$f(t)=(t^2,t^3)$ on $[0,1]$: no $\xi$ satisfies $f(1)-f(0)=f^{\mathrm{\prime }}(\xi )$

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

$f(b)-f(a)=f^{\mathrm{\prime }}(\xi )(b-a)\mathrm{.}$

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 $f(b)-f(a)=f^{\mathrm{\prime }}(c)(b-a)$), which is true for $m=1$ and false for $m\ge 2$. What survives is the inequality $\parallel f(b)-f(a){\parallel }_2\le M(b-a)$ of The mean value inequality: if $f:[a,b]\to {\mathbb{R}}^m$ is continuous and differentiable on $(a,b)$ with $\parallel f^{\mathrm{\prime }}{\parallel }_2\le M$, then $\parallel f(b)-f(a){\parallel }_2\le M(b-a)$, 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 $\parallel f^{\mathrm{\prime }}{\parallel }_2\le M$, then $\parallel f(b)-f(a){\parallel }_2\le M(b-a)$ to the auxiliary function $t\mapsto ⟨u,f(t)⟩$ 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 $\xi =1\mathrm{/}\iota (2)$ 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 $x\mathrm{\ne }0$ with derivative $-\iota (n)x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term) and $f^{\mathrm{\prime }}$ 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 $\parallel f^{\mathrm{\prime }}{\parallel }_2\le M$, then $\parallel f(b)-f(a){\parallel }_2\le M(b-a)$ 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.