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  "body": "\n## Statement\n\nLet $f \\colon \\R \\to \\R$ be an [[additive function|KF719SZD5T]], that is, $f(x + y) = f(x) + f(y)$ for all $x, y \\in \\R$.\nIf $f$ is [[continuous|0X3WGX95FE]], then\n$$f(x) = c x \\quad \\text{for all } x \\in \\R, \\qquad \\text{where } c = f(1).$$\n\n## Remarks\n\nAdditivity gives $f(nx) = n f(x)$ for integers $n$, and then $f(q) = q f(1)$ for every rational $q = a/b$, since $b f(a/b) = f(a) = a f(1)$.\nThe rationals are [[dense|0BSJSX7ME2]] in $\\R$, so every real $x$ is a limit of rationals $q_k$, and continuity gives $f(x) = \\lim f(q_k) = \\lim c q_k = c x$.\nFor example, if a continuous $f$ satisfies $f(x + y) = f(x) + f(y)$ and $f(1) = 2$, then $f(x) = 2x$, so $f(\\pi) = 2\\pi$.\nContinuity cannot be dropped.\nA basis of $\\R$ as a vector space over $\\Q$ exists by the axiom of choice, and with it one builds additive functions that are not of the form $cx$.\nWeaker hypotheses suffice, such as continuity at a single point, monotonicity on an interval, or Lebesgue measurability.\nThe same argument shows that a continuous additive map between real normed spaces is $\\R$-linear.\n\n## Explanations\n\n### L1 human\n\nIf a function turns sums into sums and has no jumps, it must be a straight line through the origin.\nAdditivity alone fixes the function on all fractions, and continuity fills in the remaining points.\nThis is why a rule like \"the cost of two orders together is the sum of their costs\", together with continuity, forces a fixed unit price.\n\n### L2 human\n\nThis is the regular solution of [[Cauchy's functional equation|ZBMEVWN100]], and many other classical functional equations reduce to it.\nFor instance, a continuous $g \\colon \\R \\to (0, \\infty)$ with $g(x + y) = g(x) g(y)$ satisfies the hypothesis after taking logarithms, so $g(x) = a^x$ with $a = g(1)$.\nIn linear algebra terms, the theorem says that a continuous homomorphism of the group $(\\R, +)$ to itself is automatically $\\R$-linear.\n",
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