Relaxed graph-area functional ($A(u)=\int \sqrt{1+|\nabla u{|}^2}dx+|D^{s}u|(\Omega )$)

For a BV space function with $Du=\nabla udx+D^{s}u$, relaxed graph area is $A(u)=\int \sqrt{1+|\nabla u{|}^2}dx+|D^{s}u|(\Omega )$. Its dual supremum uses the coupled constraint ${\phi }_0^2+|\phi {|}^2\le 1$ in the pairing $\int ({\phi }_0+u\mathrm{div}\phi )$. It is convex and lower semicontinuous in $L^1$. Independent bounds on the two test fields define a different functional.

For an assumed $W^{1,1}$ minimizer of $\alpha A(u)+\frac{1}{2}\Vert u-g{\Vert }_2^2$, compactly supported variations give $u-g-\alpha \mathrm{div}(\nabla u/\sqrt{1+|\nabla u{|}^2})=0$ in distributions. If the regularizer is instead the total variation seminorm, a bounded calibration field replaces the quotient and handles zero gradients.

The independent constraints $|{\phi }_0|\le 1$ and $|\phi |\le 1$ separate the dual supremum into $|\Omega |+|Du|(\Omega )$. They do not define square-root graph area. For $|\nabla u|=1$ the box-constrained value is $2|\Omega |$, while the relaxed graph-area functional is $\sqrt{2}|\Omega |$.