How smooth is the first derivative (in the distribution sense) of a Lipschitz function? Taking difference quotients and testing against an $L^1$ function, we see that $Df$ is in $L^\infty$. In ${\mathbb R}^1$ the converse is true, thanks to the persistence of the formula

$f(x+h) - f(x) = \int_0^1 f'(x+th) dt~ h$

(Proof: convolve with a mollifier)

However, if $f : {\mathbb R}^n \to {\mathbb R}$ is Lipschitz, then by the same argument, its derivative has a restriction to any line which is in $L^\infty$ of that line (more precisely, the tangential component of the derivative restricts). Ordinarily, one cannot restrict a distribution sensibly to lower dimensional subsets (straight lines requiring even more regularity than curves), or at least if you can because its primitive restricts, I don't know of any reason to expect the restriction to have any semblance of regularity.

For $n > 1$, is there a nice Banach space in which the derivative of a Lipschitz function belongs whose elements are smoother than just $L^\infty$?


Lipschitz functions are exactly $W^{1,\infty}$ (See 'Sobolev space' on wikipedia - under other examples and perhaps the bit about absolute continuity on lines). This means the short answer to your question is no.

| cite | improve this answer | |
  • $\begingroup$ Can you point me to a proof? I was under this impression that there were non-Lipschitz $W^{1,\infty}$ functions in dimensions greater than 1. $\endgroup$ – Phil Isett Mar 5 '11 at 1:33
  • $\begingroup$ Never mind, I have found a proof in Evans. $\endgroup$ – Phil Isett Mar 5 '11 at 1:47
  • 4
    $\begingroup$ Of course, in higher dimensions the gradient can not be just any vector-valued $L^\infty$-function $f = (f_i)$, since it satisfies the distributional identity $ \dfrac{\partial f_i}{\partial x_j} = \dfrac{\partial f_j}{\partial x_i}. $ So you don't get the whole of $L^\infty$, only those functions that satisfy this identity. $\endgroup$ – Mark Peletier Jul 22 '11 at 12:47
  • $\begingroup$ Mark -- I think your remark here was the (fairly obvious but nonetheless fundamental) thing I was failing to realize. It's just that this particular system of PDE does not exactly bestow upon its solutions any additional regularity. $\endgroup$ – Phil Isett Aug 25 '11 at 20:26

Every Lipschitz function is absolutely continuous. Consequently, its derivative exists and is uniformly bounded almost everywhere. The Lipschitz constant is just the $L^\infty$ norm of the derivative.

If you want a Banach space of smoother functions, then just define it. For example, let $X$ be the space of Lipschitz functions on $\mathbb R^n$ with integrable derivatives: $$X = \{ f :~ \nabla f \in L^1 \cap L^\infty \}.$$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.