Let $k(t)$ be any function absolutely integrable over $(-\infty,+\infty)$, let $$K(u)=\int_{-\infty}^{+\infty}k(t) e^{-uti}dt$$ and let $$f(t)=\sum_n a_n e^{-\lambda_n t i}, \ \ \ \lambda_n\in\mathbb R$$ Then $$\int_{-\infty}^{+\infty}k(t) f(t) e^{\lambda_n ti}dt=\sum_m a_m K(\lambda_m-\lambda_n)$$ Prove that, taking $n=\nu$ in the previous equation, we deduce $$\left|\int_{-\infty}^{+\infty}k(t) f(t) e^{\lambda_n ti}dt\right|\geq |a_\nu K(0)| -|a_\nu|\sum_{m\neq\nu} \left|K(\lambda_m-\lambda_\nu)\right|$$ where $|a_\nu|$ is the greatest $|a_n|$.
Any suggestion please?
By reverse triangle inequality
$$\left|\int_{-\infty}^{\infty}k\left(t\right)f\left(t\right)e^{\lambda_{\nu}it}dt\right|=\left|\sum_{m}a_{m}K\left(\lambda_{m}-\lambda_{\nu}\right)\right|=\left|a_{\nu}K\left(0\right)+\sum_{m\neq\nu}a_{m}K\left(\lambda_{m}-\lambda_{\nu}\right)\right|\geq\left|a_{\nu}K\left(0\right)\right|-\left|\sum_{m\neq\nu}a_{m}K\left(\lambda_{m}-\lambda_{\nu}\right)\right|\geq\left|a_{\nu}K\left(0\right)\right|-\sum_{m\neq\nu}\left|a_{m}\right|\left|K\left(\lambda_{m}-\lambda_{\nu}\right)\right|\geq\left|a_{\nu}K\left(0\right)\right|-\left|a_{\nu}\right|\sum_{m\neq\nu}\left|K\left(\lambda_{m}-\lambda_{\nu}\right)\right|.$$