Gradient iff conditions on vector-valued Lipschitz function on $R^n$ Suppose that $f\in C^1(\mathbb{R})$.
$f\colon \mathbb{R}\to \mathbb{R}$ is Lipschitz if and only if $|f'|$ is bounded on $\mathbb{R}$.
Is it still true in multidimensional and vector-valued situation?
Say, $f\colon \mathbb{R}^n \to \mathbb{R}$. Is it true that $||\nabla f||_2$ bounded $\Longleftrightarrow f$ is Lipschitz?
Say, $f\colon \mathbb{R}^n \to \mathbb{R}^m$. Is it true that $||Jf||_2$ bounded $\Longleftrightarrow f$ is Lipschitz, where $Jf$ is the Jacobian of $f$ and $\|\,\|_2$ is matrix induced norm in L2 sense?
If not, what are some good iff/if conditions by using gradients?

Example, let $f(\mathbf{x}) = x_1x_2$, where $\mathbf{x} = [x_1\,x_2]^T\in \mathbb{R}^2$. Is this $f(\mathbf{x})$ Lipschitz? It intuitively looks like Lipschitz, but its gradient is unbounded.
 A: The idea that a function is Lipschitz iff its grandient is bounded is correct, but to state rigorously there are a couple of technical details that you have to take into account, which naturally lead to the notion of Sobolev space $W^{k,p}$. Indeed, one can show that if $U$ is a bounded domain with Lipschitz boundary then a function is Lipschitz if and only if it belongs to $$W^{1,\infty}(U):=\left\{u\in L^1_{\text{loc}}(U):u\in L^\infty(U), \nabla u\in L^\infty(U)\right\}$$where $\nabla u$ is understood in the distributional sense. In particular, the norm in your first question should be the $L^\infty$ norm instead of the euclidean one.
I recomend Evans' "Partial Differential Equations'', chapter 5 in general, section 5.8.2, b) for this questions in particular and the references given there for further investigations.
Hope it helps.
A: In your example $f(xy)=xy,$ $f$ is not Lipschitz. Proof: $|f(t,t)-f(0,0)|=t^2,$ and
$$\frac{t^2}{|(t,t)-(0,0)|}\to \infty$$
as $t\to \infty.$
But yes, it is true that if $f:\mathbb R^n\to \mathbb R^m,$ then $f$ is Lipschitz iff $\|Jf(x)\|_2$ is bounded. Let's take $m=1$ for simplicity. To prove $f$ Lipschitz implies $\|Jf(x)\|_2$ is bounded, note the latter is bounded iff each $D_kf(x)$ is bounded. If $D_kf(x)$ is not bounded, then there is a sequence $x_j$ such that $|D_kf(x_j)|>j.$  That implies
$$\left |\frac{f(x_j+he_k)-f(x_j)}{h}\right| > j\text {  for small positive } h.$$
That violates the Lipschitz condition.
I'll leave the other direction and the case $m>1$ to you for now.
