Let $B \subset \mathbb{R}^n$ be an open ball and let $f:B\rightarrow \mathbb{R}^n$ be a differentiable mapping.
Prove that $$\sup_{x,y\in B, x\neq y} \frac{|f(x)-f(y)|}{|x-y|}=\sup_{x\in B} ||D_f(x)||_{OP}$$
I've tried using the definitions of the operator norm and the differential, but it just doesn't work. I've also tried using the fact that the operator norm of $A_{n\times n}$ is the sqrt of the maximal real eigenvalue of $A^t A$. Does anyone please may help? Thanks in advance.
Denote the LHS and RHS by symbols $$m\equiv \sup_{x,y\in B, x\neq y} \frac{|f(x)-f(y)|}{|x-y|}$$ $$b\equiv \sup_{x\in B} ||D_f(x)||_{OP}$$
Then we need to show $m \leq b$ and $b\leq m$. Of these $b \leq m$ is obvious (if you have $b>m$, pick some point $x_0$ where the operator norm is greater than $m$ and pick another point $y$ close enough to $x_0$ to get a contradiction immediately). So I shall focus on the part where we need to show $m \leq m$.
Suppose for contradiction that $m > b$. Then $\exists$ distinct $ x, y \in B$ such that $$|f(y)-f(x)| > b |x-y| \label{a}\tag{1}$$ Because the ball $B$ is convex, we can construct the straight line path $c$ in $B$ from $x$ to $y$, i.e., a smooth curve $c:I \to B$ such that $c(0)=x$ and $c(1)=y$ where $I$ denotes the interval $[0,1]$.
Let $\vec{k}(t)$ denote the tangent vector $c’(t)$ to the curve $c(t)$. Then $\vec{k}$ is a constant vector for all $t$ and $|\vec{k}|=|y-x|$ (why?).
Then consider the function $f \circ c:I \to \Bbb R^n$. We have $$f(y)-f(x)=(f\circ c)(1)-(f\circ c)(0)=\int_I (f\circ c)’=\int_I (Df)_{c(t)} \cdot c’(t)$$ where the last equality is the chain rule. So then
$$|f(y)-f(x)|= \left | \int_I (Df)_{c(t)} \cdot c’(t) \right |\leq \int_I |(Df)_{c(t)} \cdot c’(t)| \leq \int_I ||(Df)_{c(t)}||_{OP} \cdot |c’(t)|\leq \int_I b |\vec{k}| = b|\vec{k}| = b|y-x| $$
Which gives you a contradiction to ($\ref{a}$). Therefore, $m \leq b$.