Gibbs overshoot at a piecewise C^1 jump

Put $m:=(f(x_0+)+f(x_0-))\mathrm{/}2$, and define $\tilde{f}(x):=m$ when $x-x_0\in \mathbb{Z}$ and $\tilde{f}(x):=f(x)$ otherwise. Membership in $x_0+\mathbb{Z}$ is invariant under integer translation, so $\tilde{f}$ is one-periodic; on each period it is piecewise $C^1$ and has midpoint value $m$ at the jump represented by $x_0$. The exceptional set $x_0+\mathbb{Z}=\{x_0+n:n\in \mathbb{N}\}\cup \{x_0-n:n\in \mathbb{N}\}$ is countable and hence Lebesgue null by Every at most countable subset of ${\mathbb{R}}^n$ is Lebesgue null; in particular ${\lambda }_1(\mathbb{Q})=0$. Thus $f$ and $\tilde{f}$ agree almost everywhere, so after multiplication by any character their integrals agree by Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree. Hence they have the same Fourier coefficients and the same partial sums in the normalization of Period-one Fourier coefficients, partial sums, and convolution on the torus. They also have the same one-sided limits and the same jump $J$.

Apply [F1] to the $2\pi$-periodic function $g(y):=\tilde{f}(y\mathrm{/}(2\pi ))$ at $y_0:=2\pi x_0$. Its Fourier coefficients and partial sums correspond exactly to the period-one coefficients and partial sums of $\tilde{f}$ under the substitution $y=2\pi x$, while the source offset $2\pi \mathrm{/}(2N+1)$ becomes $1\mathrm{/}(2N+1)$. Therefore $S_N\tilde{f}\text{\hspace{0.17em}⁣}(x_0+\frac{1}{2N+1})⟶f(x_0+)+J(\frac{\mathrm{Si}(\pi )}{\pi }-\frac{1}{2})\mathrm{.}$ By step 1.1 the same limit holds for $S_{N}f$.

Since $\mathrm{Si}(\pi )\approx 1.85194$, one has $\frac{\mathrm{Si}(\pi )}{\pi }-\frac{1}{2}\approx \mathrm{0.08949.}$ Thus when $J>0$ the limiting value in step 2.1 lies above the right limit by the claimed fixed amount.