We define the $d$-dimensional torus by $\mathbb T^d = \mathbb R^d/\mathbb Z^d$, and define the Fourier transform $\hat f\in\ell^\infty(\mathbb Z^d)$ of functions $f\in L^1(\mathbb T^d)$ as $$ \hat f(k) = \int_{\mathbb T^d}\! f(x)e^{-2\pi i x\cdot k}\,\mathrm dx. $$
Now, suppose that $f\ge0$ is smooth, and $\hat f(0)=\int_{\mathbb T^d}\! f\,\mathrm dx=1$, i.e. $\,f\,\mathrm dx$ is a probability measure. A basic $L^1$ bound gives that $|\hat f(k)|\le 1$ for all $k\in\mathbb Z^d$, but can we go further and say that $|\hat f(k)|<1$ for all $k\ne0$?
We can reformulate the problem as follows. If we define $f^{*t}$ to be the $t$-fold convolution of $f$ on $\mathbb T^d$, then it is clear that $f^{*t}\to\phi$ as $t\to\infty$ for some trigonometric polynomial $\phi$, with convergence in every $H^s$ norm. Is it necessarily true that $\phi=1$?
The problem can even be reformulated in a third way. Suppose $(X)_{t=0}^\infty$ is a discrete time homogeneous Markov chain on $\mathbb T^d$, with isotropic and smooth transition kernel, i.e. the transition kernel $(m_x)_{x\in\mathbb T^d}$ satisfies $m_x = (\cdot+x)_\#\mu$ for $\mu(\mathrm dx)=f(x)\,\mathrm dx$ with $f$ smooth. Then does $X_t\xrightarrow{t\to\infty}1$ in law for any initial condition?
The answer is yes, and it's true for $f\in L^1$ (assuming the other condtions hold). Take $d=1$ for simplicity. Suppose $k\ne 0$ and $|\hat {f} (k)| = 1.$ Then
$$\left |\frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-ikx}\,dx\right | = 1 = \frac{1}{2\pi}\int_0^{2\pi}|f(x)e^{-ikx}|\,dx.$$
This implies $f(x)e^{-ikx} = c|f(x)e^{-ikx}| = cf(x)$ a.e. for some constant $c,|c|=1.$ This gives $e^{-ikx}=c$ on a set of positive measure, contradiction.