Lipschitz $\implies |f'(x)|\le d$

Question

Let $U\subset \mathbb R^m$ be a convex subset and $f:U\to \mathbb R^n$ be a differentiable function and I would like to prove the following equivalence:

$$|f'(x)|\le d\ \text{for every $x\in U$}\Leftrightarrow |f(x)-f(y)|\le d|x-y|\ \text{for every $x, y\in U$}.$$

The $\Rightarrow$ is easy I used the Mean Value Theorem with the fact $U$ is convex. I'm having troubles to prove the converse. I had tried to use this fact $|f'(x)\cdot u|=|f(x+u)-f(u)-r(u)|$ and the triangle inequality to prove the converse and surprisingly I didn't have any progress. Are there some facts I'm missing to use?

Answer 1

Fix $x\in U$ and let $\epsilon > 0$ be arbitrary. By differentiability, there exists $\delta > 0$ such that for $\|h\| < \delta$, we have:

$$\|Df(x)h\| - \|f(x+h) - f(x)\| \le \|f(x+h) - f(x) - Df(x)h\| < \epsilon \|h\|$$

Hence $\|Df(x)h\| < (d+\epsilon)\|h\|$, for all $\|h\| < \delta$. Now let $z \neq 0$, and $h_z = \frac{\delta}{2 \|z\|}z$. Then $\|Df(x) h_z\| < (d+\epsilon)\|h_z\|$, hence $\|Df(x)z\| < (d+\epsilon)\|z\|$. Thus, we get $\|Df(x)z\| \le (d+\epsilon)\|z\|$ for all $z \in \Bbb R^m$, hence $\|Df(x)\| \le (d+\epsilon)$. This is true for all $\epsilon >0$, hence $\|Df(x)\| \le d$.

Answer 2

Pick some $h$ and define $\phi(t) = f(x+th)$, then $\|{\phi(t)-\phi(0) \over t}\| \le \| { f(x+th)-f(x) \over t} \| \le d \|h\|$.

Since $\lim_{t \to 0} {\phi(t)-\phi(0) \over t} = f'(x)h$, we see that $\|f'(x)h \| \le d \|h\|$ for all $h$. Setting $h= f'(x)^T$ shows that $\|f'(x)\| \le d$.