Prove that $f$ is Lipschitz iff its derivative is bounded 
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!
 A: 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 
$$
|t|<\delta\quad\Longrightarrow\quad
\frac{1}{t}\|\,f(c+t\xi)-f(c)\|>2+\frac{\varepsilon}{2}\quad\Longrightarrow\quad
\|\,f(c+t\xi)-f(c)\|>\left(2+\frac{\varepsilon}{2}\right)\|t\xi\|.
$$
A: 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.
A: 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.
