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The distributional derivative of this indicator function is $e_1{\mathcal{H}}^1$ on the vertical segment $x_1=1/2$, so its total variation seminorm on a domain is one. This is singular with respect to planar Lebesgue measure, whereas Sobolev space weak derivatives in $W^{1,1}$ have integrable densities. Its bounded-variation space norm is $1/2+1=3/2$.

For $1<a\le 2$, the radial function $f(x)=(1+|x|{)}^{-a}$ belongs to $L^2({\mathbb{R}}^2)$ and has finite distributional total variation seminorm on a domain, but is not in $L^1({\mathbb{R}}^2)$. Cutting it off outside radius $T$ gives bounded-variation space functions tending to $f$ in $L^2$, with variation tending to that of $f$: the added jump costs $2\pi T(1+T{)}^{-a}\to 0$. Thus the penalty equal to TV on $BV\cap L^2$ and infinity elsewhere is not sequentially lower semicontinuous in $L^2$. The closed homogeneous bounded-variation space extension avoids this domain issue.

In the $L^2$ formulation of total variation denoising on the entire plane, one may allow finite distributional total variation seminorm on a domain without requiring global $L^1$ membership. Its extended penalty is a supremum of continuous test-function pairings and hence is sequentially lower semicontinuous. This domain can be larger than $BV({\mathbb{R}}^2)\cap L^2$, since the usual bounded-variation space includes an $L^1$ condition.

On a bounded domain of positive measure, the constant functions $u_k\equiv k$ satisfy $\mathrm{TV}(u_k)=0$ while $\Vert u_k{\Vert }_{BV}=k|\Omega |\to \infty$. Thus the total variation seminorm on a domain is neither coercive in the $BV$ norm nor strictly convex. Additional control of the mean can remove this obstruction; the Poincaré inequality for total variation supplies the needed bound on suitable connected domains.

Write $\mathrm{TV}(u)=|Du|(\Omega )$ for the total variation seminorm on a domain. For a locally integrable real function its dual definition isThe bounded-variation space consists of functions $u\in L^1(\Omega )$ with finite $|Du|(\Omega )$, equipped with $\Vert u{\Vert }_{BV}=\Vert u{\Vert }_1+|Du|(\Omega )$. In the equivalent distributional derivative description, $Du$ is a finite vector-valued Radon measure and $|Du|$ is its total variation measure.

To establish completeness of the bounded-variation space, let $(u_n)$ be Cauchy in this norm. Completeness of $L^1$ gives $u_n\to u$ in $L^1$. Given $\epsilon >0$, choose $N$ so that $\Vert u_n-u_m{\Vert }_{BV}<\epsilon$ whenever $m,n\ge N$. For fixed $n\ge N$, the given lower semicontinuity yieldsSince $\Vert u_n-u_m{\Vert }_1\to \Vert u_n-u{\Vert }_1$, we obtain $\Vert u_n-u{\Vert }_1+|D(u_n-u)|(\Omega )\le \epsilon$. In particular $u_n-u$ has finite variation; the triangle inequality then puts $u$ in the BV space. The same bound proves convergence in the full norm, so this is a Banach space.

For the disk data, put $B=B(0,R)$ and assume $\alpha >0$. The exact total variation denoising of a disk isHere $(s{)}_{+}=\max (s,0)$ is the positive part. A total variation calibration certifies global optimality, including competitors that are not radial or piecewise constant. Define the bounded vector fieldIt has $|z|\le 1$. Its normal component is continuous across the circle, so the distributional divergence has no extra boundary measure. Direct differentiation gives $q=\mathrm{div}z=(2/R){\chi }_B$. The dual definition implies $\int vq\le \mathrm{TV}(v)$: for the standard $BV\cap L^2$ domain one can cut $z$ off at radius $L$, with the error bounded by $C{L}^{-1}{\int }_{L<|x|<2L}|v|\to 0$, and then smooth the test field. In the larger homogeneous bounded-variation space, the same error is bounded by $C\Vert v{\Vert }_{L^2(L<|x|<2L)}\to 0$. Thus both usual whole-plane formulations give the same certificate.

