Prove that $f$ is Lipschitz iff its derivative is bounded

Question

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!

Answer 1

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 $$ |t|<\delta\quad\Longrightarrow\quad \frac{1}{t}\|\,f(c+t\xi)-f(c)\|>2+\frac{\varepsilon}{2}\quad\Longrightarrow\quad \|\,f(c+t\xi)-f(c)\|>\left(2+\frac{\varepsilon}{2}\right)\|t\xi\|. $$

Answer 2

HINT: Use the definition of the derivative $$\lim_{y\to 0} \frac{\|f(x+y)-f(x)-Df(x)(y)\|}{\|y\|} = 0$$ together with the reverse triangle inequality.

Answer 3

Suppose that there exists $x\in U$ such that $\|D(f)(x)\|>M>2$, since the function $f$ is continuously differentiable, there exists a neighborhood $V$ of $x$ such that for every $y\in V$, $\|Df(y)\|>M$. You can assume that $V=B(x,r)$ a ball. Consider $c_t(x)=x+tu$ where $t\in [0,r/2]$ and $u\in \mathbb{R}^n$ such that $\|u\|=1$.

Write $g(t)=f(c(t))$. There exists $t_0\in (0,r/2)$ such that:

$\|g(r/2)-g(0)\|=\|f(x+(r/2)u)-f(x)\|=\|g'(t_0)\|\|(x+(r/2)u)-x\|=$

$\|Df(x+t_0)(u)\|\|(x+(r/2)u-x\|$.

Since $\|Df(x+t_0u)\|>M>2$, you can find $u_0$ with $\|u_0\|=1$ and $\|Df(x+t_0)(u_0)\|>M$. This implies that $\|f(x+(r/2)u_0)-f(x)\|=\|Df(x+t_0)(u_0)\|\|(x+(r/2)u_0)-x\|$. Contradiction.