Fejer and Poisson Summability of Fourier Series

Expanding the average in [L1] gives ${\sum }_{m=0}^N{D}_m(t)={\sum }_{m=0}^N{\sum }_{\mathrm{\mid }k\mathrm{\mid }\le m}{e}_k(t)={\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}(N+1-\mathrm{\mid }k\mathrm{\mid })e_k(t)\mathrm{.}$ On the other hand, ${\mid {\sum }_{j=0}^N{e}_j(t)\mid }^2={\sum }_{j=0}^N{\sum }_{\mathrm{\ell }=0}^N{e}_{j-\mathrm{\ell }}(t)={\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}(N+1-\mathrm{\mid }k\mathrm{\mid })e_k(t)\mathrm{.}$ Therefore $F_N(t)=\frac{1}{N+1}{\mid {\sum }_{j=0}^N{e}_j(t)\mid }^2\mathrm{.}$

If $t\mathrm{\notin }\mathbb{Z}$, the finite geometric-series formula gives ${\sum }_{j=0}^N{e}_j(t)={\sum }_{j=0}^N{e}^{2\pi ijt}=e^{\pi iNt}\frac{\sin ((N+1)\pi t)}{\sin (\pi t)},$ so step 1.1 yields the displayed square formula. This proves $F_N(t)\ge 0$ for every $t\mathrm{\notin }\mathbb{Z}$, and at integers the same formula extends by continuity to $F_N(0)=N+1$. Also [L1] gives ${\int }_0^1{F}_N(t)dt=\frac{1}{N+1}{\sum }_{m=0}^N{\int }_0^1{D}_m(t)dt=1.$

For $t\in [\delta ,1-\delta ]$, one has $\mathrm{\mid }\sin (\pi t)\mathrm{\mid }\ge \sin (\pi \delta )$. Using step 2.1 and $\mathrm{\mid }\sin ((N+1)\pi t)\mathrm{\mid }\le 1$ therefore gives $F_N(t)\le \frac{1}{(N+1){\sin }^2(\pi \delta )}\mathrm{.}$ Integrating over an interval of length at most $1$ yields ${\int }_{\delta }^{1-\delta }{F}_N(t)dt\le \frac{1}{(N+1){\sin }^2(\pi \delta )}\to 0.$

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 $L^p([0,1])$.

Let $\epsilon >0$. Because $f$ is continuous on the compact interval $[0,1]$ and one-periodic, it is uniformly continuous modulo $1$. Choose $\delta \in (0,1\mathrm{/}2]$ such that $\mathrm{\mid }f(x-t)-f(x)\mathrm{\mid }<\epsilon \mathrm{/}2$ whenever $x\in \mathbb{R}$ and $t\in [0,\delta ]\cup [1-\delta ,1]$.

For every $x\in \mathbb{R}$, subtract $f(x)$ inside the integral from [L1]: $\mathrm{\mid }{\sigma }_{N}f(x)-f(x)\mathrm{\mid }\le {\int }_0^1\mathrm{\mid }f(x-t)-f(x)\mathrm{\mid }{F}_N(t)dt\mathrm{.}$ Split the integral into the near set $[0,\delta ]\cup [1-\delta ,1]$ and the far set $[\delta ,1-\delta ]$. By step 1.1 and the positivity from [L2], the near part is at most $\epsilon \mathrm{/}2$. The far part is at most $2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\delta }^{1-\delta }{F}_N(t)dt\mathrm{.}$

By [L2], choose $N_0$ so large that $2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\delta }^{1-\delta }{F}_N(t)dt<\epsilon \mathrm{/}2$ for all $N\ge N_0$. Then step 2.1 gives ${\sup }_{x\in \mathbb{R}}\mathrm{\mid }{\sigma }_{N}f(x)-f(x)\mathrm{\mid }<\epsilon (N\ge N_0)\mathrm{.}$ Since $\epsilon$ was arbitrary, the convergence is uniform.

Let $\epsilon >0$. By [L3], choose $\delta \in (0,1\mathrm{/}2]$ so that ${\int }_{-h}^h\mathrm{\mid }f(x-t)-f(x)\mathrm{\mid }dt<\frac{\epsilon }{4}h(0<h\le \delta )\mathrm{.}$ For $0<h\le \delta$, put $G(h):={\int }_0^h(\mathrm{\mid }f(x+t)-f(x)\mathrm{\mid }+\mathrm{\mid }f(x-t)-f(x)\mathrm{\mid })dt\mathrm{.}$ Then $G(h)<\epsilon h\mathrm{/}4$ for $0<h\le \delta$.

