CommeUnJeu · L2 MP

Vector-valued functions of a real variable

⌚ ~59 min ▢ 7 blocks ✓ 16 exercises ➣ Prerequisites : Normed vector spaces, Limits and continuity in a normed space, Differentiability, Integration on a segment, Limits and continuity

The previous chapters studied a real variable through real numbers. Here the variable stays real but the value becomes a vector: a vector function is a map $f \colon I \to E$ from an interval $I$ of $\mathbb{R}$ to a finite-dimensional normed vector space $E$. A planar or spatial curve, a moving point, a time-dependent matrix --- all are vector functions.
The guiding idea is simple: everything is read componentwise. Fix a basis $\mathcal{B} = (e_1, \dots, e_p)$ of $E$; then $f = f_1 e_1 + \cdots + f_p e_p$, where the component functions $f_i \colon I \to \mathbb{K}$ are ordinary scalar functions of a real variable. Differentiability, the integral and the Taylor formulas of $f$ are defined and computed through the $f_i$ --- and, because a change of basis is a constant invertible linear map, the results do not depend on the basis chosen.
One contrast with a real function dominates the chapter. A real function, continuous on $[a,b]$ and differentiable on $\mathopen]a,b\mathclose[$, has a mean value theorem: an equality $f(b) - f(a) = (b-a) f'(c)$ for some $c$ strictly between $a$ and $b$. The equality is special to real-valued functions: a function valued in any space $E$ other than $\mathbb{R}$ --- a $\mathbb{C}$-valued or a genuinely vector-valued function --- has no general equality mean value theorem; the derived vector $f'(t_0)$ is a vector, and the equality form has no counterpart. What survives are the inequalities: the mean value inequality and the Taylor--Lagrange inequality. The chapter is built toward them.
Standing notation. Throughout, $I$ is an interval of $\mathbb{R}$, $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$, and $E$ is a $\mathbb{K}$-vector space of finite dimension $p$, with a norm $\norme{\cdot}$. A vector function is a map $f \colon I \to E$; in a basis $\mathcal{B} = (e_1, \dots, e_p)$ its component functions are $f_1, \dots, f_p \colon I \to \mathbb{K}$. The derived vector is $f'(t_0)$, the $k$-th derivative $f^{(k)}$. At an endpoint of $I$, « differentiable » means one-sidedly differentiable. The notions of norm and convergence are those of Normed vector spaces; limits and continuity of a vector function are those of Limits and continuity in a normed space.

I Differentiability

I.1 Derivative at a point and on an interval

The difference quotient of a vector function is $\bigl(f(t) - f(t_0)\bigr)/(t - t_0)$ --- a vector divided by a scalar, hence a vector. When it has a limit as $t \to t_0$, that limit is again a vector, the derived vector. As announced, the whole notion reduces to the component functions.

Definition — Differentiability

Let $f \colon I \to E$ and $t_0 \in I$. The function $f$ is differentiable at $t_0$ if the difference quotient $\dfrac{f(t) - f(t_0)}{t - t_0}$ has a limit in $E$ as $t \to t_0$, $t \ne t_0$. That limit is then the derived vector $f'(t_0)$. Equivalently, for $h$ such that $t_0 + h \in I$, $$ f(t_0 + h) = f(t_0) + h\, f'(t_0) + o(h), $$ where the remainder is $o(|h|)$ as $h \to 0$. The function $f$ is differentiable on $I$ if it is differentiable at every point of $I$; the map $f' \colon t \mapsto f'(t)$ is then its derivative. At an endpoint of $I$ the limit, and the expansion, are one-sided.

Example — The helix

The helix $f \colon \mathbb{R} \to \mathbb{R}^3$, $f(t) = (\cos t, \sin t, t)$, is differentiable on $\mathbb{R}$. Its difference quotient tends componentwise (as the Proposition below justifies) to $f'(t) = (-\sin t, \cos t, 1)$: the derived vector is the velocity of the moving point.

Proposition — Differentiability componentwise

Let $f \colon I \to E$ with components $f_1, \dots, f_p$ in a basis $\mathcal{B}$, and let $t_0 \in I$. Then $f$ is differentiable at $t_0$ if and only if each component $f_i$ is, and in that case $$ \textcolor{colorprop}{f'(t_0) = \sum_{i=1}^p f_i'(t_0)\, e_i}. $$

Proof

The difference quotient is itself read componentwise: $$ \frac{f(t) - f(t_0)}{t - t_0} = \sum_{i=1}^p \frac{f_i(t) - f_i(t_0)}{t - t_0}\, e_i. $$ A limit in the finite-dimensional space $E$ is taken componentwise (recalled from Limits and continuity in a normed space). Hence the quotient has a limit as $t \to t_0$ if and only if each component quotient $\bigl(f_i(t) - f_i(t_0)\bigr)/(t - t_0)$ has a limit, that is if and only if each $f_i$ is differentiable at $t_0$; and the limit is then $\sum_i f_i'(t_0)\, e_i$.

