Dini pointwise convergence criterion for Fourier series

Put $a(t):=f(x+t)+f(x-t)-2s$. By [L3], $S_{N}f(x)-s={\int }_0^{\delta }a(t)D_N(t)dt+{\int }_{\delta }^{1\mathrm{/}2}a(t)D_N(t)dt\mathrm{.}$

For $0<t\le \delta <1\mathrm{/}2$, one has $0<\pi t<\pi \mathrm{/}2<2$, so [L4] gives $\sin (\pi t)\ge \pi t\mathrm{/}3$. Define ${\psi }_0(t):={\mathbf{1}}_{(0,\delta ]}(t)\frac{a(t)e^{i\pi t}}{\sin (\pi t)}\mathrm{.}$ Then $\mathrm{\mid }{\psi }_0(t)\mathrm{\mid }\le \frac{3}{\pi }{\mathbf{1}}_{(0,\delta ]}(t)\frac{\mathrm{\mid }a(t)\mathrm{\mid }}{t},$ so the hypothesis makes ${\psi }_0\in L^1(\mathbb{T})$. Using [L1], ${\int }_0^{\delta }a(t)D_N(t)dt=\mathrm{Im}{\int }_0^1{\psi }_0(t)e^{2\pi iNt}dt=\mathrm{Im}{\widehat{\psi }}_0(-N)\mathrm{.}$

Define ${\psi }_1(t):={\mathbf{1}}_{[\delta ,1\mathrm{/}2]}(t)\frac{a(t)e^{i\pi t}}{\sin (\pi t)}\mathrm{.}$ Since $\sin (\pi t)$ is bounded away from $0$ on $[\delta ,1\mathrm{/}2]$ and $a\in L^1([\delta ,1\mathrm{/}2])$, one has ${\psi }_1\in L^1(\mathbb{T})$. Again [L1] turns the second integral in step 1.1 into ${\int }_{\delta }^{1\mathrm{/}2}a(t)D_N(t)dt=\mathrm{Im}{\widehat{\psi }}_1(-N)\mathrm{.}$

By [L2], both ${\widehat{\psi }}_0(-N)$ and ${\widehat{\psi }}_1(-N)$ tend to $0$. Steps 2.1 and 2.2 therefore make both integrals in step 1.1 tend to $0$. Hence $S_{N}f(x)-s\to 0$, or equivalently $S_{N}f(x)\to s$.