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TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)
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On an interval II, for ff continuous on II and differentiable at every interior point: f0f' \ge 0 throughout gives ff nondecreasing, f>0f' > 0 gives ff increasing, f0f' \le 0 and f<0f' < 0 give the two decreasing forms; conversely a nondecreasing ff has f0f' \ge 0 and a nonincreasing ff has f0f' \le 0 wherever it is differentiable, and no strict converse is claimed

Statement

Let IRI \subseteq \mathbb{R} be order-convex (Intervals of R\mathbb{R}: the nine order-convex forms, nondegeneracy, and length), let f:IRf : I \to \mathbb{R} be continuous on II (Continuity of f:ARf : A \to \mathbb{R} at a point of AA and on AA: the ε\varepsilon-δ\delta condition, its agreement with limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c) at a limit point, and continuity at an isolated point) and differentiable at every point of II interior to II (Interior, closure, boundary and exterior of a subset of R\mathbb{R}, The derivative f(c)=limxcf(x)f(c)xcf'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} of f:ARf : A \to \mathbb{R} at a point cAc \in A that is a limit point of AA, and differentiability on a set). The words nondecreasing, increasing, nonincreasing and decreasing are those of Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of R\mathbb{R}, with the dictionary to monotone sequences, in which increasing is the strict notion.

  1. If f(x)0f'(x) \ge 0 at every interior point xx of II, then ff is nondecreasing on II.
  2. If f(x)>0f'(x) > 0 at every interior point xx of II, then ff is increasing on II.
  3. If f(x)0f'(x) \le 0 at every interior point xx of II, then ff is nonincreasing on II.
  4. If f(x)<0f'(x) < 0 at every interior point xx of II, then ff is decreasing on II.

Conversely, with no continuity hypothesis and no hypothesis at any other point:

  1. If f:IRf : I \to \mathbb{R} is nondecreasing on II and differentiable at a point cIc \in I that is a limit point of II, then f(c)0f'(c) \ge 0; if ff is nonincreasing and differentiable at such a cc, then f(c)0f'(c) \le 0.

No strict converse is claimed here, and none is true. Claim 5 gives the weak inequality only, and it cannot be improved: an increasing function may have a vanishing derivative at a point. That failure is recorded separately, as a false statement later on this page, with its witness worked out on the companion page. Reading claim 2 backwards is the single most common misuse of this theorem, and this statement does not license it.

Claims 1 to 4 need the interval; claim 5 does not. The forward direction runs through the mean value theorem on a segment joining two points of II, so order-convexity is essential. Claim 5 is a statement about one point and uses only that the difference quotients have a constant sign.

Facts & Assumptions

Given: An order-convex IRI \subseteq \mathbb{R} and a function f:IRf : I \to \mathbb{R}; for claims 1 to 4 also that ff is continuous on II and differentiable at every interior point of II, with the stated sign condition; for claim 5 that ff is monotone on II and differentiable at a limit point cIc \in I of II.

[L1]

Mean value theorem (The mean value theorem, as the case g(x)=xg(x) = x of Cauchy's: for ff continuous on [a,b][a,b] with a<ba < b and differentiable on (a,b)(a,b) there is c(a,b)c \in (a,b) with f(b)f(a)=f(c)(ba)f(b) - f(a) = f'(c)(b-a)): for u<vu < v and h:[u,v]Rh : [u,v] \to \mathbb{R} continuous on [u,v][u,v] and differentiable at every point of (u,v)(u,v), there is c(u,v)c \in (u,v) with h(v)h(u)=h(c)(vu)h(v) - h(u) = h'(c)(v-u).

[L2]

Order-convexity (Intervals of R\mathbb{R}: the nine order-convex forms, nondegeneracy, and length): u,vIu, v \in I with uvu \le v gives [u,v]I[u,v] \subseteq I; and for u<vu < v in II every x(u,v)x \in (u,v) is interior to II, since Nε(x)(u,v)IN_{\varepsilon}(x) \subseteq (u,v) \subseteq I for ε:=min{xu, vx}>0\varepsilon := \min\{x-u,\ v-x\} > 0 (The ε\varepsilon-neighbourhood and the punctured ε\varepsilon-neighbourhood of a point of R\mathbb{R}, Interior, closure, boundary and exterior of a subset of R\mathbb{R}).

[L3]

Difference quotient and restriction of the domain (The derivative f(c)=limxcf(x)f(c)xcf'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} of f:ARf : A \to \mathbb{R} at a point cAc \in A that is a limit point of AA, and differentiability on a set): differentiability of hh at pp means that q(x):=(h(x)h(p))/(xp)q(x) := (h(x)-h(p))/(x-p) on A{p}A \setminus \{p\} has limit h(p)h'(p); if BAB \subseteq A, if pBp \in B is a limit point of BB and if h:ARh : A \to \mathbb{R} is differentiable at pp, then hBh|_B is differentiable at pp with the same derivative; and every point of an order-convex set with at least two elements is a limit point of it (Limit point, isolated point, adherent point, derived set, and dense subset of R\mathbb{R}).

[L5]

Monotone vocabulary (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of R\mathbb{R}, with the dictionary to monotone sequences): ff is nondecreasing on II when f(x)f(y)f(x) \le f(y) for all x,yIx, y \in I with xyx \le y; increasing when f(x)<f(y)f(x) < f(y) for all x<yx < y; nonincreasing and decreasing are the two conditions with the inequalities on the values reversed.