Proposition — A differentiable function is continuous

If $f \colon I \to E$ is differentiable at $t_0$, then $f$ is continuous at $t_0$.

Proof

By the order-$1$ expansion, $f(t_0 + h) = f(t_0) + h\, f'(t_0) + o(h)$. As $h \to 0$, both $h\, f'(t_0)$ and $o(h)$ tend to $0_E$, so $f(t_0 + h) \to f(t_0)$: the function $f$ is continuous at $t_0$.

Example — A matrix-valued function

The rotation matrix $R \colon \mathbb{R} \to \mathcal{M}_2(\mathbb{R})$, $$ R(t) = \begin{pmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{pmatrix}, $$ is a vector function valued in the four-dimensional space $\mathcal{M}_2(\mathbb{R})$. Differentiated entrywise, it is differentiable on $\mathbb{R}$ with $R'(t) = \begin{pmatrix} -\sin t & -\cos t \\ \cos t & -\sin t \end{pmatrix} = R\bigl(t + \tfrac{\pi}{2}\bigr)$.

Method — Differentiate a vector function

To differentiate $f \colon I \to E$:

fix a basis, write the component functions $f_1, \dots, f_p$, differentiate each as an ordinary scalar function, and recombine $f' = \sum_i f_i'\, e_i$;
or, when $f$ is built from simpler vector functions, apply the operation rules of the next subsection rather than returning to the components.

Geometrically, the difference quotient is the secant vector $f(t) - f(t_0)$ rescaled; as $t \to t_0$ it turns into the derived vector $f'(t_0)$, which --- when it is non-zero --- is tangent to the curve traced by $f$.

Skills to practice

Differentiating a vector function

I.2 Operations on differentiable functions

Every differentiation rule of a real function has a vector counterpart, proved by reading the rule componentwise. Two cases are genuinely new --- and central to all that follows: composition with a linear map, and composition with a bilinear map.

Proposition — First operations

Let $f, g \colon I \to E$ be differentiable on $I$.

For $\alpha, \beta \in \mathbb{K}$, $\alpha f + \beta g$ is differentiable and $(\alpha f + \beta g)' = \alpha f' + \beta g'$.
For a differentiable scalar function $\mu \colon I \to \mathbb{K}$, the product $\mu \cdot f$ is differentiable and $(\mu \cdot f)' = \mu' f + \mu f'$.
For a differentiable $\varphi \colon J \to I$ ($J$ an interval of $\mathbb{R}$), $f \circ \varphi$ is differentiable on $J$ and $(f \circ \varphi)' = \varphi' \cdot (f' \circ \varphi)$.

Proof

Each statement is read componentwise and reduces to a known rule for scalar functions of a real variable (recalled from Dérivabilité).

The components of $\alpha f + \beta g$ are $\alpha f_i + \beta g_i$; differentiating, $(\alpha f_i + \beta g_i)' = \alpha f_i' + \beta g_i'$, which recombine into $\alpha f' + \beta g'$.
The components of $\mu \cdot f$ are the products $\mu f_i$; the scalar product rule gives $(\mu f_i)' = \mu' f_i + \mu f_i'$, which recombine into $\mu' f + \mu f'$.
The components of $f \circ \varphi$ are $f_i \circ \varphi$; the chain rule gives $(f_i \circ \varphi)' = \varphi' \cdot (f_i' \circ \varphi)$, which recombine into $\varphi' \cdot (f' \circ \varphi)$.

Proposition — Composition with a linear map

Let $F$ be a normed vector space and $u \in \mathcal{L}(E,F)$ a linear map ($E$ is the standing finite-dimensional space). If $f \colon I \to E$ is differentiable, then $u \circ f$ is differentiable and $(u \circ f)' = u \circ f'$.

