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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.

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

2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$
    – anon
    Oct 28, 2020 at 18:09
  • $\begingroup$ 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). $\endgroup$
    – hamath
    Oct 28, 2020 at 18:20
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    $\begingroup$ 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 $u\in L^\infty(U)$? $\endgroup$
    – anon
    Oct 29, 2020 at 14:33
  • $\begingroup$ 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? $\endgroup$
    – anon
    Oct 29, 2020 at 14:35
  • $\begingroup$ 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. $\endgroup$
    – hamath
    Oct 29, 2020 at 17:34
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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.

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