always the "best" approximation?
QUESTION [7 upvotes]: In Stein and Sharkachi's Introductory Fourier Analysis book, in Chapter 3, we're given the "Best Approximation Theorem".  
The theorem implies (and almost states) that if $f \in L^2 (0,2\pi)$, then the "best" way to approximate $f$ by a sequence of trigonometric polynomials is precisely by the partial sums of its Fourier series. 
Rigorously speaking,  if $a_N(\theta) = \sum_{|n| \leq N} b_n e^{in\theta}$ is any other sequence of trigonemtric polynomials, then $$||f-S_N(f)||_{L^2} \leq ||f-a_N||_{L^2} $$ 
My question is the following, is this true in any other norms? The most natural norms would to start with would be , of course, the $L^p$ norms.  I'm just curious how "canonical" are Fourier series in the approximation .

REPLY [9 votes]: No: the Fourier series is only best for the $L^2$ norm.  For a very simple example, consider the function $f$ defined by $f(x)=0$ for $x\in [0,c]$ and $f(x)=1$ for $x\in[c,2\pi]$ (for some $c\in [0,2\pi]$) and consider just the first term of the Fourier expansion, which is $a=1-\frac{c}{2\pi}$.  This constant $a$ minimizes the $L^2$ norm $\|f-a\|_2$.  On the other hand, if we wanted to choose a constant $b$ which minimizes the $L^1$ norm $\|f-b\|_1$, we should choose $b=0$ if $c>\pi$ and $b=1$ if $c<\pi$ (for instance, if $c>\pi$, increasing $b$ above $0$ always just increases the $L^1$ norm since the integral on $[0,c]$ contributes more than the integral on $[c,2\pi]$).  So for $c\neq 0,\pi,2\pi$, we can get a better approximation than the Fourier expansion in the $L^1$ norm.
(More generally, the best constant approximation to a real-valued function is the mean value if you use the $L^2$ norm, but the median value if you use the $L^1$ norm.)