Can we bound $\sum_{i=1}^ke^{-{\rm i}2\pi\langle\omega,\:x_i\rangle}$ by $\sum_{i=1}^ke^{-{\rm i}2\pi\langle\omega,\:x_i\rangle}\hat g(a_i\omega)$?

Question

Let $x_1,\ldots,x_k\in[0,1)^d$, $a_1,\ldots,a_d>0$ and $$f(\omega):=\sum_{i=1}^ke^{-{\rm i}2\pi\langle\omega,\:x_i\rangle}\hat g(a_i\omega)\;\;\;\text{for }\omega\in\mathbb R^d,$$ where $g(x)=e^{-(x/\sigma)^2}$ for some $\sigma>0$.

Suppose $|f(\omega)|<\varepsilon$ for all $\omega\in\mathbb R^d$ with $0<\|\omega\|<\delta$. Can we show that $\left|\sum_{i=1}^ke^{-{\rm i}2\pi\langle\omega,\:x_i\rangle}\right|$ is also bounded by something multiplied by $\varepsilon$? This is clearly true when all $a_i$ are equal to the same value, since we can then pull the $\hat\varphi$ out.