Proof

Fix a basis $\mathcal{B} = (e_1, \dots, e_p)$ of $E$ and write $f = \sum_{i=1}^p f_i\, e_i$. By linearity of $u$, $$ u \circ f = \sum_{i=1}^p f_i \cdot u(e_i), $$ where each $u(e_i)$ is a fixed vector of $F$. Fix $t_0 \in I$. For $t \ne t_0$ the difference quotient of the $i$-th term is $$ \frac{f_i(t)\, u(e_i) - f_i(t_0)\, u(e_i)}{t - t_0} = \frac{f_i(t) - f_i(t_0)}{t - t_0}\, u(e_i), $$ and as $t \to t_0$ it tends to $f_i'(t_0)\, u(e_i)$, because $f_i$ is differentiable at $t_0$ and multiplying the scalar quotient by the fixed vector $u(e_i)$ is continuous ($\norme{\lambda\, u(e_i) - \lambda'\, u(e_i)} = |\lambda - \lambda'|\, \norme{u(e_i)}$). Summing the $p$ terms, $u \circ f$ is differentiable at $t_0$ and $$ \begin{aligned} (u \circ f)'(t_0) = \sum_{i=1}^p f_i'(t_0)\, u(e_i) &= u\!\left( \sum_{i=1}^p f_i'(t_0)\, e_i \right) && \text{(linearity of $u$)}\\ &= u(f'(t_0)) && \text{(componentwise formula $f' = \sum_i f_i'\, e_i$)}. \end{aligned} $$ The argument is a direct difference-quotient computation: no continuity of $u$ and no finite-dimensionality of $F$ are used.

Proposition — Composition with a bilinear map

Let $F, G$ be finite-dimensional normed vector spaces, $H$ a normed vector space, and $B \colon F \times G \to H$ a bilinear map. If $f \colon I \to F$ and $g \colon I \to G$ are differentiable on the same interval $I$, then $B(f,g) \colon t \mapsto B(f(t), g(t))$ is differentiable and $$ \textcolor{colorprop}{B(f,g)' = B(f', g) + B(f, g')}. $$

Proof

Fix a basis $(a_1, \dots, a_q)$ of $F$ and a basis $(b_1, \dots, b_r)$ of $G$; write $f = \sum_i f_i\, a_i$ and $g = \sum_j g_j\, b_j$. By bilinearity of $B$, $$ B(f,g) = B\!\left( \sum_i f_i\, a_i,\ \sum_j g_j\, b_j \right) = \sum_{i,j} (f_i\, g_j) \cdot B(a_i, b_j), $$ a linear combination of the products of the scalar function $f_i\, g_j$ by the fixed vector $B(a_i, b_j) \in H$. Each $f_i\, g_j$ is a product of two differentiable scalar functions, hence differentiable with $(f_i\, g_j)' = f_i'\, g_j + f_i\, g_j'$ (recalled from Dérivabilité). For each pair $(i,j)$, the difference quotient of $(f_i\, g_j) \cdot B(a_i,b_j)$ is the scalar quotient of $f_i\, g_j$ times the fixed vector $B(a_i,b_j)$, so --- as in the previous proof --- it tends to $(f_i\, g_j)'(t_0) \cdot B(a_i,b_j)$. Summing the finitely many terms, $B(f,g)$ is differentiable and $$ \begin{aligned} B(f,g)' &= \sum_{i,j} (f_i'\, g_j + f_i\, g_j') \cdot B(a_i, b_j) && \text{(differentiate each term)}\\ &= \sum_{i,j} f_i'\, g_j\, B(a_i, b_j) + \sum_{i,j} f_i\, g_j'\, B(a_i, b_j) && \text{(split the sum)}\\ &= B(f', g) + B(f, g') && \text{(re-fold by bilinearity of $B$)}. \end{aligned} $$ No continuity of $B$ is invoked; $F$ and $G$ are taken finite-dimensional only so that $f$ and $g$ have the finite component decompositions used above.

Example — Inner product and cross product

The bilinear-map rule covers two everyday cases. On a real Euclidean space $(E, \langle\cdot\mid\cdot\rangle)$ --- finite-dimensional, so the bilinear-map Proposition applies --- the inner product is an $\mathbb{R}$-bilinear map, so for $f, g \colon I \to E$ differentiable, $$ \langle f \mid g\rangle' = \langle f' \mid g\rangle + \langle f \mid g'\rangle. $$ In $\mathbb{R}^3$ the cross product $\wedge$ is bilinear, so $(f \wedge g)' = f' \wedge g + f \wedge g'$. Both are one and the same rule, applied to two different bilinear maps.

Method — Choose the right operation rule

Read the vector function as a combination of simpler ones and pick the matching rule: a linear combination, a scalar function times a vector function ($\mu \cdot f$), a reparametrisation ($f \circ \varphi$), a linear map applied ($u \circ f$), or a bilinear map applied ($B(f,g)$ --- inner product, cross product, matrix product). Returning to the components is always available as a fallback.

