= 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 $\|f_n-g\|_2$ 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 $\int f\operatorname{div}\zeta$ is weakly continuous because $\operatorname{div}\zeta\in L^2$. 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. Hence
$$
\boxed{\text{a unique }f_*\in BV(\Omega)\cap L^2(\Omega)\text{ exists, and }a\le f_*\le b\text{ a.e.}}
$$
No compactness theorem for the $BV$ embedding is required for this weak-$L^2$ argument.
Back to article page