[L6]

Order arithmetic (Sign rules for products and monotonicity of multiplication, Inverses of positives are positive, and reciprocation reverses order, Ordered field): for reals ss and tt with t>0t > 0, s>0s > 0 gives st>0st > 0, s<0s < 0 gives st<0st < 0 and s=0s = 0 gives st=0st = 0, so by trichotomy s0s \ge 0 gives st0st \ge 0 and s0s \le 0 gives st0st \le 0; a nonzero real and its inverse have the same sign, so a quotient s/ts/t with s0s \ge 0 and t>0t > 0, or with s0s \le 0 and t<0t < 0, is 0\ge 0, and a quotient with s0s \le 0 and t>0t > 0, or with s0s \ge 0 and t<0t < 0, is 0\le 0.

[L7]

Limits preserve the non-strict order (If fgf \le g on a punctured neighbourhood of cc then limflimg\lim f \le \lim g, non-strictly): if h1,h2h_1, h_2 are functions on a set DD having cc as a limit point, if both limits at cc exist and if h1h2h_1 \le h_2 at every xDx \in D with 0<xc<η0 < |x - c| < \eta for some real η>0\eta > 0, then limxch1(x)limxch2(x)\lim_{x \to c} h_1(x) \le \lim_{x \to c} h_2(x). The constant function 00 on DD has limit 00 at cc (The ε\varepsilon-δ\delta limit limxcf(x)=L\lim_{x \to c} f(x) = L of f:ARf : A \to \mathbb{R} at a limit point cc of AA).

Proof

technique · direct
1.1

If II has at most one element then all four of the conditions in [L5] hold on II vacuously or trivially, since there is no pair x<yx < y in II, and claims 1 to 4 are immediate. So assume II has at least two elements, and let u,vIu, v \in I with u<vu < v be arbitrary.

givenL5
1.2

Claim 5. Let ff be nondecreasing on II and differentiable at a limit point cIc \in I of II, and let q(x):=(f(x)f(c))/(xc)q(x) := (f(x)-f(c))/(x-c) on I{c}I \setminus \{c\}, so limxcq(x)=f(c)\lim_{x \to c} q(x) = f'(c) by [L3]. For xIx \in I with x>cx > c one has f(x)f(c)f(x) \ge f(c) by [L5], so the numerator is 0\ge 0 while the denominator xcx - c is >0> 0, and [L6] gives q(x)0q(x) \ge 0. For xIx \in I with x<cx < c one has f(x)f(c)f(x) \le f(c), so the numerator is 0\le 0 while xc<0x - c < 0, and [L6] again gives q(x)0q(x) \ge 0. So the constant function 00 is q\le q at every point of I{c}I \setminus \{c\}, in particular at every such point with 0<xc<10 < |x - c| < 1; both functions have limits at the limit point cc of I{c}I \setminus \{c\}, namely 00 and f(c)f'(c), so [L7] gives 0f(c)0 \le f'(c). The nonincreasing case is the same argument with both inequalities on the values reversed, which makes q0q \le 0 throughout and hence f(c)0f'(c) \le 0.

L3L5L6L7
2.1

By [L2] the segment [u,v][u,v] is contained in II and is nondegenerate. The restriction f[u,v]f|_{[u,v]} is continuous on [u,v][u,v] by [L4]; and for x(u,v)x \in (u,v) the point xx is interior to II by [L2], so ff is differentiable at xx, while xx is a limit point of [u,v][u,v] by [L3], so f[u,v]f|_{[u,v]} is differentiable at xx with (f[u,v])(x)=f(x)(f|_{[u,v]})'(x) = f'(x).

step 1.1L2L3L4
3.1

By step 2.1 the function f[u,v]f|_{[u,v]} satisfies the hypotheses of [L1] on [u,v][u,v], so fix c(u,v)c \in (u,v) with f(v)f(u)=f(c)(vu)f(v) - f(u) = f'(c)\,(v-u); and vu>0v - u > 0 since u<vu < v.

step 2.1L1choose
4.1

If f(x)0f'(x) \ge 0 at every interior point of II then in particular f(c)0f'(c) \ge 0, so f(v)f(u)=f(c)(vu)0f(v)-f(u) = f'(c)(v-u) \ge 0 by [L6], that is f(u)f(v)f(u) \le f(v). If f(x)>0f'(x) > 0 at every interior point then f(c)>0f'(c) > 0 and the same product is >0> 0, that is f(u)<f(v)f(u) < f(v).

step 3.1L6
4.2

If f(x)0f'(x) \le 0 at every interior point then f(c)0f'(c) \le 0 and f(v)f(u)0f(v)-f(u) \le 0 by [L6], that is f(u)f(v)f(u) \ge f(v). If f(x)<0f'(x) < 0 at every interior point then f(c)<0f'(c) < 0 and f(v)f(u)<0f(v)-f(u) < 0, that is f(u)>f(v)f(u) > f(v).

step 3.1L6
5.1

The pair u<vu < v in II was arbitrary, so steps 4.1 and 4.2 establish exactly the four conditions of [L5]: for the two non-strict ones the case u=vu = v is the trivial equality f(u)=f(u)f(u) = f(u), and the two strict ones are conditions on pairs u<vu < v only. Claims 1 to 4 are proved.

step 1.1step 4.1step 4.2L5
6.1

Claims 1 to 4 are step 5.1 and claim 5 is step 1.2.

step 1.2step 5.1

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