Total variation denoising

Example

Let $\Omega \subseteq \R^n$ be a bounded open set, let $f \in L^2(\Omega)$ be a noisy image, and let $\lambda > 0$. For $u \in L^2(\Omega)$ let $$TV(u) = \sup \Big\{ \int_\Omega u \operatorname{div} \varphi \, dx : \varphi \in C^1_c(\Omega; \R^n), \ \abs{\varphi(x)} \le 1 \text{ for all } x \Big\},$$ the total variation of $u$, which equals $\int_\Omega \abs{\nabla u} \, dx$ for smooth $u$. Total variation denoising, in the model of Rudin, Osher and Fatemi, replaces $f$ by the minimizer of $$E(u) = TV(u) + \frac{\lambda}{2} \int_\Omega (u - f)^2 \, dx, \qquad u \in L^2(\Omega).$$ The functional $E$ is lower semicontinuous, coercive and strictly convex on the Lp space $L^2(\Omega)$, so it has exactly one minimizer.

A worked case on $\Omega = (0, 1)$: let $h > 0$ and $f = h$ on $(0, \tfrac12)$, $f = 0$ on $(\tfrac12, 1)$, a single edge of height $h$. Assume $\lambda \ge 4/h$ and set $\delta = 2/\lambda$. The minimizer is $u = h - \delta$ on $(0, \tfrac12)$ and $u = \delta$ on $(\tfrac12, 1)$.

Check: let $w(x) = 2x$ on $[0, \tfrac12]$ and $w(x) = 2 - 2x$ on $[\tfrac12, 1]$. Then $\abs{w} \le 1$, $w(0) = w(1) = 0$, $w' = \lambda (f - u)$, and $\int_0^1 u w' \, dx = h - 2\delta = TV(u)$. For every $v$ one has $TV(v) \ge \int_0^1 v w' \, dx$, and expanding the squares gives $E(v) - E(u) \ge \frac{\lambda}{2} \int_0^1 (v - u)^2 \, dx \ge 0$.