Let $$f(x,y)=\sum_{n\in\mathbb{Z}\backslash\{0\}}\frac{1}{n}e^{2\pi i(xn+yn^2)} $$ Is it true that $\|f\|_{L^{\infty}(\mathbb{R}^2)}<\infty$? i.e. is $f$ essentially bounded?

The answer is yes. Fix $x,y$, and write $e(\alpha) := e^{2\pi i \alpha}$.

Using a Littlewood-Paley partition of unity and the triangle inequality, we may bound

$$ |f(x,y)| \leq \sum_N a_N$$ where $N$ ranges over powers of two, $$ a_N := \left|\sum_{n \in {\bf Z} \backslash 0} \psi( \frac{n}{N}) \frac{1}{n} e(x n + yn^2)\right|, $$ and $\psi$ is a suitable even bump function supported on (say) $\pm [1/4,4]$. (Actually, if one wished, one could replace the smooth cutoff $\psi(\frac{n}{N})$ here by a restriction $N \leq |n| < 2N$, and the arguments below would all go through essentially unchanged.) By the triangle inequality, we have $a_N = O(1)$ uniformly in $N$.

Fix $0 < \delta \leq 1$. We will show that there are only $O( \log \frac{1}{\delta} )$ values of $N$ for which $a_N \geq \delta$, which on taking $\delta=2^{-k}$ for natural numbers $k$ and summing gives the desired bound $\sum_N a_N = \sum_{k=1}^\infty O(k 2^{-k}) = O(1)$.

By discarding all the small $N$, we may assume that $N \geq C \delta^{-C}$ for a large absolute constant $C$.

Suppose that $a_N \geq \delta$, then by summation by parts we have $|\sum_{n \in I} e( xn + yn^2 )| \gg \delta N$ for some interval $I$ in $[-10N,10N]$. Applying Weyl sum estimates (see e.g. Exercise 16 of my blog notes) and taking common denominators this implies that there are rational numbers $a/q, b/q$ with $q = O( \delta^{-O(1)})$ such that $|x - a/q| \ll\delta^{-O(1)} / N$ and $|y-b/q| \ll \delta^{-O(1)} / N^2$ (we allow $a,b,q$ to have common factors). As we are assuming $N \geq C \delta^{-C}$ for a large $C$, this forces the rationals $a/q, b/q$ to be depend only on $\delta$ and not on $N$ (since two distinct rationals $a/q,a'/q'$ with $q,q' = O(\delta^{-O(1)})$ will be separated from each other by too far of a distance).

We now have

$$ N |x-a/q| + N^2 |y-b/q| \ll \delta^{-O(1)}.$$

Suppose that in fact we had

$$ N |x-a/q| + N^2 |y-b/q| \leq C^{-1} \delta^C$$

for a large absolute constant $C$. Then we can approximate $e(xn+yn^2)$ by $e((an+bn^2)/q)$ with acceptable error and conclude that $$ \left|\sum_n \psi(\frac{n}{N}) \frac{1}{n} e( (an+bn^2)/q )\right| \gg \delta N.$$ As the function $e( (an+bn^2)/q )$ is periodic with period $q = O( \delta^{-O(1)})$, which is much smaller than $N$, one split into arithmetic progressions mod $q$ and approximate Riemann sums by Riemann integrals (crudely upper bounding the mean of $e((an+bn^2)/q)$ in magnitude by $1$) and obtain $$ \left|\int_{\bf R} \psi(\frac{t}{N}) \frac{1}{t}\ dt\right| \gg \delta N.$$ But the integrand is odd and so the integral vanishes, a contradiction.

Thus we have $$ \delta^{O(1)} \ll N |x-a/q| + N^2 |y-b/q| \ll \delta^{-O(1)}$$ and so (since $a,b,q$ do not depend on $N$) there are only $O(\log \frac{1}{\delta})$ powers of two $N$ for which $a_N \geq \delta$, as claimed.

Prof. Tao's answer is excellent. I also found two research papers answering the question so I list them below as complementary reference:

G.I.Arkhipov and K.I.Oskolkov,

*On a special trigonometric series and its applications*, 1989 Math. USSR Sb. 62 145 Link to the article: http://iopscience.iop.org/0025-5734/62/1/A10 See**Theorem 1**.E.S.Stein and S.Wainger,

*Discrete analogues of singular Radon transform*, Bulletin of AMS 1990 Link to the article: http://www.ams.org/journals/bull/1990-23-02/S0273-0979-1990-15973-7/S0273-0979-1990-15973-7.pdf See the**Lemma**in Section 6

The key of their results is that the upper bound depends only on the degree (not the coefficients, i.e. $x$ and $y$ in the question) of the polynomial.

isessentially bounded: $f(x,0) = (1-2x)\pi i$ for $0 < x < 1$ (a "sawtooth wave") $\endgroup$ – Robert Israel Jun 18 '15 at 4:20