# 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.

• $R$ should be $\mathbb R.$
– zhw.
Oct 28, 2020 at 18:09

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.

• Thanks! Two questions: 1. Do we really need weak derivative if $f\in C^1(R)$? 2. Do we have to use bounded domain, or there is no go to prove a global Lipschitz?
– anon
Oct 28, 2020 at 18:09
• 1. if $f\in C^1(\mathbb{R})$ then it has a weak derivative which coincides with the usual one (that's a nice exercise for you to try). 2. If you're in an arbitrary open set you get locally Lipschitz is equivalent to $W^{1,\infty}_{\text{loc}}$ (which is naturally defined from the previous comment). Oct 28, 2020 at 18:20
• Questions: Isn't $W^{1,\infty}(U):=\left\{u\in L^1_{\text{loc}}(U): \nabla u\in L^\infty(U)\right\}$ the correct Sobolev space? Why is there additional $u\in L^\infty(U)$?
– anon
Oct 29, 2020 at 14:33
• Another question: what do we mean that a Jacobian by using weak derivatives? Note that in Sobolev space, $L^\infty$ norm is essential norm. It means that we can not say a Jacobian with $L^\infty$ bounded weak derivatives gives bounded Lipschitz constant everywhere. Where am I going wrong?
– anon
Oct 29, 2020 at 14:35
• Regarding the first question, the correct way to define the Sobolev spaces is the space of all measurable functions $u$ in $U$ such that $$\|u\|_{L^\infty(U)}+\|\nabla u\|_{L^\infty(U)}<\infty$$ where the second norm is the sum of the norms of each component of $\nabla u$ (although there are equivalent definitions). That's how you get a Banach space and the related properties. Oct 29, 2020 at 17:34

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.