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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 $2^{\aleph_0}$.

How to resolve this contradiction?

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 $x$
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  • $\begingroup$ Why assume the Continuum Hypothesis? $\endgroup$ – Jonas Meyer Sep 30 '14 at 21:02
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Sep 30 '14 at 21:02
  • $\begingroup$ @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 $\aleph_0$. $\endgroup$ – Bas1l Sep 30 '14 at 21:23
  • $\begingroup$ @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 $\aleph_1$ :-) $\endgroup$ – Bas1l Sep 30 '14 at 21:28
  • $\begingroup$ Sorry, I prefer reading $2^{\aleph_0}$ where appropriate, instead of imagining things. :-) $\endgroup$ – Asaf Karagila Sep 30 '14 at 21:53
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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).

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  • $\begingroup$ Thanks a lot! Still, one point is unclear to me. Seems that it is sufficient to specify the function f(x) only at some countable set of values x. Say, x = T arctan(m/n) (so we map countable rational numbers onto an interval [0, T]). Values at the other points are interpolated. Let's define another function g(x) which has zeros at unspecified x. As far as there is an uncountable amount of unspecified x, the difference between functions <g-f,g-f> will be determined by unspecified x and will not be zero. So defining a function at countable x does not define an equivalence class? $\endgroup$ – Bas1l Sep 30 '14 at 21:12
  • $\begingroup$ @Mike Earnest: Can you elaborate on which countably many choices you are making in your last sentence? $\endgroup$ – Samuel Sep 30 '14 at 22:04
  • $\begingroup$ @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. $\endgroup$ – Mike Earnest Oct 1 '14 at 1:53
  • $\begingroup$ @Samuel Edited my post to explain the last sentence $\endgroup$ – Mike Earnest Oct 1 '14 at 1:54
  • $\begingroup$ @MikeEarnest, i agree with you (that specifying a function at countably many values does not determine its equivalence class). In principle, i was writing exactly about it in my previous comment. What bothers me, is that by doing so we provide the same "amount of information" (countably many independent real values) as in the Fourier series, so it shall be enough. $\endgroup$ – Bas1l Oct 1 '14 at 21:30

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