[proofplan] The proof is a direct unpacking of the definition of the convex subdifferential. If the zero vector is a subgradient at $x_*$, the subgradient inequality says that every value $f(y)$ lies above $f(x_*)$, which is exactly global minimality. Conversely, if $x_*$ is a global minimiser, then the same inequality holds with the linear term given by the zero vector, so the zero vector satisfies the defining condition for membership in $\partial f(x_*)$. [/proofplan] [step:Derive global minimality from the zero subgradient] Assume $0_{\mathbb{R}^n} \in \partial f(x_*)$. By the definition of the convex subdifferential, for every $y \in \mathbb{R}^n$, \begin{align*} f(y) \ge f(x_*) + 0_{\mathbb{R}^n} \cdot (y - x_*). \end{align*} Since $0_{\mathbb{R}^n} \cdot (y - x_*) = 0$, this becomes \begin{align*} f(y) \ge f(x_*). \end{align*} Because this inequality holds for every $y \in \mathbb{R}^n$, the point $x_*$ minimises $f$ over $\mathbb{R}^n$. [guided] Assume $0_{\mathbb{R}^n} \in \partial f(x_*)$. The definition of the convex subdifferential says that a vector $p \in \mathbb{R}^n$ belongs to $\partial f(x_*)$ exactly when \begin{align*} f(y) \ge f(x_*) + p \cdot (y - x_*) \end{align*} for every $y \in \mathbb{R}^n$. Here the subgradient is the vector $p = 0_{\mathbb{R}^n}$, so the defining inequality gives \begin{align*} f(y) \ge f(x_*) + 0_{\mathbb{R}^n} \cdot (y - x_*) \end{align*} for every $y \in \mathbb{R}^n$. The Euclidean [inner product](/page/Inner%20Product) with the zero vector is zero, hence \begin{align*} f(y) \ge f(x_*) \end{align*} for every $y \in \mathbb{R}^n$. This is precisely the statement that no point in $\mathbb{R}^n$ has smaller $f$-value than $x_*$. Therefore $x_*$ minimises $f$ over $\mathbb{R}^n$. [/guided] [/step] [step:Recognise the zero vector as a subgradient at a global minimiser] Assume that $x_*$ minimises $f$ over $\mathbb{R}^n$. Then for every $y \in \mathbb{R}^n$, \begin{align*} f(y) \ge f(x_*). \end{align*} Since $0_{\mathbb{R}^n} \cdot (y - x_*) = 0$, this is equivalent to \begin{align*} f(y) \ge f(x_*) + 0_{\mathbb{R}^n} \cdot (y - x_*). \end{align*} Thus the vector $0_{\mathbb{R}^n}$ satisfies the defining inequality for membership in $\partial f(x_*)$. Hence \begin{align*} 0_{\mathbb{R}^n} \in \partial f(x_*). \end{align*} [/step] [step:Conclude the equivalence] The first step proves that $0_{\mathbb{R}^n} \in \partial f(x_*)$ implies that $x_*$ minimises $f$ over $\mathbb{R}^n$. The second step proves the converse implication. Therefore $x_*$ minimises $f$ over $\mathbb{R}^n$ if and only if \begin{align*} 0_{\mathbb{R}^n} \in \partial f(x_*). \end{align*} [/step]