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What are some conditions that ensure that a function $f(x) : \mathbb{R} \to \mathbb{R}$ which is in $L^1_{loc}$ and almost everywhere differentiable (in the classical sens ) with derivative in $L^1_{loc}$ has its derivative equal to its weak derivative (its derivative in the sens of distributions) ie :

$$ \forall \phi \in C^{\infty}_c(\mathbb{R}) ; \int f'\phi = - \int f \phi'$$

For example this is false for the characteristic function $\xi_{[0;1]} $which weak derivative is the difference of two dirac measures while its classical derivative is almost everywhere 0.

It works on the other hand for $C^1$ functions.

Does it work for Lipschitz functions ? Are there some necessary conditions ?

Edit : It does work for lipshitz function thx to the dominated convergence theorem

incas
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1 Answers1

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The necessary and sufficient condition is that $f$ should be absolutely continuous. This is a version of the fundamental theorem of calculus, and can be found in most graduate-level real analysis books.

Nate Eldredge
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    Does the existence of a weak derivative of any function ensures it to be differentiable almost everywhere? – Dark_Knight Apr 24 '17 at 22:01
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    If its weak derivative is $L^1_{loc}$, then yes. By letting your test function approach $1_{[a,b]}$, you get $f(b)-f(a) = \int_a^b f'$ for all $a,b$, and it follows by the above-mentioned FTC that $f$ is absolutely continuous and a.e. differentiable. Of course, any $L^1_{loc}$ function has a weak derivative as a distribution, and this says nothing about its pointwise differentiability. – Nate Eldredge Apr 24 '17 at 22:39
  • Isn't the characteristic function of the rational numbers has $0$ as weak derivative, but even nowhere continuous? – David Lingard Aug 01 '19 at 22:00
  • @DavidLingard: Yes, of course. What I should have said is that if $f$ is a function whose weak derivative is $L^1_{\text{loc}}$ then $f$ is almost everywhere equal to a function which is absolutely continuous and a.e. differentiable, and whose classical derivative equals its weak derivative almost everywhere. – Nate Eldredge Aug 03 '19 at 05:24
  • @NateEldredge Ah ok. Is it true that in the theory of PDEs all it matters is having a smooth representative? – David Lingard Aug 03 '19 at 09:53
  • @DavidLingard: This is indeed a condition that's commonly studied. I couldn't say whether it's all that matters. – Nate Eldredge Aug 03 '19 at 15:20