Proof of The Fundamental Theorem of Calculus and the Mean Value Inequality for Vector- and Matrix-Valued Maps

For a map $x$ into $\mathbb{R}^{q}$ the component maps are written $x_{k}$, and for a map $A$ into $\mathcal{M}_{p\times q}(\mathbb{R})$ the entry maps are written $A_{ij}$, as in Derivative of a Vector-Valued or Matrix-Valued Map on an Interval (i) and Derivative of a Vector-Valued or Matrix-Valued Map on an Interval (ii). By these two clauses, a map into $\mathbb{R}^{q}$ is differentiable at a point if and only if each component map is, and then its derivative there has the derivatives of the component maps as components; likewise for matrices with entries. By Riemann Integral of a Vector-Valued or Matrix-Valued Map (ii) and Riemann Integral of a Vector-Valued or Matrix-Valued Map (iv), $\int_{u}^{v}x$ is defined if and only if every $\int_{u}^{v}x_{k}$ is, and then $\bigl(\int_{u}^{v}x\bigr)_{k}=\int_{u}^{v}x_{k}$; likewise for matrices with entries. The values of a map into $\mathbb{R}^{q}$, respectively $\mathcal{M}_{p\times q}(\mathbb{R})$, lie in $\mathbb{R}^{q}$, respectively $\mathcal{M}_{p\times q}(\mathbb{R})$ (Values of a Map between Classes, and Distances in a Metric Space (iii)), so their components and entries are defined. Vectors and matrices are equal when their components, respectively entries, agree (Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (iii), Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations (iii)), and differences are formed componentwise and entrywise (Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (iv), Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations (iv)). In (i), (ii) and (iv) we write out the vector case; the matrix case is the same argument applied to the entries $A_{ij}$, in (i) with Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (ii) in place of Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (i). Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations is applied with $(X,d)=(\mathbb{R},d_{\mathbb{R}})$ and with $S$ the domain of the map concerned.

Proof of (i). As $a<b$, also $a\le b$ (Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (v)), so the interval $[a,b]$ contains two distinct points by Basic Facts about Intervals of the Real Line (v), so Rules for the Riemann Integral of Vector- and Matrix-Valued Maps (i), with $[a,b]$ in place of $I$, shows that $\int_{c}^{t}x$ is defined for every $t\in[a,b]$. Fix $k\in[q]$. Then $X_{k}(t)=\int_{c}^{t}x_{k}$ for $t\in[a,b]$. By Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v), in the form of Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (i) with $S=[a,b]$, $x_{k}$ is continuous on $[a,b]$, so it is bounded and integrable on $[a,b]$ by A Continuous Function on a Closed Interval is Riemann Integrable (i). Let $F:[a,b]\to\mathbb{R}$ be the map of Fundamental Theorem of Calculus, Part I: the Indefinite Integral is Lipschitz, Continuous, and Differentiable at Points of Continuity (i) for $f=x_{k}$, so $F(t)=\int_{a}^{t}x_{k}$, and let $\gamma=\int_{c}^{a}x_{k}$. By Additivity of the Riemann Integral over Adjacent Intervals, Subintervals and Oriented Integrals, and Integrals of Restrictions (iii), for every $t\in[a,b]$,

$$X_{k}(t)=\int_{c}^{a}x_{k}+\int_{a}^{t}x_{k}=\gamma+F(t),$$

so $X_{k}$ is the sum of $F$ and the constant map $[a,b]\to\mathbb{R}$ with value $\gamma$ (Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (i)). Let $t\in[a,b]$. Since $x_{k}$ is continuous on $[a,b]$, it is continuous at $t$ relative to $[a,b]$ (Continuous Map Between Metric Spaces (ii)), that is, continuous at $t$ in the sense of The Real Line: Standing Notation and Background for Calculus on Intervals (vi), and Fundamental Theorem of Calculus, Part I: the Indefinite Integral is Lipschitz, Continuous, and Differentiable at Points of Continuity (iv) shows that $F$ is differentiable at $t$ with $F'(t)=x_{k}(t)$, and by Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers (i) and Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers (iv) $X_{k}$ is differentiable at $t$ with $X_{k}'(t)=x_{k}(t)$. Hence $X$ is differentiable at $t$ and $X'(t)$ has components $x_{k}(t)$, that is, $X'(t)=x(t)$; as $t$ is arbitrary, $X$ is differentiable (Derivative of a Vector-Valued or Matrix-Valued Map on an Interval (iii)).

