Specify what it means for a Fourier expansion to recover data

Try

Imagine a periodic step taking value −1 on one half-period and 1 on the other. Smooth trigonometric polynomials cannot reproduce its jump exactly. Sketch what you expect near the jump as more modes are added. Does the transition narrow, does the largest overshoot disappear, and does the average squared error decrease? These are different questions.

Every Fourier coefficient is an integral. Changing a function at a single point leaves all coefficients unchanged. Therefore no theorem based only on those coefficients can recover an arbitrarily assigned value at that point.

Predict

Suppose a piecewise smooth periodic function approaches −2 from the left and 4 from the right at a jump. Predict the Fourier-series limit there. Then assign the function value 100 exactly at that point. Explain whether your prediction should change.

Experiment

The Dirichlet convergence theorem gives the midpoint, 1, under the usual piecewise C¹ assumptions with one-sided limits. The point assignment 100 does not change it.

Question	Appropriate statement
At a continuity point	Limit equals the function value
At this jump	Limit equals 1
In L² over the period	Squared integral error tends to zero
Uniformly across a genuine jump	Impossible for continuous partial sums

Explain the last row using the theorem that a uniform limit of continuous functions is continuous. This argument requires no estimate of the precise overshoot size.

Predict before changing the average

Does positive averaging remove the jump, or only the overshoot? For the continuous corner, does averaging necessarily improve the error at every fixed degree?

ProfileContinuous periodic corner |x|Jump: sign(sin x), midpoint value0 Degree N0141664Reset comparison

Gray: target; dashed red: S_N; solid navy: σ_N, the average of S_0 through S_N. Shared axes cover x∈[−π,π], y∈[−1.5,3.5]. Jump endpoints and zero use the midpoint assignment0; there is no gray vertical connector across a jump.

On257 fixed sample points: maximum partial-sum error 0.156086; maximum mean error 0.495616. These are sampled maxima, not certified suprema. A narrowing jump transition can fall between samples; the proof below determines the actual uniform conclusion.

At x=0: target 0, S_N=0.156086, σ_N=0.495616. All coefficients are analytic examples, not estimated from these257samples. Degree≤64; one comparison uses at most8,224 odd-mode evaluations. State is temporary, with no saved mastery or animation.

Repair the claim

“The curve stays in range, so the jump converges uniformly” is false. Positivity controls range; uniform limits of continuous functions cannot have a jump. “The mean always has smaller error than S_N” is also false: at N=1 for |x|, the origin values are π/2−4/π versus π/2−2/π. Positive averaging gives a general convergence theorem, not best approximation at each finite degree.

Understand

For periodic L² functions, completeness of the trigonometric system implies convergence of orthogonal projections in L². Parseval’s identity reads (1/π)∫|f|² = a₀²/2 + Σ(aₙ² + bₙ²) for real coefficients. We now prove the missing trigonometric-density mechanism. Orthogonality alone does not prove completeness.

Build a positive averaging kernel

Work on the circle of circumference 2π, with normalized integration dμ(t)=dt/(2π). Let S_N be the Fourier polynomial containing frequencies −N through N. Its average σ_N=(S_0+⋯+S_N)/(N+1) weights frequency k by 1−|k|/(N+1). Count how many of the N+1 sums contain that frequency to obtain this weight; do not confuse the average with S_N itself.

The corresponding kernel has a useful square representation:

$$F_N(t)=\frac{1}{N+1}\left|\sum_{j=0}^{N}e^{ijt}\right|^2=\sum_{k=-N}^{N}\left(1-\frac{|k|}{N+1}\right)e^{ikt}.$$

To prove the equality, expand the square as a double sum. For fixed difference k=j−ℓ there are N+1−|k| pairs. The square proves nonnegativity. Integrating the finite expansion leaves only the constant coefficient, so ∫F_N dμ=1. At t=0 the value is N+1. Away from multiples of 2π, the finite geometric sum also gives

$$F_N(t)=\frac{\sin^2((N+1)t/2)}{(N+1)\sin^2(t/2)}.$$

For any fixed 0<δ≤π and δ≤|t|≤π, its value is at most 1/((N+1)sin²(δ/2)). Therefore the normalized mass outside the δ-neighborhood is at most that same quantity and tends to zero. The kernel’s height grows while its total mass stays one and its distant mass shrinks. Positivity and unit mass alone would not guarantee this concentration: the constant kernel1 satisfies both but does not concentrate.

