normalized: $0\le \sigma_N(\omega)\le 1$, and $0\le F_N^{(\alpha)}(\omega)\le 1$.

\[thm:spectraltest\] Let ${\mathcal{F}}$, $\rho$ and $\sigma_N(\omega)$ be as in Definition \[definition:generalspectraltest\]. Then

1.  \[item:sigmaN-1\] The quantity $\sigma_N(\omega)$ is a maximum.

2.  \[item:sigmaN-2\] The sequence $\omega$ is uniformly distributed modulo one if and only if $$\lim_{N\to \infty} \sigma_N(\omega)= 0.$$

The proof generalizes the arguments in [@Hel98a Sec. 5.2]. For an arbitrary positive integer $K$, let $$A_{N,K}= \sup \left\{
       \rho({\ensuremath{\mathbf k}}) \left| S_N( \xi_\mathbf{k}, \omega) \right|:
        0<||\mathbf{k}|| \le K  \right\},$$ and $$B_{N,K}= \sup \left\{
        \rho({\ensuremath{\mathbf k}}) \left| S_N( \xi_\mathbf{k}, \omega) \right|:
        ||\mathbf{k}|| > K   \right\}.$$ Clearly, $$\label{eqn:sigmaN}
\sigma_N(\omega) = \max \left\{ A_{N,K}, B_{N,K}\right\}.$$ We have $\sigma_N(\omega)> 0$. Otherwise, all terms $S_N(\xi_{\ensuremath{\mathbf k}}, \omega)$, ${\ensuremath{\mathbf k}}\in \Lambda^*$, would be equal to zero, which is impossible. Hence, there exists $\delta > 0$ such that $\delta < \sigma_N(\omega)$, and an index $K_0 = K_0(\delta)\in \mathbb{N}$ such that for all $\mathbf{k}$ with $ ||\mathbf{k}||> K_0$ we have $\rho({\ensuremath{\mathbf k}}) < \delta$. This implies $$B_{N,K_0}
\le
\sup \left\{
        \rho(\mathbf{k}): ||\mathbf{k}|| > K_0 \right\}
\le \delta < \sigma_N(\omega).$$ Hence, $\sigma_N(\omega) = A_{N,K_0}$. We observe that the set $\{\mathbf{k}: ||\mathbf{k}|| \le K_0  \}$ is finite. In this context, we recall that all norms on ${\ensuremath{\mathbb{R}}}^s$ are equivalent. This proves (\[item:sigmaN-1\]).

In 