Cesaro summability implies Abel summability

Let $S_N:=S_{N}f(x)$ and ${\sigma }_N:={\sigma }_{N}f(x)$. Since $S_N=(N+1){\sigma }_N-N{\sigma }_{N-1}(N\ge 0,{\sigma }_{-1}:=0),$ one has ${\sum }_{N=0}^{\mathrm{\infty }}{r}^N{S}_N=(1-r){\sum }_{N=0}^{\mathrm{\infty }}(N+1)r^N{\sigma }_N$ for $0\le r<1$. On the other hand, the definition of $A_{r}f(x)$ in [L1] gives $A_{r}f(x)=(1-r){\sum }_{N=0}^{\mathrm{\infty }}{r}^N{S}_N=(1-r{)}^2{\sum }_{N=0}^{\mathrm{\infty }}(N+1)r^N{\sigma }_N\mathrm{.}$

Put $w_N(r):=(1-r{)}^2(N+1)r^N(N\ge 0)\mathrm{.}$ These weights are nonnegative, and ${\sum }_{N=0}^{\mathrm{\infty }}{w}_N(r)=(1-r{)}^2{\sum }_{N=0}^{\mathrm{\infty }}(N+1)r^N=1.$ Therefore step 1.1 rewrites the Abel mean as $A_{r}f(x)-s={\sum }_{N=0}^{\mathrm{\infty }}{w}_N(r)({\sigma }_N-s)\mathrm{.}$

Let $\epsilon >0$. Choose $N_0$ so large that $\mathrm{\mid }{\sigma }_N-s\mathrm{\mid }<\epsilon \mathrm{/}2$ for all $N\ge N_0$, and put $M:={\max }_{0\le N<N_0}\mathrm{\mid }{\sigma }_N-s\mathrm{\mid }\mathrm{.}$ By step 2.1, $\mathrm{\mid }{A}_{r}f(x)-s\mathrm{\mid }\le M{\sum }_{N=0}^{N_0-1}{w}_N(r)+\frac{\epsilon }{2}{\sum }_{N=N_0}^{\mathrm{\infty }}{w}_N(r)\mathrm{.}$ The tail sum is at most $\epsilon \mathrm{/}2$, and for each fixed $N$ one has $w_N(r)\to 0$ as $r\uparrow 1$, so the finite initial sum is $<\epsilon \mathrm{/}(2M)$ for $r$ close enough to $1$ when $M>0$, and is already $0$ when $M=0$. Hence $\mathrm{\mid }{A}_{r}f(x)-s\mathrm{\mid }<\epsilon$ for all $r$ sufficiently close to $1$. Therefore $A_{r}f(x)\to s$.