Uniform approximation: choose δ before N

For continuous periodic f, expand the finite kernel inside its integral to get σ_N f(x)=∫f(x−t)F_N(t)dμ(t). This identity needs only finite interchange, not convergence of an infinite Fourier series. Let M=max|f|. Uniform continuity on the circle makes |f(x−t)−f(x)| at most ω(δ), uniformly in x when |t|<δ, with ω(δ) tending to zero. Split the integral into near and far parts:

$$\|\sigma_N f-f\|_\infty\leq\omega(\delta)+\frac{2M}{(N+1)\sin^2(\delta/2)}.$$

The near part uses positivity and mass at most one; the far part uses the difference bound2M and concentration. Given a desired error, first choose δ to make the first term small, then choose N to make the second small. Both choices work for every x. This proves uniform convergence of these trigonometric polynomials and hence uniform density in continuous periodic functions. Endpoint agreement matters: continuity on [−π,π] without equal endpoint values does not give continuity on the circle.

For f(x)=|x| extended periodically, circular distance makes it 1-Lipschitz and M=π. With δ=π/3, the bound is π/3+8π/(N+1): valid but too coarse for a small desired tolerance. Repair it by choosing a smaller δ first and then increasing N sufficiently. A poor bound is not a failure of the theorem. Finite-degree means need not beat ordinary partial sums, as the origin comparison in the workbench shows.

Complete the L² argument without assuming Parseval

Use the existing Lebesgue regularity and continuous approximation construction for the density step on the circle. More explicitly, truncate and quantize an L² function to a bounded simple function. Approximate each indicator by a continuous separator between a compact subset and an open superset of arbitrarily small measure difference. The distance construction works in the circular metric, making the separators periodic; its squared error is bounded by the measure difference. Triangle inequalities give continuous periodic g arbitrarily close to f in L². This step uses Lebesgue regularity already taught there, not trigonometric completeness.

Now choose a trigonometric polynomial p uniformly close to g by the proved kernel construction. For normalized measure, its L² error is at most its uniform error. Thus trigonometric polynomials are dense in L². S_N is the finite-dimensional orthogonal projection, so if p has degree at most N, Pythagoras gives ||f−S_N f||₂≤||f−p||₂. Choose p first, then every N at least its degree works. This proves L² convergence of the ordinary partial sums, even though the density proof used averages.

Finally finite orthogonality gives ||f||₂²=||f−S_N f||₂²+Σ from k=−N to N of |f̂(k)|². Let N grow; the error just proved tends to zero, giving Parseval. If all coefficients vanish, f=0 as an L² equivalence class, hence almost everywhere, not necessarily at every assigned point. This reasoning does not establish uniform convergence of ordinary partial sums for all continuous data, pointwise convergence for all L² data, or differentiated-series convergence.

The MIT18.103 Fall2013 Fourier Series2 notes supply the primary theorem reference. Their average ends at S_(N−1) and divides by N; this lesson consistently ends at S_N and divides by N+1. The square/counting proof and examples above use that convention throughout.

Pointwise convergence needs different information. The Dirichlet kernel representation and cancellation against one-sided smoothness prove convergence to the mean of the one-sided limits for piecewise C¹ data. This page states that theorem and its proof mechanism; it does not assert pointwise convergence for every L² function.

Near a jump, Fourier partial sums exhibit Gibbs oscillation. The affected region narrows as frequency increases, while the limiting relative overshoot does not vanish. Its shrinking width is compatible with decreasing L² error. A graph at a coarse pixel scale can hide that distinction.

Term-by-term differentiation is another separate issue. Multiplying the nth coefficient by n magnifies high frequencies, so convergence of the original series does not automatically justify differentiating it. Absolute summability of appropriately weighted coefficients is a useful sufficient condition for uniform derivative convergence. Heat evolution will supply strong decay at positive time, making such operations legitimate away from the initial instant.

Apply

For a finite signal 3sin x + 2cos(2x), verify Parseval directly by integration and by coefficient squares. Then describe a discontinuous signal for which L² convergence holds but uniform convergence cannot. In a numerical report, label whether your error is pointwise, maximum, or integrated squared error; do not substitute one conclusion for another.

Continue

Sturm–Liouville theory extends orthogonal modes beyond ordinary sines and cosines. Boundary conditions and the weight in the inner product determine which orthogonality statement is valid.