Dirichlet Kernel Localisation and Pointwise Fourier Convergence

Expanding the convolution against the finite sum [L1, L2, algebra] gives ${\int }_0^{1}f(x-t)D_N(t)dt={\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}{\int }_0^{1}f(x-t)e^{2\pi ikt}dt\mathrm{.}$

For each $k$, substitute $u=x-t$. Then ${\int }_0^{1}f(x-t)e^{2\pi ikt}dt=e^{2\pi ikx}{\int }_{x-1}^{x}f(u)e^{-2\pi iku}du\mathrm{.}$ The integrand $u\mapsto f(u)e^{-2\pi iku}$ is one-periodic, so its integral over $[x-1,x]$ equals its integral over $[0,1]$, namely $\widehat{f}(k)$. Therefore ${\int }_0^{1}f(x-t)e^{2\pi ikt}dt=\widehat{f}(k)e_k(x)\mathrm{.}$

Summing step 2.1 over $\mathrm{\mid }k\mathrm{\mid }\le N$ yields ${\int }_0^{1}f(x-t)D_N(t)dt={\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}\widehat{f}(k)e_k(x)=S_{N}f(x)\mathrm{.}$ By [L1], the integral is also $(f\ast D_N)(x)$.

Pair the terms with indices $k$ and $-k$ in [L1]. This gives $D_N(t)=1+{\sum }_{k=1}^N(e^{2\pi ikt}+e^{-2\pi ikt})=1+2{\sum }_{k=1}^N\cos (2\pi kt)\mathrm{.}$ Hence $D_N$ is even. If $t=m\in \mathbb{Z}$, then every exponential equals $1$, so $D_N(m)=2N+1$.

Assume $t\mathrm{\notin }\mathbb{Z}$ and put $z=e^{2\pi it}\mathrm{\ne }1$. Then [L1, algebra] $D_N(t)=z^{-N}{\sum }_{j=0}^{2N}{z}^j=z^{-N}\frac{1-z^{2N+1}}{1-z}\mathrm{.}$ Rewriting numerator and denominator with half-angle factors yields $D_N(t)=\frac{e^{-\pi i(2N+1)t}-e^{\pi i(2N+1)t}}{e^{-\pi it}-e^{\pi it}}=\frac{\sin ((2N+1)\pi t)}{\sin (\pi t)}\mathrm{.}$

The triangle inequality applied to [L1] gives $\mathrm{\mid }{D}_N(t)\mathrm{\mid }\le 2N+1$ for every $t$. If $t\mathrm{\notin }\mathbb{Z}$, step 1.2 and $\mathrm{\mid }\sin ((2N+1)\pi t)\mathrm{\mid }\le 1$ give $\mathrm{\mid }{D}_N(t)\mathrm{\mid }\le \frac{1}{\mathrm{\mid }\sin (\pi t)\mathrm{\mid }}\mathrm{.}$

Write $s={\sum }_{j=1}^m{a}_j{\mathbf{1}}_{[x_{j-1},x_j)}$ for a partition $0=x_0<\cdots <x_m=1$. For $k\mathrm{\ne }0$, [L1, algebra] $\widehat{{\mathbf{1}}_{[x_{j-1},x_j)}}(k)={\int }_{x_{j-1}}^{x_j}{e}^{-2\pi ikt}dt=\frac{e^{-2\pi ik{x}_{j-1}}-e^{-2\pi ik{x}_j}}{2\pi ik}\mathrm{.}$ Therefore $\mid \widehat{{\mathbf{1}}_{[x_{j-1},x_j)}}(k)\mid \le \frac{1}{\pi \mathrm{\mid }k\mathrm{\mid }}\mathrm{.}$

By linearity, $\widehat{s}(k)={\sum }_{j=1}^m{a}_j\widehat{{\mathbf{1}}_{[x_{j-1},x_j)}}(k)\mathrm{.}$ Step 1.1 then gives $\mathrm{\mid }\widehat{s}(k)\mathrm{\mid }\le \frac{1}{\pi \mathrm{\mid }k\mathrm{\mid }}{\sum }_{j=1}^m\mathrm{\mid }{a}_j\mathrm{\mid }\mathrm{.}$ Taking $C:={\pi }^{-1}{\sum }_{j=1}^m\mathrm{\mid }{a}_j\mathrm{\mid }$ proves the displayed bound.

Since $C\mathrm{/}\mathrm{\mid }k\mathrm{\mid }\to 0$ as $\mathrm{\mid }k\mathrm{\mid }\to \mathrm{\infty }$, step 2.1 yields $\widehat{s}(k)\to 0$.