Skills to practice

Applying the operation rules

I.3 Functions of class $\mathcal{C}^k$

Differentiating again and again produces the iterated derivatives. A function whose iterated derivatives exist up to a given order, the last one continuous, is said to be of a corresponding class. Like differentiability itself, the class is read componentwise: applying the componentwise Proposition of \S1.1 at each order, $f$ is of class $\mathcal{C}^k$ exactly when every component $f_i$ is, and then $f^{(j)} = \sum_i f_i^{(j)} e_i$ for $0 \le j \le k$.

Definition — Class $\mathcal{C}^k$

Let $k \in \mathbb{N}^*$. A function $f \colon I \to E$ is of class $\mathcal{C}^k$ on $I$ if it is $k$ times differentiable on $I$ and its $k$-th derivative $f^{(k)}$ is continuous. Of class $\mathcal{C}^0$ simply means continuous. The function is of class $\mathcal{C}^\infty$ if it is of class $\mathcal{C}^k$ for every $k$.

Example — A function of class $\mathcal{C}^\infty$

The function $f \colon \mathbb{R} \to \mathbb{R}^3$, $f(t) = (\mathrm{e}^t,\, t^2,\, \cos t)$, is of class $\mathcal{C}^\infty$: each component $\mathrm{e}^t$, $t^2$, $\cos t$ is of class $\mathcal{C}^\infty$ on $\mathbb{R}$, and by the componentwise reading the iterated derivatives of $f$ exist and are continuous at every order.

Proposition — Leibniz formula

Let $n \in \mathbb{N}$. If $\mu \colon I \to \mathbb{K}$ and $f \colon I \to E$ are both of class $\mathcal{C}^n$, then $\mu \cdot f$ is of class $\mathcal{C}^n$ and $$ \textcolor{colorprop}{(\mu \cdot f)^{(n)} = \sum_{k=0}^n \binom{n}{k} \mu^{(k)}\, f^{(n-k)}}. $$

Proof

By induction on $n$.

Initialisation. For $n = 0$, $\mu$ and $f$ of class $\mathcal{C}^0$ are continuous, so $\mu \cdot f$ is continuous (a scalar function times a vector function), i.e. of class $\mathcal{C}^0$, and the formula reads $\mu \cdot f = \mu \cdot f$.
Heredity. Assume the formula for the order $n$. Let now $\mu$ and $f$ be of class $\mathcal{C}^{n+1}$; they are in particular of class $\mathcal{C}^n$, so the order-$n$ formula applies to them. Each term $\mu^{(k)} f^{(n-k)}$ is a scalar function times a vector function, both of class $\mathcal{C}^1$ (since $\mu, f$ are of class $\mathcal{C}^{n+1}$), so by the scalar-times-vector rule of \S1.2 it is differentiable with $\bigl(\mu^{(k)} f^{(n-k)}\bigr)' = \mu^{(k+1)} f^{(n-k)} + \mu^{(k)} f^{(n-k+1)}$. Differentiating the order-$n$ formula, $$ \begin{aligned} (\mu \cdot f)^{(n+1)} &= \sum_{k=0}^n \binom{n}{k} \mu^{(k+1)} f^{(n-k)} + \sum_{k=0}^n \binom{n}{k} \mu^{(k)} f^{(n-k+1)} && \text{(differentiate each term)}\\ &= \sum_{k=1}^{n+1} \binom{n}{k-1} \mu^{(k)} f^{(n+1-k)} + \sum_{k=0}^n \binom{n}{k} \mu^{(k)} f^{(n+1-k)} && \text{(shift the first index)}\\ &= \sum_{k=0}^{n+1} \binom{n+1}{k} \mu^{(k)} f^{(n+1-k)} && \text{(Pascal's rule)}. \end{aligned} $$

The induction is complete. As $\mu$ and $f$ are of class $\mathcal{C}^n$, all the derivatives $\mu^{(k)}$ and $f^{(n-k)}$ for $0 \le k \le n$ are continuous; the formula then shows that $(\mu \cdot f)^{(n)}$ is continuous, so $\mu \cdot f$ is of class $\mathcal{C}^n$.

Method — Establish a class and an iterated derivative

To show $f \colon I \to E$ is of class $\mathcal{C}^k$, check that each component is of class $\mathcal{C}^k$ as a scalar function; the iterated derivative is then $f^{(k)} = \sum_i f_i^{(k)} e_i$. For a product $\mu \cdot f$, apply the Leibniz formula rather than differentiating $k$ times by hand.

Skills to practice

Computing an iterated derivative

II Integration on a segment

II.1 The integral of a continuous vector function

The integral of a continuous vector function is built, like everything else, from the components: integrate each component function, then recombine. The one point that needs care is that the result does not depend on the basis --- and that, we will see, is a fact of linear algebra.

