TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)

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The mean value inequality: if $f : [a,b] \to \mathbb{R}^m$ is continuous and differentiable on $(a,b)$ with $\lVert f'\rVert_2 \le M$ , then $\lVert f(b)-f(a)\rVert_2 \le M(b-a)$

Statement

Let $m \in \mathbb{N}$ with $m \ge 1$ , let $a, b \in \mathbb{R}$ with $a < b$ , and let $f : [a,b] \to \mathbb{R}^{m}$ be continuous on $[a,b]$ and differentiable at every point of $(a,b)$ as a function on $[a,b]$ (Vector-valued functions $f : A \to \mathbb{R}^m$ , 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 $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length). Let $M \in \mathbb{R}$ with $M \ge 0$ satisfy

$\lVert f'(t)\rVert_2 \;\le\; M \qquad \text{for every } t \in (a,b).$

Then

$\lVert f(b) - f(a)\rVert_2 \;\le\; M\,(b-a).$

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 $a \le b$ and $f : [a,b] \to \mathbb{R}^m$ integrable when $a<b$ , $\bigl\lVert\int_a^b f\bigr\rVert_2 \le \int_a^b \lVert f\rVert_2$ ; for $a<b$ , $\lVert f\rVert_2$ is integrable. If $f : [a,b] \to \mathbb{R}^m$ is differentiable with integrable $f'$ then $\int_a^b f' = f(b)-f(a)$ ; and a bounded derivative makes $f$ Lipschitz records the comparison between the two routes.

The equality form is not asserted, and for $m \ge 2$ it is false. There need be no $\xi \in (a,b)$ with $f(b)-f(a) = f'(\xi)(b-a)$ ; the companion page carries a differentiable witness on $[0,1]$ . The $\xi$ 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 $t \mapsto \langle u, f(t)\rangle$ , not of $f$ .

Facts & Assumptions

Given: A natural $m \ge 1$ , reals $a < b$ , a function $f : [a,b] \to \mathbb{R}^{m}$ continuous on $[a,b]$ and differentiable on $(a,b)$ , a real $M \ge 0$ bounding $\lVert f'\rVert_2$ on $(a,b)$ , the vector $u := f(b)-f(a) \in \mathbb{R}^{m}$ , and the real-valued function $\varphi : [a,b] \to \mathbb{R}$ , $\varphi(t) := \langle u, f(t)\rangle$ .

[L1]

The inner product is bilinear and symmetric, $\langle w,w\rangle = \lVert w\rVert_2^{2}$ , and $\langle u,w\rangle = \sum_{i<m}u_iw_i$ (The Euclidean inner product $\langle x,y\rangle = \sum_{k<n} x_k y_k$ on $\mathbb{R}^n$ , The $p$ -norms $\lVert x\rVert_p$ for rational $p \ge 1$ , and $\lVert x\rVert_\infty$ ).

[L2]

Cauchy-Schwarz: $|\langle u,w\rangle| \le \lVert u\rVert_2\lVert w\rVert_2$ , and $t \le |t|$ (Cauchy-Schwarz $\lvert\langle x,y\rangle\rvert \le \lVert x\rVert_2\lVert y\rVert_2$ with its equality case, the triangle inequality for $\lVert\cdot\rVert_2$ , the parallelogram law and polarisation, Basic properties of the absolute value); $\lVert w\rVert_2 \ge 0$ , and $\lVert w\rVert_2 = 0$ exactly when $w = 0$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[L3]

Componentwise continuity and differentiability: $f$ is continuous at a point exactly when every $f_i$ is, and differentiable at a point exactly when every $f_i$ is, with $f'(t)_i = f_i'(t)$ (A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions clause 1, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ of $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ ).

[L4]

For a real domain, the metric notion of continuity and the notion of Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point agree (Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$ , continuity and uniform continuity of $f : A \to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace clause 1).

[L5]

Algebra of continuous real functions: sums and scalar multiples of functions continuous at a point are continuous there (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function clause 1).

[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 $(\alpha w)'(c) = \alpha w'(c)$ (Sums, scalar multiples, products and quotients: $(f+g)'(c) = f'(c) + g'(c)$ , $(\alpha f)'(c) = \alpha f'(c)$ , $(fg)'(c) = f'(c)g(c) + f(c)g'(c)$ , and $(f/g)'(c) = \bigl(f'(c)g(c) - f(c)g'(c)\bigr)/g(c)^{2}$ when $g(c) \ne 0$ 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 $\psi$ continuous on $[a,b]$ with $a<b$ and differentiable on $(a,b)$ there is $\xi \in (a,b)$ with $\psi(b)-\psi(a) = \psi'(\xi)(b-a)$ (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 $c \in (a,b)$ with $f(b) - f(a) = f'(c)(b-a)$ ).

