Recall again 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:
$$
\begin{align*}
S_Nf(x) = (f * D_N)(x) = \frac{1}{2\pi}\int_0^{2\pi}f(y)\,D_N(x-y)\,dy.
\end{align*}
$$
We hoped that the sequence of functions
$$
\begin{align*}
f*D_1,\quad f*D_2,\quad f*D_3,\quad\ldots
\end{align*}
$$
converges to the original function \(f\). That is, \(f*D_N \longrightarrow f\) as \((N\to\infty)\). However, we discovered that the good-kernel argument cannot guarantee convergence, because \(D_N\) is not a good kernel.
Fejér kernel
The next thing we want to try is to instead of trusting one partial sum, we will average several partial sums. Hence, define
$$
\begin{align*}
\sigma_N(f)(x) = \frac{S_0f(x)+S_1f(x)+\cdots+S_{N-1}f(x)}{N}.
\end{align*}
$$
Using \(S_nf=f*D_n\), this becomes
$$
\begin{align*}
\sigma_N(f)(x) = \frac{(f*D_0)(x)+(f*D_1)(x)+\cdots+(f*D_{N-1})(x)}{N}.
\end{align*}
$$
Because convolution is linear, we can move the average inside:
$$
\begin{align*}
\sigma_N(f)(x) = f* \left( \frac{D_0+D_1+\cdots+D_{N-1}}{N} \right)(x).
\end{align*}
$$
So we have replaced the problematic single kernel \(D_N\) with an average of Dirichlet kernels. This new averaged kernel will be called the Fejér kernel. Formally
Fejér kernel
Let
\[
F_N(x) = \frac{D_0(x)+D_1(x)+\cdots+D_{N-1}(x)}{N}.
\]
The averaged Fourier sum can be written compactly as
\[
\sigma_N(f)(x)=(f*F_N)(x).
\]
Hence the convergence of \(\sigma_N(f)\) now depends on whether \(F_N\) behaves like a good kernel. Next, we want to derive its closed form
$$
\begin{align*}
\boxed{
F_N(x)=\frac{\sin^2(Nx/2)}{N\sin^2(x/2)}
}
\end{align*}
$$
To derive this, as a reminder
$$
\begin{align*}
S_nf(x)=\sum_{k=-n}^{n}\widehat f(k)e^{ikx}.
\end{align*}
$$
Substitute the Fourier coefficient into \(S_nf(x)\):
$$
\begin{align*}
S_nf(x) = \sum_{k=-n}^{n} \left( \frac1{2\pi}\int_{-\pi}^{\pi}f(y)e^{-iky}\,dy
\right)e^{ikx}.
\end{align*}
$$
Move the finite sum inside the integral:
$$
\begin{align*}
S_nf(x) = \frac1{2\pi}\int_{-\pi}^{\pi} f(y)\sum_{k=-n}^{n}e^{ik(x-y)} \,dy.
\end{align*}
$$
The expression inside the sum is Dirichlet kernel which we have derived a couple of lectures ago.
$$
\begin{align*}
D_n(t)=\sum_{k=-n}^{n}e^{ikt}.
\end{align*}
$$
Now substitute this into \(F_N(x)=\frac1N\sum_{n=0}^{N-1}D_n(x)\) to get
$$
\begin{align*}
F_N(x) = \frac1N\sum_{n=0}^{N-1}\sum_{k=-n}^{n}e^{ikx}.
\end{align*}
$$
The next step is to reverse the order of summation. For a fixed (k), the term (e^{ikx}) occurs whenever
$$
\begin{align*}
n\ge |k|.
\end{align*}
$$
Among \(n=0,\ldots,N-1\), this happens \(N-|k|\) times. Therefore,
$$
\begin{align*}
\boxed{
F_N(x)
=
\sum_{k=-(N-1)}^{N-1}
\left(1-\frac{|k|}{N}\right)e^{ikx}
}.
\end{align*}
$$
Fejér kernel is a Good Kernel
Theorem
The sequence \(\{F_N\}_{N=1}^{\infty}\) is a family of good kernels.
Consequently,
\[
(f * F_N)(x) \longrightarrow f(x)
\]
whenever \(x\) is a continuity point of \(f\). Moreover, if \(f\) is continuous, then the convergence is uniform.
Recall the conditions for a good kernel from last time
- \(\frac{1}{2\pi}\int_{0}^{2\pi}K_n(x)\,dx=1\) 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 \(\delta>0\),\(\lim_{N\to\infty}\int_{\delta\leq x\leq 2\pi-\delta}|K_n(x)|\,dx=0.\)
Proof
We must verify these three properties for \(F_N\). Recall that
$$
\begin{align*}
F_N(x)=\frac{D_0(x)+D_1(x)+\cdots+D_{N-1}(x)}{N}. \tag{1}
\end{align*}
$$
Each Dirichlet kernel has normalized integral \(1\) Why? because the Dirichlet kernel is
$$
\begin{align*}
D_n(x)=\sum_{k=-n}^{n}e^{ikx}
=1+2\sum_{k=1}^{n}\cos(kx).
