Fejer means converge in L^p for 1 <= p < infinity

Let $g$ be any one-periodic member of $L^p([0,1])$. By [L1] and the positivity and unit mass from [L2], Jensen's inequality gives $\mathrm{\mid }{\sigma }_{N}g(x){\mathrm{\mid }}^p={\mid {\int }_0^{1}g(x-t)F_N(t)dt\mid }^p\le {\int }_0^1\mathrm{\mid }g(x-t){\mathrm{\mid }}^p{F}_N(t)dt\mathrm{.}$ Integrating in $x$ over $[0,1]$ and using one-periodicity yields $\mathrm{\parallel }{\sigma }_{N}g{\mathrm{\parallel }}_{L^p([0,1])}\le \mathrm{\parallel }g{\mathrm{\parallel }}_{L^p([0,1])}\mathrm{.}$ Applying this to $g-h$ shows $\mathrm{\parallel }{\sigma }_{N}g-{\sigma }_{N}h{\mathrm{\parallel }}_{L^p([0,1])}\le \mathrm{\parallel }g-h{\mathrm{\parallel }}_{L^p([0,1])}\mathrm{.}$

If $u$ is continuous and one-periodic, then [L3] gives ${\sup }_{x\in \mathbb{R}}\mathrm{\mid }{\sigma }_{N}u(x)-u(x)\mathrm{\mid }\to 0.$ Hence $\mathrm{\parallel }{\sigma }_{N}u-u{\mathrm{\parallel }}_{L^p([0,1])}\le {\sup }_{x\in \mathbb{R}}\mathrm{\mid }{\sigma }_{N}u(x)-u(x)\mathrm{\mid }\to 0.$

Let $\epsilon >0$. Because $\mathrm{\mid }f{\mathrm{\mid }}^p$ is integrable on $[0,1]$, choose $a\in (0,1\mathrm{/}6)$ so that ${\int }_0^a\mathrm{\mid }f(x){\mathrm{\mid }}^{p}dx+{\int }_{1-a}^1\mathrm{\mid }f(x){\mathrm{\mid }}^{p}dx<(\epsilon \mathrm{/}6{)}^p\mathrm{.}$ Define $F_a:\mathbb{R}\to \mathbb{C}$ by $F_a(x)=f(x)$ for $x\in [a,1-a]$ and $F_a(x)=0$ otherwise. Then [L4] gives $g\in C_c(\mathbb{R})$ with $\mathrm{\parallel }{F}_a-g{\mathrm{\parallel }}_{L^p(\mathbb{R})}<\epsilon \mathrm{/}6.$ Let ${\eta }_a$ be the piecewise linear cutoff that is $0$ on $(-\mathrm{\infty },a\mathrm{/}2]\cup [1-a\mathrm{/}2,\mathrm{\infty })$, $1$ on $[a,1-a]$, and linear on $[a\mathrm{/}2,a]$ and $[1-a,1-a\mathrm{/}2]$. Put $h:={\eta }_{a}g$. Then $h\in C_c((0,1))$ and, because ${\eta }_a=1$ on $[a,1-a]$ where $F_a$ is supported, $\mathrm{\parallel }h-F_a{\mathrm{\parallel }}_{L^p(\mathbb{R})}\le \mathrm{\parallel }g-F_a{\mathrm{\parallel }}_{L^p(\mathbb{R})}<\epsilon \mathrm{/}6.$ Now periodize $h$ by $u(x):={\sum }_{m\in \mathbb{Z}}h(x-m)\mathrm{.}$ Since $\mathrm{supp}(h)$ is a compact subset of $(0,1)$, at most one summand is nonzero at each $x$, so $u$ is continuous and one-periodic. On $[0,1]$ only the $m=0$ summand can contribute, hence $u=h$ there. Therefore $\mathrm{\parallel }f-u{\mathrm{\parallel }}_{L^p([0,1])}\le \mathrm{\parallel }f-F_a{\mathrm{\parallel }}_{L^p([0,1])}+\mathrm{\parallel }{F}_a-h{\mathrm{\parallel }}_{L^p(\mathbb{R})}<\epsilon \mathrm{/}3.$

Choose $u$ as in step 1.3. Then $\mathrm{\parallel }{\sigma }_{N}f-f{\mathrm{\parallel }}_{L^p([0,1])}\le \mathrm{\parallel }{\sigma }_N(f-u){\mathrm{\parallel }}_{L^p([0,1])}+\mathrm{\parallel }{\sigma }_{N}u-u{\mathrm{\parallel }}_{L^p([0,1])}+\mathrm{\parallel }u-f{\mathrm{\parallel }}_{L^p([0,1])}\mathrm{.}$ Step 1.1 bounds the first term by $\mathrm{\parallel }f-u{\mathrm{\parallel }}_{L^p([0,1])}<\epsilon \mathrm{/}3$, and step 1.2 makes the middle term $<\epsilon \mathrm{/}3$ for all large $N$. Thus $\mathrm{\parallel }{\sigma }_{N}f-f{\mathrm{\parallel }}_{L^p([0,1])}<\epsilon$ for all large $N$. Since $\epsilon >0$ was arbitrary, ${\sigma }_{N}f\to f$ in $L^p([0,1])$.