Let \(f\) be continuous and \(2\pi\)-periodic. Last time we were given the convolution
$$
\begin{align*}
(f*g_N)(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x-y)g_N(y)\,dy
\end{align*}
$$
and we wanted to show that the sequence of functions
$$
\begin{align*}
f*g_1,\quad f*g_2,\quad f*g_3,\quad\ldots
\end{align*}
$$
converges to the original function \(f\). That is, \(f*g_N \longrightarrow f\) as \((N\to\infty)\). We were able to construct a specific function \(g_N\) defined as
$$
\begin{align*}
g_N(y)=
\begin{cases}
\pi N, & |y|\leq \dfrac{1}{N},\\[6pt]
0, & |y|>\dfrac{1}{N},
\end{cases}
\end{align*}
$$
and we were able to show that
$$
\begin{align*}
(f*g_N)(x) = \frac N2\int_{x-1/N}^{x+1/N}f(t)\,dt,
\end{align*}
$$
which is the average of \(f\) over a small interval centered at \(x\). As \(N\to\infty\), that interval shrinks to \(x\), so the average approaches \(f(x)\). Because \(f\) is continuous and periodic, then it is uniformly continuous (recall every continuous function on a compact interval is uniformly continuous and also recall uniformly continuous means, \(|x-y|<\delta\) implies \(|f(x)-f(y)|<\varepsilon\)). This implies that this convergence is uniform. In other words,
$$
\begin{align*}
\lim_{N\to\infty} \sup_{x\in[-\pi,\pi]} \left|(f*g_N)(x)-f(x)\right|=0.
\end{align*}
$$
Next, we look at what is need for a family of functions to be good kernels.
Conditions for a good Kernel
Definition
Let \(K_N\) be a sequence of periodic Reimann integrable functions on \([\pi,\pi]\). We say that a family of functions \(K_N\) is a
good kernel if
-
\(
\frac{1}{2\pi}\int_{0}^{2\pi}K_n(x)\,dx=1\) for every \(n\in\mathbb N\)
- \( \sup_{n \geq 1}
\frac{1}{2\pi}\int_{0}^{2\pi}|K_n(x)|\,dx<\infty,\)
- For every \(\delta>0\), \(\lim_{N\to\infty} \int_{\delta\leq x\leq 2\pi-\delta}|K_n(x)|\,dx=0.
\)
Note that the third property is saying that the total area outside the interval \((-\delta,\delta)\) becomes arbitrarily small as \(N\to\infty\).
Next we see why those conditions imply convergence in the next theorem
Theorem
Let \(f\) be a Riemann integrable function on \(\mathbb{T}\), and let
\(\{K_n\}\) be a family of good kernels. Then
\[
\lim_{n\to\infty}(f*K_n)(x)=f(x)
\]
whenever \(f\) is continuous at \(x\). If \(f\) is continuous, then the
convergence is uniform.
Note, recall that uniform continuity means:
$$
\begin{align*}
\forall \varepsilon>0,\ \exists\delta>0
\quad\text{such that for all }x,t,
\end{align*}
$$
$$
\begin{align*}
|t-x|<\delta \implies |f(t)-f(x)|<\varepsilon.
\end{align*}
$$
While \(f\) continuous only, means that for every fixed \(x\) and every \(\varepsilon>0\), there exists a \(\delta\) that may depend on both \(x\) and \(\varepsilon\), so \(\delta=\delta(x,\varepsilon)\). That is, fix a point \(x\). Then \(f\) is continuous at \(x\) if
$$
\begin{align*}
\forall \varepsilon>0,\ \exists \delta>0
\end{align*}
$$
such that, for every \(t\),
$$
\begin{align*}
|t-x|<\delta
\quad\Longrightarrow\quad
|f(t)-f(x)|<\varepsilon.
\end{align*}
$$
Proof
Let \(\varepsilon>0\). By property 2, choose \(B>0\) such that
$$
\begin{align*}
\boxed{
\frac1{2\pi}\int_{-\pi}^{\pi}|K_n(y)|\,dy\le B
\quad\text{for every }n.
