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Piecewise C^1 Fourier series converges to midpoint values

Statement

Assume the Axiom of Countable Choice.

Let f:R→R be one-periodic. Assume there is a partition 0=x0<x1<⋯<xm=1 such that, for each j, the restriction of f to (xj−1,xj) extends to a C1 function on [xj−1,xj]. Then for every x∈R,

SNf(x)→f(x+)+f(x−)2as N→∞.

Facts & Assumptions

Given: The Axiom of Countable Choice, a one-periodic real function f and a partition 0=x0<⋯<xm=1 such that each restriction f∣(xj−1,xj) extends to a C1 function on [xj−1,xj].

[L1]

Assuming the Axiom of Countable Choice, a one-periodic bounded-variation function satisfies the Dirichlet-Jordan convergence theorem (Dirichlet-Jordan pointwise convergence).

[L2]

A C1 function on a compact interval is of bounded variation (C1 implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation).

Proof

technique · direct
1.1

For each j, let fj denote the C1 extension of f∣(xj−1,xj) to [xj−1,xj]. By [L2], each fj has bounded variation on its interval. Summing those finitely many variations and adding the finitely many endpoint jumps shows that the one-period representative of f has bounded variation on [0,1], including the periodic seam between 1− and 0+.

L2givenalgebra
2.1

Apply [L1] to that one-period bounded-variation representative. It yields SNf(x)→f(x+)+f(x−)2 for every x∈R, which is the claimed midpoint-value convergence.

L1step 1.1∎

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