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I've got the following problem in Calculus of Variations, about perimeters with density.

Let $g \in C^1(\mathbb{R})$ be even and strictly convex. Define the perimeter with density of a measurable subset $E$ of $\mathbb{R}$ by: $$ P(E)=\sup \left\{ \int_E (\varphi(x)e^{g(x)})' dx: \varphi \in C^1_c(\mathbb{R}), \| \varphi\|_\infty\leq1 \right\} $$
and the volume with density of $E$ by: $$ V(E)=\int_Ee^{g(x)} dx . $$ I have to study the problem: $$ \min\{P(E) : E \subseteq \mathbb{R} \text{ measurable with } V(E)=1 \}. $$ Now, I was able to prove that the following problem: $$ \min\{P(E) : E \subseteq \mathbb{R} \text{ interval with } V(E)=1 \}. $$ admits a unique minimizer which is a $0$-symmetric interval (this problem actually boils down to a one-variable minimization problem). My question is how to deal with the general case; my idea was to prove that the minimizing interval is actually a global minimizer, but I don't even know if this is the case. I thought this because maybe for this notion of perimeter still holds the structure theorem for which a finite perimeter set in $\mathbb{R}$ is equivalent to a finite union of intervals; but I'm not even sure of that. Does anyone have suggestions, references, ideas on how to do this? Thank you!

18cyclotomic
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  • I throw an idea, I am not even sure that it works: with the "standard" notion of perimeter one generally starts from a minimizing sequence in order to use the compactness theorem for $BV$ functions (cfr. Theorem 3.23 in L. Ambrosio, N. Fusco, D. Pallara - Functions of Bounded Variation and Free Discontinuity Problems); maybe you can, mutatis mutandis, adapt the idea (trying to understand if a similar compactness theorem still holds)... – gangrene May 02 '17 at 17:20
  • Actually I need to prove that the minimizing interval is a global minimizer (a.e.) so I need to prove that sets of finite perimeter are a.e. a finite union of intervals (from which I can conclude) also with this notion of "weighed" perimeter; I'm searching for a proof of this. – 18cyclotomic May 02 '17 at 17:52
  • Well but it doesn't seem (at least to me) obvious that the interval which solves $$ \min { P(E) , : , E \subseteq \mathbb{R} \text{ interval with } V(E)=1 }$$ is still a solution of $$ \min { P(E) , : , E \subseteq \mathbb{R} \text{ measurable with } V(E)=1 }.$$As long as we know the latter problem may not even have a solution, from which my suggestion: try to show that at least a solution exists (and again, nothing we know about the possible uniqueness of the possible solution)... – gangrene May 02 '17 at 18:01
  • The point is, if it is true that any set with finite perimeter is a.e a finite union of intervals, then yes, the minimizing interval is a global minimizer. – 18cyclotomic May 02 '17 at 18:12

1 Answers1

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Claim: The problem $$\min\{P(E) : E \subseteq \mathbb{R} \text{ measurable with } V(E)=1 \}\tag{1}$$ is equivalent to $$\min\{P(E) : E \subseteq \mathbb{R} \text{ measurable and bounded with } V(E)\ge 1 \}\tag{2}$$ For one thing, an unbounded set $E$ with finite volume have infinite perimeter, as its boundary $\partial E$ must contain a sequence of points diverging to infinity. So the restriction of boundedness is nothing new.

Also, if $E$ is a bounded set with $V(E)>1$, then the rescaled sets $\lambda E=\{\lambda x:x\in E\}$, $0<\lambda\le 1$, have volume that decreases to $0$ as $\lambda\to 0$, and is a continuous function of $0$. Thus there exists $\lambda_0\in (0,1)$ such that $V(\lambda_0E)=1$. Since the perimeter also decreases under rescaling, we have $P(\lambda_0E)<P(E)$. The claim is proved.

Given any measurable bounded set $E$ as in (2), let $I$ be its convex hull, i.e., the smallest interval containing $E$. Since $E\subset I$ and $\partial I\subset \partial E$, it follows that (2) is equivalent to $$\min\{P(E) : E \subseteq \mathbb{R} \text{ is a bounded interval with } V(E)\ge 1 \}\tag{3}$$ Finally, the aformentioned rescaling argument shows that (3) is equivalent to
$$\min\{P(E) : E \subseteq \mathbb{R} \text{ is an interval with } V(E) = 1 \}\tag{4}$$

Remark: when the volume constraint is replaced by inequality, it is important to insist that the set be bounded; otherwise $\mathbb R$ itself would be the minimizer, having empty boundary.

  • Thank you. Do you see any way to prove that if E has finite perimeter in this sense, then it actually has finite perimeter in the traditional sense? – 18cyclotomic May 12 '17 at 12:50