Use the perimeter identity $\mathrm{TV}({\chi }_B)=2\pi R$ and $|B|=\pi R^2$. If $c=1-2\alpha /R>0$, then $\int c{\chi }_{B}q=2\pi Rc=\mathrm{TV}(c{\chi }_B)$ and $u_{\ast }-g+\alpha q=0$. Consequently every competitor $v$ satisfiesIf $\alpha \ge R/2$, replace $z$ by $z_{\alpha }=(R/(2\alpha ))z$. Its norm is still at most one, its divergence is $q_{\alpha }={\chi }_B/\alpha$, and equality in the calibration holds at $u_{\ast }=0$. The identical comparison proves optimality and uniqueness of zero, including the threshold $\alpha =R/2$. The only general results used are completeness of $L^1$, lower semicontinuity of variation, the indicator-perimeter identity, the distributional integration-by-parts/dual variation formula and the quadratic norm identity. For $\alpha =0$, the unique squared-error minimizer is simply $g$.

Use a bounded BV extension operator for the bounded Lipschitz domain. Extend $u_k$ to $w_k\in BV({\mathbb{R}}^n)$, supported in one fixed compact set, with $\Vert w_k{\Vert }_{BV({\mathbb{R}}^n)}\le C_{\Omega }\Vert u_k{\Vert }_{BV(\Omega )}$. Such an extension is obtained by reflection in Lipschitz boundary charts, a partition of unity and a fixed cutoff function. A naive zero extension must also charge its boundary-trace jump; it cannot discard that contribution.

Let $M$ bound these norms and let $w_{k,ϵ}=w_k\ast {\rho }_{ϵ}$ be their mollifications. The BV mollification error estimate givesThe variation measure here is on ${\mathbb{R}}^n$. The printed lemma's $|Dw|(\Omega )$ needs this correction unless it also assumes all the derivative mass lies in $\Omega$: a nonconstant bump supported outside $\overline{\Omega }$ disproves its literal wording. The correct estimate follows by averaging the BV translation estimate $\Vert w(\cdot -h)-w{\Vert }_1\le |h||Dw|({\mathbb{R}}^n)$ over a mollifier supported in $|h|\le ϵ$.

For each fixed $ϵ>0$, convolution gives uniform boundsThe mollifications have common compact support and are uniformly equicontinuous. The Arzelà-Ascoli theorem supplies a uniformly convergent subsequence at each scale $ϵ=1/m$. A diagonal subsequence converges at every one of those scales. For two late members of that subsequence,First choose large $m$, then late $k,\ell$; the sequence is a Cauchy sequence in $L^1$. Its limit $w$ belongs to the BV space because the total variation seminorm on a domain is sequentially lower semicontinuous. Uniform derivative bounds and integration by parts give $D{w}_k\stackrel{\ast }{\rightharpoonup }Dw$. Restriction to $\Omega$ proves the required weak-star convergence in BV.

For real $u\in L^1(\Omega )$, its total variation seminorm on a domain isThe vector fields have compact support inside $\Omega$, so no boundary term is charged. Taking both $\xi$ and $-\xi$ makes this equivalent to a supremum of absolute pairings. When finite, the distributional derivative $Du$ is a finite vector-valued Radon measure, and the displayed supremum is its total variation norm of a measure. For a smooth function it equals ${\int }_{\Omega }|\nabla u{|}_{2}dx$.

The function of bounded variation on a domain belongs to the Banach spaceThe L1 norm is needed because variation alone vanishes on constant functions and is only a seminorm. No smoothness of $u$ is part of this definition.

Let $B=B(0,R)$, with $R>0$. Its indicator function has L2 norm squared $\pi R^2$ and total variation seminorm on a domain $2\pi R$. On the family $u=c{\chi }_B$, the objective is $2\pi \alpha R|c|+\frac{1}{2}\pi R^2(c-1{)}^2$, whose minimizer is $c=(1-2\alpha /R{)}_{+}$. To prove global optimality, a total variation calibration is needed.

