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!
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.