If $\varphi\in\mathcal{S}(\mathbb{R}):\int_\mathbb{R}\varphi dx=1$, does it then follow that $\sum_{k}\hat{\varphi}(k) \leq 2$?

Question

Let $\varphi\in\mathcal{S}(\mathbb{R})$ be a Schwartz test signal such that $\int_{\mathbb{R}}\varphi(x)dx = 1$. Does it then follow that the sum of the Fourier coefficients of $\varphi$ are at most one, or in general at most twice the integral $I = \int_{\mathbb{R}}\varphi(x)dx$, if we did not assume normality? We know that the Fourier coefficients are given by $\hat{\varphi}(k)\equiv \int_{\mathbb{R}}\varphi(x)e^{-2\pi i k x}dx$ and $\varphi(x) = \sum_{k\in\mathbb{Z}}\hat{\varphi}(k)e^{2\pi i kx}$. But I don't see an immediate way to conclude anything about the sum of the said coefficients. What could be done in this case?

Answer 1

Given $\varphi\in S(\Bbb{R})$ the Fourier series of the $1$-periodization is $$\sum_n \varphi(x+n) = \sum_k\hat{\varphi}(k)e^{2\pi i kx}$$ So take $\varphi(x) = m e^{-\pi m^2 x^2}$ to get that $\int_\Bbb{R} \varphi(x)dx=1$ and $$ \sum_k\hat{\varphi}(k) = \sum_n \varphi(n) \ge \varphi(0) = m$$