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The fundamental lemma of calculus of variations tells us,

suppose $f\in L_{loc}^1\Omega\;$ and $$\int_\Omega f\varphi=0,\forall\varphi\in C^\infty_0\Omega$$ Then $f=0\text{ a.e.}$

My question is again suppose that $f\in L_{loc}^1\Omega\;$ but this time $$\int_\Omega fD^i\varphi=0,\forall\varphi\in C^\infty_0\Omega,and \; for\;all \;i$$ do we have $f=C\text{ a.e.}$?

This question is related to A variant of the fundamental lemma of calculus of variation, but in there the background is set in distributions, which I'm not familiar.

RobPratt
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  • Your condition means that $f$ has weak derivatives equal to $0$, from which we can conclude $f$ is a.e. constant. – Koobe May 12 '23 at 02:54
  • Actually, I wanted to prove this so that I could say if f has weak derivatives equal to 0, then we can conclude f is a.e. constant. (I know there are other ways to prove this statement, I just wonder if I could prove it this way). Is it possible to prove this directly? Like you prove the fundamental lemma of calculus of variations. – min May 12 '23 at 04:01
  • Yes, this is true. It is surprisingly difficult to prove. It does not follow directly from the fundamental lemma. – daw May 12 '23 at 06:11

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