Let $f\colon \mathbf{R} \to \mathbf{R}$ be given by $f(x) = 1$ if $x \geq 0$ and otherwise $f(x) =0$.
Is there a good characterization of the minimizer of $$ \lVert f - g\rVert_{L^1} $$ over $g\colon\mathbf{R} \to \mathbf{R}$ where $g$ is convex, non-decreasing, and satisfies $f \leq g$?
There is no convex function $g$ for which $\|f-g\|_1$ is finite, even if we forget the restriction $f\le g$.
Indeed, the integrability of $f-g$ implies that $\liminf |f-g|=0$ at both infinities, specifically $$ \liminf_{x\to -\infty} |g(x)| = 0 \quad \text{and}\quad \liminf_{x\to -\infty} |g(x)-1| = 0 $$ By virtue of convexity these $\liminf$s are actual limits: $$ \lim_{x\to -\infty} g(x) = 0 \quad \text{and}\quad \lim_{x\to -\infty} g(x) = 1 $$ This forces $g$ to be a bounded nonconstant convex function on $\mathbb{R}$, and there are no such things.