Define $g:\mathbb{R}\to \mathbb{R}$ by $g(t)=f(t)$ for $0\le t<1$ and $g(t)=0$ otherwise. Then $g\in L^1(\mathbb{R})$. By [L2], choose a finite linear combination of interval indicators $u$ with ${\int }_{\mathbb{R}}\mathrm{\mid }g(t)-u(t)\mathrm{\mid }dt<\epsilon \mathrm{.}$

Restrict $u$ to $[0,1)$ and call the restriction $s$. Intersecting each interval in $u$ with $[0,1)$ produces only finitely many subintervals, so $s$ is a step function on $[0,1)$. Since $g=f$ on $[0,1)$, ${\int }_0^1\mathrm{\mid }f(t)-s(t)\mathrm{\mid }dt={\int }_0^1\mathrm{\mid }g(t)-u(t)\mathrm{\mid }dt\le {\int }_{\mathbb{R}}\mathrm{\mid }g(t)-u(t)\mathrm{\mid }dt<\epsilon \mathrm{.}$

Step 2.1 is exactly the claimed density statement on one period.

By [L3], choose a one-period step function $s$ with ${\int }_0^1\mathrm{\mid }f(t)-s(t)\mathrm{\mid }dt<\epsilon \mathrm{/}3.$

By [L2], choose $K\ge 1$ such that $\mathrm{\mid }k\mathrm{\mid }\ge K$ implies $\mathrm{\mid }\widehat{s}(k)\mathrm{\mid }<\epsilon \mathrm{/}3$.

For every integer $k$, [L1, algebra] $\mathrm{\mid }\widehat{f}(k)-\widehat{s}(k)\mathrm{\mid }\le {\int }_0^1\mathrm{\mid }f(t)-s(t)\mathrm{\mid }dt<\epsilon \mathrm{/}3.$

If $\mathrm{\mid }k\mathrm{\mid }\ge K$, then step 2.1 and step 1.2 give $\mathrm{\mid }\widehat{f}(k)\mathrm{\mid }\le \mathrm{\mid }\widehat{f}(k)-\widehat{s}(k)\mathrm{\mid }+\mathrm{\mid }\widehat{s}(k)\mathrm{\mid }<\frac{2\epsilon }{3}<\epsilon \mathrm{.}$

Since $\epsilon >0$ was arbitrary, step 3.1 is exactly $\widehat{f}(k)\to 0$ as $\mathrm{\mid }k\mathrm{\mid }\to \mathrm{\infty }$.

By [L1], $S_{N}f(x)-s={\int }_0^1(f(x-t)-s)D_N(t)dt,$ because [L2] gives ${\int }_0^1{D}_N(t)dt=1$.

Split the integral in step 1.1 at $1\mathrm{/}2$ and substitute $u=1-t$ on $[1\mathrm{/}2,1]$. Since $f$ is one-periodic and $D_N(1-u)=D_N(-u)=D_N(u)$ by [L2], this yields $S_{N}f(x)-s={\int }_0^{1\mathrm{/}2}(f(x-t)+f(x+t)-2s)D_N(t)dt\mathrm{.}$

The first displayed formula is step 2.1 with the two summands reordered. Replacing $D_N(t)$ by the closed form from [L3] gives the second displayed formula.

Apply [L3] to $h:=f-g$ with $s=0$. Since $h(y)=0$ for almost every $y\in (x-\delta ,x+\delta )$, the integrand vanishes for almost every $t\in (0,\delta )$, so $S_{N}h(x)={\int }_{\delta }^{1\mathrm{/}2}(h(x+t)+h(x-t))D_N(t)dt\mathrm{.}$

Define a one-period function $\psi$ on $[0,1)$ by $\psi (t):={\mathbf{1}}_{[\delta ,1\mathrm{/}2]}(t)\frac{(h(x+t)+h(x-t))e^{i\pi t}}{\sin (\pi t)}\mathrm{.}$ Because $\sin (\pi t)$ is bounded away from $0$ on $[\delta ,1\mathrm{/}2]$ and $h\in L^1(\mathbb{T})$, one has $\psi \in L^1(\mathbb{T})$. Using [L1], step 1.1 becomes $S_{N}h(x)=\mathrm{Im}{\int }_0^1\psi (t)e^{2\pi iNt}dt=\mathrm{Im}\widehat{\psi }(-N)\mathrm{.}$

By [L2], $\widehat{\psi }(-N)\to 0$ as $N\to \mathrm{\infty }$. Step 2.1 therefore gives $S_{N}h(x)\to 0$. Since $h=f-g$, this is exactly $S_{N}f(x)-S_{N}g(x)\to 0.$

Put $a(t):=f(x+t)+f(x-t)-2s$. By [L3], $S_{N}f(x)-s={\int }_0^{\delta }a(t)D_N(t)dt+{\int }_{\delta }^{1\mathrm{/}2}a(t)D_N(t)dt\mathrm{.}$

