E11 Packet separation and the Gibbs phenomenon
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A Fourier partial sum is global. When the target has a jump, the partial sum must approximate that jump using continuous trigonometric polynomials. This creates the classical Gibbs oscillation. Standard accounts and recovery methods appear in [Zygmund2002], [GottliebShu1997].
Standard fact E11.1: Classical square-wave Gibbs behavior
Let \(q(x)=\tfrac12\operatorname{sign}(\sin x)\), whose jump at zero has size one. For \(n\geq1\), its partial sum through the first \(n\) positive odd frequencies is
At \(x_n=\pi/(2n)\),
where \(\operatorname{Si}(z)=\int_0^z\sin(t)/t\,dt\) and \(G=\operatorname{Si}(\pi)/\pi-1/2>0\). Thus a nonzero overshoot persists along points approaching the jump. Scaling by a real amplitude \(a\) gives an overshoot or undershoot of magnitude \(|a|G\).
For the limit, put \(u_k=(k+\tfrac12)/n\). The identity
is a midpoint Riemann sum, since the integrand has continuous value \(\pi\) at zero. This is the normalized classical example in [DLMFGibbs], Section 6.16(i), equations 6.16.1–6.16.4.
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The square-wave example supplies a classical comparison. The following retained-packet identities do not require a general overshoot theorem. They change the approximation problem because the discontinuity is no longer represented by the Fourier partial sum.
Theorem E11.2: Uniform convergence with exact jump packets
Let \(f=s+j\) on a compact domain, where \(j\) is any bounded singular packet. Suppose \(Q_ms\to s\) uniformly. Then
converges uniformly to \(f\), even when \(f\) is discontinuous:
There is no error proportional to the jump size because the jump is shared exactly by \(f\) and \(\widetilde f_m\).
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Proof of Theorem E11.2
Apply Theorem E7.5 in the supremum norm.
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Theorem E11.3: Estimated packet Gibbs bound
Let \(Q_m\) be linear and uniformly bounded on \(L^p\), with
For \(f=s+j\) and an estimated packet \(\widehat j_m\), define
Then
Consequently, smooth approximation and packet recovery may be improved independently. No claim of Gibbs removal is justified unless both terms are controlled in the norm under consideration.
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Proof of Theorem E11.3
This is Theorem E7.8 with the uniform operator bound.
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Corollary E11.4: Boundary band error
Let \(\Gamma\) be the singular boundary and assume
for small \(\delta\). For finite \(p\),
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Proof of Corollary E11.4
The error equals \(s-Q_ms\). Bound it by its supremum and multiply by the measure of the boundary tube.
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Example E11.5: Square wave Fourier tail
Let
on the circle and let \(s\) be smooth. If \(Q_m\) is a Fourier partial sum, then \(Q_m(s+j)\) has the ordinary oscillatory boundary layer. In contrast,
has error exactly \(s-Q_ms\). The visual comparison in Figure illustrates this change in approximation target. It is not a new Fourier summability theorem.
