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Hello,

While studying Sobolev spaces, the following question came to my mind. Any help in this direction is appreciated.

QUESTION

Let $U\subseteq\mathbb{R}^n$ be open. Does there exist a function $f\in L^1_{\text{loc}}(U)$ such that

1) the classical derivative $Df$ exists everywhere in $U$.

2) $f$ is weakly differentiable in $U$. Let us write $D_w f$ to denote the weak derivative of $f$.

3) $Df\neq D_w f$, on a set of positive measure.

Note that, we are assuming the existence of both the derivatives. I'm aware of examples where one exists while other one does not.

The problem seems to be related to the question of validity of integration by parts for functions that are only differentiable.

Thank you.

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2  
In what sense do we assume weak differentiability of $f$? As a distribution, measure, in a Sobolev space, etc.? –  Daniel Spector Dec 17 '12 at 15:22
    
The way it is defined in the context of Sobolev spaces –  tatin Dec 17 '12 at 18:42
1  
... but $f$ is not assumed to lie in a Sobolev space with one degree of differentiability, but is instead merely assumed to lie in $L^1_{loc}$. In such a case, the weak derivative is a priori only defined as a distribution, which among other things means that hypothesis (3) is not well-formed without an additional hypothesis on either the weak derivative (to interpret it as a function or measure) or the strong derivative (to interpret it as a measure or distribution). ... –  Terry Tao Dec 17 '12 at 20:41
2  
... if one assumes Df is locally integrable (so that it becomes a distribution), then the answer is "no" in one dimension at least; see Proposition 25 of my notes terrytao.wordpress.com/2010/10/16/… . If instead one assumes that $D_w f$ is locally integrable, then the answer is again "no", from the Lebesgue differentiation theorem (Theorem 6, ibid). I haven't thought the higher dimensional case through, but perhaps one can leverage the 1D case using Fubini's theorem. –  Terry Tao Dec 17 '12 at 20:47
    
The definition of the weak derivative used here presupposes that it is locally integrable (by the looks of it), rather than just being a distribution. So $f$ must be of bounded variation on line segments and Lebesgue's differentiation theorem applies as mentioned by Terry Tao. –  George Lowther Dec 17 '12 at 23:54

1 Answer 1

up vote 2 down vote accepted

Suppose $f \in W^{1,1}_{loc}(U)$. Then no, since for such an $f$, we have that $Df$ exists and the approximate limit

$ap\lim_{y\to x} \frac{f(x)-f(y)-Df(x)(x-y)}{|x-y|} = 0$

exists for almost every $x$, while from assuming classical differentiability we have

$\lim_{y\to x} \frac{f(x)-f(y)-\nabla f(x)(x-y)}{|x-y|} = 0$

exists for every $x \in U$. In particular, the classical differential is a candidate for the approximate differential, and so $Df=\nabla f$ wherever the two exist, and hence in $U$ up to a set of measure zero.

http://www.encyclopediaofmath.org/index.php/Approximate_differentiability

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According to the distribution definition of the weak derivative of an $L^1_{loc}$ function, you have with $\phi\in C^\infty_c(U)$ $$ \langle \nabla_w f,\phi\rangle_{\mathscr D'(U),\mathscr D(U)}=-\int f \nabla \phi dx. $$ On the other hand you assume that the function $f$ is differentiable on $U$ with gradient $\nabla f$. Now the formulation of your question is not really meaningful: $\nabla_w f$ is defined weakly whereas $\nabla f$ has a point wise definition. So the meaning of equality or difference on a set of positive measure does not mean anything. –  Bazin Dec 17 '12 at 20:58
    
Let me say it again: I'm sorry for the confusion. I have had the impression that the term "weak derivative" is used when the distributional derivative is a $L^1_{loc}$ function. I thought that this is a standard convention but, as it seems, I'm wrong. –  tatin Dec 18 '12 at 2:47
    
From my question and your response, I understood that you meant a weak derivative which was actually a function (since if you are studying this for the first time, the subtleties you encounter are often overlooked until later when you realize it is a game and a dance of being careful and having the right idea). My above response is in this context, and is a proof that they must agree, up to a set of Lebesgue measure zero. Implicit in my proof is a result that Sobolev functions are approximately differentiable, which uses essentially the Lebesgue differentiation theorem. –  Daniel Spector Dec 18 '12 at 10:00

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