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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 $L^\infty$?

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2 Answers 2

up vote 5 down vote accepted

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.

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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. –  Phil Isett Mar 5 '11 at 1:33
    
Never mind, I have found a proof in Evans. –  Phil Isett Mar 5 '11 at 1:47
2  
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. –  Mark Peletier Jul 22 '11 at 12:47
    
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. –  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: $$X = \{ f :~ \nabla f \in L^1 \cap L^\infty \}.$$

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