}
\end{align*}
$$
Next, because \(f\) is continuous on a compact interval \([-\pi,\pi]\), then \(f\) is uniformly continuous. Therefore, there exists some \(\delta>0\) such that
$$
\begin{align*}
|t-x|<\delta \quad\Longrightarrow\quad |f(t)-f(x)|< \frac{\varepsilon}{2B}.
\end{align*}
$$
Next, we show that \(f\) is bounded. Since \(f\) is continuous on the compact interval \([-\pi,\pi]\), there exists some number \(M>0\) such that
$$
\begin{align*}
|f(x)|\le M
\qquad\text{for every }x\in[-\pi,\pi]. \tag{1}
\end{align*}
$$
Equivalently, we can define
$$
\begin{align*}
M=\max_{x\in[-\pi,\pi]}|f(x)|.
\end{align*}
$$
Now, we want to show that
$$
\begin{align*}
|(f*K_n)(x)-f(x)| < \epsilon \tag{2}
\end{align*}
$$
Using the definition of convolution,
$$
\begin{align*}
(f*K_n)(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x-y)K_n(y)\,dy.
\end{align*}
$$
Therefore (2) becomes
$$
\begin{align*}
|(f*K_n)(x)-f(x)| &= \left| \frac{1}{2\pi} \left( \int_{-\pi}^{\pi}f(x-y)K_n(y)\,dy \right) -f(x) \right| \tag{3}
\end{align*}
$$
Now, by property 1 of a good kernel:
$$
\begin{align*}
\boxed{
\frac1{2\pi}\int_{-\pi}^{\pi}K_n(y)\,dy=1.}
\end{align*}
$$
Multiplying this equation by \(f(x)\) gives
$$
\begin{align*}
f(x) = \frac1{2\pi}\int_{-\pi}^{\pi}f(x)K_n(y)\,dy.
\end{align*}
$$
Hence (3) becomes
$$
\begin{align*}
|(f*K_n)(x)-f(x)| &= \left| \frac1{2\pi}\int_{-\pi}^{\pi}f(x-y)K_n(y)\,dy -
\frac1{2\pi}\int_{-\pi}^{\pi}f(x)K_n(y)\,dy \right| \\
&= \frac1{2\pi} \left| \int_{-\pi}^{\pi} (f(x-y) -f(x) ) K_n(y)\,dy \right| \\
&\leq \frac1{2\pi}\int_{-\pi}^{\pi}|f(x-y)-f(x)|\,|K_n(y)|\,dy.
\end{align*}
$$
Next, we split the integral into two regions:
$$
\begin{align*}
[-\pi,\pi] = \{|y|<\delta\} \cup \{\delta\le |y|\le\pi\}.
\end{align*}
$$
Therefore,
$$
\begin{align*}
|(f*K_n)(x)-f(x)| &\le \frac1{2\pi}\int_{|y|<\delta}
|f(x-y)-f(x)|\,|K_n(y)|\,dy \\
&\quad+ \frac1{2\pi}\int_{\delta\le |y|\le\pi} |f(x-y)-f(x)|\,|K_n(y)|\,dy. \tag{4}
\end{align*}
$$
First, consider the near part, where \(|y|<\delta\). Set \(t=x-y\). Then
$$
\begin{align*}
|t-x|=|(x-y)-x|=|y|<\delta.
\end{align*}
$$
By uniform continuity we had the bound
$$
\begin{align*}
|f(x-y)-f(x)|< \frac{\varepsilon}{2B}.
\end{align*}
$$
Hence
$$
\begin{align*}
\frac1{2\pi}\int_{|y|<\delta}
|f(x-y)-f(x)|\,|K_n(y)|\,dy &\le \frac1{2\pi}\int_{|y|<\delta} \frac{\varepsilon}{2B} |K_n(y)| \,dy \\
&\le \frac{\varepsilon}{2B}. \left( \frac1{2\pi}\int_{|y|<\delta} |K_n(y)| \,dy \right) \\
&\leq \frac{\varepsilon}{2}
\end{align*}
$$
where the last inequality happens by property 2 of a good kernel where
$$
\begin{align*}
\boxed{
\frac1{2\pi}\int_{-\pi}^{\pi}|K_n(y)|\,dy\le B.}
\end{align*}
$$
Since \(\{|y|<\delta\}\) is only part of \([-\pi,\pi]\),
$$
\begin{align*}
\frac1{2\pi}\int_{|y|<\delta}|K_n(y)|\,dy\le B.
