[proofplan] We prove both directions of the equivalence between strong duality and the existence of a supporting hyperplane to the value function $\phi$ at $b$. For the backward direction, a supporting hyperplane provides a slope $\lambda$ such that $\phi(b) + \lambda^\top(c - b) \leq \phi(c)$ for all $c$. Minimising both sides over the parametrisation $c = h(x)$ identifies the left-hand side with $g(\lambda)$, and weak duality forces equality. For the forward direction, an optimal dual multiplier $\lambda^*$ achieving $g(\lambda^*) = \phi(b)$ directly furnishes the supporting hyperplane. [/proofplan] [step:$(\Leftarrow)$: A supporting hyperplane at $b$ implies strong duality] Suppose $\phi$ has a supporting hyperplane at $b$. By definition, there exists a vector $\lambda \in \mathbb{R}^m$ such that \begin{align*} \phi(b) + \lambda^\top(c - b) \leq \phi(c) \quad \text{for all } c \in \mathbb{R}^m. \end{align*} Rearranging, $\phi(b) \leq \phi(c) - \lambda^\top(c - b)$ for all $c$. Taking the infimum over $c \in \mathbb{R}^m$ on the right-hand side: \begin{align*} \phi(b) \leq \inf_{c \in \mathbb{R}^m} \bigl[\phi(c) - \lambda^\top(c - b)\bigr]. \end{align*} We now rewrite the right-hand side in terms of the Lagrangian. For each $c$, $\phi(c) = \inf_{x \in X(c)} f(x)$ where $X(c) = \{x \in X : h(x) = c\}$. Substituting and using $h(x) = c$ for $x \in X(c)$: \begin{align*} \inf_{c \in \mathbb{R}^m} \inf_{x \in X(c)} \bigl[f(x) - \lambda^\top(h(x) - b)\bigr] = \inf_{x \in X} \bigl[f(x) - \lambda^\top(h(x) - b)\bigr] = \inf_{x \in X} L(x, \lambda) = g(\lambda), \end{align*} where the first equality holds because $\bigcup_{c \in \mathbb{R}^m} X(c) = X$ (every $x \in X$ belongs to $X(h(x))$), and optimising first over $c$ then over $x \in X(c)$ is the same as optimising over all $x \in X$. Therefore $\phi(b) \leq g(\lambda)$. By [Weak Duality](/theorems/2549), $g(\lambda) \leq \phi(b)$. Combining: $\phi(b) = g(\lambda)$, which is strong duality. [guided] Suppose $\phi$ has a supporting hyperplane at $b$: there exists $\lambda \in \mathbb{R}^m$ with \begin{align*} \phi(b) + \lambda^\top(c - b) \leq \phi(c) \quad \text{for all } c \in \mathbb{R}^m. \end{align*} What does this say geometrically? The graph of $\phi$ lies on or above the affine function $c \mapsto \phi(b) + \lambda^\top(c - b)$, which touches $\phi$ at $c = b$. This is the definition of a supporting hyperplane to the epigraph of $\phi$. Rearranging the supporting condition gives $\phi(b) \leq \phi(c) - \lambda^\top(c - b)$ for all $c$. The right-hand side depends on $c$, so we can tighten the bound by taking the infimum: \begin{align*} \phi(b) \leq \inf_{c \in \mathbb{R}^m} \bigl[\phi(c) - \lambda^\top(c - b)\bigr]. \end{align*} Now we connect this to the dual function. Recall that $\phi(c) = \inf_{x \in X(c)} f(x)$ where $X(c) = \{x \in X : h(x) = c\}$. For any $x \in X(c)$, we have $h(x) = c$, so $\lambda^\top(c - b) = \lambda^\top(h(x) - b)$. The double infimum over $c$ and $x \in X(c)$ can be combined into a single infimum over $x \in X$, because the sets $\{X(c)\}_{c \in \mathbb{R}^m}$ partition $X$ (every $x \in X$ satisfies $h(x) = c$ for exactly one $c$): \begin{align*} \inf_{c \in \mathbb{R}^m} \inf_{x \in X(c)} \bigl[f(x) - \lambda^\top(h(x) - b)\bigr] = \inf_{x \in X} \bigl[f(x) - \lambda^\top(h(x) - b)\bigr] = \inf_{x \in X} L(x, \lambda) = g(\lambda). \end{align*} We therefore have $\phi(b) \leq g(\lambda)$. But [Weak Duality](/theorems/2549) gives $g(\lambda) \leq \phi(b)$ for any dual feasible $\lambda$. The two inequalities force $\phi(b) = g(\lambda)$. The dual attains its supremum at $\lambda$, and the duality gap is zero. This is strong duality. [/guided] [/step] [step:$(\Rightarrow)$: Strong duality implies a supporting hyperplane at $b$] Suppose strong duality holds: there exists $\lambda^* \in Y$ with $g(\lambda^*) = \phi(b)$. We claim the affine function $\alpha(c) := \phi(b) + \lambda^{*\top}(c - b)$ is a supporting hyperplane to $\phi$ at $b$. We must verify $\alpha(c) \leq \phi(c)$ for all $c \in \mathbb{R}^m$. Fix any $c \in \mathbb{R}^m$. For any $x \in X(c)$ (so $h(x) = c$): \begin{align*} g(\lambda^*) = \inf_{x' \in X} L(x', \lambda^*) \leq L(x, \lambda^*) = f(x) - \lambda^{*\top}(h(x) - b) = f(x) - \lambda^{*\top}(c - b). \end{align*} Since $g(\lambda^*) = \phi(b)$, this gives $\phi(b) \leq f(x) - \lambda^{*\top}(c - b)$, i.e., $\phi(b) + \lambda^{*\top}(c - b) \leq f(x)$. Taking the infimum over $x \in X(c)$: \begin{align*} \phi(b) + \lambda^{*\top}(c - b) \leq \inf_{x \in X(c)} f(x) = \phi(c). \end{align*} This holds for all $c$, so $\alpha$ is a supporting hyperplane to $\phi$ at $b$. [guided] Suppose strong duality holds, so there exists $\lambda^* \in Y$ with $g(\lambda^*) = \phi(b)$. We need to construct a supporting hyperplane. The natural candidate is the affine function $\alpha(c) := \phi(b) + \lambda^{*\top}(c - b)$. Note that $\alpha(b) = \phi(b)$, so $\alpha$ touches $\phi$ at $b$. We need to verify $\alpha(c) \leq \phi(c)$ for all $c$. Fix $c \in \mathbb{R}^m$ and let $x \in X(c)$ be any point with $h(x) = c$. Since $g(\lambda^*)$ is the infimum of the Lagrangian over all of $X$, and $x \in X(c) \subseteq X$: \begin{align*} g(\lambda^*) = \inf_{x' \in X} L(x', \lambda^*) \leq L(x, \lambda^*) = f(x) - \lambda^{*\top}(h(x) - b) = f(x) - \lambda^{*\top}(c - b), \end{align*} where we used $h(x) = c$. Since $g(\lambda^*) = \phi(b)$ by strong duality: \begin{align*} \phi(b) + \lambda^{*\top}(c - b) \leq f(x). \end{align*} This holds for every $x \in X(c)$, so the left-hand side is a lower bound on $\{f(x) : x \in X(c)\}$. Taking the infimum: \begin{align*} \phi(b) + \lambda^{*\top}(c - b) \leq \inf_{x \in X(c)} f(x) = \phi(c). \end{align*} Since $c$ was arbitrary, we have $\alpha(c) \leq \phi(c)$ for all $c \in \mathbb{R}^m$, confirming that $\alpha$ is a supporting hyperplane to $\phi$ at $b$. The dual multiplier $\lambda^*$ is the slope of this hyperplane — it encodes the rate at which the optimal value changes as the constraint right-hand side is perturbed. [/guided] [/step]