Exercise 8.5.7

Question

\begin{enumerate}[label=(\alph*)] 
\item First, argue why the integral involving $q_x(u)$ tends to zero as $N \to \infty$.
\item The first integral is a little more subtle because the function $p_x(u)$ has the $\sin(u/2)$ term in the denominator. Use the fact that $f$ is differentiable at $x$ (and a familiar limit from calculus) to prove that the first integral goes to zero as well.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item This is a direct result of the Riemann-Lebesgue Lemma (Theorem 8.5.2).
\item We would like to show that $p_x(u)$ is continuous, and the only place this is not automatically true is when $\sin(u/2) = 0$; this only occurs when $u = 0$. Strictly, $p_x(u)$ isn't even defined here, but if $\lim_{u \to 0} p_x(u)$ is well defined, then we can simply define $p_x(0) = \lim_{u \to 0} p_x(u)$ and be on our merry way. We have
$$
\lim_{u \to 0} p_x(u) = \lim_{u \to 0}\frac{f(x + u) - f(x)}{u} \frac{u}{\sin\left(\frac{u}{2}\right)} \cos\left(\frac{u}{2}\right)
= f'(x) \frac{2}{\cos\left(\frac{u}{2}\right)} \cos\left(\frac{u}{2}\right)
= 2f'(x)
$$
where we have used L'Hospital's rule in the second term. We can now conclude that $p_x(u)$ is effectively continuous, for the purposes of applying the Riemann-Lebesgue Lemma, and thus the integral goes to zero.
 \end{enumerate}

Exercise 8.5.7

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Exercise 8.5.7

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