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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. – Siminore Nov 25 '12 at 8:55
    
@ Siminore : Does the null function you said mean the constant function with value zero? – Darry Nov 25 '12 at 11:10
    
Can weak derivative $Du\in C^\alpha$ imply $u\in C^{1,\alpha}$? $\alpha\in(0,1)$. – Darry Nov 25 '12 at 11:17
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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. – Lukas Geyer Nov 26 '12 at 2:13
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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: math.ucsd.edu/~bdriver/231-02-03/Lecture_Notes/… – Siminore Nov 26 '12 at 12:08

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