## Is the derivative of a Lipschitz function better than L^\infty

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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 $L^\infty$?

-

## 2 Answers

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.

-
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 2011 at 1:33
Never mind, I have found a proof in Evans. – Phil Isett Mar 5 2011 at 1:47

1
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 2011 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 2011 at 20:26

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

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: `$$X = \{ f :~ \nabla f \in L^1 \cap L^\infty \}.$$`

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9396450519561768, "perplexity_flag": "head"}