 Solution

Assume $g\in L^2(\Omega )$, as required for the finite-valued data fidelity; extend the total variation seminorm on a domain by $+\infty$ outside $BV(\Omega )$. The zero function is a finite-energy competitor. A minimizing sequence $f_n$ has bounded $\Vert f_n-g{\Vert }_2$ and bounded variation, hence bounded $L^2$ norm. Take a weakly convergent subsequence in $L^2$, with limit $f$. For each fixed test vector field $\zeta$, the pairing $\int f\mathrm{div}\zeta$ is weakly continuous because $\mathrm{div}\zeta \in L^2$. Its supremum has weak lower semicontinuity. The squared Hilbert space norm also has weak lower semicontinuity. Thus the direct method in the calculus of variations gives a minimizer in $BV\cap L^2$.

The variation term is convex, and the squared fidelity term is strictly convex. Their sum is strictly convex on its effective domain, so the minimizer is unique.

Let $T(t)=\min (b,\max (a,t))$. The bounded-variation contraction under clipping gives $|D(T\circ f)|(\Omega )\le |Df|(\Omega )$. If $g(x)\in [a,b]$, then $|T(f(x))-g(x)|\le |f(x)-g(x)|$, strictly whenever $f(x)\notin [a,b]$. A positive-measure set outside the interval would therefore strictly decrease the fidelity integral, while not increasing variation, contradicting minimality. HenceNo compactness theorem for the $BV$ embedding is required for this weak-$L^2$ argument.