Proof of The Mean Value Theorem and Inequality on Convex Open Sets

Throughout, the arithmetic and order of $\mathbb{R}$ are used through The Structure of ℝ as a Field and an Ordered Ring (i), The Structure of ℝ as a Field and an Ordered Ring (ii), Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (v), Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (ix), Rules of Arithmetic and Order in an Ordered Field (iii) and The Quotient 1/x, Order Rules for Sums, and Nesting of Closed Intervals in ℝ (i); norms and distances in $\mathbb{R}^{n}$ through Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (ii), Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (v), Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (vi) and Standard Basis Vectors of ℝ^q: Coordinates, Orthonormality and Unit Length (iii); and the metric spaces $(\mathbb{R}^{n},d_{E})$ and $(\mathbb{R},d_{\mathbb{R}})$ through $\mathbb{R}^n$ with the Euclidean Distance Is a Complete Metric Space with Coordinatewise Convergence (i) and The Absolute Value Metric on the Real Numbers (i).

The segment. Every clause is proved by restricting $f$ to the segment from $x$ to $y$. Let $\gamma:[0,1]\to\mathbb{R}^{n}$,

$$\gamma(s)=(1-s)x+sy=x+s(y-x).$$

By Segments and Convex Subsets of $\mathbb{R}^n$ (i) and Segments and Convex Subsets of $\mathbb{R}^n$ (ii), $\gamma(s)\in[x,y]\subseteq U$ for every $s\in[0,1]$, and $\gamma(0)=x$, $\gamma(1)=y$. The interval $[0,1]$ contains two distinct points, and $\gamma$ is differentiable at every $s\in[0,1]$, relative to $[0,1]$, with

$$\gamma'(s)=y-x:$$

$\gamma$ is the constant function $x$ plus the function $s\mapsto\varphi(s)\,(y-x)$ with $\varphi(s)=s$, where $\varphi'=1$ by Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers (ii), so this follows from Rules for Derivatives of Vector- and Matrix-Valued Maps (ix), Rules for Derivatives of Vector- and Matrix-Valued Maps (iii) and Rules for Derivatives of Vector- and Matrix-Valued Maps (ii).

The mean value theorem ((i)). Let $\varphi=f\circ\gamma:[0,1]\to\mathbb{R}$. By the chain rule along a curve (Chain Rule for Differentiable Functions Between Euclidean Spaces (iv)), $\varphi$ is differentiable at every $s\in[0,1]$, with

$$\varphi'(s)=\nabla f\bigl(\gamma(s)\bigr)\cdot(y-x),$$

and it is therefore continuous on $[0,1]$ (The Increment Form of the Derivative, and Differentiability Implies Continuity (ii)). The one-variable mean value theorem (Mean Value Theorem on a Closed Real Interval and on Any Interval (i)) on $[0,1]$ yields $t\in(0,1)$ with $\varphi(1)-\varphi(0)=\varphi'(t)(1-0)$. With $z=\gamma(t)=(1-t)x+ty$, a point of $U$ by the previous paragraph, this reads

$$f(y)-f(x)=\nabla f(z)\cdot(y-x).$$

The mean value inequality ((ii)). Let $\psi=f\circ\gamma:[0,1]\to\mathbb{R}^{m}$. By Chain Rule for Differentiable Functions Between Euclidean Spaces (iii), $\psi$ is differentiable at every $s\in[0,1]$ with $\psi'(s)=Df(\gamma(s))\,(y-x)$, hence continuous on $[0,1]$ (Rules for Derivatives of Vector- and Matrix-Valued Maps (i)), and by the operator-norm bound (Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (ii)) and the hypothesis on $M$,

$$\lVert\psi'(s)\rVert\le\bigl\lVert Df\bigl(\gamma(s)\bigr)\bigr\rVert\,\lVert y-x\rVert\le M\,\lVert y-x\rVert\qquad(s\in[0,1]).$$

The mean value inequality for vector-valued functions of one variable (The Fundamental Theorem of Calculus and the Mean Value Inequality for Vector- and Matrix-Valued Maps (v)), applied with the constant $M\lVert y-x\rVert$, gives

$$\lVert f(y)-f(x)\rVert=\lVert\psi(1)-\psi(0)\rVert\le M\,\lVert y-x\rVert\cdot|1-0|=M\,\lVert y-x\rVert .$$

The real-valued inequality ((iii)). Take $z$ from (i). The Cauchy--Schwarz inequality (The Cauchy-Schwarz Inequality on $\mathbb{R}^n$ (i)) and the hypothesis on $M$ give

$$|f(y)-f(x)|=|\nabla f(z)\cdot(y-x)|\le\lVert\nabla f(z)\rVert\,\lVert y-x\rVert\le M\,\lVert y-x\rVert .$$

Vanishing derivative ((iv), (v)). If $Df(z)=0_{m\times n}$ for every $z\in U$, then $\lVert Df(z)\rVert=0$ (Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (iv)), so (ii) with $M=0$ gives $\lVert f(y)-f(x)\rVert\le0$; as the norm is nonnegative (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (i)), $\lVert f(y)-f(x)\rVert=0$, hence $f(y)-f(x)=0$ by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (iii), and $f(x)=f(y)$. Likewise, if $\nabla f(z)=0$ for every $z\in U$, then $\lVert\nabla f(z)\rVert=0$ by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (iii), so (iii) with $M=0$ gives $|f(y)-f(x)|\le0$, and again $f(x)=f(y)$.