Exercise 8.5.10

Question

\begin{enumerate}[label=(\alph*)] \item Show that $$\sigma_N(x) = \frac{1}{\pi} \int_{-\pi}^\pi f(u+x) F_N(u) du$$ \item Graph the function $F_N(u)$ for several values of $N$. Where is $F_N$ large, and where is it close to zero? Compare this function to the Dirichlet kernel $D_N(u)$. Now, prove that $F_N \to 0$ uniformly on any set of the form $\{u : \left|u\right| \geq \delta\}$, where $\delta>0$ is fixed (and $u$ is restricted to the interval $(-\pi, \pi]$.) \item Prove that $\int_{-\pi}^\pi F_N(u) du = \pi$. \item To finish the proof of Fejér's Theorem, first choose a $\delta > 0$ so that $$\left|u\right| < \delta \text{\quad implies \quad} \left|f(x+u) - f(x) \right| < \epsilon.$$ Set up a single integral that represents the difference $\sigma_N(x) - f(x)$ and divide this integral into sets where $\left|u\right| \leq \delta$ and $\left|u\right| \geq \delta$. Explain why it is possible to make each of these integrals sufficiently small, independently of the choice of $x$. \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item Note that $D_0(u) = 1/2$. We have
$$\begin{aligned}
\sigma_N(x) &= \frac{1}{N+1} \sum_{n=0}^N S_n(x) \
&= \frac{1}{N+1} \sum_{n=0}^N \frac{1}{\pi}\int_{-\pi}^\pi f(u+x) D_N(u) du \
&= \frac{1}{\pi} \int_{-\pi}^\pi f(u+x) \left(\frac{1}{N+1}\right) \left( \frac{1}{2} + \sum_{n=1}^N D_N(u) \right) du \
&= \frac{1}{\pi} \int_{-\pi}^\pi f(u+x) F_N(u) du \
\end{aligned}$$
\item $F_N(u)$ is large when $u$ is close to zero, and close to zero everywhere else. In contrast, $D_N(u)$ continues to oscillate with large amplitude away from $u=0$. To show $F_N \to 0$ uniformly when $\left|u\right| \geq \delta$, note that under this condition,
$$\left|F_N(u)\right| = \left| \frac{1}{2(N+1)} \left(\frac{\sin \left((N+1) \frac{u}{2}\right)}{\sin \frac{u}{2}}\right) \right| \leq \left| \frac{1}{2(N+1) \sin \frac{\delta}{2}} \right| $$
which approaches zero as $N \to \infty$; hence $F_N(u) \to 0$ uniformly when $\left|u\right| \geq \delta$.
\item Recall Fact 3 from earlier stating $\int_{-\pi}^\pi D_N(\theta) d\theta = \pi$.
$$\int_{-\pi}^\pi F_N(u) du = \frac{1}{N+1}\int_{-\pi}^\pi \sum_{n=0}^N D_n(u) du = \frac{1}{N+1} (N+1)\pi = \pi$$

\newcommand{\intpi}{\int_{-\pi}^\pi}
\item $$\begin{aligned}
    \sigma_N(x) - f(x) &= \frac{1}{\pi}\intpi f(u+x) F_N(u) du - \frac{1}{\pi} \intpi f(x) F_N(u) du\
&= \frac{1}{\pi}\intpi \left(f(u+x) - f(x)\right) F_N(u) du
\end{aligned}
$$
For the set where $|u| \leq \delta$,
$$ \begin{aligned}
    \left|\sigma_N(x) - f(x)\right| \leq \frac{1}{\pi}\intpi \left|f(u+x) - f(x)\right| F_N(u) du\leq \frac{\epsilon}{\pi} \intpi F_N(u) = \epsilon
\end{aligned} $$
and since $\epsilon$ is chosen freely, this part of the integral can be made arbitrarily small.

For the set where $|u| \geq \delta$, let $M$ be a bound on $f(x)$. Then
$$ \begin{aligned}
    \left|\sigma_N(x) - f(x)\right|
    &\leq \frac{1}{\pi}\intpi \left|f(u+x) - f(x)\right| F_N(u) du \leq \frac{2M}{\pi}\intpi F_N(u) du
\end{aligned} $$
Now since $F_N \to 0$ uniformly, we can choose $N$ large enough so that
$$\intpi F_N(u) < \frac{\epsilon \pi}{2M}$$
again bringing the integral arbitrarily small.
 \end{enumerate}

Exercise 8.5.10

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Exercise 8.5.10

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