Completeness Of The Trigonometric System

The Circle and Its Trigonometric System

We identify the circle with \(\mathbb{T}=\mathbb{R}/2\pi\mathbb{Z}\), and use normalized Lebesgue measure \(dm(x)=dx/(2\pi)\) on a period. Thus \(m(\mathbb{T})=1\). The space \(L^2(\mathbb{T},m)\) consists of complex-valued measurable functions with finite \(\int_{\mathbb{T}}|f|^2\,dm\), identified when they agree \(m\)-a.e.; its inner product is \(\langle f,g\rangle=\int_{\mathbb{T}}f\overline g\,dm\). All equalities between \(L^2\) functions below are understood a.e. A trigonometric polynomial is a finite linear combination of the functions \(e_n(x)=e^{inx}\), \(n\in\mathbb{Z}\).

Direct integration gives \(\langle e_j,e_k\rangle=\int_{\mathbb{T}}e^{i(j-k)x}\,dm(x)\), which is \(1\) if \(j=k\) and \(0\) otherwise. Hence \((e_n)_{n\in\mathbb{Z}}\) is an orthonormal system. To prove it is complete, it suffices to show that its finite linear span is dense in \(L^2(\mathbb{T},m)\). We will construct approximations using Fejér kernels: nonnegative trigonometric polynomials whose mass concentrates near \(0\).

Fejér Kernels

To obtain the expansion, multiply \(\sum_{j=0}^N e^{ijt}\) by its complex conjugate. For each integer \(k\) with \(|k|\le N\), there are exactly \(N+1-|k|\) pairs of indices \(j,\ell\in\{0,\ldots,N\}\) with \(j-\ell=k\). Dividing by \(N+1\) gives the stated coefficient. The constant coefficient is \(1\), while every nonconstant exponential integrates to \(0\) over \(\mathbb{T}\). Consequently, \(\int_{\mathbb{T}}K_N\,dm=1\).

The kernels concentrate near the origin in the following precise sense: for every \(\delta\in(0,\pi)\), their integral over \(\{t:\delta\le |t|\le\pi\}\) tends to zero. Indeed, the finite geometric sum gives, when \(t\not\equiv0\) modulo \(2\pi\),

For \(\delta\le |t|\le\pi\), \(|\sin(t/2)|\ge\sin(\delta/2)>0\) and \(|\sin((N+1)t/2)|\le1\). Therefore \(K_N(t)\le ((N+1)\sin^2(\delta/2))^{-1}\), so

Approximation in \(L^2\)

For \(f\in L^2(\mathbb{T},m)\), define its \(N\)-th Fejér mean by convolution, \(\sigma_N f(x)=\int_{\mathbb{T}}K_N(t)f(x-t)\,dm(t)\). This is well-defined for a.e. \(x\); the finite expansion below also gives a trigonometric-polynomial representative. Write \(\widehat f(k)=\int_{\mathbb{T}}f(y)e^{-iky}\,dm(y)\). This coefficient is finite: Cauchy–Schwarz and \(m(\mathbb{T})=1\) give \(\int|f|\,dm\le\|f\|_2\). Substituting the finite expansion of \(K_N\) and changing variables \(y=x-t\) yields \(\sigma_N f(x)=\sum_{|k|\le N}(1-|k|/(N+1))\widehat f(k)e^{ikx}\). In particular, each Fejér mean is a trigonometric polynomial.

Proof. We first establish continuity of translations in \(L^2\): if \((\tau_t f)(x)=f(x-t)\), then \(\|\tau_t f-f\|_2\to0\) as \(t\to0\) on the circle. Continuous functions are dense in \(L^2(\mathbb{T},m)\). To recall why, simple functions are dense by the construction of the Lebesgue integral. For a measurable set \(E\), regularity of Lebesgue measure on the compact circle gives a compact \(C\) and an open \(U\) with \(C\subseteq E\subseteq U\) up to null-set changes and \(m(U\setminus C)\) arbitrarily small. A continuous function taking values in \([0,1]\), equal to \(1\) on \(C\) and \(0\) outside \(U\), approximates \(\chi_E\) in \(L^2\): its squared error is at most \(1\) and is supported, up to a null set, in \(U\setminus C\). Such a function exists on the circle, for example by separating the disjoint closed sets \(C\) and \(\mathbb{T}\setminus U\) with the distance functions. Approximating the finitely many indicator functions in a simple function proves the density claim.

