This answer shares an intriguing inequality:
Let $\nu \subseteq \mathbb{R}^n$ be open and convex and $f:\nu \to \mathbb{R}^m$ be differentiable, then $$\Vert f(x) - f(y) \Vert \leq \left(\sup_{\xi \in (x,y)} \Vert Df(\xi) \Vert \right) \Vert x -y \Vert$$
Here's my proof attempt:
We begin by noticing that $\Vert x \Vert = \sup \{\langle x,y \rangle : \Vert y \Vert = 1\}$ where $\langle ,\rangle $ denotes the usual inner product given by $\langle x,x\rangle = \Vert x \Vert^2$
Fix $x,y \in \mathbb{R}^n$ and $v \in \mathbb{R}^m$ and consider $g:\mathbb{R} \to \mathbb{R}$ given by $g(t) = \langle f(tx+(1-t)y), v\rangle $
Since $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$, by Mean Value Theorem:
$$g(1)-g(0) = g'(c) \text{ for some } c \in (0,1)$$ $$\implies \langle f(x)-f(y), v\rangle = \langle Df(cx+(1-c)y)(x-y), v\rangle $$
Set $z = cx+(1-c)y$ and we get
\begin{align*} \Vert f(x)-f(y) \Vert &= \sup \{\langle f(x)-f(y), v\rangle : \Vert v \Vert = 1\} \\ &= \sup \{\langle Df(z)(x-y), v\rangle : \Vert v \Vert = 1\} \\ \end{align*}
By Cauchy-Schwarz inequality,
$$\langle Df(z)(x-y), v\rangle \leq \Vert Df(z)(x-y)\Vert \leq \Vert Df(z)\Vert.\Vert x-y \Vert$$
where $\Vert Df(z)\Vert$ denotes the operator norm.
Now clearly, $\Vert Df(z) \Vert \leq \sup_{\xi \in (x,y)} \Vert Df(\xi) \Vert$ as $z \in (x,y)$
Therefore, $\sup_{\xi \in (x,y)} \Vert Df(\xi) \Vert. \Vert x-y \Vert$ is an upper bound for $\{\langle Df(z)(x-y), v\rangle: \Vert v \Vert = 1\}$ and hence,
$$\Vert f(x) - f(y) \Vert \leq \left(\sup_{\xi \in (x,y)} \Vert Df(\xi) \Vert \right) \Vert x -y \Vert$$
My first question is: is my proof valid and correct? Next, here we have a fixed $z$ so why are we weakening the inequality by maximizing $Df(z)$ over the interval $(x,y)$? I mean, $\Vert Df(z) \Vert. \Vert x-y \Vert$ is also an upper bound so doesn't this hold true :
$$\Vert f(x) - f(y) \Vert \leq \Vert Df(z) \Vert . \Vert x -y \Vert \text{ for some fixed } z \in (x,y)$$