Proof of (ii). By Derivative of a Vector-Valued or Matrix-Valued Map on an Interval (iv) and Derivative of a Vector-Valued or Matrix-Valued Map on an Interval (i), the $k$-th component of $x'(s)$ is $x_{k}'(s)$ for $s\in I$, so the $k$-th component map of $x'$ is the derivative map $x_{k}'$ of Derivative of a Real Function on an Interval (iv), and $\int_{u}^{v}x_{k}'$ is defined for every $k\in[q]$. As $x$ is differentiable, every $x_{k}$ is differentiable at every point of $I$, that is, differentiable on $I$ (Derivative of a Real Function on an Interval (iii)), with derivative $x_{k}'$. By Rules for Oriented Riemann Integrals of Real-Valued Maps: Sums, Scalar Multiples, Constants, Continuous Maps, Additivity and Evaluation by an Antiderivative (vi), applied to $x_{k}$, $\int_{u}^{v}x_{k}'=x_{k}(v)-x_{k}(u)$, the $k$-th component of $x(v)-x(u)$; this gives the clause.

Proof of (iii). Let $u,v\in I$. Since $x'$ is continuous on $I$, $\int_{u}^{v}x'$ is defined by Rules for the Riemann Integral of Vector- and Matrix-Valued Maps (i), applied to $x'$, and (ii) gives $\int_{u}^{v}x'=x(v)-x(u)$. The matrix case is the same, with the matrix statements of Rules for the Riemann Integral of Vector- and Matrix-Valued Maps (i) and (ii).

Proof of (iv). Fix $k\in[q]$. For every $t\in I$, $x_{k}$ is differentiable at $t$ with $x_{k}'(t)$ the $k$-th component of $x'(t)=0$, which is $0$ (Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (v)), and $x_{k}$ is continuous at $t$ relative to $I$ by The Increment Form of the Derivative, and Differentiability Implies Continuity (ii); so $x_{k}$ is continuous on $I$ (Continuous Map Between Metric Spaces (ii)). By The Sign and Size of the Derivative: Monotonicity, Constancy and Lipschitz Bounds (v), $x_{k}(u)=x_{k}(v)$ for all $u,v\in I$; hence $x(u)=x(v)$. For matrices, the entries of $0_{p\times q}$ are $0$ by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations (v).

Proof of (v), vectors. Let $u,v\in I$. If $u=v$, then $\lVert x(u)-x(u)\rVert=\lVert0\rVert=0=M\,|u-u|$ by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (iii). If $v<u$, the case $u<v$ with $u$ and $v$ exchanged gives $\lVert x(u)-x(v)\rVert\le M\,|u-v|$, and $\lVert x(v)-x(u)\rVert=\lVert-(x(u)-x(v))\rVert=\lVert x(u)-x(v)\rVert$ by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (v), while $|u-v|=|v-u|$. So let $u<v$, and let $z=x(v)-x(u)$ and $\varphi:I\to\mathbb{R}$, $t\mapsto z\cdot x(t)$ (Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (i) and Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (iii), with the map of The Summation Map on Real-Valued Maps over a Finite Set, and the Dot Product, Euclidean Norm and Euclidean Distance as Maps (ii)). The constant map $I\to\mathbb{R}^{q}$, $t\mapsto z$, and $x$ are continuous on $I$, so $\varphi$ is continuous on $I$ by Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (iv) and Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v). At every inner point $t$ of $I$, $x$ is differentiable at $t$, so $\varphi$ is differentiable at $t$ with $\varphi'(t)=z\cdot x'(t)$, by Rules for Derivatives of Vector- and Matrix-Valued Maps (x). By Mean Value Theorem on a Closed Real Interval and on Any Interval (ii) there is $\xi\in I$ with $u<\xi<v$ and $\varphi(v)-\varphi(u)=\varphi'(\xi)\,(v-u)$; $\xi$ is an inner point of $I$. Hence $\lVert x'(\xi)\rVert\le M$ by hypothesis, and $0\le\lVert x'(\xi)\rVert$ by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (i); so $M\ge\lVert x'(\xi)\rVert\ge0$. By the difference form of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite (iii), and by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (i) for the last step,