\end{align*}
$$
If we integrate it over one full period:
$$
\begin{align*}
\int_0^{2\pi}D_n(x)\,dx = \int_0^{2\pi}1\,dx + 2\sum_{k=1}^{n}\int_0^{2\pi}\cos(kx)\,dx.
\end{align*}
$$
We know every cosine completes an integer number of periods, so
$$
\begin{align*}
\int_0^{2\pi}\cos(kx)\,dx=0.
\end{align*}
$$
Consequently,
$$
\begin{align*}
\int_0^{2\pi}D_n(x)\,dx=2\pi.
\end{align*}
$$
After applying the normalization factor \(1/(2\pi)\),
$$
\begin{align*}
\frac{1}{2\pi}\int_0^{2\pi}D_n(x)\,dx=1. \tag{2}
\end{align*}
$$
Now, let’s re-write \((1)\) to use \((2)\):
$$
\begin{align*}
\frac{1}{2\pi}\int_0^{2\pi}F_N(x)\,dx
&=
\frac{1}{2\pi}\int_0^{2\pi}
\frac{D_0(x)+\cdots+D_{N-1}(x)}{N}\,dx \\[4pt]
&=
\frac{1}{N} \frac{1}{2\pi}\int_0^{2\pi}
D_0(x)+\cdots+D_{N-1}(x)\,dx \\[4pt]
&=
\frac{1}{N} \left[ \frac{1}{2\pi} \int_0^{2\pi}
D_0(x)dx+\cdots+ \frac{1}{2\pi} \int_0^{2\pi} D_{N-1}(x)\,dx \right] \\[4pt]
&=
\frac1N\sum_{n=0}^{N-1}
\left(
\frac{1}{2\pi}\int_0^{2\pi}D_n(x)\,dx
\right) \\[4pt]
&= \frac1N\sum_{n=0}^{N-1}1 = \frac1N\cdot N=1.
\end{align*}
$$
Thus \(F_N\) satisfies property (1):
$$
\begin{align*}
\boxed{\frac{1}{2\pi}\int_0^{2\pi}F_N(x)\,dx=1.} \tag{3}
\end{align*}
$$
Next we verify property (2). Recall the closed form:
$$
\begin{align*}
F_N(x)=\frac{\sin^2(Nx/2)}{N\sin^2(x/2)}.
\end{align*}
$$
Both squared terms are nonnegative, so \(F_N(x)\geq 0\). However, we need to consider \(x = 0\) since \(F_N(0)=\frac{\sin^2(0)}{N\sin^2(0)}=\frac{0}{0}\). In that case, we can use the original definition
$$
\begin{align*}
F_N(0)=\frac{D_0(0)+D_1(0)+\cdots+D_{N-1}(0)}{N}.
\end{align*}
$$
where
$$
\begin{align*}
D_j(0) = \sum_{k=-j}^{j}e^{ik0} =\sum_{k=-j}^{j}1 =2j+1.
\end{align*}
$$
Hence
$$
\begin{align*}
F_N(0)
&=\frac1N\sum_{j=0}^{N-1}(2j+1)\\
&=\frac{1+3+5+\cdots+(2N-1)}{N} = \frac{N^2}{N} = N.
\end{align*}
$$
So \(F_N(x)\geq0\) everywhere. This implies that \(|F_N(x)|=F_N(x)\). Therefore by (3),
$$
\begin{align*}
\frac1{2\pi}\int_0^{2\pi}|F_N(x)|\,dx = \frac1{2\pi}\int_0^{2\pi}F_N(x)\,dx =1,
\end{align*}
$$
Thus property (2) holds with the constant \(C=1\):
$$
\begin{align*}
\boxed{\frac1{2\pi}\int_0^{2\pi}|F_N(x)|\,dx=1.}
\end{align*}
$$
Next, we verify property \(3\), fix any \(\delta>0\). First note that the numerator in \(F(N)\) is \(\sin^2(Nx/2)\). Since \(-1 \leq \sin(\theta) \leq 1\) and since sine is squared in the numerator then \(\sin^2(\theta) \leq 1\). Taking \(\theta=Nx/2\), then
$$
\begin{align*}
\sin^2(Nx/2)\leq1.
