Bounded-variation contraction under clipping

Let $T(t)=\min (b,\max (a,t))$, $a\le b$, and let $u\in BV(\Omega )$. The variation contractionfollows from the one-Lipschitz scalar composition rule, by smooth approximation and lower semicontinuity. To conclude global membership $T\circ u\in BV(\Omega )$, one must also retain its L1 norm.

That integrability is automatic when $\Omega$ has finite Lebesgue measure. It also holds on an arbitrary domain when $T(0)=0$, equivalently $0\in [a,b]$, since $|T(u)|\le |u|$. Without either condition it can fail: $u=0$ on $\mathbb{R}$ belongs to BV space, but clipping it to $[1,2]$ gives the nonintegrable constant one. Its derivative variation is zero, which does not repair the missing global integrability.

For total variation denoising on a bounded domain with data in $[a,b]$, clipping cannot increase either the variation penalty or the pointwise squared error. Uniqueness of the minimizer therefore forces range preservation. The variation inequality and this integrability qualification describe different parts of the argument.

For a scalar BV space function, clipping to $[-M,M]$ partitions its coarea formula for BV functions into middle and tail levels. The clipped function has the original superlevel sets at heights in $(-M,M)$. The positive tail levels of $u-T_{M}u$ correspond to original heights above $M$, and its negative tail levels to heights below $-M$. Thus their total variation seminorms add exactly. This is stronger than the triangle inequality and is specific to scalar monotone clipping. It supports residual-preserving clipping of an ROF minimizer.