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

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 $f' : [a,b] \to \mathbb{R}^{m}$ is integrable. Then $\int_a^b f' \;=\; f(b) - f(a).$
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 $\lVert f'(t)\rVert_2 \le M$ for every $t \in (a,b)$. Then $\lVert f(t) - f(s)\rVert_2 \;\le\; M\,|t-s| \qquad \text{for all } s,t \in [a,b],$ 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: $d(x,y) = |x-y|$ 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 $d_1$, $d_2$, $d_\infty$ are metrics on it).

• Two routes to the mean value inequality, and why the other one was taken. When $f'$ happens to be integrable, clause 1 together with 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 gives $\lVert f(b)-f(a)\rVert_2 = \Bigl\lVert\int_a^b f'\Bigr\rVert_2 \le \int_a^b \lVert f'\rVert_2 \le M(b-a),$ 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 $\lVert f'\rVert_2 \le M$, then $\lVert f(b)-f(a)\rVert_2 \le M(b-a)$ 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 $f'$ integrable? Continuity of $f'$ 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 $G' = f$ 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 $|f'| \le M$ 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 $I$, 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 $\int_b^a f := -\int_a^b f$).