Lecture 5: Examples of Good Kernels

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 but that was an assigned homework problem so skipping for now.

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