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There is a minor range issue in the printed bound: for $\epsilon >1$ and positive $C$, the upper bound is below one. We prove the intended result for $0<\epsilon \le 1$; a valid formulation for every $\epsilon >0$ replaces the upper bound by $\max \{1,{\epsilon }^{-C}\}$. In fact the argument below gives $C=32$. All auxiliary estimates are proved here.

For $0<\epsilon \le 1/2$, let $m=\lceil {\epsilon }^{-2}\rceil$, and use the Fejér kernelExpanding the square proves the identity and nonnegativity. The finite geometric series formula and $|\sin \pi t|\ge 2\Vert t\Vert$ give $F_m(t)\le (4m\Vert t{\Vert }^2{)}^{-1}$ off the integers. In particular $F_m(t)\le 1/4$ whenever $\Vert t\Vert \ge \epsilon$.

Suppose, for a contradiction, that no $1\le n\le N$ has $\Vert \alpha n^2\Vert <\epsilon$. Summing the Fejér kernel along the quadratic sequence and separating its constant term givesThe coefficients on the right sum to $m-1$. Also $|S_{-k}|=|S_k|$. Hence some $1\le k<m$ has $|S_k|\ge \eta N$, where $\eta =1/(2m)$.

We need only an elementary Van der Corput inequality for finite scalar sequences. For $|z_n|\le 1$, extended by zero outside $[N]$, each term of $S={\sum }_n{z}_n$ occurs in exactly $H$ windows of length $H$. Applying the Cauchy-Schwarz inequality to the window sums and expanding their squares gives, for $H\le N$,Consequently $|S{|}^2\le 2N^2/H+(4N/H){\sum }_{h=1}^{H-1}|C_h|$. Take $H=\lceil 8{\eta }^{-2}\rceil$. If $|S|\ge \eta N$, some $1\le h<H$ must have $|C_h|\ge {\eta }^{2}N/8$; otherwise the displayed upper bound is less than $3{\eta }^2{N}^2/4$.

For the large quadratic exponential sum just found, put $z_n=e(k\alpha n^2)$. Its quadratic exponential sum has multiplicative derivativeThus $C_h$ is a finite geometric series. Its absolute value is at most $1/(2\Vert 2kh\alpha \Vert )$, unless that distance is zero, in which case the desired estimate is automatic. It follows thatSet $n_{\ast }=2kh$. The distance to the nearest integer satisfies $\Vert \ell x\Vert \le \ell \Vert x\Vert$ for a positive integer $\ell$, by multiplying a nearest integer to $x$. ThereforeThis proves the needed quantitative quadratic recurrence once $N$ is chosen polynomially in ${\epsilon }^{-1}$.

For explicit bookkeeping, $m\le 2{\epsilon }^{-2}$ and $H\le 33m^2$. Choose $N=\lfloor {\epsilon }^{-32}\rfloor$. For $0<\epsilon \le 1/2$ we have $N>33792{\epsilon }^{-11}\ge 8mH/({\eta }^2\epsilon )$ and $N\ge H$, while $2mH\le 528{\epsilon }^{-6}\le N$. Hence $n_{\ast }\le N$ and $\Vert \alpha n_{\ast }^2\Vert <\epsilon$, contradicting our supposition. The estimates have substantial slack even at $\epsilon =1/2$.

For $1/2<\epsilon \le 1$, simply take $n=1$, since $\Vert \alpha \Vert \le 1/2<\epsilon$ and $1\le {\epsilon }^{-32}$. For the intended range, the conclusion isFor $\epsilon >1$, the same choice $n=1$ proves the corrected all-range version.