[L8]

Laws of finite sums and induction (Laws of finite sums and finite products, Finite sums and finite products, by recursion, The principle of mathematical induction).

[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 $f_i$ is continuous on $[a,b]$ in the sense of Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point and differentiable at every point of $(a,b)$ , with $f'(t)_i = f_i'(t)$ .

L3L4

1.2

$\varphi(t) = \sum_{i<m}u_i f_i(t)$ by the coordinate formula for the inner product.

1.3

$\varphi(b)-\varphi(a) = \langle u, f(b)\rangle - \langle u, f(a)\rangle = \langle u, f(b)-f(a)\rangle = \langle u,u\rangle = \lVert u\rVert_2^{2}$ , by bilinearity.

1.4

By Cauchy-Schwarz and the bound on $\lVert f'\rVert_2$ , $\langle u,f'(\xi)\rangle \le \bigl|\langle u,f'(\xi)\rangle\bigr| \le \lVert u\rVert_2\lVert f'(\xi)\rVert_2 \le \lVert u\rVert_2 M$ .

1.5

If $u = 0$ then $\lVert u\rVert_2 = 0$ while $M(b-a) \ge 0$ , so the conclusion holds.

L2L9

2.1

By induction on $p \le m$ , each partial sum $t \mapsto \sum_{i<p}u_i f_i(t)$ is continuous on $[a,b]$ and differentiable on $(a,b)$ with derivative $\sum_{i<p}u_i f_i'(t)$ : 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 $\varphi$ is continuous on $[a,b]$ , differentiable at every point of $(a,b)$ , and $\varphi'(t) = \sum_{i<m}u_i f_i'(t) = \langle u, f'(t)\rangle$ for $t \in (a,b)$ .

step 1.2step 2.1L1

4.1

By the mean value theorem applied to $\varphi$ there is $\xi \in (a,b)$ with $\varphi(b)-\varphi(a) = \varphi'(\xi)(b-a)$ .

step 3.1L7

5.1

Combining steps 1.3 and 4.1, $\lVert u\rVert_2^{2} = \langle u, f'(\xi)\rangle\,(b-a)$ .

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 $\lVert u\rVert_2^{2} \le \lVert u\rVert_2\,M\,(b-a)$ .

step 5.1step 1.4L9

7.1

If $u \ne 0$ then $\lVert u\rVert_2 > 0$ , and multiplying step 6.1 by the positive real $1/\lVert u\rVert_2$ gives $\lVert u\rVert_2 \le M(b-a)$ .

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 $\lVert f(b)-f(a)\rVert_2 \le M(b-a)$ .

step 1.5step 7.1∎

Remarks

What the auxiliary function buys. The scalar mean value theorem produces a point at which one real function has its average slope. Applying it to $\varphi = \langle u, f\rangle$ for the particular $u = f(b)-f(a)$ turns that into a statement about $f$ , at the cost of the equality becoming an inequality. The loss is not an artefact of the proof: the equality form is genuinely false for $m \ge 2$ , and the companion page's curve $t \mapsto (t^{2},t^{3})$ on $[0,1]$ is a differentiable witness.
The bound is sharp. No constant smaller than $\sup\lVert f'\rVert_2$ works in general; the companion page exhibits a curve for which the inequality is an equality.
The case split at $u = 0$ is where the statement would otherwise be incomplete, exactly as in For $a \le b$ and $f : [a,b] \to \mathbb{R}^m$ integrable when $a<b$ , $\bigl\lVert\int_a^b f\bigr\rVert_2 \le \int_a^b \lVert f\rVert_2$ ; for $a<b$ , $\lVert f\rVert_2$ is integrable. When $f(b) = f(a)$ the left-hand side is $0$ and nothing is divided.
$M \ge 0$ is a hypothesis, not a deduction. It follows from $\lVert f'(t)\rVert_2 \le M$ at any single $t \in (a,b)$ , and $(a,b)$ is nonempty here because $a<b$ ; it is stated anyway so that the conclusion reads as a genuine bound, in the style of If $f$ is continuous on an interval $I$ and $|f'| \le M$ at every interior point, then $|f(x) - f(y)| \le M|x-y|$ for all $x,y \in I$ , so $f$ is Lipschitz with constant $M$ and uniformly continuous on $I$ .

Depends on

Used by

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Sources

Mean value theorem (Wikipedia) (standard reference, not scraped)
Vector-valued function (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, Section 8.4 (standard reference, not scraped)

The mean value inequality: if f:[a,b]→Rmf : [a,b] \to \mathbb{R}^mf:[a,b]→Rm is continuous and differentiable on (a,b)(a,b)(a,b) with ∥f′∥2≤M\lVert f'\rVert_2 \le M∥f′∥2​≤M, then ∥f(b)−f(a)∥2≤M(b−a)\lVert f(b)-f(a)\rVert_2 \le M(b-a)∥f(b)−f(a)∥2​≤M(b−a)