Piecewise-constant Mumford–Shah problem (${\min }_{\nabla u=0}\{\Vert u-g{\Vert }_2^2+\beta {\mathcal{H}}^1(J_u)\}$)

The infinite gradient-weight limit restricts to SBV space image signals with $\nabla u=0$ almost everywhere. Their energy is fidelity plus jump length, represented by constants on a Caccioppoli partition. Internal perimeter is counted once by $\frac{1}{2}{\sum }_i\mathrm{Per}(E_i;\Omega )$; adjacent equal-valued regions can be merged.

Suppose two constant data intensities differ by $d$, have areas $a$ and $A-a$, and share an internal boundary of length $L$. Keeping the boundary costs $\beta L$ with zero fidelity; merging the two regions costs $a(A-a)d^2/A$ after fitting their common mean. The displayed inequality makes splitting cheaper among these two candidate geometries. This comparison does not prove either geometry is globally optimal among all partitions.

When grey values are fixed but region labels vary, a smooth interface satisfies $\beta \kappa =(c_j-g{)}^2-(c_i-g{)}^2$. The curvature uses the normal pointing out of region $i$, positive on an outward-oriented circle. It balances the change in data cost against the first variation of interface length. Fixing a full spatial image signal instead fixes its jumps and leaves no such boundary-relocation freedom.

At a freely moving regular junction of three smooth image edges with equal isotropic length weights, the endpoint first variation requires the sum of their outward unit tangent vectors to vanish. Three unit vectors with zero sum have pairwise angles of $120$ degrees. Different interface weights instead give a weighted force balance, and constrained or singular junctions require separate hypotheses.

For a fixed positive-area region $E_i$, its least-squares grey value is $c_i=|E_i{|}^{-1}{\int }_{E_i}g$. The minimum fidelity equals $\int g^2-{\sum }_i({\int }_{E_i}g{)}^2/|E_i|$. This eliminates the grey values before optimizing the partition geometry.