For $0<t\le \delta <1\mathrm{/}2$, one has $0<\pi t<\pi \mathrm{/}2<2$, so [L4] gives $\sin (\pi t)\ge \pi t\mathrm{/}3$. Define ${\psi }_0(t):={\mathbf{1}}_{(0,\delta ]}(t)\frac{a(t)e^{i\pi t}}{\sin (\pi t)}\mathrm{.}$ Then $\mathrm{\mid }{\psi }_0(t)\mathrm{\mid }\le \frac{3}{\pi }{\mathbf{1}}_{(0,\delta ]}(t)\frac{\mathrm{\mid }a(t)\mathrm{\mid }}{t},$ so the hypothesis makes ${\psi }_0\in L^1(\mathbb{T})$. Using [L1], ${\int }_0^{\delta }a(t)D_N(t)dt=\mathrm{Im}{\int }_0^1{\psi }_0(t)e^{2\pi iNt}dt=\mathrm{Im}{\widehat{\psi }}_0(-N)\mathrm{.}$

Define ${\psi }_1(t):={\mathbf{1}}_{[\delta ,1\mathrm{/}2]}(t)\frac{a(t)e^{i\pi t}}{\sin (\pi t)}\mathrm{.}$ Since $\sin (\pi t)$ is bounded away from $0$ on $[\delta ,1\mathrm{/}2]$ and $a\in L^1([\delta ,1\mathrm{/}2])$, one has ${\psi }_1\in L^1(\mathbb{T})$. Again [L1] turns the second integral in step 1.1 into ${\int }_{\delta }^{1\mathrm{/}2}a(t)D_N(t)dt=\mathrm{Im}{\widehat{\psi }}_1(-N)\mathrm{.}$

By [L2], both ${\widehat{\psi }}_0(-N)$ and ${\widehat{\psi }}_1(-N)$ tend to $0$. Steps 2.1 and 2.2 therefore make both integrals in step 1.1 tend to $0$. Hence $S_{N}f(x)-s\to 0$, or equivalently $S_{N}f(x)\to s$.

For $0<t<\delta$, the triangle inequality and the hypotheses give $\mathrm{\mid }f(x+t)+f(x-t)-2f(x)\mathrm{\mid }\le \mathrm{\mid }f(x+t)-f(x)\mathrm{\mid }+\mathrm{\mid }f(x-t)-f(x)\mathrm{\mid }\le 2C{t}^{\alpha }\mathrm{.}$

Therefore ${\int }_0^{\delta }\frac{\mathrm{\mid }f(x+t)+f(x-t)-2f(x)\mathrm{\mid }}{t}dt\le 2C{\int }_0^{\delta }{t}^{\alpha -1}dt=\frac{2C}{\alpha }{\delta }^{\alpha }<\mathrm{\infty }\mathrm{.}$

Applying [L1] with $s=f(x)$ and using step 2.1 gives $S_{N}f(x)\to f(x)$.

By [L1], write $u=P_u-N_u$ with $P_u(0)=N_u(0)=0$ and $P_u,N_u$ nondecreasing. Let $p:={\inf }_{0<t\le \delta }{P}_u(t)={\lim }_{t\downarrow 0}{P}_u(t),q:={\inf }_{0<t\le \delta }{N}_u(t)={\lim }_{t\downarrow 0}{N}_u(t)\mathrm{.}$ Since $u(t)\to 0$, one has $p-q=0$. If $p=q>0$, define $\tilde{P}(0)=\tilde{N}(0)=0$ and $\tilde{P}(t)=P_u(t)-p$, $\tilde{N}(t)=N_u(t)-p$ for $t>0$. Then $\tilde{P},\tilde{N}$ are again nondecreasing, nonnegative, normalized, and satisfy $u=\tilde{P}-\tilde{N}$, but $\tilde{P}(t)<P_u(t)$ for every $t>0$, contradicting the minimality in [L1]. Hence $p=q=0$, so $P_u(t)\to 0,N_u(t)\to 0,{\mathrm{Var}}_{[0,t]}(u)=P_u(t)+N_u(t)\to 0$ as $t\downarrow 0$ by [L2].

