Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?


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.]

share|improve this question
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

4 Answers 4

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 "$0=1$".

share|improve this answer
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

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

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

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


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

share|improve this answer
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
@Tom, x^2 is not in H1(R) :) –  Pietro Majer May 18 '10 at 19:17
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

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.

share|improve this answer

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

share|improve this answer
Your answers are in dimension 1, that is, exactly when the answer you disagree with is correct ;) –  Jung Wen Chen Oct 28 '13 at 21:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.