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  "body": "\n## Proof\n\nWe use the mean value theorem twice: once to show that $(f_n)$ is uniformly Cauchy, and once to control difference quotients uniformly. No continuity of the derivatives is needed.\n\n*Setup.*\n\n1. Let $a \u003c b$. At an interior point, $f_n'$ is the [[derivative|0F1WR9C83K]]. At $a$ and $b$, it is the [[one-sided derivative|0HXEX0DPRA]]. So for each $y \\in [a, b]$, $f_n'(y)$ is the limit of $\\frac{f_n(t) - f_n(y)}{t - y}$ as $t \\to y$ with $t \\in [a, b]$, $t \\ne y$.\n2. Each $f_n$ is continuous on $[a, b]$. Indeed, fix $y$ and put $K = |f_n'(y)| + 1$. By the limit in step 1 with tolerance $1$, there is $\\delta_0 \u003e 0$ such that the quotient has absolute value at most $K$ when $0 \u003c |t - y| \u003c \\delta_0$. So $|f_n(t) - f_n(y)| \\le K |t - y|$ for these $t$. Given $\\varepsilon \u003e 0$, this is less than $\\varepsilon$ when $|t - y| \u003c \\min(\\delta_0, \\varepsilon / K)$. By the [[epsilon-delta form|0W8YHHGCJG]], $f_n$ is continuous at $y$.\n3. Let $n, m \\ge 1$, put $h = f_n - f_m$, and let $s \\ne t$ in $[a, b]$. By step 2, $h$ is continuous on the closed interval between $s$ and $t$. By [[linearity of differentiation|0M0TKFRX2A]], $h$ is differentiable inside it with $h' = f_n' - f_m'$.\n4. By the [[mean value theorem|0BARQY52N5]], there is $\\xi$ strictly between $s$ and $t$ with $h(t) - h(s) = (f_n'(\\xi) - f_m'(\\xi))(t - s)$.\n5. Let $\\eta \u003e 0$. Since $f_n' \\to g$ [[uniformly|08FPABDQ4C]], there is $N(\\eta)$ with $|f_n'(x) - g(x)| \u003c \\frac{\\eta}{2}$ for all $n \\ge N(\\eta)$ and all $x \\in [a, b]$.\n6. By step 5 and the [[triangle inequality|03HF5PMY5T]], $|f_n'(x) - f_m'(x)| \u003c \\eta$ for all $n, m \\ge N(\\eta)$ and all $x \\in [a, b]$.\n\n*The sequence $(f_n)$ converges uniformly.*\n\n7. Let $\\varepsilon \u003e 0$. The sequence $(f_n(x_0))$ converges, so it is Cauchy, since [[convergent sequences are Cauchy|0QY87R9CY3]].\n8. By steps 6 and 7, there is $N$ such that for all $n, m \\ge N$, we have $|f_n(x_0) - f_m(x_0)| \u003c \\frac{\\varepsilon}{2}$ and $|f_n' - f_m'| \u003c \\frac{\\varepsilon}{2(b - a)}$ on $[a, b]$.\n9. Let $n, m \\ge N$, $h = f_n - f_m$, and $x \\in [a, b]$. By step 4 with $s = x_0$ and $t = x$ (if $x \\ne x_0$), and step 8,\n$$|h(x)| \\le |h(x) - h(x_0)| + |h(x_0)| \u003c \\frac{\\varepsilon}{2(b - a)} |x - x_0| + \\frac{\\varepsilon}{2} \\le \\varepsilon.$$\n10. So $(f_n)$ is [[uniformly Cauchy|01XAY17QA3]] on $[a, b]$.\n11. For each $x$, the sequence $(f_n(x))$ is Cauchy by step 9. By the [[Cauchy criterion|0ZA4AX4MW3]], it converges. Let $f(x)$ be its limit.\n12. Fix $n \\ge N$ and $x$. Letting $m \\to \\infty$ in $|f_n(x) - f_m(x)| \u003c \\varepsilon$ gives $|f_n(x) - f(x)| \\le \\varepsilon$, since [[limits preserve non-strict inequalities|0N32V0E3XJ]].\n13. Since $\\varepsilon$ was arbitrary and $N$ does not depend on $x$, step 12 shows that $f_n \\to f$ uniformly on $[a, b]$.\n\n*The limit $f$ is differentiable with $f' = g$.*\n\n14. Fix $y \\in [a, b]$. For $t \\in [a, b]$ with $t \\ne y$, let $\\varphi_n(t) = \\frac{f_n(t) - f_n(y)}{t - y}$ and $\\varphi(t) = \\frac{f(t) - f(y)}{t - y}$. By step 11, $\\varphi_n(t) \\to \\varphi(t)$ for each such $t$.\n15. Let $\\eta \u003e 0$ and $n, m \\ge N(\\eta)$. By step 4 with $s = y$, and step 6, $|\\varphi_n(t) - \\varphi_m(t)| = |f_n'(\\xi) - f_m'(\\xi)| \u003c \\eta$ for each $t \\ne y$.\n16. Letting $m \\to \\infty$ in step 15 gives $|\\varphi_n(t) - \\varphi(t)| \\le \\eta$ for all $n \\ge N(\\eta)$ and all $t \\ne y$, again since limits preserve non-strict inequalities.\n17. Let $\\varepsilon \u003e 0$, put $\\eta = \\frac{\\varepsilon}{3}$, and fix $n = N(\\eta)$. By step 5, $|f_n'(y) - g(y)| \u003c \\eta$.\n18. By step 1 and the [[epsilon-delta form of a limit|0GBZDAFPX8]], there is $\\delta \u003e 0$ with $|\\varphi_n(t) - f_n'(y)| \u003c \\eta$ for all $t \\in [a, b]$ with $0 \u003c |t - y| \u003c \\delta$.\n19. For such $t$, steps 16 to 18 and the triangle inequality give\n$$|\\varphi(t) - g(y)| \\le |\\varphi(t) - \\varphi_n(t)| + |\\varphi_n(t) - f_n'(y)| + |f_n'(y) - g(y)| \u003c 3\\eta = \\varepsilon.$$\n20. So $\\varphi(t) \\to g(y)$ as $t \\to y$ within $[a, b]$. This means that $f$ is differentiable at $y$, one-sided if $y$ is an endpoint, with $f'(y) = g(y)$.\n21. Since $y$ was arbitrary, $f_n \\to f$ uniformly on $[a, b]$, $f$ is differentiable, and $f' = g$.\n\n## Idea\n\nThe mean value theorem turns a uniform bound on $f_n' - f_m'$ into a bound on the differences of $f_n - f_m$, both between two points and in difference quotients. Convergence at one point $x_0$ anchors the functions, so they converge uniformly. The difference quotients converge uniformly too, which lets us swap the limit in $n$ with the limit $t \\to y$.\n",
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