Recall that we have started with the partial sum of the Fourier series
$$
\begin{align*}
S_Nf(x)=\sum_{n=-N}^{N}\widehat f(n)e^{inx}.
\end{align*}
$$
Recall that our main question was: as we include more frequencies, that is as \(N\to\infty\), does the partial sum \(S_Nf(x)\) approach \(f(x)\)? We then re-expressed the partial sum in terms of the Dirichlet Kernel:
$$
\begin{align*}
S_Nf(x)=\frac{1}{2\pi}\int_0^{2\pi}f(y)\,D_N(x-y)\,dy.
\end{align*}
$$
This new identity in fact has a name and is called a convolution. More generally
$$
\begin{align*}
(f*g)(x)=\frac{1}{2\pi}\int_0^{2\pi}f(y)\,g(x-y)\,dy.
\end{align*}
$$
Convolutions have many general properties that we went over in here. Hence the question of convergence becomes
Does \(f*D_N\) approach \(f\) as \(N\) grows?
Constructing a \(g_N\) that works
But we before study this this specific question where we use \(D_N\), could we choose some other functions \(g_N\) such that
$$
\begin{align*}
(f*g_N)(x)\longrightarrow f(x)?
\end{align*}
$$
In other words, what would make a good choice of \(g_N\) to converge to \(f\)? Recall that
$$
\begin{align*}
(f*g_N)(x)=\frac1{2\pi}\int_{-\pi}^{\pi}f(x-y)\,g_N(y)\,dy.
\end{align*}
$$
Fix \(x\). Notice now that when \(y\) is close to \(0\), \(x-y\) is close to \(x\). So if \(f\) is continuous at \(x\), then
$$
\begin{align*}
f(x-y)\approx f(x)\qquad\text{when }y\approx0.
\end{align*}
$$
That suggests choosing \(g_N(y)\) to put its weight near \(y=0\). We might think that we should set \(g(0)=1\) at only zero. But that doesn’t work since if \(g\) was \(1\) only at \(0\) and was zero everywhere else, the integral would be zero. Hence we want \(g\) to put its weight in a small interval around zero, with total weight
$$
\begin{align*}
\frac{1}{2\pi}\int_{-\pi}^{\pi}g(y)\,dy=1.
\end{align*}
$$
So it’s total weight \(1\) and not setting the height to be \(1\) at only \(0\). So the goal is
- Put the weight near zero.
- Make the total weight equal to \(1\):
\[
\frac1{2\pi}\int_{-\pi}^{\pi}g_N(y)\,dy=1.
\]
To do this, we first choose the narrow interval \(\left[-\frac1N,\frac1N\right]\). Then we define \(g_N\) to have an unknown height \(c_N\):
$$
\begin{align*}
g_N(y)=
\begin{cases}
c_N, & -\dfrac1N\le y\le\dfrac1N,\\[6pt]
0, & \text{otherwise}.
\end{cases}
\end{align*}
$$
Then we figure out what \(c_N\) should be by noting
$$
\begin{align*}
\frac1{2\pi}\int_{-\pi}^{\pi}g_N(y)\,dy=1.
\end{align*}
$$
Because the rectangle has height \(c_N\) and width \(\frac{2}{N}\). Then
$$
\begin{align*}
\frac1{2\pi}c_N\frac2N &= 1 \\
c_N &= \frac{2\pi}{1} \frac{N}{2} = \pi N.
\end{align*}
$$
Therefore, \(c_N=\pi N\). as illustrated below
Note above that \(N\) gives us a different function \(g_N\) and therefore a different convolution:
$$
\begin{align*}
f*g_1,\quad f*g_2,\quad f*g_3,\quad\ldots
\end{align*}
$$
We are asking whether
$$
\begin{align*}
(f*g_N)(x)\longrightarrow f(x)
\qquad\text{as }N\to\infty.
\end{align*}
$$
As \(N\) grows:
- the interval \([-1/N,1/N]\) shrinks toward \(0\).
- therefore \(x-y\) gets closer to \(x\).
- so \(f(x-y)\) gets closer to \(f(x)\), assuming \(f\) is continuous at \(x\).
Next, we see how the \(f * g_N\) converges.
Showing that \(f * g_N\) Converges
Now let’s substitute \(g_N\) back into the convolution:
$$
\begin{align*}
(f*g_N)(x)
=\frac1{2\pi}\int_{-\pi}^{\pi}f(x-y)g_N(y)\,dy.