\end{align*}
$$
Therefore, \((4)\) becomes
$$
\begin{align*}
|(f*K_n)(x)-f(x)|
&\le \frac{\varepsilon}{2} + \frac1{2\pi}\int_{\delta\le |y|\le\pi}
|f(x-y)-f(x)|\,|K_n(y)|\,dy. \tag{5}
\end{align*}
$$
Now recall by \((1)\) we have the bound \(|f(z)|\le M\) for every \(z\). By the triangle inequality
$$
\begin{align*}
|f(x-y)-f(x)|
\le |f(x-y)|+|f(x)|
\le M+M=2M.
\end{align*}
$$
so in the second integral (in \((5)\)) we can replace \(|f(x-y)-f(x)|\) with the upper bound \(2M\). So \((5)\) becomes
$$
\begin{align*}
|(f*K_n)(x)-f(x)| &\le \frac{\varepsilon}{2} + \frac1{2\pi}\int_{\delta\le |y|\le\pi} 2M\,|K_n(y)|\,dy \\
&= \frac{\varepsilon}{2} + \frac{M}{\pi}\int_{\delta\le |y|\le\pi} \,|K_n(y)|\,dy \\
\end{align*}
$$
Now, property 3 says that for every fixed \(\delta>0\),
$$
\begin{align*}
\boxed{\frac1{2\pi} \int_{\delta\le |y|\le\pi}|K_n(y)|\,dy \longrightarrow 0 \qquad\text{as }n\to\infty.
}
\end{align*}
$$
Therefore, there exists \(N\) such that whenever \(n \geq N\),
$$
\begin{align*}
\frac{M}{\pi}
\int_{\delta\le |y|\le\pi}|K_n(y)|\,dy
<
\frac{\varepsilon}{2}.
\end{align*}
$$
Hence for sufficiently large \(n\),
$$
\begin{align*}
|(f*K_n)(x)-f(x)| &\le \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \\
\end{align*}
$$
which means \(f*K_n\longrightarrow f\) uniformly.
The Dirichlet's Kernel is not a Good Kernel
Now recall that we defined
The \(N\)-th partial sum of the Fourier series can be written as
$$
\begin{align*}
\boxed{S_Nf(x)=(f*D_N)(x).}
\end{align*}
$$
The Dirichlet kernel is
$$
\begin{align*}
\boxed{
D_N(x) = \sum_{n=-N}^{N}e^{inx} = \frac{\sin\left(\left(N+\frac12\right)x\right)}{\sin(x/2)}.}
\end{align*}
$$
Hence
$$
\begin{align*}
S_Nf(x) = (f*D_N)(x)
=
\frac1{2\pi}\int_{-\pi}^{\pi}
f(x-y)D_N(y)\,dy.
\end{align*}
$$
Specifically the Dirichlet’s kernel fails property 2. It is unbounded. To derive the bound, note that
$$
\begin{align*}
\frac1{2\pi}\int_{-\pi}^{\pi}|D_N(x)|\,dx
&=
\frac1{2\pi}\int_{-\pi}^{\pi}
\frac{\left|\sin\left(\left(N+\frac12\right)x\right)\right|}
{|\sin(x/2)|}\,dx \\
&=
\frac1{\pi}\int_{0}^{\pi}
\frac{\left|\sin\left(\left(N+\frac12\right)x\right)\right|}
{\sin(x/2)}\,dx.
\end{align*}
$$
For \(0 < x \le \pi\), we use
$$
\begin{align*}
\sin(x/2)\le \frac{x}{2}.
\end{align*}
$$
Taking reciprocals reverses the inequality:
$$
\begin{align*}
\frac1{\sin(x/2)}\ge \frac{2}{x}.
\end{align*}
$$
Therefore,
$$
\begin{align*}
\frac1{2\pi}\int_{-\pi}^{\pi}|D_N(x)|\,dx
\ge
\frac{2}{\pi}\int_0^\pi
\frac{\left|\sin\left(\left(N+\frac12\right)x\right)\right|}{x}\,dx.
\end{align*}
$$
... TODO
References
- SFSU Fourier Analysis by Professor Chun-Kit Lai
- Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi