= Solution
Work in the real <Hilbert space> $L^2(\mathbb R^2)$, with $g\in L^2$ and $\alpha>0$. The penalty $J$ is a <proper convex function>: it is finite at zero, and the <bounded-variation space> domain is <convex>. Its <subdifferential> at a finite-penalty $u$ consists of $q\in L^2$ satisfying $J(v)\ge J(u)+\langle q,v-u\rangle$ for every $v\in L^2$.
If $q=(g-u)/\alpha\in\partial J(u)$, expand the quadratic term and use the <subgradient inequality>:
$$
\alpha J(v)+\tfrac12\|v-g\|_2^2-\alpha J(u)-\tfrac12\|u-g\|_2^2\ge\alpha\langle q,v-u\rangle+\langle u-g,v-u\rangle+\tfrac12\|v-u\|_2^2=\tfrac12\|v-u\|_2^2.
$$
Thus $u$ is the unique minimizer.
Conversely, let $u$ be a minimizer. Its penalty is finite since comparison with zero gives a finite objective. For any $v$ with $J(v)<\infty$, set $w_t=u+t(v-u)$, where $0<t\le1$. <Convexity> gives $J(w_t)-J(u)\le t(J(v)-J(u))$. Minimality and quadratic expansion then imply
$$
0\le\alpha\bigl(J(v)-J(u)\bigr)+\langle u-g,v-u\rangle+\tfrac t2\|v-u\|_2^2.
$$
Letting $t\downarrow0$ yields $J(v)\ge J(u)+\langle(g-u)/\alpha,v-u\rangle$. The inequality is automatic if $J(v)=\infty$. Hence
$$
\boxed{u\text{ minimizes }\alpha J+\tfrac12\|\mathord\cdot-g\|_2^2\ \Longleftrightarrow\ \frac{g-u}{\alpha}\in\partial J(u).}
$$
This also follows from the <subdifferential sum rule>, since the quadratic term is everywhere <continuous> and <differentiable>, and the <subgradient optimality condition>. The direct proof above needs no unproved existence theorem.
\b[On the entire plane, the printed global BV domain is not closed in the L2 geometry.] The usual $BV(\mathbb R^2)$ definition includes an $L^1$ condition. For $1<a\le2$, $f(x)=(1+|x|)^{-a}$ belongs to $L^2$, has finite distributional variation $2\pi a\int_0^\infty r(1+r)^{-a-1}\,dr$, and fails to belong to $L^1$. The truncated $f_T=f\chi_{B(0,T)}$ lies in $BV\cap L^2$ and tends to $f$ in $L^2$. Its variation is the interior variation plus $2\pi T(1+T)^{-a}$, and tends to a finite limit. Thus $J(f_T)$ stays bounded but the literal $J(f)=\infty$, establishing <failure of L2 closure of the global BV domain>. One must not infer universal minimizer existence from an inapplicable closed-penalty <proximal operator> theorem. The standard closed extension uses the <homogeneous bounded-variation space>, allowing finite distributional variation without global $L^1$. The optimality equivalence just proved is valid for the literal penalty whenever a minimizer exists; the next datum has an explicit certified minimizer in its domain.
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