[proofplan] We construct a subgradient at $x$ using a supporting hyperplane argument on the epigraph. The epigraph $\operatorname{epi}(f) = \{(z, y) \in \mathbb{R}^d \times \mathbb{R} : y \geq f(z)\}$ is a closed convex set, and the point $(x, f(x))$ lies on its boundary. We approach $(x, f(x))$ from outside the epigraph via a sequence of points, use the [Projection Theorem](/theorems/1985) to produce outer normals, extract a convergent subsequence by the Bolzano--Weierstrass theorem, and show that the limiting normal vector defines a subgradient. [/proofplan]

[step:Establish that the epigraph is closed and convex, and construct an approaching sequence] The epigraph of $f$ is \begin{align*} C := \operatorname{epi}(f) = \{(z, y) \in \mathbb{R}^d \times \mathbb{R} : y \geq f(z)\}. \end{align*} Since $f$ is convex and real-valued on $\mathbb{R}^d$, $C$ is convex: for $(z_1, y_1), (z_2, y_2) \in C$ and $t \in [0,1]$, \begin{align*} f((1-t)z_1 + tz_2) \leq (1-t)f(z_1) + tf(z_2) \leq (1-t)y_1 + ty_2, \end{align*} so $((1-t)z_1 + tz_2, (1-t)y_1 + ty_2) \in C$. Moreover, $C$ is closed: a finite-valued convex function on $\mathbb{R}^d$ is continuous (this is a standard consequence of local boundedness of convex functions), so if $(z_k, y_k) \in C$ and $(z_k, y_k) \to (z_0, y_0)$, then $y_0 = \lim_k y_k \geq \lim_k f(z_k) = f(z_0)$ by continuity, hence $(z_0, y_0) \in C$. The point $(x, f(x))$ lies on the boundary of $C$. Define the sequence \begin{align*} w_k := \left(x,\; f(x) - \frac{1}{k}\right) \in \mathbb{R}^{d+1}, \quad k = 1, 2, 3, \ldots \end{align*} Since $f(x) - 1/k < f(x)$, each $w_k \notin C$, and $w_k \to (x, f(x))$ as $k \to \infty$. [/step]

[step:Apply the Projection Theorem to each $w_k$ to obtain outer normal vectors] For each $k$, since $C$ is closed and convex in $\mathbb{R}^{d+1}$ and $w_k \notin C$, the [Projection Theorem](/theorems/1985) guarantees the existence of a unique projection $\pi_C(w_k) \in C$. The variational characterisation (part (i) of the Projection Theorem) states that for all $w \in C$: \begin{align*} (w_k - \pi_C(w_k))^\top(w - \pi_C(w_k)) \leq 0. \end{align*} Define $v_k := w_k - \pi_C(w_k)$. Since $w_k \notin C$ but $\pi_C(w_k) \in C$, we have $v_k \neq 0$. The variational inequality becomes: for all $w \in C$, \begin{align*} v_k^\top w \leq v_k^\top \pi_C(w_k) \leq v_k^\top w_k, \end{align*} where the second inequality uses $v_k^\top \pi_C(w_k) = v_k^\top(w_k - v_k) = v_k^\top w_k - \|v_k\|_2^2 \leq v_k^\top w_k$. Normalise: define $\hat{v}_k := v_k / \|v_k\|_2$. Then for all $w \in C$: \begin{align*} \hat{v}_k^\top w \leq \hat{v}_k^\top w_k. \end{align*} [/step]

[step:Extract a convergent subsequence of the normalised normals] The sequence $(\hat{v}_k)_{k \geq 1}$ lies in the unit sphere $\{u \in \mathbb{R}^{d+1} : \|u\|_2 = 1\}$, which is compact (closed and bounded in $\mathbb{R}^{d+1}$). By the Bolzano--Weierstrass theorem, there exists a subsequence $(\hat{v}_{k_j})_{j \geq 1}$ converging to some $\hat{v} \in \mathbb{R}^{d+1}$ with $\|\hat{v}\|_2 = 1$. Write $\hat{v} = (-\tilde{g}, \alpha)$ where $\tilde{g} \in \mathbb{R}^d$ and $\alpha \in \mathbb{R}$. Passing to the limit in $\hat{v}_{k_j}^\top w \leq \hat{v}_{k_j}^\top w_{k_j}$, and using $w_{k_j} \to (x, f(x))$: for all $w = (z, y) \in C$, \begin{align*} -\tilde{g}^\top z + \alpha y \leq -\tilde{g}^\top x + \alpha f(x). \end{align*} [/step]

