= Solution
Write $\operatorname{TV}(u)=|Du|(\Omega)$ for the <total variation seminorm on a domain>. For a <locally integrable> real function its dual definition is
$$
|Du|(\Omega)=\sup\left\{\int_\Omega u\,\operatorname{div}\varphi\,dx:\varphi\in C_c^1(\Omega;\mathbb R^2),\ |\varphi(x)|\le1\right\}.
$$
The <bounded-variation space> consists of functions $u\in L^1(\Omega)$ with finite $|Du|(\Omega)$, equipped with $\|u\|_{BV}=\|u\|_1+|Du|(\Omega)$. 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 $\varepsilon>0$, choose $N$ so that $\|u_n-u_m\|_{BV}<\varepsilon$ whenever $m,n\ge N$. For fixed $n\ge N$, the given <lower semicontinuity> yields
$$
|D(u_n-u)|(\Omega)\le\liminf_{m\to\infty}|D(u_n-u_m)|(\Omega).
$$
Since $\|u_n-u_m\|_1\to\|u_n-u\|_1$, we obtain $\|u_n-u\|_1+|D(u_n-u)|(\Omega)\le\varepsilon$. 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> is
$$
\boxed{u_*=(1-2\alpha/R)_+\chi_B.}
$$
Here $(s)_+=\max(s,0)$ is the <positive part>. A <total variation calibration> certifies global optimality, including competitors that are not radial or piecewise constant. Define the bounded <vector field>
$$
z(x)=\begin{cases}x/R,&|x|\le R,\\Rx/|x|^2,&|x|>R.\end{cases}
$$
It has $|z|\le1$. Its normal component is <continuous> across the circle, so the <distributional divergence> has no extra boundary measure. Direct differentiation gives $q=\operatorname{div}z=(2/R)\chi_B$. The dual definition implies $\int vq\le\operatorname{TV}(v)$: for the standard $BV\cap L^2$ domain one can cut $z$ off at radius $L$, with the error bounded by $C L^{-1}\int_{L<|x|<2L}|v|\to0$, and then smooth the test field. In the larger <homogeneous bounded-variation space>, the same error is bounded by $C\|v\|_{L^2(L<|x|<2L)}\to0$. Thus both usual whole-plane formulations give the same certificate.
Use the <perimeter> identity $\operatorname{TV}(\chi_B)=2\pi R$ and $|B|=\pi R^2$. If $c=1-2\alpha/R>0$, then $\int c\chi_Bq=2\pi Rc=\operatorname{TV}(c\chi_B)$ and $u_*-g+\alpha q=0$. Consequently every competitor $v$ satisfies
$$
\begin{aligned}
E(v)-E(u_*)&\ge\alpha\langle q,v-u_*\rangle+\langle u_*-g,v-u_*\rangle+\tfrac12\|v-u_*\|_2^2\\
&=\tfrac12\|v-u_*\|_2^2\ge0.
\end{aligned}
$$
If $\alpha\ge R/2$, 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 $u_*=0$. 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 $g$.
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