= Solution
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 $2\pi\alpha R|c|+\tfrac12\pi R^2(c-1)^2$, whose minimizer is $c=(1-2\alpha/R)_+$. To prove global optimality, a <total variation calibration> is needed.
Define the bounded radial <vector field>
$$
z(x)=\begin{cases}x/R,&|x|\le R,\\Rx/|x|^2,&|x|>R.\end{cases}
$$
It satisfies $|z|\le1$ and has a <continuous> normal component across $\partial B$. Its <distributional divergence> therefore has no boundary measure, and direct differentiation yields $q=\operatorname{div}z=(2/R)\chi_B\in L^2$. 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 $|z|\le1$. The dual definition therefore gives $\langle q,v\rangle\le J(v)$ for every finite-penalty $v$, and trivially for all other $v$.
Moreover $\langle q,\chi_B\rangle=(2/R)\pi R^2=2\pi R=J(\chi_B)$. 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 proves
$$
\boxed{u(x)=\left(1-\frac{2\alpha}{R}\right)_+\chi_{B(0,R)}(x).}
$$
The 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>.
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