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  "body": "\n## Definition\n\nLet $f \\colon \\R \\to \\C$ be a [[continuous function|0X3WGX95FE]].\nLet $\\mu$ range over the nonzero complex [[Radon measures|HZNEQQBCW4]] on $\\R$ with compact [[support|FQTXV6BQZG]].\nThe function $f$ is *mean-periodic* if the [[convolution|2Z29JN3VES]] of $f$ with some such $\\mu$ vanishes:\n$$(f * \\mu)(x) = \\int_{\\R} f(x - y) \\, d\\mu(y) = 0 \\quad \\text{for all } x \\in \\R.$$\n\n## Remarks\n\nGive $C(\\R)$ the topology of uniform convergence on compact sets.\nIts dual space is the space of compactly supported complex Radon measures.\nBy 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 $C(\\R)$.\nSome authors allow $\\mu$ to be a compactly supported distribution instead of a measure.\n\nExample: if $f$ is continuous with period $p \u003e 0$, take $\\mu = \\delta_0 - \\delta_p$.\nThen $(f * \\mu)(x) = f(x) - f(x - p) = 0$, so $f$ is mean-periodic.\nSimilarly $f(x) = e^{\\lambda x}$ with $\\lambda \\in \\C$ satisfies $f * (\\delta_0 - e^{\\lambda} \\delta_1) = 0$.\n\nNon-example: $f(x) = e^{x^2}$ is not mean-periodic.\nSuppose $f * \\mu = 0$ with $\\mu$ compactly supported, and let $\\nu$ be the measure $e^{y^2} d\\mu(y)$.\nThen $(f * \\mu)(x) = e^{x^2} \\int e^{-2xy} \\, d\\nu(y)$, so $F(z) = \\int e^{-2zy} \\, d\\nu(y)$ vanishes for real $z$.\nSince $\\nu$ has compact support, $F$ is an entire function, so $F = 0$ on $\\C$.\nOn the imaginary axis $F$ is the Fourier transform of $\\nu$, so $\\nu = 0$ and $\\mu = 0$.\n",
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