# $f \in L^1(\mathbb R), f>0$ then $|\hat f(y)| < \hat f(0), y \ne 0$

Suppose $f$ is a strictly positive function in $L^1(\mathbb R)$. Show $$|\hat f(y)| < \hat f(0) \text{, for all } y \ne 0.$$

Using monotonicity of the integral, I can show $|\hat f(y)| \le \hat f(0)$.
I don't see how to make the inequality strict.

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If $\widehat f(0)=|\widehat f(y)|$ for some $y\neq 0$, let $\theta$ such that $|\widehat f(y)|=e^{i\theta}\widehat f(y)$. Then $$\int_{\Bbb R}f(t)dt=\int_{\Bbb R}f(t)e^{i(ty+\theta})dt,$$ which gives $$\int_{\Bbb R}f(t)(1-e^{i(ty+\theta)})dt=0=\int_{\Bbb R}f(t)(1-\cos(ty+\theta))-i\int_{\Bbb R}f(t)\sin(ty+\theta)dt.$$ In particular $$\int_{\Bbb R}\underbrace{f(t)(1-\cos(ty+\theta))}_{\geq 0}dt=0,$$ so $f(t)(1-\cos(ty+\theta))=0$ almost everywhere. As $f>0$ almost everywhere, we get a contradiction.

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