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Sep 14, 2020 at 17:27 comment added Pietro Majer Excuse me, what result do you mean?
Sep 14, 2020 at 0:17 comment added Ben Ciotti Ok, so in that case, could the result be shown? Perhaps using the additional assumptions?
Sep 13, 2020 at 8:52 comment added Daniele Tampieri @BenCiotti, as a matter of fact, substituting $C^1_0(\Omega)$ with $C^1_c(\Omega)$ is equivalent to say that $f$ is a locally integrable function
Sep 13, 2020 at 6:06 comment added Pietro Majer Isn't this the same as saying $f$ is $ L^1_{loc}(\Omega)$?
Sep 13, 2020 at 0:44 comment added Ben Ciotti Very nice Pietro, thank you! May I ask a followup (which essentially comes down to the distinction you raised in your first comment)? Suppose I remove the assumption that $fg\in L^1$ for all $g\in C_0^1(\Omega)$, and replace it by the assumption that $fg\in L^1$ for all $g\in C_c^1(\Omega)$. Would the result still hold? Here $C^1_c(\Omega)$ is the set of continuously differentiable functions with compact support in $\Omega$.
Sep 12, 2020 at 23:23 comment added Pietro Majer Note that no assumption on $\partial\Omega$ are needed
Sep 12, 2020 at 22:31 history answered Pietro Majer CC BY-SA 4.0