[proofplan] We identify the finite real sequence with its empirical measure on the torus $\mathbb R/\mathbb Z$. The Erdős-Turán discrepancy inequality bounds the deviation of this empirical measure on every interval by a truncation error and finitely many nonzero Fourier coefficients. We then rewrite the positive-frequency estimate in the symmetric form appearing in the statement, using conjugation to see that the negative-frequency exponential sums have the same absolute values as the positive-frequency ones. [/proofplan]

[step:Apply the interval form of the Erdős-Turán inequality]Let $\mathbb T:=\mathbb R/\mathbb Z$, and let $\mu_{\mathbb T}$ denote normalized Haar probability measure on $\mathbb T$. For an interval $I\subset \mathbb T$, define the empirical interval-mass map \begin{align*} A_I:\mathbb R^N&\to[0,1] \end{align*} by \begin{align*} A_I(x_1,\dots,x_N):=\frac{1}{N}\#\{1\le n\le N:x_n\bmod 1\in I\}. \end{align*} We use the same-run theorem [citetheorem:9093], the Erdős-Turán discrepancy inequality. Its statement applies to $N,H\in\mathbb N$, a real tuple $x_1,\dots,x_N$, the additive character $e(t)=\exp(2\pi i t)$, the torus $\mathbb R/\mathbb Z$ with normalized Haar probability measure, and the interval-mass maps $A_I$ just defined. These are exactly the present hypotheses and definitions, with $\mu_{\mathbb T}$ playing the role of the Haar measure denoted $\mu$ there. Therefore there is an absolute constant $C_0>0$ such that, for every interval $I\subset\mathbb T$, \begin{align*} |A_I(x_1,\dots,x_N)-\mu_{\mathbb T}(I)|\le C_0\left(\frac{1}{H}+\sum_{h=1}^{H}\frac{1}{h}\left|\frac{1}{N}\sum_{n=1}^{N}e(hx_n)\right|\right). \end{align*}[/step]

[guided]We first translate the discrepancy problem into a statement about empirical measures on the torus. Define $\mathbb T:=\mathbb R/\mathbb Z$, and let $\mu_{\mathbb T}$ be normalized Haar probability measure on $\mathbb T$. For an interval $I\subset\mathbb T$, define the empirical interval-mass map \begin{align*} A_I:\mathbb R^N&\to[0,1] \end{align*} by \begin{align*} A_I(x_1,\dots,x_N):=\frac{1}{N}\#\{1\le n\le N:x_n\bmod 1\in I\}. \end{align*} The external input is the same-run theorem [citetheorem:9093], the Erdős-Turán discrepancy inequality. We record exactly what is being used: if $N,H\in\mathbb N$, $x_1,\dots,x_N\in\mathbb R$, $e(t)=\exp(2\pi i t)$, and $\mathbb R/\mathbb Z$ is equipped with normalized Haar probability measure, then every interval $I\subset\mathbb R/\mathbb Z$ satisfies an estimate of the form \begin{align*} |A_I(x_1,\dots,x_N)-\mu_{\mathbb T}(I)|\le C_0\left(\frac{1}{H}+\sum_{h=1}^{H}\frac{1}{h}\left|\frac{1}{N}\sum_{n=1}^{N}e(hx_n)\right|\right) \end{align*} for an absolute constant $C_0>0$. We verify the hypotheses one by one. The present theorem assumes $N,H\in\mathbb N$ with $N\ge 1$ and $H\ge 1$, so the integer-size hypotheses are satisfied. It assumes $x_1,\dots,x_N\in\mathbb R$, so the required finite real tuple is available. The additive character $e:\mathbb R\to\mathbb C$ has already been defined by $e(t)=\exp(2\pi i t)$, matching the cited theorem. Finally, the map $A_I$ above is exactly the empirical interval-mass map appearing in the cited theorem, and $\mu_{\mathbb T}$ is the normalized Haar probability measure required there. Thus the cited theorem gives the displayed estimate for every interval $I\subset\mathbb T$. This is the quantitative Weyl estimate before taking the supremum over intervals. The right-hand side does not depend on $I$, so it is already uniform over all intervals.[/guided]

