# Why coefficients of Fourier series are countable, though the initial periodic function is described with an uncountable set of points

Coefficients in the Fourier series for any periodic square-integrable function $f(x)$ form a countable (though infinite) set, i.e., they have cardinality $\aleph_0$. As far as Fourier exponents form a basis in the corresponding Hilbert space $L^2([0,T])$, we may say that any basis of this space is countable.

On the other hand, the initial function $f(x)$ is described by an infinite and uncountable set of values—for each $x$. This set has cardinality $2^{\aleph_0}$.

P.S. May the resolution be connected to the fact that $f(x)$ may be discontinuous only at the set of points with zero measure to belong to the Hilbert space $L^2([0,T])$? So we actually need a countable number of function values to deduce the values at missing points, so that the deduced function has zero difference—in the sense of the norm in $L^2([0,T])$—from the initial function and thus has the same Fourier series? We may construct countable $x$ by mapping rational values to the interval $[0,T]$.

To sum it up:

1. why $f(x) \in L^2([0,T])$ is initially described with uncountable set of values, but $L^2([0,T])$ has a countable basis
2. is it because we may restore a function with the same Fourier series (coordinates in a basis) through initial function values at countable $x$
• Why assume the Continuum Hypothesis? – Jonas Meyer Sep 30 '14 at 21:02
• What set has size $\aleph_1$? The real numbers? That's not true. Well, it's consistently false anyway. $2^{\aleph_0}\geq\aleph_1$ but it's consistent that $2^{\aleph_0}>\aleph_1$. That's what Cohen showed in 1963 when he invented forcing. – Asaf Karagila Sep 30 '14 at 21:02
• @JonasMeyer, you are right. It is not necessary here. Probably i should have written $\mathfrak c$ (cardinality of the continuum), which anyway is larger than $\aleph_0$. – Bas1l Sep 30 '14 at 21:23
• @AsafKaragila, yes, i meant the cardinality of real numbers and assumed the continuum hypothesis. As mentioned, it is not necessary here and may safely be avoided. Please read $2^{\aleph_0}$ instead of $\aleph_1$ :-) – Bas1l Sep 30 '14 at 21:28
• Sorry, I prefer reading $2^{\aleph_0}$ where appropriate, instead of imagining things. :-) – Asaf Karagila Sep 30 '14 at 21:53

The problem, as you hinted at, is that for $L_2[0,T]$ to be a Hilbert space, you have to let its elements be almost everywhere equivalence classes of functions on $[0,T]$, not the functions themselves. Otherwise, the positive definiteness axiom of Hilbert spaces $(\langle f,f\rangle=0\iff f=0$) would be violated. Though you make continuum ($2^{\aleph_0}$) many choices when specifying a square integrable function on $[0,T]$ (at each $t$, you choose $f(t)$), you only make countable many such choices to specify its equivalence class (for each $\sin(kx),\cos(kx)$, you choose its coefficient in the Fourier series).
• @Basil Specifying a function at countably many values does not determine its equivalence class. Let $f(x)=1_{\mathbb{Q}}$, so $f(x)=1$ iff $x$ is rational, and $g(x)=1$. These are not equivalent, since $f=0$ a.e, but they agree on a countable dense set. – Mike Earnest Oct 1 '14 at 1:53