The problem is: Let $f \in L^{1}(R) \cap C(R)$ . Supose that $f$ is positive. Show that $|\hat{f}(\xi)| < |\hat{f}(0)|$ for all $\xi \neq 0$.
My idea: By the definition of the Fourier transform we have $$|\hat{f}(\xi)| \leq |\hat{f}(0)|$$
But i am not seeing a contradiction if i have the equality for $\xi \neq 0$.
Someone can give me a hint? thanks in advance
Since $f \geqslant 0$, we have $\hat{f}(0) \geqslant 0$. For $\xi \neq 0$, pick an $\alpha \in \mathbb{R}$ so that $e^{-i\alpha}\hat{f}(\xi) \geqslant 0$. Then
$$\begin{align} \hat{f}(0) - \lvert \hat{f}(\xi)\rvert &= \hat{f}(0) - \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(t) e^{-i(\alpha + t\xi)}\,dt\\ &= \hat{f}(0) - \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(t)\cos (\alpha + t\xi)\,dt\\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(t)\left(1-\cos (\alpha + t\xi)\right)\,dt. \end{align}$$
Since $1 - \cos (\alpha + t\xi) \geqslant 0$ everywhere, and has only isolated zeros, we have, due to the non-negativity and continuity of $f$
$$\int_{-\infty}^\infty f(t)\left(1-\cos (\alpha + t\xi)\right)\,dt = 0 \iff f \equiv 0.$$
So if $f \in L^1(\mathbb{R}) \cap C(\mathbb{R})$, $f \geqslant 0$ and $f \not\equiv 0$, we have the strict inequality
$$\lvert \hat{f}(\xi)\rvert < \hat{f}(0)$$
for all $\xi\neq 0$.