\end{align*}
$$
But since \(g_N(y)=0\) outside \([-1/N,1/N]\), we can shorten the integral to be
$$
\begin{align*}
(f*g_N)(x)
=\frac1{2\pi}\int_{-1/N}^{1/N}f(x-y)(\pi N)\,dy.
\end{align*}
$$
Pulling the constant \(\pi N\) outside gives
$$
\begin{align*}
\boxed{
(f*g_N)(x) =\frac N2\int_{-1/N}^{1/N}f(x-y)\,dy.
}
\end{align*}
$$
Now make the substitution \(t = x-y\), then
$$
\begin{align*}
\frac{dt}{dy} =\frac{d}{dy}(x-y) =0-1=-1.
\end{align*}
$$
Hence \(dt=-dy\). The bounds change:
$$
\begin{align*}
y=-\frac1N \quad\Rightarrow\quad t=x+\frac1N,
\end{align*}
$$
$$
\begin{align*}
y=\frac1N \quad\Rightarrow\quad t=x-\frac1N.
\end{align*}
$$
Reversing the bounds removes the minus sign, giving
$$
\begin{align*}
\boxed{
(f*g_N)(x)=\frac N2\int_{x-1/N}^{x+1/N}f(t)\,dt.}
\end{align*}
$$
Observe now that The interval \(\left[x-\frac1N,x+\frac1N\right]\) has length
$$
\begin{align*}
\left(x+\frac1N\right)-\left(x-\frac1N\right)=\frac2N.
\end{align*}
$$
and since we’re dividing
The average value of \(f\) over an interval is
$$
\begin{align*}
\frac{1}{\text{length}}\int_{\text{interval}}f(t)\,dt.
\end{align*}
$$
But \(\frac1{\text{length}}=\frac1{2/N}=\frac N2\). Hence
$$
\begin{align*}
\boxed{(f*g_N)(x)=
\text{the average value of \(f\) on }
\left[x-\frac1N,x+\frac1N\right].}
\end{align*}
$$
Now as \(N \to \infty\), the interval \(\left[x-\frac1N,x+\frac1N\right]\) shrinks to the single point \(x\). If \(f\) happens to be continuous at \(x\), then \(f(t)\) must approach \(f(x)\) (Because of continuity, otherwise we can’t know). Therefore, the average value
$$
\begin{align*}
\frac N2\int_{x-1/N}^{x+1/N}f(t)\,dt
\end{align*}
$$
is averaging numbers that are all close to \(f(x)\).
$$
\begin{align*}
\boxed{(f*g_N)(x)\longrightarrow f(x).}
\end{align*}
$$
Formalizing The Idea
Lemma (Shrinking Averages)
Let \(f:\mathbb R\to\mathbb C\) be continuous and \(2\pi\)-periodic. Define
\[
g_N(y)=
\begin{cases}
\pi N, & |y|\leq \dfrac{1}{N},\\[6pt]
0, & |y|>\dfrac{1}{N},
\end{cases}
\]
extended \(2\pi\)-periodically. Then
\[
(f*g_N)(x)
=
\frac N2\int_{x-1/N}^{x+1/N}f(t)\,dt,
\]
and \(f*g_N\) converges to \(f\) uniformly as \(N\to\infty\). Equivalently,
\[
\lim_{N\to\infty}
\sup_{x\in[-\pi,\pi]}
\left|(f*g_N)(x)-f(x)\right|=0.
\]
Proof
Let \(\varepsilon>0\). Our goal is to find \(N_0\) such that whenever \(N\ge N_0\),
$$
\begin{align*}
\left|
\frac N2\int_{x-1/N}^{x+1/N}f(t)\,dt-f(x)
\right|<\varepsilon.
\end{align*}
$$
For uniform convergence, the same \(N_0\) must work for every \(x\in[-\pi,\pi]\). So
We want to compare the average of \(f(t)\) with \(f(x)\):
$$
\begin{align*}
\left|
\frac N2\int_{x-1/N}^{x+1/N}f(t)\,dt-f(x) = \frac N2\int_{x-1/N}^{x+1/N}f(t)\,dt- (1 \cdot f(x))
\right|.
\end{align*}
$$
Now observe that \(1\) can be also written as
$$
\begin{align*}
\frac N2\int_{x-1/N}^{x+1/N}1\,dt = \frac N2\left(\frac2N\right) =1,
\end{align*}
$$
Therefore,
$$
\begin{align*}
\left|
\frac N2\int_{x-1/N}^{x+1/N}f(t)\,dt-f(x)
\right| &= \left| \frac N2\int_{x-1/N}^{x+1/N}f(t)\,dt - \left( \frac N2\int_{x-1/N}^{x+1/N}1\,dt
\right)f(x) \right|
\end{align*}
$$
But now observe that
$$
\begin{align*}
f(x)=\frac N2\int_{x-1/N}^{x+1/N}f(x)\,dt.
