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  "body": "\n## Proof\n\n1. Since $f$ is [[Lipschitz|0QNVQ4VD9T]], there is a real $K \\ge 0$ with $\\abs{f(x) - f(y)} \\le K \\abs{x - y}$ for all $x, y \\in D$.\n2. Let $\\varepsilon \u003e 0$. Put $\\delta = \\frac{\\varepsilon}{K + 1}$. Then $\\delta \u003e 0$, since $K + 1 \u003e 0$.\n3. Let $x, y \\in D$ with $\\abs{x - y} \u003c \\delta$. By step 1, $\\abs{f(x) - f(y)} \\le K \\abs{x - y} \\le K \\delta$.\n4. Since $K \u003c K + 1$ and $\\delta \u003e 0$, we have $K \\delta \u003c (K + 1) \\delta = \\varepsilon$. So $\\abs{f(x) - f(y)} \u003c \\varepsilon$.\n5. The number $\\delta$ depends only on $\\varepsilon$, not on $x$ or $y$. So $f$ is [[uniformly continuous|0Q4DR37DVE]].\n\n## Idea\n\nThe Lipschitz bound controls the change of $f$ by a fixed multiple of the change of $x$. So one $\\delta$, a fixed fraction of $\\varepsilon$, works at every point. Using $K + 1$ instead of $K$ avoids division by zero when $K = 0$.\n",
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  "name": "Proof that Lipschitz functions are uniformly continuous",
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