Let $f:\mathbb{R}\longrightarrow\mathbb{C}$ be a $2\pi-$periodic function. Denote $e_{k}(x):=e^{ikx}$, and then we know that the Dirichlet Kernel has forms $$D_{n}(x):=\dfrac{1}{2\pi}\sum_{k=-n}^{n}e_{k}(x)=\dfrac{1}{2\pi}\dfrac{\sin(n+\frac{1}{2})x}{\sin(\frac{1}{2}x)}.$$
It plays a vital role in the pointwise convergence of the partial sum $$S_{n}(f):=\sum_{k=-n}^{n}c_{k}(f)e_{k}(x),$$ where $c_{k}(f)$ is the Fourier Coefficient of $f$, that is $$c_{k}(f):=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-ikx}dx.$$
We want to know that when $S_{n}(f)(x)\longrightarrow f(x)$ pointwisely, as $n\longrightarrow\infty$.
Then we need:
[Riemann-Lebesgue Lemma:]If $f\in L^{1}(\mathbb{R})$, then $$\int_{\mathbb{R}}f(x)e^{i\lambda x}dx\longrightarrow 0,\ \text{as}\ \lambda\in\mathbb{R},\ |\lambda|\longrightarrow\infty.$$
And the statement of the pointwise convergence is a corollary:
[Corollary] If $f\in L^{1}(\mathbb{S})$ satisfying $$\int_{-\delta}^{\delta}\dfrac{|f(x+y)-f(x)|}{|y|}dy<\infty\ \text{for some}\ \delta,$$ then $S_{n}(f)(x)\longrightarrow f(x)$ as $n\longrightarrow\infty$.
I tried to prove this corollary but I got stuck near the end.
Below is my proof:
To study the convergence, as usual we want to study $S_{n}(f)(x)-f(x)$, so let's compute: $$S_{n}(f)(x)-f(x)=(D_{n}*f)(x)-f(x)=\int_{-\pi}^{\pi}D_{n}(x-y)f(y)dy-f(x).$$
Note that using orthogonality, we have $$\int_{-\pi}^{\pi}D_{n}(y)dy=\int_{-\pi}^{\pi}\dfrac{1}{2\pi}\sum_{k=-n}^{n}e_{k}(y)dy=\dfrac{1}{2\pi}\sum_{k=-n}^{n}\int_{-\pi}^{\pi}e_{k}(y)dy=0+\dfrac{1}{2\pi}\cdot 2\pi=1.$$
Using this result, we are able to rewrite $S_{n}(f)(x)-f(x)$ as $$S_{n}(f)(x)-f(x)=\int_{-\pi}^{\pi}D_{n}(x-y)f(y)dy-\int_{-\pi}^{\pi}D_{n}(y)f(x)dy,$$ and using the commutativity of convolution $(D_{n}*f)(x)=(f*D_{n})(x)$,
\begin{align*}
\text{the above equation}&=\int_{-\pi}^{\pi}f(x-y)D_{n}(y)dy-\int_{-\pi}^{\pi}D_{n}(y)f(x)dy\\
&=\int_{-\pi}^{\pi}\Big(f(x-y)-f(x)\Big)D_{n}(y)dy.
\end{align*}
Now plug in the formula of $D_{n}(y)$, we have
\begin{align*}
|S_{n}(f)(x)-f(x)|&=\Big|\int_{-\pi}^{\pi}\dfrac{f(x-y)-f(x)}{\sin(\frac{1}{2}y)}\sin(n+\dfrac{1}{2})ydy\Big|\\
&=\Big|\int_{-\pi}^{\pi}\dfrac{f(x+y)-f(x)}{\sin(\frac{1}{2}y)}\sin(n+\dfrac{1}{2})ydy\Big|\\
&\leq\int_{-\pi}^{\pi}\Big|\dfrac{f(x+y)-f(x)}{\sin(\frac{1}{2}y)}\Big|\ \cdot \ \Big|\sin(n+\dfrac{1}{2})y\Big|dy,
\end{align*}
where the second equality was obtained by replacing $y\mapsto -y$, and two negative signs coming from $\sin$ function canceled and $-dy$ does not matter since we are in the absolute value.
I got stuck here with two problems:
$(1)$ My goal was to argue that $$g(y):=\dfrac{f(x+y)-f(x)}{\sin(\frac{1}{2}y)}\in L^{1}(-\pi, \pi)$$ using the hypothesis, but I don't know how to pass the denominator $\sin(\frac{1}{2}y)$ to $y$ in the hypothesis, and I don't know how to extend the integral from $-\delta$ to $\delta$, to the integral from $-\pi$ to $\pi$.
$(2)$ If $(1)$ is satisfied, then we can apply Riemann-Lebesgue Lemma, but how could I connect $\sin(n+\dfrac{1}{2})y$ to $e^{i(n+\frac{1}{2})y}$?
Thank you so much!
Best Answer
In order:
(1) Notice that you can rewrite $$ \frac{f(x+y) - f(x)}{\sin(\frac12 y)} = \frac{f(x+y) - f(x)}{y} \cdot \frac{y}{\sin(\frac12 y)} $$ The second term in approximately 1 and smooth when $y\in (-\delta,\delta)$ for $\delta$ sufficiently small, and so is harmless.
(2) $\sin(x) = \frac1{2i} (e^{ix} - e^{-ix})$. So controlling the integral against $\exp$ via Riemann-Lebesgue also controls integrals against $\sin$.
(3) Notice that for $y\not\in [-\delta,\delta]$, as $n\to \infty$ the Dirichlet kernel $D_n(y)$ converges uniformly to zero.
As a general principle with a lot of these types of Fourier/harmonic analysis computations: your goal is to take advantage of the fact that $D_n$ is an "approximation to the identity", which morally speaking says that as $n\to \infty$ it starts looking more and more like the Dirac $\delta$. So you should always immediately split your integral into two parts, the first localized to where the kernel becomes "singular", and where you get rid of the singularity with some sort of "integration by parts" or "cancellation", and the second for the rest of the space where the kernel converges to zero.
If you can internalize this trick, you will have no problem understanding even more complicated constructions like Calderon-Zygmund theory.