\end{align*}
$$
Hence we can insert \(f(x)\) inside the integral to get
$$
\begin{align*}
\left|
\frac N2\int_{x-1/N}^{x+1/N}f(t)\,dt-f(x)
\right| &= \left| \frac N2\int_{x-1/N}^{x+1/N}f(t)\,dt - \frac N2\int_{x-1/N}^{x+1/N}f(x)\,dt
\right|\\
&= \left| \frac N2\int_{x-1/N}^{x+1/N} \bigl(f(t)-f(x)\bigr)\,dt \right|.
\end{align*}
$$
By the triangle inequality
$$
\begin{align*}
\left| \frac N2 \int_{x-1/N}^{x+1/N}
\bigl(f(t)-f(x)\bigr)\,dt \right| \leq \frac N2 \int_{x-1/N}^{x+1/N} |f(t)-f(x)|\,dt. \tag{1}
\end{align*}
$$
Because \(f\) is continuous and periodic, it is uniformly continuous. Therefore, for the \(\varepsilon>0\) chosen at the start, there exists some \(\delta>0\) such that whenever \(|t-x|<\delta\), then
$$
\begin{align*}
|f(t)-f(x)|<\varepsilon.
\end{align*}
$$
This is good but what we need is the entire right hand side to be less than \(\epsilon\). So we want the entire interval \(\left[x-\frac1N,x+\frac1N\right]\) to fit in \((x - \delta, x + \delta)\). To do this, we want to choose an integer \(N_0\) satisfying
$$
\begin{align*}
N_0>\frac1\delta.
\end{align*}
$$
Then, whenever \(N\geq N_0\),
$$
\begin{align*}
\frac1N\leq\frac1{N_0}<\delta.
\end{align*}
$$
We claim this choice will make the entire interval \(\left[x-\frac1N,x+\frac1N\right]\) to fit in \((x - \delta, x + \delta)\). To see this, for every \(t\) in the integration interval, we have
$$
\begin{align*}
x-\frac1N\leq t &\leq x+\frac1N \\
-\frac1N\leq t-x &\leq\frac1N \\
|t-x| &\leq\frac1N \\
\end{align*}
$$
Hence \(|t-x|\leq\frac1N<\delta\). Thus we have shown
$$
\begin{align*}
\left[x-\frac1N,x+\frac1N\right]
\subseteq
(x-\delta,x+\delta)
\qquad\text{for every }N\geq N_0.
\end{align*}
$$
Therefore, by uniform continuity,
$$
\begin{align*}
|f(t)-f(x)|<\varepsilon
\quad\text{for every }
t\in\left[x-\frac1N,x+\frac1N\right].
\end{align*}
$$
Let’s substitute the bound back in (1) to get
$$
\begin{align*}
\left|(f*g_N)(x)-f(x)\right|
&\leq \frac N2 \int_{x-1/N}^{x+1/N}
|f(t)-f(x)|\,dt\\
&<
\frac N2
\int_{x-1/N}^{x+1/N}
\varepsilon\,dt
=\frac N2\varepsilon\left(\frac2N\right) =\varepsilon.
\end{align*}
$$
Therefore, for every \(N\geq N_0\),
$$
\begin{align*}
\left|(f*g_N)(x)-f(x)\right|<\varepsilon.
\end{align*}
$$
This is precisely the definition of
$$
\begin{align*}
(f*g_N)(x)\longrightarrow f(x).
\end{align*}
$$
Conditions for a good Kernel
Lemma (Good Kernel)
Let \(K_N\) be a sequence of periodic Reimann integrable functions on \([\pi,\pi]\). We say that \(K_N\) is a
good kernel if
-
\(
\frac{1}{2\pi}\int_{-\pi}^{\pi}K_N(x)\,dx=1\) for every \(N\in\mathbb N\)
-
\(
\sup_{N\in\mathbb N}
\frac{1}{2\pi}\int_{-\pi}^{\pi}|K_N(x)|\,dx<\infty,
\)
- For every \(\delta>0\),
\(
\lim_{N\to\infty}
\int_{\delta\leq |x|\leq\pi}|K_N(x)|\,dx=0.
\)
References
- SFSU Fourier Analysis by Professor Chun-Kit Lai
- Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi