I'm stuck with this exercise:
Let $\Omega$ be bounded (open) Lipschitz domain; let also $ g \in L^1(\Omega)$ be fixed and consider the functional $I: BV(\Omega) \to \mathbb{R}$ defined by \begin{equation} I(f) := \vert Df \vert (\Omega) + \int_{\Omega} \sqrt{1+(f-g)^2} d\mathscr{L}^n \end{equation} Prove thath $I$ admits a minimizer in $BV(\Omega)$.
Here $\vert Df \vert(\Omega)$ is the total variation of the measure $Df$.
My idea is to consider a minimizing sequence and then I want to use the compactness. But I don't have details in order to achive this plan. Can you formalize my idea?
In particular, let me call $F = \{ I(f) \in \mathbb{R} : f\in BV(\Omega) \} \subset \mathbb{R}$. Since $\mathbb{R}$ is complete then there exists an infimum $a \geq 0$ ($I(f) \geq 0$). By properties of infimum we can consider a minimizing seqeunece $F_k \in BV(\Omega)$ such that $I(F_k) \to a$. By compactness theorem there exists $F \in BV(\Omega)$ such that $F_{k_j} \to F$ in $L^1(\Omega)$. How can I show that such $F$ is the minimizer?