[step:Show that $\alpha < 0$ and extract a subgradient]Taking $z = x$ and letting $y \to +\infty$ (since $(x, y) \in C$ for all $y \geq f(x)$), the left-hand side becomes $-\tilde{g}^\top x + \alpha y$. For this to remain bounded above by $-\tilde{g}^\top x + \alpha f(x)$ as $y \to +\infty$, we require $\alpha \leq 0$. Suppose $\alpha = 0$. Then the inequality becomes $-\tilde{g}^\top z \leq -\tilde{g}^\top x$ for all $z \in \mathbb{R}^d$ (since every $(z, f(z)) \in C$), i.e., $\tilde{g}^\top(z - x) \geq 0$ for all $z \in \mathbb{R}^d$. Taking $z = x - \tilde{g}$ gives $-\|\tilde{g}\|_2^2 \geq 0$, hence $\tilde{g} = 0$. But then $\hat{v} = (0, 0)$, contradicting $\|\hat{v}\|_2 = 1$. Therefore $\alpha < 0$.[/step]

[guided]Why must $\alpha$ be strictly negative? The vector $\hat{v} = (-\tilde{g}, \alpha)$ is a unit outward normal to the epigraph at the boundary point $(x, f(x))$. The epigraph extends upward (in the $y$-direction) without bound, so any outward normal must have a non-positive last component. If the last component were zero, the normal would lie entirely in the "horizontal" subspace $\mathbb{R}^d \times \{0\}$, meaning the supporting hyperplane is vertical. But the epigraph of a finite-valued convex function on all of $\mathbb{R}^d$ is "wide" enough that no vertical hyperplane can support it (because $f$ is defined everywhere, the epigraph projects onto all of $\mathbb{R}^d$). This forces $\alpha < 0$.[/guided]

[step:Rearrange the inequality to obtain $g \in \partial f(x)$]Since $\alpha < 0$, define $g := \tilde{g}/|\alpha| = -\tilde{g}/\alpha \in \mathbb{R}^d$. (Note that $\alpha < 0$, so $|\alpha| = -\alpha$.) Divide the inequality $-\tilde{g}^\top z + \alpha y \leq -\tilde{g}^\top x + \alpha f(x)$ by $|\alpha| = -\alpha > 0$ (which flips the sign of the $\alpha$ terms): \begin{align*} \frac{\tilde{g}^\top}{\alpha} z - y \leq \frac{\tilde{g}^\top}{\alpha} x - f(x), \end{align*} which rearranges to \begin{align*} f(x) - g^\top x \leq y - g^\top z. \end{align*} This holds for all $(z, y) \in C$, i.e., for all $z \in \mathbb{R}^d$ and all $y \geq f(z)$. Taking $y = f(z)$: \begin{align*} f(x) + g^\top(z - x) \leq f(z) \quad \text{for all } z \in \mathbb{R}^d. \end{align*} This is exactly the subgradient inequality, so $g \in \partial f(x)$. In particular, $\partial f(x) \neq \varnothing$.[/step]

[guided]The full argument is a geometric one: we separate the point $(x, f(x))$ from the interior of the epigraph using a supporting hyperplane. The supporting hyperplane has equation $-\tilde{g}^\top z + \alpha y = -\tilde{g}^\top x + \alpha f(x)$, and the entire epigraph lies on one side. The crucial observation is that the "vertical" component $\alpha$ of the normal must be strictly negative (the hyperplane is not vertical), which allows us to solve for the slope $g = \tilde{g}/|\alpha|$ and recover the subgradient inequality. This argument generalises: any convex function $f$ that is proper (not identically $+\infty$) and lower semicontinuous has non-empty subdifferential at every point of its effective domain. The proof here specialises to the case where $f$ is finite-valued on all of $\mathbb{R}^d$, which avoids issues with relative interiors and effective domains.[/guided]