$$\varphi(v)-\varphi(u)=z\cdot x(v)-z\cdot x(u)=z\cdot\bigl(x(v)-x(u)\bigr)=z\cdot z=\lVert z\rVert^{2}.$$

By Rules of Arithmetic and Order in an Ordered Field (x) and The Cauchy-Schwarz Inequality on $\mathbb{R}^n$ (i), and since $0\le\lVert z\rVert$ (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (i)) and $\lVert x'(\xi)\rVert\le M$,

$$\varphi'(\xi)=z\cdot x'(\xi)\le|z\cdot x'(\xi)|\le\lVert z\rVert\,\lVert x'(\xi)\rVert\le\lVert z\rVert\,M .$$

Multiplying by $v-u>0$ (Rules of Arithmetic and Order in an Ordered Field (vi)) and using $\varphi(v)-\varphi(u)=\varphi'(\xi)(v-u)$ gives the chain

$$\lVert z\rVert^{2}=\varphi'(\xi)\,(v-u)\le\lVert z\rVert\,\lVert x'(\xi)\rVert\,(v-u)\le\lVert z\rVert\,M\,(v-u).$$

If $\lVert z\rVert=0$, then $\lVert z\rVert=0\le M(v-u)$ because $M\ge0$; otherwise multiplying by $\lVert z\rVert^{-1}>0$ (Rules of Arithmetic and Order in an Ordered Field (viii), Rules of Arithmetic and Order in an Ordered Field (vi)) gives $\lVert z\rVert\le M(v-u)=M\,|v-u|$.

Proof of (v), matrices. Let $u,v\in I$ and $w\in\mathbb{R}^{q}$, and let $y:I\to\mathbb{R}^{p}$, $t\mapsto A(t)w$ (Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (i) and Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (iii), with the map of The Matrix Product, Matrix-Vector Product, Transpose, Trace, Identity Matrix and Standard Basis Vectors Exist and Are Unique (ii)). Then $y$ is continuous on $I$ by Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (iv) and Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v), the constant map $t\mapsto w$ being continuous by the former. At every inner point $t$ of $I$, $y$ is differentiable at $t$ with $y'(t)=A'(t)w$ by Rules for Derivatives of Vector- and Matrix-Valued Maps (x), and $\lVert y'(t)\rVert\le\lVert A'(t)\rVert\,\lVert w\rVert\le M\,\lVert w\rVert$ by Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (ii) and $0\le\lVert w\rVert$ (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (i)). The vector case, with $p$ in place of $q$ and $M\lVert w\rVert$ in place of $M$, gives $\lVert A(v)w-A(u)w\rVert\le M\,\lVert w\rVert\,|v-u|$. By Matrix-Vector Product (i), Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations (iv) and Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (ii), for $i\in[p]$,

$$\bigl((A(v)-A(u))w\bigr)_{i}=\sum_{j=1}^{q}\bigl(A_{ij}(v)w_{j}-A_{ij}(u)w_{j}\bigr)=\sum_{j=1}^{q}A_{ij}(v)w_{j}-\sum_{j=1}^{q}A_{ij}(u)w_{j},$$

so $(A(v)-A(u))w=A(v)w-A(u)w$. Hence $\lVert(A(v)-A(u))w\rVert\le\bigl(M\,|v-u|\bigr)\lVert w\rVert$ for every $w\in\mathbb{R}^{q}$, and Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (iii) gives $\lVert A(v)-A(u)\rVert\le M\,|v-u|$.