Dirichlet-Jordan pointwise convergence

By [L3], the one-sided limits $f(x+)$ and $f(x-)$ exist. Put $s:=\frac{f(x+)+f(x-)}{2}\mathrm{.}$ Choose $\delta \in (0,1\mathrm{/}2)$ and define, on $[0,\delta ]$, $u_{+}(0):=0,u_{-}(0):=0,$ $u_{+}(t):=f(x+t)-f(x+)\text{and}{u}_{-}(t):=f(x-t)-f(x-)(0<t\le \delta )\mathrm{.}$ Because translations and reflections preserve bounded variation on a compact interval, and changing a function at one point preserves bounded variation, both $u_{+}$ and $u_{-}$ have bounded variation on $[0,\delta ]$. The cited one-sided-limit result [L3] gives $u_{\pm }(t)\to 0$ as $t\downarrow 0$, and by construction $u_{\pm }(0)=0$.

Applying [L1] with the value $s$ from step 1.1 yields $S_{N}f(x)-s={\int }_0^{\delta }(u_{+}(t)+u_{-}(t))D_N(t)dt+{\int }_{\delta }^{1\mathrm{/}2}(f(x+t)+f(x-t)-2s)D_N(t)dt\mathrm{.}$

By [L2] applied to $u_{+}$ and to $u_{-}$, and then using [L5], ${\int }_0^{\delta }{u}_{+}(t)D_N(t)dt\to 0,{\int }_0^{\delta }{u}_{-}(t)D_N(t)dt\to 0.$ Hence the first integral in step 2.1 tends to $0$.

Define $\psi (t):={\mathbf{1}}_{[\delta ,1\mathrm{/}2]}(t)\frac{(f(x+t)+f(x-t)-2s)e^{i\pi t}}{\sin (\pi t)}\mathrm{.}$ Since $\sin (\pi t)$ is bounded away from $0$ on $[\delta ,1\mathrm{/}2]$ and the numerator is integrable there, $\psi \in L^1(\mathbb{T})$. By [L5], the second integral in step 2.1 equals $\mathrm{Im}{\int }_0^1\psi (t)e^{2\pi iNt}dt=\mathrm{Im}\widehat{\psi }(-N),$ so it tends to $0$ by [L4].

Steps 3.1 and 3.2 make both integrals in step 2.1 tend to $0$. Therefore $S_{N}f(x)-s\to 0$, which is exactly $S_{N}f(x)\to \frac{f(x+)+f(x-)}{2}\mathrm{.}$