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  "body": "\n## Proof\n\nWe prove the two directions separately. For $u \\ne v$ in $I$ write $s(u, v) = \\dfrac{f(v) - f(u)}{v - u}$, the slope of the chord. Note that $s(u, v) = s(v, u)$.\n\n*Convex implies $f'$ increasing.*\n\n1. Assume $f$ is [[convex|0VV0HSW1ZS]]. Let $x \u003c y$ in $I$; we show $f'(x) \\le f'(y)$.\n2. For each $t$ with $x \u003c t \u003c y$, the [[slope inequalities|0BY5FTHEEE]] give $s(x, t) \\le s(x, y) \\le s(t, y)$.\n3. By the definition of the [[derivative|0F1WR9C83K]], $s(x, t) = \\dfrac{f(x + h) - f(x)}{h}$ with $h = t - x$. So $s(x, t) \\to f'(x)$ as $t \\to x^+$.\n4. Suppose $f'(x) \u003e s(x, y)$. Put $\\varepsilon = f'(x) - s(x, y)$. By step 3 there is $t \\in (x, y)$ with $s(x, t) \u003e f'(x) - \\varepsilon = s(x, y)$.\n5. This contradicts step 2. So $f'(x) \\le s(x, y)$.\n6. Likewise $s(t, y) = \\dfrac{f(y + h) - f(y)}{h}$ with $h = t - y \u003c 0$. So $s(t, y) \\to f'(y)$ as $t \\to y^-$.\n7. Suppose $f'(y) \u003c s(x, y)$. With $\\varepsilon = s(x, y) - f'(y)$, step 6 gives $t \\in (x, y)$ with $s(t, y) \u003c s(x, y)$. This contradicts step 2.\n8. So $f'(x) \\le s(x, y) \\le f'(y)$. Hence $f'$ is increasing on $I$.\n\n*$f'$ increasing implies convex.*\n\n9. Assume $f'$ is increasing on $I$. Let $u, v \\in I$ and $\\lambda \\in [0, 1]$. We show $f(\\lambda u + (1 - \\lambda) v) \\le \\lambda f(u) + (1 - \\lambda) f(v)$.\n10. If $u = v$, or $\\lambda \\in \\set{0, 1}$, both sides are equal. Exchanging $u$ with $v$ and $\\lambda$ with $1 - \\lambda$ leaves the claim unchanged. So we may assume $u \u003c v$ and $0 \u003c \\lambda \u003c 1$.\n11. Let $p = \\lambda u + (1 - \\lambda) v$. Then $p - u = (1 - \\lambda)(v - u) \u003e 0$ and $v - p = \\lambda (v - u) \u003e 0$. So $u \u003c p \u003c v$.\n12. Since $I$ is an [[interval|0TK8Y3J5RW]], $[u, v] \\subseteq I$. The function $f$ is differentiable on $I$, hence [[continuous|07ZMX4ANJS]] on $I$.\n13. By the [[mean value theorem|0BARQY52N5]] on $[u, p]$ and on $[p, v]$, there are $c_1 \\in (u, p)$ and $c_2 \\in (p, v)$ with $s(u, p) = f'(c_1)$ and $s(p, v) = f'(c_2)$.\n14. Since $c_1 \u003c c_2$ and $f'$ is increasing, $s(u, p) \\le s(p, v)$.\n15. By step 11 this reads $\\dfrac{f(p) - f(u)}{(1 - \\lambda)(v - u)} \\le \\dfrac{f(v) - f(p)}{\\lambda (v - u)}$.\n16. Multiplying by $\\lambda (1 - \\lambda)(v - u) \u003e 0$ gives $\\lambda (f(p) - f(u)) \\le (1 - \\lambda)(f(v) - f(p))$.\n17. Rearranging, $f(p) \\le \\lambda f(u) + (1 - \\lambda) f(v)$. So $f$ is convex.\n\n## Idea\n\nFor a convex function, the slope of a chord over $[x, y]$ lies between the tangent slopes at $x$ and at $y$, since chord slopes increase. Conversely, the mean value theorem turns chord slopes into values of $f'$. If $f'$ is increasing, chords to the left of a point are less steep than chords to the right, and that is convexity.\n",
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