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 $L^\infty$?



Is the derivative of a Lipschitz function better than L^\infty
Lipschitz functions are exactly $W^{1,\infty}$ (See '[Sobolev space](http://en.wikipedia.org/wiki/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.


