= Solution
If the <Strong Lagrangian property> holds with multiplier $\lambda$, its infimum identity implies, for every $x\in X$,
$$
f(x)-\lambda^T(h(x)-b)\geq\phi(b).
$$
For any feasible right-hand side $u$, take the infimum over $h(x)=u$ to obtain
$$
\boxed{\phi(u)\geq\phi(b)+\lambda^T(u-b).}
$$
The inequality also holds at infeasible $u$, where $\phi(u)=+\infty$, and it is equality at $u=b$. Thus $r=\phi(b)+\lambda^T(u-b)$ is the required <non-vertical supporting hyperplane of a value function>. This proves the converse and completes the equivalence. Neither implication requires that $X$ be a <convex set>.
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