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  "body": "\n## Proof\n\nWe multiply by an integrating factor and show that the product is non-increasing.\n\n1. Let $B(t) = \\int_a^t \\beta(s)\\, ds$ for $t \\in I$. Since $\\beta$ is continuous, $B$ is continuous on $I$ and $B'(t) = \\beta(t)$ for $t \\in I^\\circ$, by the [[first fundamental theorem of calculus|063H5KJC69]].\n2. By the [[chain rule|0RXX2PT1MQ]] and [[the derivative of the exponential|0EX9GFJ0B3]], $t \\mapsto e^{-B(t)}$ is differentiable on $I^\\circ$ with derivative $-\\beta(t) e^{-B(t)}$. Also $e^{-B(t)} \u003e 0$, because [[the exponential function is positive|0P9BNBW2W3]].\n\nDifferential form.\n\n3. Let $v(t) = u(t)\\, e^{-B(t)}$ for $t \\in I$. Then $v$ is continuous on $I$.\n4. By the [[product rule|0V9A43H7WV]] and step 2, for $t \\in I^\\circ$ we have $v'(t) = \\left( u'(t) - \\beta(t) u(t) \\right) e^{-B(t)}$.\n5. The hypothesis gives $u'(t) - \\beta(t) u(t) \\le 0$, and $e^{-B(t)} \u003e 0$. So $v'(t) \\le 0$ on $I^\\circ$.\n6. Then $-v$ has a non-negative derivative on $I^\\circ$, so $-v$ is [[increasing|026BJBQNGJ]] on $I$. Hence $v(t) \\le v(a) = u(a)$ for each $t \\in I$, since $B(a) = 0$.\n7. Multiplying by $e^{B(t)} \u003e 0$ gives $u(t) \\le u(a)\\, e^{B(t)}$, which is the differential form.\n\nIntegral form.\n\n8. Fix $t \\in I$. If $t = a$, the hypothesis gives $u(a) \\le \\alpha(a) = \\alpha(a) e^{B(a)}$. So let $t \u003e a$, and put $c = \\alpha(t)$.\n9. For $s \\in [a, t]$ we have $\\alpha(s) \\le c$, because $\\alpha$ is non-decreasing. So $u(s) \\le c + S(s)$, where $S(s) = \\int_a^s \\beta(r) u(r)\\, dr$.\n10. The function $S$ is continuous on $[a, t]$, and $S'(s) = \\beta(s) u(s)$ for $s \\in (a, t)$, by the first fundamental theorem of calculus.\n11. Since $\\beta(s) \\ge 0$, step 9 gives $\\beta(s) u(s) \\le c\\, \\beta(s) + \\beta(s) S(s)$ for $s \\in [a, t]$.\n12. Let $q(s) = e^{-B(s)} \\left( S(s) + c \\right)$ for $s \\in [a, t]$. By the product rule and step 2, $q'(s) = e^{-B(s)} \\left( \\beta(s) u(s) - \\beta(s) S(s) - c\\, \\beta(s) \\right)$ on $(a, t)$.\n13. By step 11 the bracket in step 12 is at most $0$, so $q'(s) \\le 0$ on $(a, t)$. As in step 6, $q(t) \\le q(a) = c$, since $S(a) = 0$ and $B(a) = 0$.\n14. Multiplying $q(t) \\le c$ by $e^{B(t)} \u003e 0$ gives $S(t) + c \\le c\\, e^{B(t)}$.\n15. By step 9 with $s = t$, $u(t) \\le c + S(t) \\le c\\, e^{B(t)} = \\alpha(t) \\exp\\left( \\int_a^t \\beta(s)\\, ds \\right)$. This is the integral form.\n\n## Idea\n\nThe function $e^{B}$ with $B' = \\beta$ solves $v' = \\beta v$. Dividing $u$ by it turns the growth condition into the statement that a function has a non-positive derivative, so it can only decrease. In the integral form the same trick is applied to the integral term $S$, after the monotone $\\alpha$ is frozen at its value at $t$.\n",
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