Definition — Integral of a continuous vector function

Let $f \colon [a,b] \to E$ be continuous, with $a \le b$, and let $\mathcal{B} = (e_1, \dots, e_p)$ be a basis of $E$. The integral of $f$ on $[a,b]$ is $$ \int_a^b f = \sum_{i=1}^p \left( \int_a^b f_i \right) e_i, $$ each $f_i \colon [a,b] \to \mathbb{K}$ being continuous (continuity is componentwise), so that the component integrals --- the integral of a continuous $\mathbb{K}$-valued function --- exist. This defines $\int_\alpha^\beta f$ whenever $\alpha \le \beta$; for an arbitrary pair $\alpha, \beta$ of points of an interval on which $f$ is continuous we extend it by the oriented-integral convention $$ \int_\beta^\alpha f = - \int_\alpha^\beta f \qquad (\alpha \le \beta), $$ so that $\int_\alpha^\beta f$ is now defined for the two endpoints in either order (and equals $0_E$ when $\alpha = \beta$).

Proposition — The integral does not depend on the basis

Let $f \colon [a,b] \to E$ be continuous, with $a \le b$. For every linear form $\varphi$ on $E$, $$ \varphi\!\left( \int_a^b f \right) = \int_a^b (\varphi \circ f). $$ Consequently the vector $\int_a^b f$ does not depend on the basis used to define it.

Proof

Let $\varphi$ be a linear form on $E$. In the basis $\mathcal{B}$, $\varphi(x) = \sum_i \varphi(e_i)\, x_i$ for $x = \sum_i x_i e_i$, so $\varphi \circ f = \sum_i \varphi(e_i)\, f_i$, a fixed $\mathbb{K}$-linear combination of the component functions. By linearity of $\varphi$ and of the $\mathbb{K}$-valued integral, $$ \begin{aligned} \varphi\!\left( \int_a^b f \right) = \sum_{i=1}^p \varphi(e_i) \int_a^b f_i &= \int_a^b \sum_{i=1}^p \varphi(e_i)\, f_i && \text{(linearity of the integral)}\\ &= \int_a^b (\varphi \circ f). \end{aligned} $$ Now let $\mathcal{B}'$ be a second basis and $J'$ the vector $\sum_j (\int_a^b f_j')\, e_j'$ obtained from the components in $\mathcal{B}'$. The identity just proved holds equally for $J'$: $\varphi(J') = \int_a^b (\varphi \circ f)$ for every linear form $\varphi$. Hence $\varphi\bigl(\int_a^b f\bigr) = \varphi(J')$ for every linear form $\varphi$. Two vectors of $E$ on which every linear form agrees are equal, so $\int_a^b f = J'$: the integral is the same in both bases.

Example — A vector integral computed componentwise

For $f \colon t \mapsto (\cos t, \sin t)$ on $[0,\pi]$, $$ \int_0^\pi f = \left( \int_0^\pi \cos t\, \mathrm{d}t,\ \int_0^\pi \sin t\, \mathrm{d}t \right) = (0,\, 2). $$

Proposition — Properties of the integral

Let $f, g$ be continuous on an interval $I$ of $\mathbb{R}$, valued in $E$, and let $a, b, c \in I$.

Linearity: for $\alpha, \beta \in \mathbb{K}$, $\int_a^b (\alpha f + \beta g) = \alpha \int_a^b f + \beta \int_a^b g$.
Chasles: $\int_a^b f = \int_a^c f + \int_c^b f$, valid for any $a, b, c \in I$ with the oriented convention.
Riemann sums: for $a \le b$, $\dfrac{b-a}{n} \displaystyle\sum_{k=1}^n f\!\left( a + k\,\dfrac{b-a}{n} \right) \xrightarrow[n \to +\infty]{} \int_a^b f$.
Triangle inequality: for $a \le b$, $\ \norme[\Big]{\displaystyle\int_a^b f} \le \displaystyle\int_a^b \norme{f}$.

Proof

Linearity, Chasles and the convergence of the Riemann sums are read componentwise: each component identity is the corresponding statement for the $\mathbb{K}$-valued integral (recalled from Intégration sur un segment), and a limit in $E$ is taken componentwise. The triangle inequality is the one genuinely new argument. Write $S_n = \frac{b-a}{n} \sum_{k=1}^n f(t_k)$ with $t_k = a + k\frac{b-a}{n}$. By the triangle inequality for a finite sum of vectors, $$ \norme{S_n} = \norme[\Big]{\frac{b-a}{n} \sum_{k=1}^n f(t_k)} \le \frac{b-a}{n} \sum_{k=1}^n \norme{f(t_k)}. $$ On the left, $S_n \to \int_a^b f$ and the norm is continuous (it is $1$-Lipschitz, recalled from Limits and continuity in a normed space), so $\norme{S_n} \to \norme[\big]{\int_a^b f}$. On the right, $\frac{b-a}{n} \sum_{k=1}^n \norme{f(t_k)}$ is a Riemann sum of the continuous real function $t \mapsto \norme{f(t)}$, hence tends to $\int_a^b \norme{f}$. Passing to the limit in the inequality gives $\norme[\big]{\int_a^b f} \le \int_a^b \norme{f}$.

