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When the weak derivative just is the strong (or classical) derivative? For instance, can we prove that weak derivate $Du\in C^\alpha$(or $C^0$) implies $u\in C^{1,\alpha}$(or $C^1$).

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    Consider the characteristic function $\chi_{\mathbb{Q}}$ of $\mathbb{Q}$, the set of rational numbers. It is easy to check that the null function is the weak derivative of $\chi_{\mathbb{Q}}$, but $\chi_{\mathbb{Q}}$ is not even continuous.2012-11-25
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    @ Siminore : Does the null function you said mean the constant function with value zero?2012-11-25
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    Can weak derivative $Du\in C^\alpha$ imply $u\in C^{1,\alpha}$? $\alpha\in(0,1)$.2012-11-25
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    A constant function is of class $C^\infty$.2012-11-25
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    @Siminore: As distributions the zero function and the characteristic function of $\mathbb{Q}$ are the same. The question, suitably interpreted is whether the fact that the weak derivative is represented by a $\mathcal{C}^\alpha$ function implies that the original distribution is represented by a $\mathcal{C}^{1+\alpha}$ function, or equivalently has a $\mathcal{C}^{1+\alpha}$ version.2012-11-26
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    @ Lukas Geyer: What you interpreted is absolutely right! But I even not need u is a general distribution, it is enough for me that consider the function u is $L^1_{loc}$. :-)2012-11-26
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    Not only,I will keep trying to prove the statement, but also expect someone can post his/her answer to me! ^-^2012-11-26
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    When it comes to equivalence classes up to set of zero measure, you cannot hope to prove esily pointwise results. There is a nice paragraph about weak and pointwise derivatives in these notes: http://www.math.ucsd.edu/~bdriver/231-02-03/Lecture_Notes/weak-derivatives.pdf2012-11-26

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