The two-pixel total-variation proximal map

The function $g$ is finite everywhere, hence proper. Its epigraph is the intersection of the two closed halfspaces $\lambda (u_2-u_1)\le s$ and $-\lambda (u_2-u_1)\le s$, so it is closed. The triangle inequality for absolute value gives convexity under [F5]. Write a candidate $u$ uniquely as $u=(a-b,a+b)$, so $a$ is its mean and $b$ its half-difference. Direct expansion gives $\mathrm{\parallel }u-v{\mathrm{\parallel }}_2^2=2(a-m{)}^2+2(b-d{)}^2$ and $g(u)=2\lambda \mathrm{\mid }b\mathrm{\mid }$. Hence the objective is $\frac{(a-m{)}^2}{t}+\frac{(b-d{)}^2}{t}+2\lambda \mathrm{\mid }b\mathrm{\mid }\mathrm{.}$ The first term is uniquely minimized at $a=m$; the remaining terms depend only on $b$.

The $b$-dependent objective is $2\lambda \mathrm{\mid }b\mathrm{\mid }+(b-d{)}^2\mathrm{/}t$, which under the scalar/one-coordinate identification is the proximal objective for the proper, closed, convex function $2\lambda \mathrm{\mid }\cdot \mathrm{\mid }$ with parameter $t\mathrm{/}2$. By [F2], its unique minimizer is $S_{(t\mathrm{/}2)(2\lambda )}(d)=S_{t\lambda }(d)$. This and $a=m$ give the displayed vector. The objective is strictly convex because of its positive quadratic term, consistent with uniqueness in [F3].