Method — Compute and bound a vector integral

Compute $\int_a^b f$ componentwise: integrate each $f_i$ and recombine. To bound the integral, use the triangle inequality $\norme[\big]{\int_a^b f} \le \int_a^b \norme{f}$ --- valid for $a \le b$ --- and then a bound on $\norme{f}$; never try to « take the norm inside » in any other way.

The Riemann sum chains the small contributions $\frac{b-a}{n} f(t_k)$ tip to tail; as $n$ grows, the chain closes onto the vector $\int_a^b f$.

Skills to practice

Integrating a vector function

II.2 The integral as a function of its upper bound

Letting the upper bound vary turns the integral into a vector function of its own. It links integration back to differentiation --- and produces the mean value inequality, the working substitute for the absent equality theorem.

Theorem — Fundamental theorem of calculus

Let $f \colon I \to E$ be continuous and $a \in I$. The function $$ F \colon x \longmapsto \int_a^x f $$ is of class $\mathcal{C}^1$ on $I$, and $F' = f$.

Proof

By the definition of the vector integral, the components of $F$ are $F_i \colon x \mapsto \int_a^x f_i$. Each $f_i$ is a continuous $\mathbb{K}$-valued function, so by the fundamental theorem of calculus for $\mathbb{K}$-valued functions (recalled from Intégration sur un segment) each $F_i$ is of class $\mathcal{C}^1$ with $F_i' = f_i$. Every component of $F$ being of class $\mathcal{C}^1$, the function $F$ is of class $\mathcal{C}^1$, and $F' = \sum_i F_i'\, e_i = \sum_i f_i\, e_i = f$.

Proposition — Integral of a derivative

If $f \colon I \to E$ is of class $\mathcal{C}^1$, then for all $a, b \in I$, $$ \int_a^b f' = f(b) - f(a). $$

Proof

Set $G \colon x \mapsto \int_a^x f'$. Since $f'$ is continuous, the fundamental theorem of calculus gives $G$ of class $\mathcal{C}^1$ with $G' = f'$. Then $(f - G)' = f' - f' = 0_E$ on the interval $I$; componentwise, each $f_i - G_i$ has zero derivative, hence is constant (recalled from Dérivabilité), so $f - G$ is a constant vector. At $x = a$, $f(a) - G(a) = f(a) - 0_E = f(a)$, so $f(x) - G(x) = f(a)$ for every $x$, that is $G(x) = f(x) - f(a)$. Taking $x = b$ gives $\int_a^b f' = G(b) = f(b) - f(a)$.

Theorem — Mean value inequality

Let $f \colon I \to E$ be of class $\mathcal{C}^1$, and suppose there is a real $M \ge 0$ with $\norme{f'(t)} \le M$ for every $t \in I$. Then $$ \textcolor{colorprop}{\forall\, a, b \in I, \qquad \norme{f(b) - f(a)} \le M\, |b - a|}. $$

Proof

Case $a \le b$. By the integral of a derivative and the triangle inequality for the integral, $$ \begin{aligned} \norme{f(b) - f(a)} = \norme[\Big]{\int_a^b f'} &\le \int_a^b \norme{f'} && \text{(triangle inequality)}\\ &\le \int_a^b M = M\,(b - a) && \text{(bound $\norme{f'} \le M$)}. \end{aligned} $$ Since $b - a = |b - a|$ here, this is the claim.
Case $b < a$. Apply the previous case to the pair $(b, a)$: $\norme{f(a) - f(b)} \le M\,(a - b)$. As $\norme{f(b) - f(a)} = \norme{f(a) - f(b)}$ and $a - b = |b - a|$, the claim holds again.

Why an inequality and not an equality? A function valued in a space $E$ other than $\mathbb{R}$ has no general equality mean value theorem. The equality $f(b) - f(a) = (b-a)\, f'(c)$ is a theorem of the real case (Dérivabilité), and it has no vector counterpart. Take the circle $f \colon t \mapsto (\cos t, \sin t)$ on $[0, 2\pi]$: it returns to its start, $f(2\pi) = f(0) = (1,0)$, yet $f'(t) = (-\sin t, \cos t)$ is never the zero vector, since $\norme{f'(t)} = 1$. No point $c$ can give $f(2\pi) - f(0) = 2\pi\, f'(c)$. Only the mean value inequality survives --- and it is enough for analysis.

