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Problem: Let $G$ be a bounded Borel set. Let $X$ be the set of finite perimeter sets in $\mathbb R^N$ and $F: X \to \mathbb R \cup \{+\infty\}$ defined as

\[ F(E)= \begin{cases} Per(E) \hspace{1,5cm} \text{if} \hspace{0,2cm} \mathcal L^N(G - E)=0,\\ +\infty \hspace{2,1cm} \text{otherwise}. \end{cases} \]

show that $F$ admits minumum.

Attempt: We want to use the Direct Method. We must chose the right topology on $X$. It is probably the topology induced by $L^1_{Loc}$. If $\chi_{E_n} \to \chi_{E}$ in $L^1_{Loc}$ we get $\mathcal L^N(G - E) \leq \mathcal L^N(G - E_n) + \mathcal L^N(E_n - E) = 0 + \epsilon = \epsilon$ thus the subset where $F(E)=Per(E)$ is closed for sequences. We need to show that $F\not\equiv +\infty$, $F$ is coercive, $F$ is lower semi-continuous for sequences. Pick a ball containing $G$, we get $F\not\equiv +\infty$. $Per(*)$ is lower semi-continuous thus so is $F$ thanks to the previous observation. I have some problems showing $F$ is coercive, because when $F(E) \leq \lambda$ of course $Per(E)\leq \lambda$, but in order to use the Compactness Theorem we need also $|\chi_E|_1\leq c$ but that could not be afforded I guess, because we have no control on the measure of a set covering $G$ with perimeter $\leq \lambda$.

Thanks.

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2 Answers 2

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The minimizer of $F$ is $L^N$-equivalent to the minimal hull containing the bounded obstacle $G$. Then the minimizer exists due to standard compactness theorem. This one-sided minimizer is also called pesudoconvex set. More references about minimal hull and pseudoconvex set include the papers by M. Miranda(1971), N. Fusco(2004) and E. Barozzi(2009).

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The isoperimetric inequality should do it for you: $$Per(E)\geq C_N \mathcal{L}^N(E)^{\frac{N-1}{N}}$$ where $C_N$ depends only on $N$.

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