= Solution
For each fixed $z$, choose the <convex perturbation function>
$$
\boxed{f_z(x,u)=k(x)+h(z+u-x).}
$$
It is jointly convex and has marginal
$$
p_z(u)=\inf_x[k(x)+h(z+u-x)]=F(z+u).
$$
Because $h$ and $k$ are nonnegative and finite everywhere,
$$
0\leq F(w)\leq k(0)+h(w)<\infty
$$
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 $z$:
$$
\boxed{\inf_x[k(x)+h(z-x)]=\max_y[-f_z^*(0,y)].}
$$
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.
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