Method — Bound an increment

To bound $\norme{f(b) - f(a)}$ for a $\mathcal{C}^1$ vector function, do not look for an equality: bound $\norme{f'}$ by a constant $M$ on the interval between $a$ and $b$, then apply the mean value inequality $\norme{f(b) - f(a)} \le M\,|b-a|$.

Skills to practice

Using the FTC and bounding with the mean value inequality

III Taylor formulas

III.1 Taylor with integral remainder and the Taylor-Lagrange inequality

The Taylor formulas approximate a function near a point by a polynomial in $x - a$, with vector coefficients $f^{(k)}(a)$. The remainder of the exact formula is a vector integral; bounding its norm gives the Taylor--Lagrange inequality --- once again an inequality, with no equality form.

Theorem — Taylor with integral remainder

Let $n \in \mathbb{N}$, let $f \colon I \to E$ be of class $\mathcal{C}^{n+1}$, and let $a, x \in I$. Then $$ f(x) = \sum_{k=0}^n \frac{(x-a)^k}{k!}\, f^{(k)}(a) + \int_a^x \frac{(x-t)^n}{n!}\, f^{(n+1)}(t)\, \mathrm{d}t. $$

Proof

Apply the Taylor formula with integral remainder for a $\mathbb{K}$-valued function (recalled from Intégration sur un segment) to each component $f_i$, which is of class $\mathcal{C}^{n+1}$: $$ f_i(x) = \sum_{k=0}^n \frac{(x-a)^k}{k!}\, f_i^{(k)}(a) + \int_a^x \frac{(x-t)^n}{n!}\, f_i^{(n+1)}(t)\, \mathrm{d}t. $$ Multiply by $e_i$ and sum over $i$. The polynomial part recombines into $\sum_{k=0}^n \frac{(x-a)^k}{k!} f^{(k)}(a)$, since $f^{(k)} = \sum_i f_i^{(k)} e_i$; the remainder part recombines into $\int_a^x \frac{(x-t)^n}{n!} f^{(n+1)}(t)\, \mathrm{d}t$, by the very definition of the vector integral.

Theorem — Taylor-Lagrange inequality

Let $n \in \mathbb{N}$, let $f \colon I \to E$ be of class $\mathcal{C}^{n+1}$, and let $a, x \in I$. On the segment with endpoints $a$ and $x$, the real-valued continuous function $\norme{f^{(n+1)}}$ attains a maximum; call it $M$. Then $$ \textcolor{colorprop}{\norme[\Big]{ f(x) - \sum_{k=0}^n \frac{(x-a)^k}{k!}\, f^{(k)}(a) } \le M\, \frac{|x-a|^{n+1}}{(n+1)!}}. $$

Proof

First, the maximum $M$ is well defined: $t \mapsto \norme{f^{(n+1)}(t)}$ is a real-valued function, continuous as the composition of the continuous $f^{(n+1)}$ with the continuous norm, on the segment with endpoints $a$ and $x$; a real function continuous on a segment is bounded and attains its bounds (extreme value theorem, recalled from Limites et continuité).
By the integral-remainder formula, the left-hand side is $\norme[\big]{R}$, where $R = \int_a^x \frac{(x-t)^n}{n!} f^{(n+1)}(t)\, \mathrm{d}t$.

Case $a \le x$. For $t \in [a,x]$, $(x-t)^n \ge 0$; by the triangle inequality for the integral, $$ \begin{aligned} \norme{R} &\le \int_a^x \frac{(x-t)^n}{n!}\, \norme{f^{(n+1)}(t)}\, \mathrm{d}t && \text{(triangle inequality)}\\ &\le \int_a^x \frac{(x-t)^n}{n!}\, M\, \mathrm{d}t && \text{(bound $\norme{f^{(n+1)}} \le M$)}\\ &= M\, \frac{(x-a)^{n+1}}{(n+1)!} && \text{(integrate the polynomial)}. \end{aligned} $$
Case $x < a$. Here $R = -\int_x^a \frac{(x-t)^n}{n!} f^{(n+1)}(t)\, \mathrm{d}t$, so $\norme{R} = \norme[\big]{\int_x^a \frac{(x-t)^n}{n!} f^{(n+1)}(t)\, \mathrm{d}t}$. For $t \in [x,a]$ one has $x - t \le 0$, hence $\bigl|(x-t)^n\bigr| = (t-x)^n$; by the triangle inequality, $$ \begin{aligned} \norme{R} &\le \int_x^a \frac{(t-x)^n}{n!}\, \norme{f^{(n+1)}(t)}\, \mathrm{d}t && \text{(triangle inequality)}\\ &\le M \int_x^a \frac{(t-x)^n}{n!}\, \mathrm{d}t = M\, \frac{(a-x)^{n+1}}{(n+1)!} && \text{(bound, then integrate)}. \end{aligned} $$

In both cases $\norme{R} \le M\, |x-a|^{n+1}/(n+1)!$ --- the announced inequality. Note. For a function valued in a space $E$ other than $\mathbb{R}$ this is an inequality and only an inequality: there is no general equality form, unlike the Taylor--Lagrange equality of the real case.

