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Off the circle and onto the line: the Fourier transform, inversion, and Plancherel

Let the period go to infinity and discrete coefficients become a continuous spectrum. Meet the Fourier transform, the Riemann–Lebesgue lemma, the inversion formula, and Plancherel's energy identity.

From coefficients to a spectrum

On the circle, frequencies are the integers and a function carries one coefficient per integer. Now let the period grow without bound. The allowed frequencies pack closer and closer until, in the limit, they fill a continuum: the discrete list of coefficients becomes a function of a real frequency ξ. That limiting object is the Fourier transform.

For an L¹ function f on ℝ we define f̂(ξ) = integral over ℝ of f(x) e^{−2πi x ξ} dx. The integral converges absolutely because |e^{−2πixξ}| = 1 and f is integrable, so f̂ is well-defined and bounded: |f̂(ξ)| ≤ ‖f‖₁ for every ξ. A first structural fact: f̂ is continuous (dominated convergence on the parameter ξ).

Riemann–Lebesgue: high frequencies fade

The Riemann–Lebesgue lemma says f̂(ξ) → 0 as |ξ| → ∞: rapid oscillation averages an integrable function to nothing. It is the continuous cousin of c_n → 0 from Bessel, and it is proved by a clean shift trick.

Goal:  f in L^1(R)  =>  f-hat(xi) -> 0 as |xi| -> infinity.

Step 1 (shift trick).  Since e^{-2pi i x xi} = -e^{-2pi i (x + 1/(2 xi)) xi}, substitute u = x + 1/(2 xi):
   f-hat(xi) = - integral f(u - 1/(2 xi)) e^{-2pi i u xi} du.
Average this with the original definition:
   f-hat(xi) = (1/2) integral [ f(x) - f(x - 1/(2 xi)) ] e^{-2pi i x xi} dx.

Step 2 (estimate).  Take absolute values (|e^{...}| = 1):
   |f-hat(xi)| <= (1/2) integral | f(x) - f(x - 1/(2 xi)) | dx = (1/2) || f - tau_h f ||_1,   h = 1/(2 xi).

Step 3 (continuity of translation in L^1).  As xi -> infinity, h -> 0, and translation is
continuous in L^1:  || f - tau_h f ||_1 -> 0.    (True for compactly supported continuous g
by uniform continuity; extend to all of L^1 by density of such g.)

Therefore |f-hat(xi)| -> 0.   QED
Riemann–Lebesgue via the half-period shift and continuity of translation in L¹.

Inversion and Plancherel

Can we rebuild f from f̂? Yes — Fourier inversion: if f and f̂ are both integrable, then f(x) = integral over ℝ of f̂(ξ) e^{+2πi x ξ} dξ at every point of continuity. The honest proof does not just plug f̂ back in (that double integral need not converge); it inserts a Gaussian damping e^{−εξ²}, which makes everything absolutely convergent, recognizes the result as convolution against a Gaussian approximate identity, and lets ε → 0. Same machinery as Fejér, new kernel.