# Continuous representatives in Sobolev Spaces

My question arise from the study of the possible extensions of Rademacher's Theorem to the Sobolev Space $W^{1,p}(\Omega)$, with $\Omega\subset \mathbb{R}^n$. In specific I'm studying the proof of the fact that any element of $W^{1,p}(\Omega)$ is differentiable almost everywhere if $p>n$.

A key result in the proof is that any element in $W^{1,p}(\Omega)$, if $p>n$, has a continuous representative. Unfortunately I did not found any reference for this result.

Moreover, looking on Wikipedia's page on Sobolev inequalities, I discovered that in my setup (which is part of the general case $k<\frac{p}{n}$) every element of $W^{1,p}(\Omega)$, if $p>n$, should be an Holder's continuous function.

I'm puzzled by that, because I always thought to the element of a Soboloev Space as class of functions, and seems to me unrealistic that any element of any class of those Sobolev spaces is Holder continuous (which, if I'm not wrong, is stated as a property that holds everywhere).

So my questions are: 1) Do you have a reference which explain why any class of functions in $W^{1,p}(\Omega)$ (for $p>n$) has a continuous representative? 2) Is it correct the result stated by Wikipedia on Holder continuity? If the answer is yes, where am I wrong in thinking Sobolev functions?

Thank you very much for your time!

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The theorem you are looking for is called sobolev embedding theorem (see here for the 3rd google hit).

It does not show that every representative of a class (an element of $W^{m,p}$) is (hölder) continuous (when you think about it a little bit, this is clearly impossible, because you can alter these functions on a set of measure zero to no longer be continuous), but rather, that there is a function representing a class, which has the desired property.

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Thank you for your answer! Actually this was my thought, but Wikipedia states it in an ambiguous way (or, at least, I misunderstood the concept). –  Giovanni De Gaetano Jul 10 '11 at 10:49
@paul: this is not true, sobolev spaces are defined as subspaces of $L^p$ and therefore they contain equivalence classes of functions and not functions as their elements. Also one does not consider derivatives in a classical sense, but weak derivatives, which are itself only defined up to alteration on a subset of measure zero. –  Alexander Thumm Jul 10 '11 at 18:15
Oop, yes, adjustments on measure 0 sets seem not to matter, although I don't know why such an adjustment would be done. Perhaps there are different ways to talk about Sobolev spaces... But to my taste I would not look at the inclusion of test functions into distributions and then declare the images to be ambiguous by any functions (integration against) which gave the 0 distribution. Nor when looking at Sobolev imbedding theorems, I would not consider altering the continuous or $C^k$ representative on a set of measure $0$. In fact, I don't think of completions as equivalence classes, usually... –  paul garrett Jul 12 '11 at 13:04
On further reflection, I do think there is some danger in dismissing things happening on sets of measure 0, since this immediately causes trouble in negatively-indexed $L^2$ Sobolev spaces, for example, despite being harmless in non-negatively indexed spaces. And there is some strange volatility to say that a Sobolev norm dominates a $C^k$ norm... if/but the corresponding limits of test functions are not guaranteed to be $C^k$, being ill-determined pointwise? I suppose the utility depends on one's context. –  paul garrett Jul 12 '11 at 13:14
... and, even if the "construction" of a completion is as equivalence classes, this isn't really the characterization. In an extreme example, I am presuming that not many people think of "real numbers" as equivalence classes, so, ambiguous up to null Cauchy sequences. And Dirac delta cannot literally be in a suitable negatively-indexed Sobolev space, because it can't literally be in the dual of a suitable positively-indexed one? Maybe I'm just using such stuff in a completely different context... –  paul garrett Jul 12 '11 at 13:55

1) In any of the books on Sobolev spaces. References are on the wiki page you cited.

2) Yes. As a toy example the space $W_2^1([0,1])$ can be taken here. Suppose $f\in W_2^1([0,1])$. Let $g(x)=\int_0^x f'(y)dy\$. Then $g'(x)=f(x)$ and $f-g\equiv C$ a.e. in $[0,1]$. From the other hand $$|\Delta g(x)|^2=\left|\int_x^{x+\Delta x}f'(y) dy\right|^2\le \left( \int_x^{x+\Delta x}dy\right) \int_x^{x+\Delta x}|f'(y)|^2 dy\le |\Delta x| \|f\|_{W_2^1}.$$ So the representative $g+C$ of the function $f$ is Hölder continuous. In the general case the original approach of Sobolev was exactly the same: to obtain an integral representaion of a function and then to use integral inequalities.

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