 Solution

For each fixed $z$, choose the convex perturbation functionIt is jointly convex and has marginalBecause $h$ and $k$ are nonnegative and finite everywhere,for every $w$. The inf-convolution is convex: for approximate decompositions of $w_1,w_2$, take their convex combination and use convexity of both costs; then let their approximation errors tend to zero. Thus $F$ is a finite convex function on all of ${\mathbb{R}}^n$, hence continuous everywhere. In particular $p_z$ is proper and finite near zero, so the qualification in the previous solution applies at every $z$:This is strong duality for a finite-valued infimal convolution.

No primal attainment was used. For example $k(x)=e^x$ and $h(x)=0$ meet all the stated assumptions, but $F(z)=0$ is approached only as $x\to -\infty$. This illustrates why a proof based on an assumed minimizing decomposition would be incomplete. The printed functions are real-valued everywhere, not merely extended-real; that distinction supplies continuity and the strong-duality qualification.