Let $f$ be a vector-valued, continuously differentiable function on an open, convex set $U$ in $\mathbb R^n$ with values in $\mathbb R^m$. Show that $\|Df(c)(y)\| \le 2\|y\|$ for all $c\in U, y\in \mathbb R^n$ iff $\|f(x)-f(z)\|\le 2\|x-z\|$ for all $x, z\in U$.
My attempt:
$(\Rightarrow)$ $\|Df(c)(y)\| \le 2\|y\|$ means that the operator norm of $Df$ is bounded by $2$, and $U$ is convex, so the conclusion follows. I had no difficulty figuring out this.
$(\Leftarrow)$ What I am stuck is to prove the converse. I attempted as follows:
Let $\gamma(t)=tx+(1-t)z$ for $0\le t\le 1$ and $g(t)=f(\gamma(t))$. Then $$\|f(x)-f(z)\|=\|g(1)-g(0)\|\le \|g'(\tilde{t})\|=\|Df(\gamma(\tilde{t}))(x-z)\|,$$ and I'm stuck here.
Does anyone have ideas? Thank you for your help!
Best Answer
Suppose that for some $c\in U$, $$ \|Df(c)\|=2+\varepsilon>0. $$ Then, there exists a unit vector $\xi\in\mathbb R^n$, (i.e., $\|\xi\|=1$), such that $$ \|Df(c)\xi\|=2+\varepsilon. $$ However, $$ Df(c)\xi=\left.\frac{d}{dt}\right|_{t=0}f(c+t\xi)=\lim_{t\to 0}\frac{f(c+t\xi)-f(c)}{t}, $$ and thus $$ 2+\varepsilon=\|Df(c)\xi\|=\lim_{t\to 0}\frac{1}{t}\|\,f(c+t\xi)-f(c)\|. $$ Hence, for some $\delta>0$, we have that $$ |t|<\delta\quad\Longrightarrow\quad \frac{1}{t}\|\,f(c+t\xi)-f(c)\|>2+\frac{\varepsilon}{2}\quad\Longrightarrow\quad \|\,f(c+t\xi)-f(c)\|>\left(2+\frac{\varepsilon}{2}\right)\|t\xi\|. $$