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Here in Wikipedia, it is said that in the one-dimensional case, it is enough to assume that the $(k-1)$-th derivative of the function $f$, is differentiable almost everywhere and is equal almost everywhere to the Lebesgue integral of its derivative.

In the other hand in the multidimensional case, it is said that we should work with derivatives in the sense of distributions, because what was used in the one dimensional case does not work.

I just want some clarification why there's such differences between the one dimensional and multidimensional case.

user165633
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1 Answers1

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The point is that weak derivatives really are the "weakest" ones that give back the original function when integrated. That is, if $f=f(x_1\ldots x_n)$, then the function $g$ is equal to the weak derivative $\frac{\partial f}{\partial x_j}$ if and only if $$\tag{1}f(x_1\ldots x_j+h \ldots x_n)-f(x_1\ldots x_j\ldots x_n)=\int_{x_j}^{x_j+h} g(x_1\ldots \tau\ldots x_n)\,d\tau.$$ Therefore one could in principle define Sobolev spaces in terms of integrals, just like in the one-dimensional case. However, due to the usual caveats of measure theory, relation (1) does not hold at all "horizontal" lines $\{x_j + h\ : h\in \mathbb{R}\}$ but only for almost all $x_j$. This generates some complications - I am no expert on that - and that is why the usual distributional approach is more popular.

For further information you might want to look for the keyword "ACL property" on some book on Sobolev spaces such as the one by G. Leoni.

P.S.: See also this answer.

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Giuseppe Negro
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  • It makes sense since in the one dimensional case all the functions in $W^{1,1}$ are absolutely continuous functions. – user165633 Jun 02 '14 at 10:58
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    @user144542: Exactly: that's because in the one dimensional case you only have one "horizontal" line. So you don't have to worry about the "almost all" caveat mentioned above. – Giuseppe Negro Jun 02 '14 at 11:09