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Is there any example of a function which is weakly differentiable but none of its versions are classically differentiable anywhere (or differentiable only on a set of measure 0) ? Thanks

Aerinmund Fagelson
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chandu1729
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3 Answers3

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In any Sobolev space where functions are capable of being infinite at a point, they are capable of being nowhere continuous.

Let $f$ be a function in some Sobolev space which is infinite at 0 (for example $f$ might be $|x|^{-1/3}$ in $H^1(\Omega)$ for $\Omega$ some open ball in $\mathbb{R}^3$), let $q_n$ be an enumeration of the rational points of $\Omega$, and let $f_h(x) = f(x-h)$. Then we can define $$g = \sum_n 2^{-n} f_{q_n}.$$ This $g$ is unbounded on any open set, but its norm is no more than double the norm of $f$, so in particular it is still weakly differentiable.

One can easily verify that the sequence of partial sums converges, and that the result is nowhere continuous, [and that you can't make it continuous by modifying on a set of measure zero].

Logan Stokols
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    Why is $g$ weakly differentiable? – chandu1729 Sep 01 '15 at 17:13
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    Because the sum is a cauchy sequence, and sobolev spaces are complete. Suppose e.g. the derivative of $f$ is $L^2$, then the $L^2$ norm of the derivative of $g$ is at most double that of $f$, and in particular it is a function. – Logan Stokols Sep 01 '15 at 17:19
  • cool example :) – Aerinmund Fagelson Jan 19 '16 at 21:29
  • @Logan Stokols...Excellent example. But I am not sure if the translation of $f(x-h)$ is possible if your domain is bounded?. $x-h$ may be outside the domain and $f$ need not be defined there. Please comment on this. – Alexander Jun 07 '20 at 04:35
  • @Alexander It's a purely notational issue. We could for example define $f_h(x)$ to be 0 if $x-h \notin \Omega$ or $f(x-h)$ if $x-h \in \Omega$. You could also define the functions $f$ and $g$ on all of $\mathbb{R}^n$ and then only consider the restriction of $g$ to $\Omega$. As long as the end result is defined, it should have the desired property. – Logan Stokols Jun 08 '20 at 18:23
  • Perfect! Thanks – Alexander Jun 08 '20 at 18:55
  • Is there a reason you picked $\mathbb{R}^3$, and not $\mathbb{R}$? – rod Sep 27 '23 at 11:05
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This is rather a comment on the one-dimensional case and on $W^{1,\infty}_\mathrm{loc}(\Omega)$ than an answer.

In one dimension, this cannot happen. Take a function $f \in L^1_{\mathrm{loc}}(0,1)$, with weak derivative $f' \in L^1_{\mathrm{loc}}(0,1)$. Then, for any $[a,b] \subset (0,1)$, you have $f' \in L^1([a,b])$. Hence, $f$ is absolutely continuous on this interval and hence almost everywhere differentiable.

On the other hand, if $f \in W^{1,\infty}_{\mathrm{loc}}(\Omega)$ for some $\Omega \subset \mathbb{R}^n$, then $f$ is Lipschitz on compact subsets of $\Omega$ and therefore almost everywhere differentiable by Rademacher's theorem.

This shows, that one has to look for counterexamples in at least two dimensions, where the weak derivative is not in $L^\infty_\text{loc}(\Omega)$ (i.e. "nowhere bounded").

gerw
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Surely, if you take a classically differentiable function and change its values on a dense (Lebesgue) null set, as the example with the characteristic function of the rationals in Wikipedia, that function would still have a weak derivative. But maybe you wanted less trivial examples?

Jeppe Stig Nielsen
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    From the point of view of the theory of weak derivatives the indicator function of the rationals is the same as the zero function. – kahen Mar 31 '13 at 16:28
  • I agree with Kahen. I just edited the question. None of its versions should have classical derivative! – chandu1729 Mar 31 '13 at 16:38
  • @kahen I agree. I was more or less quoting Wikipedia. Maybe the asker meant a function whose equivalence class (of functions identical to it almost everywhere) contains only functions that are classically nowhere differentiable. – Jeppe Stig Nielsen Mar 31 '13 at 16:40