# Strict upper bound of Fourier Transform

I've got, again, a question about the fourier transform. Actually, let's $f \in L^1(\mathbb{R})$ be a positive function ($f \geq 0$). We want to show that :

$$\forall y\neq0 \; |\hat{f}(y)| < |\hat{f}(0)|$$

Of course we have $$\forall y\neq0 \; |\hat{f}(y)| \leq |\hat{f}(0)|$$

because $|\hat{f}(0)| = ||f||_1$ but I don't figure out how to show the strict bound. We know that :

$$|\hat{f}(y)| = |\hat{f}(0)| \iff |\int_{\mathbb{R}} f(t)e^{ity} \,\mathrm{d}t| = \int_{\mathbb{R}} f(t) \,\mathrm{d}t$$ $$\iff (\int_{\mathbb{R}} f(t)\cos(ty) \,\mathrm{d}t)^2 + (\int_{\mathbb{R}} f(t)\sin(ty) \,\mathrm{d}t)^2 = (\int_{\mathbb{R}} f(t) \,\mathrm{d}t)^2$$

From here I've tried several things but nothing gave me the result.

Someone has an idea ?

Thank you !

Your last formula is useful: we can write $$\left( \int_{\mathbb{R}} g(t) \, dt \right)^2 = \int_{\mathbb{R}} \int_{\mathbb{R}} g(t)g(s) \, ds \, dt,$$ so $$\left( \int_{\mathbb{R}} f(t) \, dt \right)^2 - \left( \int_{\mathbb{R}} f(t) \sin{ty} \, dt \right)^2 - \left( \int_{\mathbb{R}} f(t) \cos{ty} \, dt \right)^2 \\ = \int_{\mathbb{R}} \int_{\mathbb{R}} (1-\cos{ty}\cos{sy}-\sin{ty}\sin{sy}) f(t)f(s) \, ds \, dt \\ = \int_{\mathbb{R}} \int_{\mathbb{R}} 2\sin^2{\left(\tfrac{1}{2}(s-t)y\right)} f(t)f(s) \, ds \, dt$$
When $y \neq 0$, $2\sin^2{\left(\tfrac{1}{2}(s-t)y\right)}$ is positive almost everywhere, so provided that $f$ is positive on some measurable set $E$, you can find a subset $E'$ of $E \times E$ so that the whole integrand is strictly positive on $E'$, and then it follows that the last integral is larger than $$\iint_{E'} 2\sin^2{\left(\tfrac{1}{2}(s-t)y\right)} f(t)f(s) \, ds \, dt > 0 .$$