Example — Taylor-Lagrange on the helix

For the helix $f(t) = (\cos t, \sin t, t)$, with $\mathbb{R}^3$ Euclidean, bound $\norme{f(x) - f(0) - x\, f'(0)}$ for $x \in \mathbb{R}$.

Answer

The function $f$ is of class $\mathcal{C}^\infty$, with $f''(t) = (-\cos t, -\sin t, 0)$, so $\norme{f''(t)} = \sqrt{\cos^2 t + \sin^2 t + 0} = 1$ for every $t$. Hence $M = \sup \norme{f''} = 1$ on any segment. Applying the Taylor--Lagrange inequality at order $n = 1$, with $a = 0$, $$ \norme{f(x) - f(0) - x\, f'(0)} \le M\, \frac{|x|^2}{2!} = \frac{x^2}{2}. $$

Method — Apply a Taylor formula

Choose the formula by the goal: the integral remainder for an exact expression of $f(x)$; the Taylor--Lagrange inequality to bound the distance from $f(x)$ to its Taylor polynomial --- compute $M = \sup \norme{f^{(n+1)}}$ on the segment; the Taylor--Young formula (next subsection) for the local behaviour up to $o((x-a)^n)$.

Skills to practice

Applying Taylor with integral remainder and Taylor-Lagrange

III.2 Taylor-Young

The Taylor--Young formula is the local version: it describes the behaviour of $f$ in the immediate neighbourhood of $a$, with a remainder negligible compared to $(x-a)^n$.

Theorem — Taylor-Young

Let $n \in \mathbb{N}$, let $f \colon I \to E$ be of class $\mathcal{C}^n$, and let $a \in I$. Then, as $x \to a$ with $x \in I$ (a one-sided neighbourhood if $a$ is an endpoint), $$ f(x) = \sum_{k=0}^n \frac{(x-a)^k}{k!}\, f^{(k)}(a) + R(x), \qquad \norme{R(x)} = o\bigl(|x-a|^n\bigr). $$ The vector remainder, written $o\bigl((x-a)^n\bigr)$, is a function $R$ whose norm is negligible compared with $|x-a|^n$.

Proof

Apply the Taylor--Young formula for a $\mathbb{K}$-valued function (recalled from Dérivabilité) to each component $f_i$, of class $\mathcal{C}^n$: $f_i(x) = \sum_{k=0}^n \frac{(x-a)^k}{k!} f_i^{(k)}(a) + \varepsilon_i(x)$ with $|\varepsilon_i(x)| = o\bigl(|x-a|^n\bigr)$. Multiplying by $e_i$ and summing, $f(x) = \sum_{k=0}^n \frac{(x-a)^k}{k!} f^{(k)}(a) + R(x)$ with $R(x) = \sum_i \varepsilon_i(x)\, e_i$. By the triangle inequality, $\norme{R(x)} \le \sum_i |\varepsilon_i(x)|\, \norme{e_i}$; each $|\varepsilon_i(x)|$ is $o\bigl(|x-a|^n\bigr)$ and the sum is finite, so $\norme{R(x)} = o\bigl(|x-a|^n\bigr)$ --- which is what $R(x) = o\bigl((x-a)^n\bigr)$ means.

Example — Taylor-Young of a planar curve

For $f \colon t \mapsto (\cos t, \sin t)$ at $a = 0$, the order-$2$ Taylor--Young expansion is obtained componentwise from $\cos t = 1 - \tfrac{t^2}{2} + o(t^2)$ and $\sin t = t + o(t^2)$: $$ f(t) = (1, 0) + t\,(0, 1) + \frac{t^2}{2}\,(-1, 0) + o(t^2). $$

Going further

This chapter differentiated and integrated a function valued in a fixed finite-dimensional space. Numerical and vector series and Sequences and series of functions study sequences and series of such functions; Linear differential equations solves differential equations whose unknown is a vector function; Matrix exponential and differential systems builds the matrix exponential $t \mapsto \exp(tA)$, the archetypal $\mathcal{C}^\infty$ vector function.

Skills to practice

Applying Taylor-Young