Exercise 6.12

Proof.

(a)

Let $a\in \mathbb{Z}$. As $\bar{\zeta }={\zeta }^{-1}$, $$\begin{array}{llll}\hfill \overline{{\psi }_a(t)}& =\overline{{\zeta }^{\mathit{at}}}={\zeta }^{-\mathit{at}}& \hfill & \\ \hfill & ={\zeta }^{a(-t)}={\zeta }^{(-a)t}& \hfill & \\ \hfill & ={\psi }_a(-t)={\psi }_{-a}(t)& \hfill & \end{array}$$

$\overline{{\psi }_a(t)}={\psi }_a(-t)={\psi }_{-a}(t)$

(b)

From Corollary of Lemma 1 : $$\frac{1}{p}\underset{a=0}{\overset{p-1}{\sum }}{\psi }_a(t-s)=\frac{1}{p}\underset{a=0}{\overset{p-1}{\sum }}{\zeta }^{a(t-s)}=\delta (t,s)$$

$$\frac{1}{p}\underset{a}{\sum }{\psi }_a(t-s)=\delta (t,s).$$