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Consider a ball B and let $f(x) \in L^1(B)$ such that $\int_B f(x) dx = 0$. Furtheremore, there exists a closed set $E \subset B$ such that $f|_E$ is Lipschitz. The standard Lipschitz extension theorem gives a function $g$ which is Lipschitz on $B$ such that $g|_E=f$.

My question is the following: Is there a construction that additionally gives $\int_B g(x) dx = 0$?

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    $\begingroup$ Assuming $E\subsetneq B$ (otherwise the problem is trivial) a cheap solution is $g(x)-{\int_Bg(y)dy\over \int_B \text{dist}(y,E)dy}\ \text{dist}(x,E)$. $\endgroup$
    – Pietro Majer
    Commented Apr 7, 2018 at 17:58
  • $\begingroup$ Thank you for the answer, that is a clever solution. I realised my question is not well stated, as the extension I am looking for is more specific. I shall update the question tomorrow to be more clear on what I am exactly looking for. $\endgroup$
    – Adi
    Commented Apr 8, 2018 at 16:32

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