Abel means converge in L^p, uniformly, and at Lebesgue points

Let $g$ be one-periodic and in $L^p([0,1])$ for some $1\le p<\mathrm{\infty }$. Using [L1] and the positivity and unit mass in [L2], Jensen's inequality gives $\mathrm{\mid }{A}_{r}g(x){\mathrm{\mid }}^p\le {\int }_0^1\mathrm{\mid }g(x-t){\mathrm{\mid }}^p{P}_r(t)dt\mathrm{.}$ Integrating in $x$ over $[0,1]$ shows $\mathrm{\parallel }{A}_{r}g{\mathrm{\parallel }}_{L^p([0,1])}\le \mathrm{\parallel }g{\mathrm{\parallel }}_{L^p([0,1])},$ and hence $\mathrm{\parallel }{A}_{r}g-A_{r}h{\mathrm{\parallel }}_{L^p([0,1])}\le \mathrm{\parallel }g-h{\mathrm{\parallel }}_{L^p([0,1])}\mathrm{.}$

Assume now that $f$ is continuous, and let $\epsilon >0$. Uniform continuity modulo $1$ gives $\delta \in (0,1\mathrm{/}2]$ such that $\mathrm{\mid }f(x-t)-f(x)\mathrm{\mid }<\epsilon \mathrm{/}2$ whenever $t\in [0,\delta ]\cup [1-\delta ,1]$. Using [L1] and [L2] exactly as in the Fejer proof yields $\mathrm{\mid }{A}_{r}f(x)-f(x)\mathrm{\mid }\le \epsilon \mathrm{/}2+2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\delta }^{1-\delta }{P}_r(t)dt\mathrm{.}$ By [L2], the far term is $<\epsilon \mathrm{/}2$ for all $r$ close enough to $1$, uniformly in $x$. Therefore $A_{r}f\to f$ uniformly.

Assume $x$ is a Lebesgue point of $f$, and let $\epsilon >0$. By [L3], choose $\delta \in (0,1\mathrm{/}2]$ so that ${\int }_{-h}^h\mathrm{\mid }f(x-t)-f(x)\mathrm{\mid }dt<\frac{\epsilon }{4}h(0<h\le \delta )\mathrm{.}$ Define $G(h):={\int }_0^h(\mathrm{\mid }f(x+t)-f(x)\mathrm{\mid }+\mathrm{\mid }f(x-t)-f(x)\mathrm{\mid })dt,$ so $G(h)<\epsilon h\mathrm{/}4$ for $0<h\le \delta$. Pairing $[0,1\mathrm{/}2]$ and $[1\mathrm{/}2,1]$ in [L1] gives $\mathrm{\mid }{A}_{r}f(x)-f(x)\mathrm{\mid }\le {\int }_0^{1\mathrm{/}2}(\mathrm{\mid }f(x+t)-f(x)\mathrm{\mid }+\mathrm{\mid }f(x-t)-f(x)\mathrm{\mid })P_r(t)dt\mathrm{.}$ Set $a_r:=\min (\delta ,1-r)$. On $(0,a_r)$, the closed form in [L2] gives $P_r(t)\le 2\mathrm{/}(1-r)$, so this interval contributes at most $\epsilon \mathrm{/}2$. If $a_r<\delta$ and $r\ge 1\mathrm{/}2$, then $\sin (\pi t)\ge 2t$ on $[a_r,\delta ]$, so [L2] gives $P_r(t)\le \frac{1-r^2}{16r{t}^2}\le \frac{1-r}{4t^2}\mathrm{.}$ The same integration-by-parts estimate as in the Fejer proof shows that $[a_r,\delta ]$ contributes at most $3\epsilon \mathrm{/}16$. Finally, $[\delta ,1\mathrm{/}2]$ contributes $o(1)$ as $r\uparrow 1$ by [L2]. Hence $A_{r}f(x)\to f(x)$.

Let $1\le p<\mathrm{\infty }$ and assume $f{\mathrm{\mid }}_{[0,1]}\in L^p([0,1])$. Let $\epsilon >0$. Repeating the construction from the Fejer $L^p$ theorem with [L5], one obtains a continuous one-periodic function $u$ such that $\mathrm{\parallel }f-u{\mathrm{\parallel }}_{L^p([0,1])}<\epsilon \mathrm{/}3.$ Then step 1.1 and the uniform convergence of step 1.2 give $\mathrm{\parallel }{A}_{r}f-f{\mathrm{\parallel }}_{L^p([0,1])}\le \mathrm{\parallel }{A}_r(f-u){\mathrm{\parallel }}_{L^p([0,1])}+\mathrm{\parallel }{A}_{r}u-u{\mathrm{\parallel }}_{L^p([0,1])}+\mathrm{\parallel }u-f{\mathrm{\parallel }}_{L^p([0,1])}<\epsilon$ for all $r$ sufficiently close to $1$. Hence $A_{r}f\to f$ in $L^p([0,1])$.

Step 1.3 proves the pointwise conclusion at every Lebesgue point, and [L4] therefore gives the almost-everywhere convergence.