Let $f\in L^1(\mathbb{R}^d)$ with $f > 0$. I would like to prove that for any $\xi \in \mathbb{R}^d \setminus\{0\}$, it holds $$|\widehat{f}(\xi) |< \widehat{f}(0).$$
Without loss of generlity, $d = 1$. According to this answer, I first prove that we have $\leq$ for any $\xi \in \mathbb{R}^d$. Now, in order to prove that strictly inequality holds, I first show, as suggested in the answer, that the statement is true for $f$ a positive step function. Omitting the constant factor, I'm struggling to show that $$ \left|\int_{\mathbb{R}} f(x) e^{-i\xi x} \, dx \right| \, < \, \int_{\mathbb{R}} f(x) \, dx .$$
My attempt: Since $f \in L^1$ is positive, we may find a sequence of step functions $(f_n)$ such that $f_n \nearrow f$ pointwise as $n\to \infty$. By the monotone convergence theorem, $$\int_{\mathbb{R}}f(x) e^{-i\xi x} \, dx \, = \, \lim_{n\to \infty} \int_{n\to \infty} f_n(x)e^{-i\xi x} \, dx $$ and $$\int_{\mathbb{R}} |f(x) e^{-i\xi x}| \, dx \, = \, \lim_{n\to \infty} \int_{\mathbb{R}} |f_n(x)e^{-i\xi x}| \, dx.$$ However, passing to the limit, the strict inequality that I have for any $f_n$ becomes no more strict.
Any suggestion how to fix that? Thanks in advance!
For $\xi\ne 0$ take $a=-arg(\hat{f}(\xi))$ so that $$| \hat{f}(\xi)|= \Re(e^{ia} \hat{f}(\xi))=\int_{\Bbb{R}^d} f(x)\cos(a-x.\xi)dx$$
$f>0$ so that $$\int_{\Bbb{R}^d} f(x)1_{\cos(a-x.\xi)<1/2} dx \ne 0$$ obtaining
$$| \hat{f}(\xi)|\le \int_{\Bbb{R}^d} f(x)1_{\cos(a-x.\xi)\ge 1/2} dx + 1/2 \int_{\Bbb{R}^d} f(x)1_{\cos(a-x.\xi)< 1/2} dx $$ $$<\int_{\Bbb{R}^d} f(x)1_{\cos(a-x.\xi)\ge 1/2} dx + \int_{\Bbb{R}^d} f(x)1_{\cos(a-x.\xi)< 1/2} dx=\hat{f}(0)$$ If you assume only that $f\ge 0,\|f\|_{L^1}\ne 0$ then $\{x\in \Bbb{R}^d,\cos(a-x.\xi)=1\}$ has measure zero so there is $r<1$ such that $\int_{\Bbb{R}^d} f(x)1_{\cos(a-x.\xi)< r} dx\ne 0$ and $\hat{f}(0)-| \hat{f}(\xi)|\ge (1-r) \int_{\Bbb{R}^d} f(x)1_{\cos(a-x.\xi)< r} dx>0 $