Fejer means converge to midpoint values at jumps

Put $m:=(f(x-)+f(x+))\mathrm{/}2$. Using [L1], split the integral over $[0,1]$ at $1\mathrm{/}2$ and substitute $u=1-t$ on $[1\mathrm{/}2,1]$. Because $F_N(1-u)=F_N(u)$, this gives ${\sigma }_{N}f(x)-m={\int }_0^{1\mathrm{/}2}(f(x-t)-f(x-)+f(x+t)-f(x+))F_N(t)dt\mathrm{.}$ Also $2{\int }_0^{1\mathrm{/}2}{F}_N(t)dt={\int }_0^1{F}_N(t)dt=1.$

Let $\epsilon >0$. Choose $\delta \in (0,1\mathrm{/}2]$ so that $\mathrm{\mid }f(x-t)-f(x-)\mathrm{\mid }<\epsilon ,\mathrm{\mid }f(x+t)-f(x+)\mathrm{\mid }<\epsilon (0<t<\delta )\mathrm{.}$ Then the interval $(0,\delta )$ contributes at most $2\epsilon {\int }_0^{1\mathrm{/}2}{F}_N(t)dt=\epsilon$ by step 1.1. The interval $[\delta ,1\mathrm{/}2]$ contributes at most $({\int }_0^1\mathrm{\mid }f(y)\mathrm{\mid }dy+{\int }_0^1\mathrm{\mid }f(y)\mathrm{\mid }dy+\frac{\mathrm{\mid }f(x-)\mathrm{\mid }+\mathrm{\mid }f(x+)\mathrm{\mid }}{2}){\sup }_{t\in [\delta ,1-\delta ]}{F}_N(t),$ which tends to $0$ by [L2].

Choose $N_0$ so large that the far contribution in step 2.1 is $<\epsilon$ for $N\ge N_0$. Then $\mathrm{\mid }{\sigma }_{N}f(x)-m\mathrm{\mid }<2\epsilon (N\ge N_0)\mathrm{.}$ Since $\epsilon$ was arbitrary, ${\sigma }_{N}f(x)\to m$.