\end{align*}
$$
Hence we have
$$
\begin{align*}
F_N(x) = \frac{\sin^2(Nx/2)}{N\sin^2(x/2)} \leq \frac{1}{N\sin^2(x/2)}. \tag{4}
\end{align*}
$$
For the denominator, we first note that \([\delta,2\pi-\delta] = [\delta,\pi]\cup[\pi,2\pi-\delta]\). If we leave the first unchanged and subtract \(2\pi\) from the second (we can do this due to periodicity), then we get \([-\pi,-\delta]\cup[\delta,\pi]\). Hence, we can also compress this to
$$
\begin{align*}
\delta\leq |x|\leq\pi.
\end{align*}
$$
and if we divide by 2, then
$$
\begin{align*}
\frac{\delta}{2} \leq \frac{|x|}{2} \leq \frac{\pi}{2}.
\end{align*}
$$
On \([0,\pi/2]\), sine is increasing. Also, squaring removes the sign when \(x<0\). Therefore,
$$
\begin{align*}
\sin^2(x/2)=\sin^2(|x|/2)
\geq\sin^2(\delta/2).
\end{align*}
$$
Hence we can update the bound in (4) to become
$$
\begin{align*}
F_N(x) = \frac{\sin^2(Nx/2)}{N\sin^2(x/2)} \leq \frac{1}{N\sin^2(x/2)} &\leq
\frac{1}{N\sin^2(\delta/2)}.
\end{align*}
$$
So away from \(x=0\), the entire Fejér kernel is bounded above by a constant times \(\frac{1}{N}\):
$$
\begin{align*}
F_N(x)\leq\frac{1}{N\sin^2(\delta/2)}
\quad\text{when }\delta\leq|x|\leq\pi.
\end{align*}
$$
Also note that \(|F_N(x)| = F_N(x)\) as we established previously. Next, we evaluate the integral
$$
\begin{align*}
\int_{\delta\leq|x|\leq\pi}|F_N(x)|\,dx
&\leq
\int_{\delta\leq|x|\leq\pi}
\frac{1}{N\sin^2(\delta/2)}\,dx\\
&=
\frac{1}{N\sin^2(\delta/2)} \int_{\delta\leq|x|\leq\pi} 1 \,dx\\
&=
\frac{2(\pi-\delta)}
{N\sin^2(\delta/2)}\\
&\leq
\frac{2\pi}
{N\sin^2(\delta/2)}.
\end{align*}
$$
Hence as \(N \rightarrow \infty\), the limit of the above expression goes to zero. Hence we have verified the three properties and so \(F_N(x)\) is a good kernel. \(\blacksquare\)
Consequences of Fejér kernel being a Good Kernel
Corollary (Uniqueness of Fourier series)
Suppose that
\(\widehat{f}(n)=0\) for every \(n\in\mathbb Z\). Then \(f(x)=0\) at every continuity point \(x\) of \(f\).
In words: if every Fourier coefficient of \(f\) is zero, then \(f\) must be zero wherever it is continuous.
Proof
Recall that
$$
\begin{align*}
S_Nf(x)=\sum_{n=-N}^{N}\widehat f(n)e^{inx}.
\end{align*}
$$
The assumption says \(\widehat f(n)=0\) for every integer \(n\). Substituting this into the partial sum gives
$$
\begin{align*}
S_Nf(x)
&=\sum_{n=-N}^{N}0\cdot e^{inx} =0.
\end{align*}
$$
Because \(S_Nf(x)=(f*D_N)(x)\). We conclude that
$$
\begin{align*}
(f*D_N)(x)=0. \tag{1}
\end{align*}
$$
for every \(N\) and every \(x\). Now, the Fejér mean is the average of the Fourier partial sums:
$$
\begin{align*}
\sigma_N(f)(x) =
\frac{S_0f(x)+S_1f(x)+\cdots+S_{N-1}f(x)}{N}.
\end{align*}
$$
But by (1), \(S_jf(x)=0\) for every \(j\). Hence
Therefore,
$$
\begin{align*}
\sigma_N(f)(x) &= \frac{0+0+\cdots+0}{N} =0.
\end{align*}
$$
We also know that the Fejér mean can be written as
$$
\begin{align*}
\sigma_N(f)(x)=(f*F_N)(x).
\end{align*}
$$
Consequently,
$$
\begin{align*}
\boxed{(f*F_N)(x)=0} \tag{2}
\end{align*}
$$
for every \(N\) and every \(x\). By the Theorem, \(\{F_N\}\) is a family of good kernels. Therefore, at every continuity point \(x\) of \(f\),
$$
\begin{align*}
(f*F_N)(x)\longrightarrow f(x).
\end{align*}
$$
But by 2, we know that \((f*F_N)(x)=0\) for every \(N\). Hence its limit must also be zero:
$$
\begin{align*}
\lim_{N\to\infty}(f*F_N)(x)=0.
\end{align*}
$$
But this implies that \(f(x) = 0\) at every continuity point of \(f\). \(\blacksquare\)
References
- SFSU Fourier Analysis by Professor Chun-Kit Lai
- Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi