If two nonnegative finite-valued <convex functions> are defined on all of a finite-dimensional space, their <infimal convolution> is finite, <convex> and continuous everywhere. <Convex perturbation duality> then gives <strong duality> and dual attainment at every argument. It does not require or imply attainment of the primal decomposition infimum.

More precisely, let $f,g:\mathbb R^n\to[0,\infty)$ be finite everywhere and write $h=f\mathbin\square g$. Nonnegativity supplies a lower bound, while the two trial splits $(x,0)$ and $(0,x)$ give
$$
0\leq h(x)\leq\min\{f(x)+g(0),\ f(0)+g(x)\}<\infty.
$$
Mixing nearly optimal splits proves <convexity>, and <continuity of a convex function> applies because the <effective domain> is the entire <finite-dimensional vector space>. These statements concern the optimized value, rather than existence of an optimal split. For the continuity theorem and dual attainment under the equality constraint $u+v=x$, see https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/db7bde2cd43e8dbaba11f6911311b87c_MIT6_253S12_summary.pdf#page=40[Bertsekas, Convex Optimization Theory: A Summary, Propositions 1.3.11 and 5.3.3(a)]. The equality has a feasible point in the whole space for every $x$, and the displayed bound makes its optimal value finite.

\b[A finite value without any optimal split.] On $\mathbb R$, take the <convex functions> $f(u)=e^u$ and $g(v)=0$. Both are nonnegative and finite everywhere, but
$$
(f\mathbin\square g)(x)=\inf_{u\in\mathbb R}e^u=0
$$
for every $x$. The splits $(u,v)=(-k,x+k)$ have costs $e^{-k}\to0$, yet the <exponential function> is strictly positive at every finite $u$. No split attains the <infimum>. The value function is even constant and <differentiable>; regularity of the value function does not prevent optimizing splits from escaping to infinity. For example, $k=1,3,6$ gives costs approximately $0.367879$, $0.049787$ and $0.002479$.

A sufficient extra condition for attainment at a fixed $x$ is that $u\mapsto f(u)+g(x-u)$ be a <coercive function>, meaning it tends to $+\infty$ as $\|u\|\to\infty$. Under the finite-valued assumptions it is a <continuous function>, so a sufficiently large closed ball contains a <global minimum> by the <extreme value theorem>. If this cost is also a <strictly convex function>, the optimal split is unique. The quadratic example above satisfies both conditions; the <exponential function> example does not satisfy <coercivity>.

Nonnegativity is sufficient, not necessary: finite lower bounds $f\geq A$, $g\geq B$ would give $h\geq A+B$ by the same argument. Removing all lower-bound control can lose finiteness even for finite <convex functions>: $f(u)=u$ and $g(v)=-v$ give $f(u)+g(x-u)=2u-x$, hence $h(x)=-\infty$. Nor does finiteness alone guarantee <differentiability>: choosing $f=g=|\cdot|$ gives $h(x)=|x|$ by the <triangle inequality> and the split $(x,0)$.


 Finite-valued infimal convolution (source code)