The Mean Value Theorem and Inequality on Convex Open Sets

In the setting of Multivariable Calculus: Notation, let $n$, $m$ and $U$ be as in Multivariable Calculus: Notation (ii), suppose that $U$ is convex, and let $x,y\in U$. (The Jacobian matrix $Df(z)$ and the gradient $\nabla f(z)$ are the derivative matrix and the derivative vector of $f$ at $z$, by The Derivative Is the Jacobian Matrix, and Is Unique (ii) and The Derivative Is the Jacobian Matrix, and Is Unique (i).)

(i) Mean value theorem. Let $f:U\to\mathbb{R}$ be differentiable on $U$. Then there is $t\in(0,1)$ such that the point $z=(1-t)x+ty$ of $U$ satisfies

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

(ii) Mean value inequality. Let $f:U\to\mathbb{R}^{m}$ be differentiable on $U$, and let $M\in\mathbb{R}$ satisfy $\lVert Df(z)\rVert\le M$ for every $z\in U$. Then

$$\lVert f(y)-f(x)\rVert\le M\,\lVert y-x\rVert.$$

(iii) Mean value inequality, real-valued. Let $f:U\to\mathbb{R}$ be differentiable on $U$, and let $M\in\mathbb{R}$ satisfy $\lVert\nabla f(z)\rVert\le M$ for every $z\in U$. Then

$$|f(y)-f(x)|\le M\,\lVert y-x\rVert.$$

(iv) Vanishing derivative. If $f:U\to\mathbb{R}^{m}$ is differentiable on $U$ with $Df(z)=0_{m\times n}$ for every $z\in U$, then $f(x)=f(y)$.

(v) Vanishing gradient. If $f:U\to\mathbb{R}$ is differentiable on $U$ with $\nabla f(z)=0$ for every $z\in U$, then $f(x)=f(y)$.