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  "body": "\n## Statement\n\nLet $T \u003e 0$, and let $\\varphi, \\psi \\colon [0, T] \\to \\C$ be Lebesgue integrable.\nSuppose that the [[convolution|FCKX8KFVGN]]\n$$(\\varphi * \\psi)(x) = \\int_0^x \\varphi(t) \\, \\psi(x - t) \\, dt$$\nis $0$ for almost every $x \\in (0, T)$.\nThen there are $\\lambda \\ge 0$ and $\\mu \\ge 0$ with $\\lambda + \\mu \\ge T$ such that $\\varphi = 0$ almost everywhere on $(0, \\lambda)$ and $\\psi = 0$ almost everywhere on $(0, \\mu)$.\n\n## Remarks\n\nEquivalently, write $\\lambda(\\varphi)$ for the largest $\\lambda$ such that $\\varphi$ vanishes almost everywhere on $(0, \\lambda)$.\nThen for integrable $\\varphi, \\psi$ on $[0, T]$ one has $\\lambda(\\varphi * \\psi) = \\lambda(\\varphi) + \\lambda(\\psi)$ whenever the right side is less than $T$.\nA standard form concerns functions or distributions with compact support on $\\R$.\nIt says that the convex hull of the support of $\\varphi * \\psi$ is the sum of the convex hulls of the supports of $\\varphi$ and $\\psi$.\nThe classical proof uses complex analysis: the Laplace or Fourier transforms of such functions are entire functions of exponential type, and the theorem reduces to growth estimates for them.\nReal-variable proofs are also known.\n\nExample: with $T = 2$ and $\\varphi = \\psi = \\mathbf{1}_{(1, 2)}$, the convolution vanishes on $(0, 2)$, and the conclusion holds with $\\lambda = \\mu = 1$, so $\\lambda + \\mu = T$ exactly.\n\nThe easy direction is the converse: if $\\varphi = 0$ on $(0, \\lambda)$ and $\\psi = 0$ on $(0, \\mu)$, then each product $\\varphi(t) \\psi(x - t)$ with $x \u003c \\lambda + \\mu$ vanishes, so $\\varphi * \\psi = 0$ on $(0, \\lambda + \\mu)$.\n\n## Explanations\n\n### L1 human\n\nConvolving two signals that start at time $0$ can only produce silence for as long as the first signal waits plus the second signal waits.\nThere is no way for the two signals to cancel each other out early.\nSo if the convolution is zero for a long time, at least one of the signals must have been silent for a correspondingly long time.\n\n### L2 human\n\nLetting $T \\to \\infty$ gives the corollary that the [[convolution ring of continuous functions on the half-line has no zero divisors|E9QNNPQD2P]].\nThis is the algebraic foundation of the [[Mikusiński operational calculus|TD5XK5AHNV]], which builds the field of fractions of that ring.\nThe support form of the theorem is used in the theory of distributions and in the uniqueness theory of partial differential equations.\n",
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