Using [L1], pair the intervals $[0,1\mathrm{/}2]$ and $[1\mathrm{/}2,1]$ exactly as in the Dirichlet symmetric-difference formula. This gives $\mathrm{\mid }{\sigma }_{N}f(x)-f(x)\mathrm{\mid }\le {\int }_0^{1\mathrm{/}2}(\mathrm{\mid }f(x+t)-f(x)\mathrm{\mid }+\mathrm{\mid }f(x-t)-f(x)\mathrm{\mid })F_N(t)dt\mathrm{.}$ Set $a_N:=\min (\delta ,(N+1{)}^{-1})$. Since step 1.1 gives $G(a_N)<\epsilon a_N\mathrm{/}4$ and the square formula in [L2] yields $F_N(t)\le N+1$, the interval $(0,a_N)$ contributes at most $\epsilon \mathrm{/}4$. If $a_N<(\delta )$, then for $t\in [a_N,\delta ]$ one has $\sin (\pi t)\ge 2t$, so [L2] gives $F_N(t)\le \frac{1}{4(N+1)t^2}\mathrm{.}$ Integration by parts with $G^{\mathrm{\prime }}(t)$ equal almost everywhere to the displayed integrand therefore gives ${\int }_{a_N}^{\delta }\frac{G^{\mathrm{\prime }}(t)}{t^2}dt=\frac{G(\delta )}{{\delta }^2}-\frac{G(a_N)}{a_N^2}+2{\int }_{a_N}^{\delta }\frac{G(t)}{t^3}dt\le \frac{3\epsilon }{4a_N},$ so the interval $[a_N,\delta ]$ contributes at most $3\epsilon \mathrm{/}16$. Consequently ${\int }_0^{\delta }(\mathrm{\mid }f(x+t)-f(x)\mathrm{\mid }+\mathrm{\mid }f(x-t)-f(x)\mathrm{\mid })F_N(t)dt\le \frac{7\epsilon }{16}$ for every $N$.

On $[\delta ,1\mathrm{/}2]$, the integrand is integrable and [L2] gives ${\sup }_{t\in [\delta ,1-\delta ]}{F}_N(t)\to 0.$ Hence ${\int }_{\delta }^{1\mathrm{/}2}(\mathrm{\mid }f(x+t)-f(x)\mathrm{\mid }+\mathrm{\mid }f(x-t)-f(x)\mathrm{\mid })F_N(t)dt\to 0.$ Choose $N_0$ so large that this far contribution is $<9\epsilon \mathrm{/}16$ for all $N\ge N_0$. Then step 2.1 yields $\mathrm{\mid }{\sigma }_{N}f(x)-f(x)\mathrm{\mid }<\epsilon (N\ge N_0)\mathrm{.}$ Thus ${\sigma }_{N}f(x)\to f(x)$.

The first claim holds at every Lebesgue point by step 3.1. Applying [L4] therefore gives ${\sigma }_{N}f(x)\to f(x)$ for almost every $x$.

Let $z=e^{2\pi it}$. By [L1], $P_r(t)=1+{\sum }_{k=1}^{\mathrm{\infty }}{r}^k{z}^k+{\sum }_{k=1}^{\mathrm{\infty }}{r}^k{z}^{-k}\mathrm{.}$ Both geometric series converge absolutely, so $P_r(t)=1+\frac{rz}{1-rz}+\frac{r{z}^{-1}}{1-r{z}^{-1}}=\frac{1-r^2}{(1-rz)(1-r{z}^{-1})}\mathrm{.}$ Since $z+z^{-1}=2\cos (2\pi t)$, this is exactly $P_r(t)=\frac{1-r^2}{1-2r\cos (2\pi t)+r^2}\mathrm{.}$

Step 1.1 shows $P_r(t)\ge 0$ because $1-2r\cos (2\pi t)+r^2=(1-r{)}^2+2r(1-\cos (2\pi t))\ge 0,$ and the numerator is positive for $r<1$. Also the constant Fourier coefficient of $P_r$ is $1$, so ${\int }_0^1{P}_r(t)dt=1.$

The limit only concerns $r\uparrow 1$, so it is enough to consider $r\in [1\mathrm{/}2,1)$. If $t\in [\delta ,1-\delta ]$, then $1-2r\cos (2\pi t)+r^2=(1-r{)}^2+4r{\sin }^2(\pi t)\ge 2{\sin }^2(\pi \delta )\mathrm{.}$ Therefore step 1.1 gives $P_r(t)\le \frac{1-r^2}{2{\sin }^2(\pi \delta )}\mathrm{.}$ As $r\uparrow 1$, the right-hand side tends to $0$, so the displayed supremum tends to $0$. Multiplying that supremum bound by the interval length at most $1$ gives the same limit for the tail integral.

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.

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 $A_{r}f(x)\to s$.

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, ${\sigma }_{N}f(x)\to m$.