Mean-periodic function

Remarks

Give $C(\R)$ the topology of uniform convergence on compact sets. Its dual space is the space of compactly supported complex Radon measures. By the Hahn–Banach theorem, $f$ is mean-periodic if and only if the closed linear span of the translates $x \mapsto f(x - a)$, $a \in \R$, is not all of $C(\R)$. Some authors allow $\mu$ to be a compactly supported distribution instead of a measure.

Example: if $f$ is continuous with period $p > 0$, take $\mu = \delta_0 - \delta_p$. Then $(f * \mu)(x) = f(x) - f(x - p) = 0$, so $f$ is mean-periodic. Similarly $f(x) = e^{\lambda x}$ with $\lambda \in \C$ satisfies $f * (\delta_0 - e^{\lambda} \delta_1) = 0$.

Non-example: $f(x) = e^{x^2}$ is not mean-periodic. Suppose $f * \mu = 0$ with $\mu$ compactly supported, and let $\nu$ be the measure $e^{y^2} d\mu(y)$. Then $(f * \mu)(x) = e^{x^2} \int e^{-2xy} \, d\nu(y)$, so $F(z) = \int e^{-2zy} \, d\nu(y)$ vanishes for real $z$. Since $\nu$ has compact support, $F$ is an entire function, so $F = 0$ on $\C$. On the imaginary axis $F$ is the Fourier transform of $\nu$, so $\nu = 0$ and $\mu = 0$.