The planar Mumford–Shah regularity conjecture and local weak-L⁴ gradient bounds

The explanation

Draw a crack along the negative x-axis in the plane, ending at the origin. Away from the crack, let u = √r sin(θ/2), where r is the distance to the origin and θ is the angle. This function jumps across the crack and is smooth elsewhere. Its slope has size 1/(2√r), which blows up at the tip. The slope is bigger than t only where r < 1/(4t²). For t = 10 that is r < 0.0025, a disc of area about 0.0000196. In general that area is π/(16t⁴), so it shrinks like t⁻⁴. Mathematicians call this weak L⁴. This crack tip is one of the shapes the theory expects.

The Mumford–Shah energy scores a pair: a function u and a closed set K where u may jump. It adds the squared slope of u off K, the length of K, and the squared distance from given data gSource 1. Mumford and Shah proposed it to split an image into regionsSource 2. Their conjecture predicts that, inside the domain, K is made of smooth arcs with only free ends and triple junctionsSource 3. Bonnet, David, Ambrosio, Fusco and Pallara built most of the regularity theory. After their classification results, one obstacle remained for blow-up limits: a possible bounded piece of the limiting edge setSource 4.

The paper claims the interior part of the conjecture for reduced absolute minimizers with bounded data. Near every point of K, the set is a smooth arc through the point, an arc ending there, or three arcs meeting at 120 degreesSource 5. Only finitely many connected pieces of K meet any region kept away from the boundarySource 6. It also deduces that the slope of u is weak L⁴ near each interior point, matching the crack tip aboveSource 7.

The key step rules out the bounded piece. Moving that piece sideways keeps its length the sameSource 8. An interaction energy then depends on the shift in a harmonic way, and minimality forces it to be constant. This makes the remaining fields zero, so deleting the piece saves length, a contradictionSource 9. A recent preprint by Deangelis also claims the classification and the interior structure, by a different argumentSource 10.

How sure is this? There is no Lean check. A computer proof checker has not looked at any part of it. The proof relies on a large body of earlier regularity theory, used as stated in the literature. No specialist review of the paper has been published.

What it doesn't show. This does not cover what happens at the boundary of the domain, or dimensions above two. It has no Lean check, and a separate recent preprint by Deangelis claims a proof by another route.