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TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)
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If continuously differentiable functions converge at one point and their derivatives converge uniformly on a closed interval, then the functions converge uniformly to a differentiable function whose derivative is the derivative limit

Statement

Let a<ba<b be reals and let fk:[a,b]Rf_k:[a,b]\to\mathbb{R} be continuously differentiable: each fkf_k is differentiable on [a,b][a,b] and each derivative fkf_k' is continuous there (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). Suppose there is c[a,b]c\in[a,b] such that the real sequence (fk(c))(f_k(c)) converges to AA, and suppose fkgf_k'\to g uniformly on [a,b][a,b]. Then there is a differentiable function f:[a,b]Rf:[a,b]\to\mathbb{R} such that

fkf  uniformly,f=g.f_k\longrightarrow f\ \text{ uniformly},\qquad f'=g.

Facts & Assumptions

Given: Reals a<ba<b, a point c[a,b]c\in[a,b], continuously differentiable functions fk:[a,b]Rf_k:[a,b]\to\mathbb{R}, convergence fk(c)Af_k(c)\to A, and uniform convergence fkgf_k'\to g.

[L1]

A uniform limit of continuous real-valued functions is continuous, and a continuous function on [a,b][a,b] is Riemann integrable (The uniform limit of continuous real-valued functions on a metric space is continuous, A continuous function on [a,b][a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion).

[L2]

Uniform convergence of integrable functions preserves integrability and the limit of the integrals (A uniform limit of Riemann-integrable functions is Riemann integrable, and its integral is the limit of their integrals).

[L3]

If s<ts<t, GG is differentiable on [s,t][s,t], and GG' is integrable there, then stG=G(t)G(s)\int_s^tG'=G(t)-G(s) (The second fundamental theorem: if GG is differentiable on [a,b][a,b] with G=fG' = f and ff is integrable, then abf=G(b)G(a)\int_a^b f = G(b)-G(a)). Restriction to a closed subinterval preserves differentiability and the derivative at its limit points, and integrability on [a,b][a,b] passes to every nondegenerate closed subinterval (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 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, Limit point, isolated point, adherent point, derived set, and dense subset of R\mathbb{R}, Intervals of R\mathbb{R}: the nine order-convex forms, nondegeneracy, and length, A function integrable on [a,b][a,b] is integrable on every closed subinterval). Finally tsh=sth\int_t^s h=-\int_s^t h and ssh=0\int_s^s h=0 whenever the displayed integrals are defined (The integral with oriented limits: aaf:=0\int_a^a f := 0 and baf:=abf\int_b^a f := -\int_a^b f).

[L6]

A uniform bound pqη|p-q|\le\eta on an interval gives uvpuvqηvu\left|\int_u^v p-\int_u^v q\right|\le\eta|v-u| (Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error).

Proof

technique · constructive
1.1

By [L7], each real-continuous derivative fkf_k' is metric-continuous. The uniform-limit clause of [L1] makes gg metric-continuous, and [L7] makes gg real-continuous. The integrability clause of [L1] therefore makes every fkf_k' and gg Riemann integrable; [L2] also gives the integrability of the uniform limit.

L1L2L7
1.2

Let ε>0\varepsilon>0. Choose N0N_0 such that fk(c)A<ε/2|f_k(c)-A|<\varepsilon/2 for kN0k\ge N_0, and choose N1N_1 such that fk(x)g(x)<ε/(2(ba+1))|f_k'(x)-g(x)|<\varepsilon/\bigl(2(b-a+1)\bigr) for kN1k\ge N_1 and all x[a,b]x\in[a,b].

givenchoose
2.1

Fix kk and x[a,b]x\in[a,b]. If c<xc<x, restrict fkf_k to [c,x][c,x]; its derivative is fkf_k' there and that derivative is integrable there by steps 1.1 and [L3], so the first clause of [L3] gives fk(x)=fk(c)+cxfkf_k(x)=f_k(c)+\int_c^x f_k'. If x<cx<c, apply that clause on [x,c][x,c] and then use orientation; if x=cx=c, use ccfk=0\int_c^c f_k'=0. Thus in every case fk(x)=fk(c)+cxfkf_k(x)=f_k(c)+\int_c^x f_k'.

step 1.1L3
2.2

Define H(x):=axgH(x):=\int_a^xg and construct f:[a,b]Rf:[a,b]\to\mathbb{R} by f(x):=A+H(x)H(c)=A+cxgf(x):=A+H(x)-H(c)=A+\int_c^xg.

step 1.1L4construct
3.1

By [L4] and [L5], HH is differentiable with H=gH'=g, and therefore the constructed function f=A+HH(c)f=A+H-H(c) is differentiable with f=gf'=g.

step 2.2L4L5
3.2

Choose NN at least as large as N0,N1N_0,N_1. For kNk\ge N and x[a,b]x\in[a,b], steps 2.2 and 2.1 with [L6] give fk(x)f(x)fk(c)A+cxfkcxg<ε/2+εxc/(2(ba+1))<ε|f_k(x)-f(x)|\le |f_k(c)-A|+\left|\int_c^x f_k'-\int_c^xg\right|<\varepsilon/2+\varepsilon|x-c|/\bigl(2(b-a+1)\bigr)<\varepsilon.

step 2.2step 2.1step 1.2L6choosealgebra
4.1

The index NN in step 3.2 serves every x[a,b]x\in[a,b], so fkff_k\to f uniformly; step 3.1 gives f=gf'=g. Thus the constructed ff has both asserted properties.

step 3.2step 3.1discharge-construct

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