For $0\le t\le \delta$, define $H_N(t):={\int }_0^t{D}_N(s)ds\mathrm{.}$ Using [L3] and the change of variables $y=(2N+1)\pi s$, $H_N(t)={\int }_0^{(2N+1)\pi t}\sin y\cdot g_N(y)dy,$ where $g_N(y):=\frac{1}{(2N+1)\pi \sin (y\mathrm{/}(2N+1))}\mathrm{.}$ If $(2N+1)\pi t\le 1$, then $0<y\mathrm{/}(2N+1)\le 1<2$, so [L5] and $\mathrm{\mid }\sin y\mathrm{\mid }\le y$ give $\mathrm{\mid }\sin y{g}_N(y)\mathrm{\mid }\le 3\mathrm{/}\pi$ on $[0,(2N+1)\pi t]$, hence $\mathrm{\mid }{H}_N(t)\mathrm{\mid }\le 3\mathrm{/}\pi$. If $(2N+1)\pi t>1$, the same estimate controls the part on $[0,1]$, while on $[1,(2N+1)\pi t]$ the function $g_N$ is positive and decreasing because $0<y\mathrm{/}(2N+1)\le \pi \mathrm{/}2$. Bonnet's theorem [L4] therefore gives a point $\xi$ with ${\int }_1^{(2N+1)\pi t}\sin y{g}_N(y)dy=g_N(1){\int }_1^{\xi }\sin ydy+g_N((2N+1)\pi t){\int }_{\xi }^{(2N+1)\pi t}\sin ydy\mathrm{.}$ The absolute value is at most $2g_N(1)+2g_N((2N+1)\pi t)\le 4g_N(1)\le \frac{12}{\pi }$ by monotonicity of $g_N$ and [L5]. Thus $\mathrm{\mid }{H}_N(t)\mathrm{\mid }\le \frac{15}{\pi }(0\le t\le \delta ,N\ge 0)\mathrm{.}$

Fix $\epsilon >0$. By step 1.1, choose $\eta \in (0,\delta )$ such that ${\mathrm{Var}}_{[0,\eta ]}(u)<\frac{\pi \epsilon }{60}\mathrm{.}$ Applying [L4] to the monotone functions $P_u$ and $N_u$ on $[0,\eta ]$, and using $P_u(0)=N_u(0)=0$, step 1.2 yields $\mid {\int }_0^{\eta }{P}_u(t)D_N(t)dt\mid \le \frac{30}{\pi }{P}_u(\eta ),\mid {\int }_0^{\eta }{N}_u(t)D_N(t)dt\mid \le \frac{30}{\pi }{N}_u(\eta )\mathrm{.}$ Therefore $\mid {\int }_0^{\eta }u(t)D_N(t)dt\mid \le \frac{30}{\pi }(P_u(\eta )+N_u(\eta ))=\frac{30}{\pi }{\mathrm{Var}}_{[0,\eta ]}(u)<\frac{\epsilon }{2}\mathrm{.}$

On $[\eta ,\delta ]$, define $\psi (t):={\mathbf{1}}_{[\eta ,\delta ]}(t)\frac{u(t)e^{i\pi t}}{\sin (\pi t)}\mathrm{.}$ Because $\sin (\pi t)$ is bounded away from $0$ on $[\eta ,\delta ]$ and $u$ is bounded on the compact interval $[\eta ,\delta ]$, one has $\psi \in L^1(\mathbb{T})$. By [L3], ${\int }_{\eta }^{\delta }u(t)D_N(t)dt=\mathrm{Im}{\int }_0^1\psi (t)e^{2\pi iNt}dt=\mathrm{Im}\widehat{\psi }(-N)\mathrm{.}$

By [L6], $\widehat{\psi }(-N)\to 0$. So step 3.1 gives ${\int }_{\eta }^{\delta }u(t)D_N(t)dt\to 0.$ Choose $N_0$ such that the absolute value of this integral is below $\epsilon \mathrm{/}2$ for every $N\ge N_0$. Combining with step 2.1 shows that, for $N\ge N_0$, $\mid {\int }_0^{\delta }u(t)D_N(t)dt\mid <\epsilon \mathrm{.}$ Since $D_N(t)=\sin ((2N+1)\pi t)\mathrm{/}\sin (\pi t)$ on $(0,\delta ]$ by [L3], the stated limit follows.

By [L3], the one-sided limits $f(x+)$ and $f(x-)$ exist. Put $s:=\frac{f(x+)+f(x-)}{2}\mathrm{.}$ Choose $\delta \in (0,1\mathrm{/}2)$ and define, on $[0,\delta ]$, $u_{+}(0):=0,u_{-}(0):=0,$ $u_{+}(t):=f(x+t)-f(x+)\text{and}{u}_{-}(t):=f(x-t)-f(x-)(0<t\le \delta )\mathrm{.}$ Because translations and reflections preserve bounded variation on a compact interval, and changing a function at one point preserves bounded variation, both $u_{+}$ and $u_{-}$ have bounded variation on $[0,\delta ]$. The cited one-sided-limit result [L3] gives $u_{\pm }(t)\to 0$ as $t\downarrow 0$, and by construction $u_{\pm }(0)=0$.