Mumford–Shah functional

Example

Let $\Omega \subseteq \R^2$ be a bounded open set, the image domain. Let $g \in L^\infty(\Omega)$ be a grey-level image, and let $\alpha, \beta > 0$ be parameters. For a relatively closed set $K \subseteq \Omega$ and a function $u \in C^1(\Omega \setminus K)$, the Mumford–Shah functional is $$E(u, K) = \int_{\Omega \setminus K} \abs{\nabla u}^2 \, dx + \alpha \int_{\Omega \setminus K} (u - g)^2 \, dx + \beta \, \mathcal{H}^1(K).$$ Here $\mathcal{H}^1$ is the one-dimensional Hausdorff measure, the length of $K$. A minimizing pair $(u, K)$ gives a smoothed image $u$ and a set of edges $K$.

Check on a simple image: let $D$ be a closed disc of radius $r$ in $\Omega$, and let $g = \mathbf{1}_D$. The pair $u = g$, $K = \partial D$ has no gradient term and no fidelity term, so $E(u, K) = 2 \pi \beta r$. The pair $u = 0$, $K = \emptyset$ has energy $\alpha \pi r^2$. So for $\alpha > 2\beta/r$ the first pair is cheaper than the second, and for $\alpha < 2\beta/r$ the second is cheaper. Large weights on fidelity favour keeping the edge, and small weights favour erasing it.