If $f \in L^1$ and $f$ is everywhere strictly positive, then does it follow that $|\widehat{f}(y)| < \widehat{f}(0)$ for $y \neq 0$? As the question title suggests, if $f \in L^1$ and $f$ is everywhere strictly positive, then does it follow that $|\widehat{f}(y)| < \widehat{f}(0)$ for $y \neq 0$?
 A: First we have $\hat{f}(0)\geq |\hat{f}(y)|$ since 
$$|\hat{f}(y)|= \bigg|\int f(x) e^{-2\pi i xy} dx \bigg| \leq \int |f(x)| | e^{-2\pi i xy}| dx = \int f(x) dx = \hat{f}(0).$$
And for strict inequality observe that 
$$\bigg|\int f(x) e^{-2\pi i xy} dx \bigg|^2 = \bigg(\int f(x)\cos(2\pi xy) dx\bigg)^2+\bigg(\int f(x)\sin(2\pi xy) dx\bigg)^2  \quad (\star)$$
Now we apply Jensen's inequality. Looking at "$f(x) dx$" as our measure, we have 
$$\bigg(\frac{1}{\int f(x) dx} \int \cos(2\pi xy) f(x)dx\bigg)^2 < \frac{1}{\int f(x) dx} \int \cos(2\pi xy)^2 f(x)dx$$
and similarly for the integral with $\sin(x)$.  
Note the equality in Jensen's inequality $\phi\left(\frac{1}{\mu(E)} \int_E g d\mu  \right) = \frac{1}{\mu(E)} \int_E \phi(g)d\mu$ holds if and only if $g$ is constant or $\phi$ is linear. 
Apply this to $(\star)$, we get 
$$\bigg|\int f(x) e^{-2\pi i xy} dx \bigg|^2<\bigg(\int f(x) dx \bigg)^2.$$
A: \begin{align}
    |\hat{f}(y)| &= \left|\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-iyx}f(x)dx\right| \\
  & \le \frac{1}{\sqrt{2\pi}}\int_{\infty}^{\infty}|f(x)|dx \\
  & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)dx =\hat{f}(0).
\end{align}
