If $f:[a,b]\to {\mathbb{R}}^m$ is differentiable with integrable $f^{\mathrm{\prime }}$ then ${\int }_a^b{f}^{\mathrm{\prime }}=f(b)-f(a)$; and a bounded derivative makes $f$ Lipschitz

Let $m\in \mathbb{N}$ with $m\ge 1$ and let $a,b\in \mathbb{R}$ with $a<b$.

1. Fundamental theorem, second part, in ${\mathbb{R}}^m$. Let $f:[a,b]\to {\mathbb{R}}^m$ be differentiable at every point of $[a,b]$ as a function on $[a,b]$ (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral), and suppose $f^{\mathrm{\prime }}:[a,b]\to {\mathbb{R}}^m$ is integrable. Then ${\int }_a^b{f}^{\mathrm{\prime }}=f(b)-f(a)\mathrm{.}$
2. A bounded derivative gives a Lipschitz function. Let $f:[a,b]\to {\mathbb{R}}^m$ be continuous on $[a,b]$ and differentiable at every point of $(a,b)$, and let $M\ge 0$ satisfy $\parallel f^{\mathrm{\prime }}(t){\parallel }_2\le M$ for every $t\in (a,b)$. Then $\parallel f(t)-f(s){\parallel }_2\le M\mathrm{\mid }t-s\mathrm{\mid }\text{for all }s,t\in [a,b],$ that is, $f$ is Lipschitz with constant $M$ as a map $([a,b],d_{\mathbb{R}})\to ({\mathbb{R}}^m,d_2)$ (Lipschitz map, $\alpha$-Hölder map for rational $0<\alpha \le 1$, and contraction, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, ${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$, and $d_1$, $d_2$, $d_{\mathrm{\infty }}$ are metrics on it).

• Two routes to the mean value inequality, and why the other one was taken. When $f^{\mathrm{\prime }}$ happens to be integrable, clause 1 together with For $a\le b$ and $f:[a,b]\to {\mathbb{R}}^m$ integrable when $a<b$, $\parallel {\int }_a^{b}f{\parallel }_2\le {\int }_a^b\parallel f{\parallel }_2$; for $a<b$, $\parallel f{\parallel }_2$ is integrable gives $\parallel f(b)-f(a){\parallel }_2=\parallel {\int }_a^b{f}^{\mathrm{\prime }}{\parallel }_2\le {\int }_a^b\parallel f^{\mathrm{\prime }}{\parallel }_2\le M(b-a),$ the last step by monotonicity of the integral. That is a second 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 $\parallel f(b)-f(a){\parallel }_2\le M(b-a)$ under an extra hypothesis. The theorem is proved the other way, from the scalar mean value theorem, precisely because it then needs no integrability at all: it applies to every differentiable $f$. The point is not academic — the companion page's witness for the failure of the equality form is differentiable and is nowhere assumed integrable.

• When is $f^{\mathrm{\prime }}$ integrable? Continuity of $f^{\mathrm{\prime }}$ suffices, each component being then continuous and hence integrable, which is the classical form of clause 1 and is what Every continuous function on an interval has a primitive; two primitives differ by a constant; and ${\int }_a^{b}f=G(b)-G(a)$ for any primitive $G$ supplies for the scalar case. Clause 1 as stated is stronger: it asks only for integrability, exactly as The second fundamental theorem: if $G$ is differentiable on $[a,b]$ with $G^{\mathrm{\prime }}=f$ and $f$ is integrable, then ${\int }_a^{b}f=G(b)-G(a)$ does.

• The scalar case of clause 2 is already published as If $f$ is continuous on an interval $I$ and $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }\le M$ at every interior point, then $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }\le M\mathrm{\mid }x-y\mathrm{\mid }$ for all $x,y\in I$, so $f$ is Lipschitz with constant $M$ and uniformly continuous on $I$, for a function on an order-convex subset of $\mathbb{R}$. Clause 2 is its ${\mathbb{R}}^m$ analogue on a closed bounded interval, and it is proved from the vector mean value inequality rather than by applying the scalar statement coordinatewise, which would give the worse constant $\iota (m)M$.

• Additivity is not needed above but is available, so that clause 1 may be applied on any subinterval and the pieces reassembled (For $a<c<b$: $f$ is integrable on $[a,b]$ if and only if it is integrable on $[a,c]$ and on $[c,b]$, and then ${\int }_a^{b}f={\int }_a^{c}f+{\int }_c^{b}f$; with the oriented form for arbitrary $a,b,c$, The integral with oriented limits: ${\int }_a^{a}f:=0$ and ${\int }_b^{a}f:=-{\int }_a^{b}f$).