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I think that $f$ is weakly differentiable if and only if there exists a sequence of smooth functions $f_n$, which converges locally to $f$ in $L^1$ and such that $f'_n$ is locally Cauchy in $L^1$. Then the weak differential is the (local) limit of $f'_n$ in $L^1$.

Is that true?

EDIT: I have deleted my wrong proof.

MikeTeX
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  • @coudy. For the sake of simplicity, let assume that the domain of $f$ is $\bf R$. I left this question informal because I think it can be dealt similarly in $\bf R^n$. – MikeTeX Mar 01 '23 at 12:28
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    The proof of the second implication is wrong: 'Conversely suppose ...' should be 'Suppose $f$ to be weakly differentiable', also the claim 'It follows easily' is wrong. – daw Mar 01 '23 at 12:41
  • @daw. Except the fact that I have messed $\to$ with $=$, I think the rest of the proof may be OK. I have edited the question and added some explanations. – MikeTeX Mar 01 '23 at 17:11
  • $\int f_n \phi'\to \int f \phi'$ does not imply convergence of the weak derivative in $L^1$. – daw Mar 01 '23 at 19:37
  • @daw. I don't understand to what your answer is addressed. I wrote $\int f_n\varphi' \to \int f \varphi'$, and since $\int f_n\varphi' = \int f'_n\varphi$, we do have $\int f'_n\varphi\to \int f\varphi' = \int \alpha\varphi$ by hypothesis. So what's wrong with this? – MikeTeX Mar 02 '23 at 07:42
  • This does not imply $f_n' \to \alpha$ in $L^1$. – daw Mar 02 '23 at 07:54
  • @daw. See my edit. – MikeTeX Mar 02 '23 at 08:37
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    Not correct. The claim is false, no matter how many proofs you attempt. Here is a counterexample: Let $g_n=\sin(n\pi x)$ on $U=(0,1)$. Then $\int g_n \phi\to0$ but $g_n$ does not converge to zero in $L^1$. – daw Mar 02 '23 at 11:20
  • @daw. No. Your $g_n\varphi$ does not tend to 0 in $L^1([0,1])$. Indeed, $\int \sin^2 (n \pi x) \to 1/(2\pi)$, because this is equal to ${1\over n\pi}\int_0^{n\pi} \sin^2 (y)= {1\over n\pi}(n\cdot 1/2)$. Now $|\sin n \pi x| > \sin ^2 n \pi x$, hence your example does not work whenever $\varphi = 1$ on $[0, 1]$ (say). – MikeTeX Mar 02 '23 at 13:35
  • you cannot choose $\phi=g_n$. $\phi$ has to be a fixed function. sorry, you are wrong. check this website, which mentions the result somewhere, see, e.g. https://math.stackexchange.com/questions/2359172/how-to-show-that-sequence-constructed-from-a-periodic-function-weakly-converges – daw Mar 02 '23 at 13:39
  • @daw. The link you provided add nothing to the question, as they want to show that $\int g_n \varphi \to 0$ and not $\int |g_n|\varphi \to 0$ as needed here. Also, I don't understand your objection "$\phi = g_n\phi$ has to be a fixed function". – MikeTeX Mar 02 '23 at 14:01
  • OK. I think I've understood the source of the confusion. I've messed things there. I will edit my question. – MikeTeX Mar 02 '23 at 14:22

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It appears the theorem is true (if my understanding is exact). Here is a pic of Gilbarg and Trudinger, Elliptic partial differential equations of second order, thm. 7.4

Gilbarg and Trudinger 7.4

MikeTeX
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