Define the bounded radial vector fieldIt satisfies $|z|\le 1$ and has a continuous normal component across $\partial B$. Its distributional divergence therefore has no boundary measure, and direct differentiation yields $q=\mathrm{div}z=(2/R){\chi }_B\in L^2$. Although $z$ is not compactly supported, multiply it by a smooth radial cutoff equal to one through radius $T$ and zero beyond $2T$. The extra divergence has magnitude $O(R/T^2)$ on an annulus of area $O(T^2)$, hence L2 norm $O(R/T)$. Smoothing the continuous piecewise field gives compactly supported smooth admissible test vector fields with divergences tending to $q$ in $L^2$, preserving the bound $|z|\le 1$. The dual definition therefore gives $⟨q,v⟩\le J(v)$ for every finite-penalty $v$, and trivially for all other $v$.

Moreover $⟨q,{\chi }_B⟩=(2/R)\pi R^2=2\pi R=J({\chi }_B)$. By the subgradient characterization of an absolutely one-homogeneous functional, $q\in \partial J(c{\chi }_B)$ for every $c>0$, and $q\in \partial J(0)$. If $0<\alpha <R/2$, choose $c=1-2\alpha /R$; then $(g-c{\chi }_B)/\alpha =q$. If $\alpha \ge R/2$, choose $u=0$; then $g/\alpha =(R/(2\alpha ))q$ belongs to $\partial J(0)$, since multiplying the dual bound by a number in $[0,1]$ preserves it. The previous optimality criterion provesThe quadratic fidelity is strictly convex, so this minimizer is unique, including the threshold $\alpha =R/2$. The disk retains its radius on the positive branch and disappears on the zero branch; this is total variation denoising of a disk.

The total variation seminorm on a domain isThe class is symmetric under negation, so no additional absolute value is needed around the integral. The bounded-variation space isEquivalently, the distributional derivative is a finite vector-valued Radon measure, and the displayed quantity is its total mass in variation. For $f\in W^{1,1}(\Omega )$, it equals ${\int }_{\Omega }|\nabla f{|}_{2}dx$. Thus jumps are allowed, with their interface contributions measured by variation.

Assume $g\in L^2(\Omega )$, as required for the finite-valued data fidelity; extend the total variation seminorm on a domain by $+\infty$ outside $BV(\Omega )$. The zero function is a finite-energy competitor. A minimizing sequence $f_n$ has bounded $\Vert f_n-g{\Vert }_2$ and bounded variation, hence bounded $L^2$ norm. Take a weakly convergent subsequence in $L^2$, with limit $f$. For each fixed test vector field $\zeta$, the pairing $\int f\mathrm{div}\zeta$ is weakly continuous because $\mathrm{div}\zeta \in L^2$. Its supremum has weak lower semicontinuity. The squared Hilbert space norm also has weak lower semicontinuity. Thus the direct method in the calculus of variations gives a minimizer in $BV\cap L^2$.

The variation term is convex, and the squared fidelity term is strictly convex. Their sum is strictly convex on its effective domain, so the minimizer is unique.

Let $T(t)=\min (b,\max (a,t))$. The bounded-variation contraction under clipping gives $|D(T\circ f)|(\Omega )\le |Df|(\Omega )$. If $g(x)\in [a,b]$, then $|T(f(x))-g(x)|\le |f(x)-g(x)|$, strictly whenever $f(x)\notin [a,b]$. A positive-measure set outside the interval would therefore strictly decrease the fidelity integral, while not increasing variation, contradicting minimality. HenceNo compactness theorem for the $BV$ embedding is required for this weak-$L^2$ argument.

For $u\in L^1(\Omega )$, the total variation seminorm on a domain isUsing $C_c^1$ instead gives the same definition. The zero test field shows nonnegativity, and $\mathrm{TV}(0)=0$, so this is a proper extended-real function. Each test field defines a linear functional of $u$; taking their supremum proves convexity, equivalentlyIt is not strictly convex: two distinct constant functions both have zero total variation, as does every convex combination of them.

