Dirichlet Kernel Localisation and Pointwise Fourier Convergence - Examples

For $0<t<1$, [L1] gives $D_N(1-t)=D_N(t)$, so ${\int }_0^1\mathrm{\mid }{D}_N(t)\mathrm{\mid }dt=2{\int }_0^{1\mathrm{/}2}\mathrm{\mid }{D}_N(t)\mathrm{\mid }dt\mathrm{.}$ Split the last integral at $1\mathrm{/}(2N+1)$. On $(0,1\mathrm{/}(2N+1)]$, [L1] and the bound $\mathrm{\mid }{D}_N(t)\mathrm{\mid }\le 2N+1$ give a contribution at most $1$. On $[1\mathrm{/}(2N+1),1\mathrm{/}2]$, one has $0<\pi t\le \pi \mathrm{/}2<2$, so [L2] implies $\sin (\pi t)\ge \pi t\mathrm{/}3$. Using [L1], $\mathrm{\mid }{D}_N(t)\mathrm{\mid }\le \frac{1}{\sin (\pi t)}\le \frac{3}{\pi t}\mathrm{.}$ Therefore ${\int }_0^1\mathrm{\mid }{D}_N(t)\mathrm{\mid }dt\le 2+\frac{6}{\pi }{\int }_{1\mathrm{/}(2N+1)}^{1\mathrm{/}2}\frac{dt}{t}\le C\log (N+1)$ for a universal $C$.

For $m=0,\dots ,N-1$, let $J_m:=[\frac{m+1\mathrm{/}6}{2N+1},\frac{m+5\mathrm{/}6}{2N+1}]\mathrm{.}$ These intervals lie in $(0,1\mathrm{/}2)$ and are disjoint. If $t\in J_m$, then $(2N+1)\pi t\in [(m+1\mathrm{/}6)\pi ,(m+5\mathrm{/}6)\pi ]$, so $\mathrm{\mid }\sin ((2N+1)\pi t)\mathrm{\mid }\ge 1\mathrm{/}2$. Also $\sin (\pi t)\le \pi t$. Hence [L1] gives $\mathrm{\mid }{D}_N(t)\mathrm{\mid }\ge \frac{1}{2\pi t}(t\in J_m)\mathrm{.}$

Integrating the lower bound from step 1.2 over each $J_m$ and summing yields ${\int }_0^{1\mathrm{/}2}\mathrm{\mid }{D}_N(t)\mathrm{\mid }dt\ge \frac{1}{2\pi }{\sum }_{m=0}^{N-1}{\int }_{J_m}\frac{dt}{t}=\frac{1}{2\pi }{\sum }_{m=0}^{N-1}\log \text{\hspace{0.17em}⁣}\left(\frac{m+5\mathrm{/}6}{m+1\mathrm{/}6}\right)\mathrm{.}$ Since $\log \text{\hspace{0.17em}⁣}\left(\frac{m+5\mathrm{/}6}{m+1\mathrm{/}6}\right)\ge \frac{2\mathrm{/}3}{m+5\mathrm{/}6}\ge \frac{2}{3(m+1)},$ one gets ${\int }_0^{1\mathrm{/}2}\mathrm{\mid }{D}_N(t)\mathrm{\mid }dt\ge \frac{1}{3\pi }{\sum }_{m=0}^{N-1}\frac{1}{m+1}\ge \frac{1}{3\pi }{\int }_1^{N+1}\frac{dt}{t}=\frac{1}{3\pi }\log (N+1)\mathrm{.}$ Using step 1.1 once more, ${\int }_0^1\mathrm{\mid }{D}_N(t)\mathrm{\mid }dt\ge \frac{2}{3\pi }\log (N+1)\mathrm{.}$

Steps 1.1 and 2.1 give the two-sided logarithmic bound.

At $t=0$, [L1] gives $D_N(0)={\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}{e}_k(0)={\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}1=2N+1,$ which agrees with the removable value recorded in [L2].

