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Is it true that the characteristic (indicator) function of a subset of Euclidean space with finite positive measure is never in the Sobolev space $H^1 = W^{1,2}$ And if so, what is the best/easiest/most elementary way to see this?

Context:

I have this on good authority (it is stated in a decent textbook). However, I have had no joy in showing this to be the case myself. Due to its placement in aforementioned textbook as the first exercise at the end of a chapter about $H^1$, it feels like it oughtn't be difficult to show, but a group of my friends and I had no luck. It is bugging me now.

[Though I am a student taking a course based on the textbook, this is not `homework'; I will not be graded on it in any way and I have attempted the problem myself.]

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In your attempt, what definition of $W^{1,2}$ did you use? –  Gerald Edgar May 18 '10 at 16:51
    
Well, those functions in $L^2$ all of whose weak first derivatives are given by $L^2$ functions. –  Spencer May 18 '10 at 17:01
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4 Answers

If it's in $H^1$ it's a.e. differentiable, with weak differential a.e. equal its differential, which for an indicator function is a.e. zero. So if you integrate any candidate for your weak derivative multiplied by a compactly supported test function you should get zero. Now if you use the right test function and the definition of a weak derivative you ought to be able to contrive a contadiction of the form "$0=1$".

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Hum, I like yours better. So I removed mine. –  Willie Wong May 18 '10 at 17:19
    
Why thankyou @willie :) –  Tom Boardman May 18 '10 at 17:27
    
Does this arguement avoid assuming that the boundary of the set is of measure zero? I might be overcomplicating things but it's something that worried me last time I thought about the problem. –  Spencer May 18 '10 at 17:39
    
Just an objection: if f is differentiable in a point, it should continuous, hence locally bounded in that point, which is not quite true in H<sup>1</sup> if n>1 (such an f may have a dense set of singularities). –  Pietro Majer May 18 '10 at 17:48
    
@Pietro- if it's in H1 it will be a.e. differentiable- even the stuff with a dense set of singularities has that going for it. Don't worry- it's definitely a theorem ;) –  Tom Boardman May 18 '10 at 17:53
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One reason is this: if $f$ is in $H^1({\mathbb R^n})$, you have $\int_{\mathbb R^n}|f(x+h)-f(x)|^2 dx\le C|h|^2$ for all $h \in{\mathbb R^n}$ . Now in the case of $f:=\chi_E$ the integral is just the $L^1$ distance, $\|\chi_E - \chi_{E-h}\|_1$. By the triangular inequality, for any positive integer m and any h as above one gets

$\|\chi_E -\chi_{E-h}\|_1 \le$

$\sum_{j=0}^{m-1}\|\chi_{E - \frac{j}{m} h } - \chi_{E-\frac {j+1}{m}h}\|_1 \le$

$Cm|h/m|^2=C|h|^2/m$,

whence $\|\chi_E - \chi_{E-h}\|_1=0$ for all $h$. This is impossible since for $|h|\to\infty$ it converges to 2meas(E).

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Any chance you could please just give extra clarification as to where the $m$'s come from? Many thanks. –  Spencer May 18 '10 at 17:41
    
@ Pietro, I'm not sure about your first inequality- what if n=1 and f(x) is x^2 ? Trying h=1 for example seems to scupper things... –  Tom Boardman May 18 '10 at 19:08
    
(I tried to edit but had some problem with TeX). Anyway: just apply the first inequality to h/m where m is a positive integer, and make m steps to estimate the distance between E and E-h. Since m is arbitrary you conclude. Is it everything OK to you? This is quite an elementary proof. –  Pietro Majer May 18 '10 at 19:14
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@Tom, x^2 is not in H1(R) :) –  Pietro Majer May 18 '10 at 19:17
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Actually for L2 functions that inequality is a characterization of H1. This stuff is quite popular in calculus of variations and regularity. To prove the inequality, do it first for smooth functions with compact support (just write f(x+h)-f(x) as integral of the derivative in the direction h, then use Cauchy-Schwartz and Fubini. You will find C=||Df||_2^2). Then pass to the limit. An analog result holds with all p>1, to characterize W^{1,p}; with p=1 you get BV. Reference: I think Brezis (Fun.Anal.) is OK. –  Pietro Majer May 18 '10 at 20:02
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A not-so-elementary way to see it is to use the theorem that if $f$ is an $H^1$ function, then for almost every line segment, the restriction of $f$ to that line segment gives an absolutely continuous function. (In fact, this is essentially sufficient as well as necessary.) Wikipedia cites Maz'ya's book Sobolev spaces, and I also found a proof in Ziemer.

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The answer posted by Tom, as written is actually not true. A function in $H^1$ will not in general be differential almost everywhere; it depends on the dimension. In one dimension however it is indeed true that $H^1$ functions are differentiable almost everywhere (they are in fact absolutely continuous). There are two ways of seeing it is not in $H^1$. The simple answer is that if you differentiate the characteristic function of say $[0,\infty)$ then you will get the dirac measure. However let me just answer your question first:

Answer 1: Take any smooth compactly supported $\phi:\mathbb{R} \to \mathbb{R}$. By definition of weak derivative we have $\int \phi g^{\prime} dx = - \int \phi^{\prime} g dx$ where I've set $g=1_{[0,\infty)}$. This would have to be true for all such $\phi$ if the weak derivative existed. Now take $\phi^{\epsilon}$ to be supported in a neighborhood $(-\epsilon,\epsilon)$ of $0$. We are making the crucial assumption that $g^{\prime}$ is an integrable and hence it follows that $\int \phi^{\epsilon} g^{\prime} \to 0$ as $\epsilon \to 0$. However, $\phi^{\epsilon}$ is smooth and so $\int \partial_x\phi^{\epsilon}(x)g(x)dx = \phi^{\epsilon}(0)$ since $\phi$ was assumed to have compact support in $(-\epsilon,\epsilon)$. Now just fix $\phi^{\epsilon}(0)=1$ and we have that $\phi^{\epsilon}(0) \to 0$ by the first integral equality. This is a clear contradiction.

Notice that in fact that this really shows that $g' dx = \delta(x)$.

Answer 2: Take $1_{[0,1]}$ instead so that it is an $L^2([0,1])$ function. This is in fact the fourier transform of a "sinc" function, $\sin(k)/k$ up to some normalization constants. If we consider the $H^1$ norm in frequency space we would need $\int_0^{\infty} |k|^2\frac{\sin(k)^2}{|k|^2} < \infty$ which is clearly false. This requires being at ease with the fourier transform so if you're not, answer 1 is probably best.

It is true in $\mathbb{R}^n$ that if $u \in W^{1,p}$ for $p > n$ then $u$ is a.e differentiable and equals a.e its weak gradient (see Evans chapter 5). This is to correct what Tom had said although perhaps we was thinking about the $n=1$ case in which case $2 > 1$.

Hope this helps! Dorian

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Your answers are in dimension 1, that is, exactly when the answer you disagree with is correct ;) –  Athanagor Wurlitzer Oct 28 '13 at 21:10
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