[step:Rewrite the positive-frequency sum in symmetric form]For each integer $h$ with $1\le |h|\le H$, define the normalized exponential sum $E_h\in\mathbb C$ by \begin{align*} E_h:=\frac{1}{N}\sum_{n=1}^{N}e(hx_n). \end{align*} Since $e(-t)=\overline{e(t)}$ for every $t\in\mathbb R$, one has \begin{align*} E_{-h}:=\frac{1}{N}\sum_{n=1}^{N}e(-hx_n)=\overline{E_h}. \end{align*} Thus $|E_{-h}|=|E_h|$, and therefore \begin{align*} \sum_{h=1}^{H}\frac{1}{h}|E_h|\le \sum_{1\le |h|\le H}\frac{1}{|h|}|E_h|. \end{align*} Consequently, for every interval $I\subset\mathbb T$, \begin{align*} |A_I(x_1,\dots,x_N)-\mu_{\mathbb T}(I)|\le C_0\left(\frac{1}{H}+\sum_{1\le |h|\le H}\frac{1}{|h|}\left|\frac{1}{N}\sum_{n=1}^{N}e(hx_n)\right|\right). \end{align*}[/step]

[guided]The interval Erdős-Turán estimate from the previous step uses only the positive frequencies $1\le h\le H$, while the theorem statement uses the symmetric range $1\le |h|\le H$. For each integer $h$ with $1\le |h|\le H$, define the normalized exponential sum $E_h\in\mathbb C$ by \begin{align*} E_h:=\frac{1}{N}\sum_{n=1}^{N}e(hx_n). \end{align*} If $1\le h\le H$, then $e(-t)=\overline{e(t)}$ for every $t\in\mathbb R$, so \begin{align*} E_{-h}=\frac{1}{N}\sum_{n=1}^{N}e(-hx_n)=\overline{E_h}. \end{align*} Taking absolute values gives $|E_{-h}|=|E_h|$. Hence the symmetric sum contains the positive-frequency sum as a sub-sum: \begin{align*} \sum_{h=1}^{H}\frac{1}{h}|E_h|\le \sum_{1\le |h|\le H}\frac{1}{|h|}|E_h|. \end{align*} Substituting this larger right-hand side into the interval estimate preserves the inequality for every interval $I\subset\mathbb T$.[/guided]

[step:Take the supremum over intervals to obtain the discrepancy bound]For $0\le a<b\le 1$, let $I_{a,b}\subset\mathbb T$ be the interval represented by $[a,b)\subset[0,1)$. Then \begin{align*} \mu_{\mathbb T}(I_{a,b})=b-a \end{align*} and \begin{align*} A_{I_{a,b}}(x_1,\dots,x_N)=\frac{1}{N}\#\{1\le n\le N:\{x_n\}\in [a,b)\}. \end{align*} Taking the supremum over all $0\le a<b\le 1$ in the estimate from the previous step gives \begin{align*} D_N(x_1,\dots,x_N)\le C_0\left(\frac{1}{H}+\sum_{1\le |h|\le H}\frac{1}{|h|}\left|\frac{1}{N}\sum_{n=1}^{N}e(hx_n)\right|\right). \end{align*} Setting $C:=C_0$ completes the proof.[/step]

[guided]We now connect the interval estimate on $\mathbb T$ with the discrepancy $D_N$ appearing in the theorem statement. Fix [real numbers](/page/Real%20Numbers) $a,b$ with $0\le a<b\le 1$, and let $I_{a,b}\subset\mathbb T$ denote the interval represented by the half-open interval $[a,b)\subset[0,1)$. Normalized Haar probability measure on $\mathbb T$ agrees with one-dimensional length on this fundamental interval, so \begin{align*} \mu_{\mathbb T}(I_{a,b})=b-a. \end{align*} The condition $x_n\bmod 1\in I_{a,b}$ is equivalent to the condition that the fractional part $\{x_n\}$ lies in $[a,b)$. Therefore the empirical interval-mass map satisfies \begin{align*} A_{I_{a,b}}(x_1,\dots,x_N)=\frac{1}{N}\#\{1\le n\le N:\{x_n\}\in [a,b)\}. \end{align*} Substituting this particular interval into the estimate proved in the previous step gives, for every $0\le a<b\le 1$, \begin{align*} \left|\frac{1}{N}\#\{1\le n\le N:\{x_n\}\in [a,b)\}-(b-a)\right|\le C_0\left(\frac{1}{H}+\sum_{1\le |h|\le H}\frac{1}{|h|}\left|\frac{1}{N}\sum_{n=1}^{N}e(hx_n)\right|\right). \end{align*} The right-hand side is independent of $a$ and $b$. Taking the supremum over all $0\le a<b\le 1$ is exactly the definition of $D_N(x_1,\dots,x_N)$, so \begin{align*} D_N(x_1,\dots,x_N)\le C_0\left(\frac{1}{H}+\sum_{1\le |h|\le H}\frac{1}{|h|}\left|\frac{1}{N}\sum_{n=1}^{N}e(hx_n)\right|\right). \end{align*} Finally set $C:=C_0$. Since $C_0$ is absolute, so is $C$, and this is the asserted bound.[/guided]