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TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)
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The mean value inequality: if f:[a,b]→Rm is continuous and differentiable on (a,b) with ∥f′∥2≤M, then ∥f(b)−f(a)∥2≤M(b−a)

Statement

Let m∈N with m≥1, let a,b∈R with a<b, and let f:[a,b]→Rm be continuous on [a,b] and differentiable at every point of (a,b) as a function on [a,b] (Vector-valued functions f:A→Rm, their limits and continuity, with the dictionary to the metric notions, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, Intervals of R: the nine order-convex forms, nondegeneracy, and length). Let M∈R with M≥0 satisfy

∥f′(t)∥2  ≤  Mfor every t∈(a,b).

Then

∥f(b)−f(a)∥2  ≤  M (b−a).

No integrability of f′ is assumed, so the theorem applies to every differentiable f; that is why it is proved from the scalar mean value theorem rather than from For a≤b and f:[a,b]→Rm integrable when a<b, ∥∫abf∥2≤∫ab∥f∥2; for a<b, ∥f∥2 is integrable. If f:[a,b]→Rm is differentiable with integrable f′ then ∫abf′=f(b)−f(a); and a bounded derivative makes f Lipschitz records the comparison between the two routes.

The equality form is not asserted, and for m≥2 it is false. There need be no ξ∈(a,b) with f(b)−f(a)=f′(ξ)(b−a); the companion page carries a differentiable witness on [0,1]. The ξ produced in the proof below depends on the fixed vector u=f(b)−f(a) and is a mean value point of the real function t↦⟨u,f(t)⟩, not of f.

Facts & Assumptions

Given: A natural m≥1, reals a<b, a function f:[a,b]→Rm continuous on [a,b] and differentiable on (a,b), a real M≥0 bounding ∥f′∥2 on (a,b), the vector u:=f(b)−f(a)∈Rm, and the real-valued function φ:[a,b]→R, φ(t):=⟨u,f(t)⟩.

[L1]

The inner product is bilinear and symmetric, ⟨w,w⟩=∥w∥22, and ⟨u,w⟩=∑i<muiwi (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, The p-norms ∥x∥p for rational p≥1, and ∥x∥∞).

[L6]

Algebra of derivatives: sums and scalar multiples of functions differentiable at a point are differentiable there, with (w+z)′(c)=w′(c)+z′(c) and (αw)′(c)=αw′(c) (Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0 clauses 1 and 2); and a differentiable function is continuous (A function differentiable at c is continuous at c).

[L7]

The mean value theorem: for ψ continuous on [a,b] with a<b and differentiable on (a,b) there is ξ∈(a,b) with ψ(b)−ψ(a)=ψ′(ξ)(b−a) (The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a)).

[L9]

Order arithmetic: b−a>0; a product of nonnegatives is nonnegative; and u>0 gives u−1>0, so an inequality may be multiplied by a positive real (Inverses of positives are positive, and reciprocation reverses order).

Proof

technique · direct
1.1

Every component fi is continuous on [a,b] in the sense of Continuity of f:A→R at a point of A and on A: the ε-δ condition, its agreement with lim⁡x→cf(x)=f(c) at a limit point, and continuity at an isolated point and differentiable at every point of (a,b), with f′(t)i=fi′(t).

L3L4
1.2

φ(t)=∑i<muifi(t) by the coordinate formula for the inner product.

L1
1.3

φ(b)−φ(a)=⟨u,f(b)⟩−⟨u,f(a)⟩=⟨u,f(b)−f(a)⟩=⟨u,u⟩=∥u∥22, by bilinearity.

L1
1.4

By Cauchy-Schwarz and the bound on ∥f′∥2, ⟨u,f′(ξ)⟩≤∣⟨u,f′(ξ)⟩∣≤∥u∥2∥f′(ξ)∥2≤∥u∥2M.

L2
1.5

If u=0 then ∥u∥2=0 while M(b−a)≥0, so the conclusion holds.

L2L9
2.1

By induction on p≤m, each partial sum t↦∑i<puifi(t) is continuous on [a,b] and differentiable on (a,b) with derivative ∑i<puifi′(t): the empty sum is the constant 0, and each successor step adds one scalar multiple of a function that is continuous and differentiable by step 1.1.

step 1.1L5L6L8
3.1

Hence φ is continuous on [a,b], differentiable at every point of (a,b), and φ′(t)=∑i<muifi′(t)=⟨u,f′(t)⟩ for t∈(a,b).

step 1.2step 2.1L1
4.1

By the mean value theorem applied to φ there is ξ∈(a,b) with φ(b)−φ(a)=φ′(ξ)(b−a).

step 3.1L7
5.1

Combining steps 1.3 and 4.1, ∥u∥22=⟨u,f′(ξ)⟩ (b−a).

step 3.1step 1.3step 4.1
6.1

Since b−a>0, multiplying the inequality of step 1.4 by b−a and using step 5.1 gives ∥u∥22≤∥u∥2 M (b−a).

step 5.1step 1.4L9
7.1

If u≠0 then ∥u∥2>0, and multiplying step 6.1 by the positive real 1/∥u∥2 gives ∥u∥2≤M(b−a).

step 6.1L2L9
8.1

The two cases of steps 1.5 and 7.1 exhaust the possibilities for u=f(b)−f(a), so ∥f(b)−f(a)∥2≤M(b−a).

step 1.5step 7.1∎

Remarks

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