Du Bois-Reymond theorem Created 2026-10-14 Updated 2026-11-05
The Du Bois-Reymond theorem says that continuity of a periodic function does not guarantee convergence of its ordinary Fourier series at every point. At any prescribed circle point , there is even a real continuous with unbounded Fourier partial sums .
For , view as bounded linear functionals on the real Banach space with its supremum norm. The Dirichlet kernel harmonic lower bound givesIf every continuous input had bounded values under all , the Uniform boundedness principle would bound these operator norms, a contradiction. Translation moves the exceptional point to any specified . The generic unbounded Fourier sums at a fixed point strengthens this existence proof to a category statement.[ref]
245C, Notes 2: The Fourier transform by Terence Tao. §2, Exercise 25, Dirichlet-kernel norm growth and the uniform boundedness argument
This does not contradict mean-square convergence: a continuous function is square-integrable, and its series still converges in . Nor does it obstruct Fejér's theorem, which averages partial sums and gives uniform convergence for every continuous periodic input. The Fejér kernel is positive and its sums are a uniform-norm contraction, but the sums themselves need not be nonnegative: for , every Fejér sum equals .
Past exam of the mathematics course of the University of Cambridge 2005 iii Paper 6 1 iv Solution Created 2026-10-03 Updated 2026-10-14
The real space is a Banach space in the uniform norm: a uniformly Cauchy sequence has a uniform pointwise limit by completeness of , and that uniform limit is continuous. Each functional from the preceding part is bounded, but their norms are unbounded.
For integers , defineEach is closed, being an intersection of inverse images of closed intervals under continuous functionals. It has empty interior. Otherwise some ball would be contained in it. For every of norm at most one, both and would belong to that ball, soTaking the supremum over would give for all , contradicting the preceding part.
Thus the are nowhere dense. The Baire category theorem says their union cannot cover . A function outside that union satisfiesso its Fourier series diverges at zero. In fact these functions form a dense subset of . This proves the relevant Uniform boundedness principle argument directly and establishes generic unbounded Fourier sums at a fixed point.