$f(t) = (t^{2}, t^{3})$ on $[0,1]$: no $\xi$ satisfies $f(1)-f(0) = f'(\xi)$

Refuted claim: if $f : [a,b] \to \mathbb{R}^{m}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there is $\xi \in (a,b)$ with

$f(b) - f(a) \;=\; f'(\xi)\,(b-a).$

That is the equality form of the mean value theorem (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 $f(b) - f(a) = f'(c)(b-a)$), which is true for $m = 1$ and false for $m \ge 2$. What survives is the inequality $\lVert f(b)-f(a)\rVert_2 \le M(b-a)$ of 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 $\lVert f(b)-f(a)\rVert_2 \le M(b-a)$, and this item is the witness showing that the inequality cannot be upgraded.

The witness. Take $m = 2$, $[a,b] = [0,1]$ and $f : [0,1] \to \mathbb{R}^{2}$ with components $f_0(t) = t^{2}$ and $f_1(t) = t^{3}$ (Integer powers $a^m$, Vector-valued functions $f : A \to \mathbb{R}^m$, their limits and continuity, with the dictionary to the metric notions).

Why this curve and not the classical one. The crispest classical witness is $t \mapsto (\cos t, \sin t)$ on $[0,2\pi]$, whose derivative has constant norm $1$ while the endpoints coincide. The trigonometric functions are introduced later in the reading order than this page, so they may not be used here; the polynomial curve above carries the same refutation with the material available. This substitution is recorded here, in the item itself, so that a reader who knows the classical example is told why it is absent rather than left to suppose that this library does not know it.

• What exactly fails. The scalar mean value theorem is applied in the proof of 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 $\lVert f(b)-f(a)\rVert_2 \le M(b-a)$ to the auxiliary function $t \mapsto \langle u, f(t)\rangle$ for the fixed vector $u = f(b)-f(a)$, and it does produce a point $\xi$. That $\xi$ is a mean value point of that real function, and it depends on $u$; there is no reason for the two coordinates to be served by one and the same point, and on this curve they are not: the first coordinate demands $\xi = 1/\iota(2)$ and the second demands $\iota(3)\xi^{2} = 1$.

• The curve is not pathological. Both components are polynomials, so $f$ is differentiable at every real (For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$; for $n = 0$ it is the constant $1$, with derivative $0$; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term) and $f'$ is continuous. No smoothness hypothesis would rescue the equality form.

• The bound of 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 $\lVert f(b)-f(a)\rVert_2 \le M(b-a)$ is not merely true but sharp, and the companion item A curve for which the mean value inequality is an equality, showing the constant cannot be improved exhibits a curve on which it is an equality. The two items together show that the failure of the equality form is not a failure of the bound.