[proofplan] The proof is a direct comparison of two definitions. The condition that $x_0$ is a global minimizer says exactly that $f(x) \ge f(x_0)$ for every $x \in C$. The condition $0 \in \partial f(x_0)$ says that the subgradient inequality holds with the zero vector, and this reduces to the same inequality. [/proofplan] [step:Unpack the zero subgradient condition as a pointwise lower bound] Let $0_{\mathbb{R}^n} \in \mathbb{R}^n$ denote the zero vector. By the relative-domain definition of the subdifferential at $x_0$, the condition $0_{\mathbb{R}^n} \in \partial f(x_0)$ is equivalent to the assertion that, for every $x \in C$, \begin{align*} f(x) \ge f(x_0)+0_{\mathbb{R}^n}\cdot (x-x_0). \end{align*} Since $0_{\mathbb{R}^n}\cdot (x-x_0)=0$ for every $x \in C$, this is equivalent to \begin{align*} f(x) \ge f(x_0) \text{ for every } x \in C. \end{align*} [guided] Let $0_{\mathbb{R}^n}$ be the zero vector in $\mathbb{R}^n$. We begin by translating the symbolic condition $0_{\mathbb{R}^n} \in \partial f(x_0)$ into the inequality it represents. By definition, a vector $g \in \mathbb{R}^n$ belongs to $\partial f(x_0)$ exactly when \begin{align*} f(x) \ge f(x_0)+g\cdot (x-x_0) \text{ for every } x \in C. \end{align*} Here we take $g=0_{\mathbb{R}^n}$. Therefore $0_{\mathbb{R}^n} \in \partial f(x_0)$ is equivalent to \begin{align*} f(x) \ge f(x_0)+0_{\mathbb{R}^n}\cdot (x-x_0) \text{ for every } x \in C. \end{align*} The dot product term vanishes because the first factor is the zero vector: \begin{align*} 0_{\mathbb{R}^n}\cdot (x-x_0)=0. \end{align*} Thus the subgradient condition with the zero vector is precisely \begin{align*} f(x) \ge f(x_0) \text{ for every } x \in C. \end{align*} This is the key point: the affine lower support inequality becomes the global minimum inequality when the supporting slope is zero. [/guided] [/step] [step:Identify the lower bound with global minimality] By definition, $x_0$ is a global minimizer of $f$ on $C$ if and only if \begin{align*} f(x_0) \le f(x) \text{ for every } x \in C. \end{align*} This is the same condition as \begin{align*} f(x) \ge f(x_0) \text{ for every } x \in C. \end{align*} The previous step showed that this condition is equivalent to $0_{\mathbb{R}^n} \in \partial f(x_0)$. Hence $x_0$ is a global minimizer of $f$ on $C$ if and only if $0 \in \partial f(x_0)$. [/step]