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

Statement

Let $m \in N$ with $m \geq 1$ and let $a, b \in R$ with $a < b$ .

Fundamental theorem, second part, in $R^{m}$ . Let $f : [a, b] \to R^{m}$ 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 $f^{'} : [a, b] \to R^{m}$ is integrable. Then $\int_{a}^{b} f^{'} = f (b) - f (a) .$
A bounded derivative gives a Lipschitz function. Let $f : [a, b] \to R^{m}$ be continuous on $[a, b]$ and differentiable at every point of $(a, b)$ , and let $M \geq 0$ satisfy $∥ f^{'} (t) ∥_{2} \leq M$ for every $t \in (a, b)$ . Then $∥ f (t) - f (s) ∥_{2} \leq M ∣ t - s ∣ for all s, t \in [a, b],$ that is, $f$ is Lipschitz with constant $M$ as a map $([a, b], d_{R}) \to (R^{m}, d_{2})$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction, The absolute value makes $R$ 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, $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it).

Facts & Assumptions

Given: A natural $m \geq 1$ , reals $a < b$ , and a function $f : [a, b] \to R^{m}$ with the hypotheses of the clause under discussion; points $s, t \in [a, b]$ .

[L1]

The vector-valued derivative and integral are componentwise: $f^{'} (c)_{i} = f_{i}^{'} (c)$ , and $f$ is integrable exactly when every $f_{i}$ is, with $(\int_{a}^{b} f)_{i} = \int_{a}^{b} f_{i}$ ; equality of two elements of $R^{m}$ is equality of all their coordinates (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, 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).

[L2]

The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = g$ and $g$ is integrable on $[a, b]$ , then $\int_{a}^{b} g = G (b) - G (a)$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[L3]

The mean value inequality on a subinterval (The mean value inequality: if $f : [a, b] \to R^{m}$ is continuous and differentiable on $(a, b)$ with $∥ f^{'} ∥_{2} \leq M$ , then $∥ f (b) - f (a) ∥_{2} \leq M (b - a)$ ): for $s < t$ , $f$ continuous on $[s, t]$ and differentiable on $(s, t)$ with $∥ f^{'} ∥_{2} \leq M$ there, $∥ f (t) - f (s) ∥_{2} \leq M (t - s)$ .

[L4]

Restricting the domain of a function preserves a limit and its value, the $ε$ - $δ$ condition then quantifying over fewer points; in particular if $f$ is differentiable at $c$ as a function on $[a, b]$ and $c$ is a limit point of $[s, t] \subseteq [a, b]$ , then the restriction of $f$ to $[s, t]$ is differentiable at $c$ with the same derivative (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ , Vector-valued functions $f : A \to R^{m}$ , their limits and continuity, with the dictionary to the metric notions, Limit point, isolated point, adherent point, derived set, and dense subset of $R$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L5]

Norms and the induced metric: $∥ w ∥_{2} \geq 0$ , $∥ - w ∥_{2} = ∥ w ∥_{2}$ , and $d_{2} (u, w) = ∥ u - w ∥_{2}$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ , Each $∥ \cdot ∥_{p}$ is a norm on $R^{n}$ , and the induced metrics are exactly $d_{1}$ , $d_{2}$ and $d_{\infty}$ of the published metric-spaces page); $d_{R} (s, t) = ∣ s - t ∣$ and $∣ t - s ∣ = ∣ s - t ∣$ (The absolute value makes $R$ 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, Basic properties of the absolute value).

[L6]

Lipschitz maps: $f$ is Lipschitz with constant $L \geq 0$ when $d_{Y} (f (x), f (x^{'})) \leq L d_{X} (x, x^{'})$ for all $x, x^{'}$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction).

Proof

technique · direct

1.1L1

Under the hypotheses of clause 1, each component $f_{i}$ is differentiable at every point of $[a, b]$ with derivative $f_{i}^{'} = (f^{'})_{i}$ , and each $(f^{'})_{i}$ is integrable on $[a, b]$ .

1.2L4

Under the hypotheses of clause 2, if $s < t$ in $[a, b]$ then $f$ restricted to $[s, t]$ is continuous on $[s, t]$ and differentiable at every point of $(s, t)$ with the same derivative, since $(s, t) \subseteq (a, b)$ and every point of $(s, t)$ is a limit point of $[s, t]$ .

2.1step 1.1L2

Applying [L2] to $G : = f_{i}$ and $g : = (f^{'})_{i}$ gives $\int_{a}^{b} (f^{'})_{i} = f_{i} (b) - f_{i} (a)$ for every $i < m$ .

2.2step 1.2L3L5

Under the hypotheses of clause 2, for $s < t$ in $[a, b]$ the mean value inequality applies on $[s, t]$ and gives $∥ f (t) - f (s) ∥_{2} \leq M (t - s) = M ∣ t - s ∣$ .

3.1step 2.1L1

The $i$ -th coordinate of $\int_{a}^{b} f^{'}$ is $\int_{a}^{b} (f^{'})_{i}$ and the $i$ -th coordinate of $f (b) - f (a)$ is $f_{i} (b) - f_{i} (a)$ ; by step 2.1 these agree for every $i < m$ , so the two vectors are equal, which is clause 1.

3.2step 2.2L5

If $s = t$ then $∥ f (t) - f (s) ∥_{2} = 0 = M ∣ t - s ∣$ ; and if $t < s$ then step 2.2 applied with the roles exchanged gives $∥ f (s) - f (t) ∥_{2} \leq M ∣ s - t ∣$ , and $∥ f (t) - f (s) ∥_{2} = ∥ - (f (s) - f (t)) ∥_{2} = ∥ f (s) - f (t) ∥_{2}$ while $∣ s - t ∣ = ∣ t - s ∣$ .

4.1step 2.2step 3.2L5L6∎

Steps 2.2 and 3.2 cover all pairs $s, t \in [a, b]$ , so $∥ f (t) - f (s) ∥_{2} \leq M ∣ t - s ∣$ always; since $d_{2} (f (t), f (s)) = ∥ f (t) - f (s) ∥_{2}$ and $d_{R} (t, s) = ∣ t - s ∣$ , this is exactly the Lipschitz condition with constant $M \geq 0$ , which is clause 2.

Remarks

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 $a \leq b$ and $f : [a, b] \to R^{m}$ integrable when $a < b$ , $∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ ; for $a < b$ , $∥ f ∥_{2}$ is integrable gives $∥ f (b) - f (a) ∥_{2} = ∥ \int_{a}^{b} f^{'} ∥_{2} \leq \int_{a}^{b} ∥ f^{'} ∥_{2} \leq M (b - a),$ the last step by monotonicity of the integral. That is a second proof of The mean value inequality: if $f : [a, b] \to R^{m}$ is continuous and differentiable on $(a, b)$ with $∥ f^{'} ∥_{2} \leq M$ , then $∥ f (b) - f (a) ∥_{2} \leq M (b - a)$ 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 $∣ f^{'} ∣ \leq M$ at every interior point, then $∣ f (x) - f (y) ∣ \leq M ∣ x - y ∣$ 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 $R$ . Clause 2 is its $R^{m}$ 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 $ι (m) M$ .
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 $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

If f:[a,b]→Rmf:[a,b]→Rm is differentiable with integrable f′f′ then ∫abf′=f(b)−f(a)∫ab​f′=f(b)−f(a); and a bounded derivative makes ff Lipschitz

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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