The function of bounded variation on a domain space isTotal variation is not coercive on $BV(\Omega )$: for $u_k\equiv k$ on a domain of positive measure, $\Vert u_k{\Vert }_{BV}=k|\Omega |\to \infty$, whereas $\mathrm{TV}(u_k)=0$. This is noncoercivity of total variation on constants; a mean-zero constraint can remove the constant obstruction.

Use the interior-domain definition of total variation seminorm on a domain, so there is no boundary-extension term at $\pm 1$. For any admissible scalar test function $\phi$,Thus $\mathrm{TV}(u_R)\le 2$.

Choose $0<\epsilon <\min \{R,1-R\}$. From the supplied smooth bump function, defineThe two supports are disjoint and inside $(-1,1)$. Each bump has maximum $e^{-1}$, so $\Vert \phi {\Vert }_{\infty }\le 1$, while $\phi (R)=1$ and $\phi (-R)=-1$. This attains the upper bound, provingEquivalently, its distributional derivative is $D{u}_R={\delta }_{-R}-{\delta }_R$, with total variation two. This is total variation of an interval indicator.

For $u\in L^1(\Omega )$, its total variation seminorm on a domain isThe corresponding bounded-variation space and its zero-mean subspace areIf $u_{\Omega }=|\Omega {|}^{-1}{\int }_{\Omega }udx$, the Poincaré inequality for total variation isIn particular, $\Vert u{\Vert }_{L^1}\le C_{\Omega }\mathrm{TV}(u)$ for $u\in B{V}_0(\Omega )$.

A proper convex function is both a convex function and a proper extended-real function. For example, the total variation seminorm on a domain is nonnegative, finite at zero, and a supremum of linear functionals, making it proper and convex.

Staircasing is the formation of approximately constant plateaux in reconstructed data, even when the original data vary smoothly. The linear growth of the total variation seminorm on a domain favors sparse spatial gradients, so preserving jumps can come at the cost of flattening gradual transitions.

Total variation measures accumulated absolute change rather than just the difference between two endpoint values. For a real or complex sequence $(a_n)$, its total variation isallowing $+\infty$. The triangle inequality gives $|a_m-a_n|\le {\sum }_{k=n}^{m-1}|a_{k+1}-a_k|$. Finite variation therefore makes the sequence Cauchy and hence convergent in $\mathbb{R}$ or $\mathbb{C}$.

The converse fails. The sequence $a_n=(-1{)}^n/n$ tends to zero, but $|a_{n+1}-a_n|=1/n+1/(n+1)$, whose sum diverges. Small late values can still accumulate infinitely much back-and-forth movement.

For a function on an interval, the total variation of a function replaces the sum over consecutive indices by a supremum over all partitions of an interval. For spatial fields, the total variation seminorm on a domain uses the distributional derivative as a measure. Total generalized variation extends that regularization idea to higher-order structure. These settings share cumulative-change control, while their definitions specify different index or spatial domains.

Total variation flow is the $L^2$ gradient flow of the total variation seminorm on a domain. Formally it is $u_t=\mathrm{div}(\nabla u/|\nabla u|)$, but the subgradient formulation handles zero gradients and jumps. Adding quadratic fidelity gives a strictly convex stationary total variation denoising problem.

Variational regularization reconstructs an unknown by balancing data fit with a penalty expressing desired structure. A standard model isThe linear operator $A$ predicts data, $f^{\delta }$ denotes observations, and the extended-real functional $J$ can impose smoothness, sparsity or hard constraints. This construction includes nonlinear reconstruction maps even when $A$ itself is linear.^[ref]

For $J(u)=\Vert u{\Vert }^2$, the model is quadratic Tikhonov regularization. For $J(u)=\mathrm{TV}(u)$, the penalty controls the total variation seminorm on a domain and can retain jumps without demanding a square-integrable classical gradient. Adding the indicator functional of a constraint set enforces constraints such as nonnegativity.

Existence, uniqueness and stability are separate conclusions. A direct method in the calculus of variations requires suitable compactness of bounded-energy sets and lower semicontinuity; convexity alone does not supply compactness. A unique minimizer at fixed $\alpha$ also does not prove convergence to the true object as noise vanishes: the regularization parameter choice must balance bias and noise.