Given \(f\in L^2\) and \(\varepsilon>0\), choose a continuous \(g\) with \(\|f-g\|_2<\varepsilon\). Translation preserves \(m\), so \(\|\tau_t(f-g)\|_2=\|f-g\|_2\). Uniform continuity of \(g\) on the compact circle gives \(\|\tau_t g-g\|_2\le\|\tau_t g-g\|_\infty\to0\). The triangle inequality then gives \(\|\tau_t f-f\|_2\le2\varepsilon+\|\tau_t g-g\|_2\). Since \(\varepsilon\) is arbitrary, translation continuity follows.

Because \(K_N\,dm\) is a probability measure, the integral form of Cauchy–Schwarz gives, for a.e. \(x\), \(|\sigma_N f(x)-f(x)|^2\le\int_{\mathbb{T}}K_N(t)|f(x-t)-f(x)|^2\,dm(t)\). Integrating in \(x\) and applying Tonelli's theorem to this nonnegative measurable integrand gives \(\|\sigma_N f-f\|_2^2\le\int_{\mathbb{T}}K_N(t)\|\tau_t f-f\|_2^2\,dm(t)\). The integrand is measurable; the inner norm is bounded by \(4\|f\|_2^2\), since \(\|\tau_t f-f\|_2\le2\|f\|_2\).

Let \(\varepsilon>0\). Translation continuity provides \(\delta\in(0,\pi)\) such that \(\|\tau_t f-f\|_2^2<\varepsilon\) whenever \(|t|<\delta\). Split the last integral into \(|t|<\delta\) and \(\delta\le|t|\le\pi\). Since \(K_N\) has integral \(1\), and by the concentration estimate already proved,

in the sense that the limit superior is at most \(\varepsilon\). This holds for every \(\varepsilon>0\), so the squared norms tend to zero. Each \(\sigma_N f\) is a trigonometric polynomial, proving density. \(\square\)

Completeness and a Common Pitfall

Proof. Suppose \(\langle f,e_n\rangle=0\) for every integer \(n\). Then \(\widehat f(n)=\langle f,e_n\rangle=0\) for every \(n\), so the finite formula for \(\sigma_N f\) gives \(\sigma_N f=0\) a.e. for every \(N\). The theorem gives \(\|f\|_2=\|f-\sigma_N f\|_2\to0\), hence \(\|f\|_2=0\) and \(f=0\) a.e. Conversely, the zero element is orthogonal to every \(e_n\). This proves the stated criterion for completeness. \(\square\)

Pitfall: confusing \(L^2\) density with uniform approximation. The half-circle indicator cannot be approximated by continuous functions with arbitrarily small essential uniform error. In fact, let \(g\) be any continuous complex-valued function and suppose its essential uniform error \(c=\|g-f\|_\infty\) were less than \(1/2\). Choose \(c'\) with \(c<c'<1/2\); then \(|g-f|\le c'\) a.e. On the interior of \([0,\pi)\), this gives \(|g-1|\le c'\) a.e., and continuity extends this bound to the boundary point \(0\). On the interior of the complementary half-circle, it gives \(|g|\le c'\) a.e., and continuity extends that bound to \(0\) as well. The triangle inequality would imply \(1=|1-0|\le|1-g(0)|+|g(0)|\le2c'<1\), a contradiction. The argument works for complex-valued \(g\); it uses absolute values, not an order on complex numbers. This obstruction to small uniform error does not prevent \(L^2\) approximation, which is the conclusion proved using Fejér kernels.