The numbers $(4N+2{)}^{-1}$ and $3(4N+2{)}^{-1}$ are not integers, so [L2] applies. Since $(2N+1)\pi \mathrm{/}(4N+2)=\pi \mathrm{/}2$ and $3(2N+1)\pi \mathrm{/}(4N+2)=3\pi \mathrm{/}2$, the numerator is respectively $1$ and $-1$. The denominators are positive because their angles lie in $(0,\pi )$. Hence the two displayed signs follow.

Since ${\int }_0^1(x-\frac{1}{2})dx=0$, [L1] gives $\widehat{f}(0)=0$. For $k\mathrm{\ne }0$, direct integration gives $\widehat{f}(k)={\int }_0^1(x-\frac{1}{2})e^{-2\pi ikx}dx=-\frac{1}{2\pi ik}\mathrm{.}$

Insert the coefficients from step 1.1 into the partial-sum formula [L1]. Pairing the $k$ and $-k$ terms gives $S_{N}f(x)=-{\sum }_{k=1}^N\frac{\sin (2\pi kx)}{\pi k}\mathrm{.}$

The sawtooth is piecewise $C^1$, hence of bounded variation on one period. If $x\mathrm{\notin }\mathbb{Z}$, choose the unique integer $m$ with $y:=x-m\in (0,1)$. Because both $f$ and its Fourier partial sums are one-periodic, one has $S_{N}f(x)=S_{N}f(y)$ and $f(x)=f(y)=y-\frac{1}{2}$. The one-sided limits at $y$ therefore both equal $f(x)$, so [L2] gives $S_{N}f(x)\to f(x)$. At an integer $x$, the one-sided limits are $-1\mathrm{/}2$ and $1\mathrm{/}2$, whose midpoint is $0$, so [L2] gives $S_{N}f(x)\to 0$.

On the interval $(-1\mathrm{/}8,1\mathrm{/}8)$ modulo $1$, the indicator $g$ vanishes, so $f=g=0$ there. Thus the hypothesis of [L1] holds at $x=0$.

Applying [L1] at $x=0$ gives $S_{N}g(0)-S_{N}f(0)\to 0$. But every Fourier coefficient of the zero function is $0$, so $S_{N}f(0)=0$ for all $N$. Therefore $S_{N}g(0)\to 0$.

Let $d(x,\mathbb{Z}):={\min }_{m\in \mathbb{Z}}\mathrm{\mid }x-m\mathrm{\mid }$ and define the one-periodic function $f(x):=\{\begin{array}{ll}0,& d(x,\mathbb{Z})=0,\\ {\displaystyle \frac{1}{\log (e\mathrm{/}d(x,\mathbb{Z}))},}& d(x,\mathbb{Z})>0.\end{array}$ Because $d(x,\mathbb{Z})\to 0$ exactly when $x$ approaches an integer and $1\mathrm{/}\log (e\mathrm{/}r)\to 0$ as $r\downarrow 0$, the function $f$ is continuous on $\mathbb{T}$.

At $x=0$ one has $f(0)=0$ and, for $0<t\le 1\mathrm{/}4$, $d(t,\mathbb{Z})=d(-t,\mathbb{Z})=t$. Hence $f(t)+f(-t)-2f(0)=\frac{2}{\log (e\mathrm{/}t)}\mathrm{.}$ Therefore ${\int }_0^{1\mathrm{/}4}\frac{\mathrm{\mid }f(t)+f(-t)-2f(0)\mathrm{\mid }}{t}dt=2{\int }_0^{1\mathrm{/}4}\frac{dt}{t\log (e\mathrm{/}t)}=\mathrm{\infty },$ because the change of variables $u=\log (e\mathrm{/}t)$ turns the integral into ${\int }_{\log (4e)}^{\mathrm{\infty }}du\mathrm{/}u$. So the hypothesis in [L1] fails at $x=0$: $f$ is continuous but does not satisfy the Dini condition there.