real analysis - Is a function globally Lipschitz continuous and $mathcal{C}^1$ if and only if it is $mathcal{C}^1$ and its total derivative is bounded?

Is $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ globally Lipschitz continuous, i.e. there exists an $L>0$ such that

$\frac{|f(x)-f(y)|}{|x-y|}\leq L$ for all $x,y\in\mathbb{R}^n$,

and $\mathcal{C}^1$ if and only if $f$ is $\mathcal{C}^1$ and its total derivative is bounded?

Based on intuition alone, I'm strongly inclined to believe that the answer is yes. However I'm having trouble coming up with a proof (probably because my grasp of multivariable calculus is far from great). Could someone give one if the statement is true, or provide a counter example if it is false?

Thanks.

Answer

If $f$ is $\mathscr{C}^1$, then $f(x) - f(y) = \int_0^1 Df(y + t(x-y)).(x-y) dt$, by the fundamental theorem of calculus.

Hence, $$\begin{aligned} \| f(x) - f(y) \| &\le& &\int_0^1 \|Df(y+t(x-y)).(x-y) \| dt& \\ &\le& &\left( \int_0^1 \| Df( y + t(x-y) )\| dt \right) \| x-y \|& \le \sup_{z \in \mathbb{R}^n} \, \|Df(z) \| \; \| x-y \| \end{aligned}$$

If $\sup_{z \in \mathbb{R}^n} \, \| Df(z) \| = C$ is finite, we get $\| f(x) - f(y) \| \le C \|x - y \|$ for all $x,y$.

Reciprocally, suppose that your function is $\mathscr{C}^1$ and that it's globally Lipschitz, with constant $C$.

Then, for all $x \in \mathbb{R}^n$, and all $h \in \mathbb{R}^n$, we know that $$Df(x).h = \lim_{t \to 0} \frac{f(x+th) - f(x)}{t}$$

But, by assumption, $\| f(x +th) - f(x) \| \le C \|th \| = C |t| \|h\|$, and we finally get $\|Df(x).h \| \le C \|h \|$ for all $h$, which by definition implies $\| Df(x) \| \le C$. Hence the total derivative is bounded all over $\mathbb{R}^n$.

All of this works also on an open set of $\mathbb{R}^n$, instead of the whole space.

Remark also that you don't need to assume $f$ to be $\mathscr{C}^1$, but only differentiable. The second part of my proof works as well, and for the first part, instead of applying fundamental theorem of calculus, you can use the mean value theorem.