Generalization of properties of subgradient for convex function defined on arbitrary open convex subsets

In Bertsekas, Convex Optimization Algorithms The following Proposition is proved.

Let $ \Phi: \mathbb{R}^{n} \to \mathbb{R}$ be a convex function. For every $ x \in \mathbb{R}^{n}$ , we have

(a) The subgradient $ \partial \Phi(x)$ is a nonempty, convex and compact set, and we have \begin{equation} \label{eq-quotient} \Phi'(x;d):= \lim \limits_{\alpha \to 0} \dfrac{\Phi(x+\alpha d)-\Phi(x)}{\alpha} =\max_{g \in \partial f(x)} g^{\intercal} d \quad \forall \ d \in \mathbb{R^{n}} \end{equation} (b) If $ \Phi$ is differentiable at $ x$ with gradient $ \nabla \Phi(x)$ , then $ \nabla \Phi(x)$ is its unique subgradient at $ x$ , and we have $ \Phi'(x;d)= \nabla \Phi(x)^{\intercal}$

What I want to show: I want to generalize the nonemptyness, closedness and compactness of the subgradient to convex functions, defined on arbitrary open convex subsets of $ \mathbb{R}^{n}$ . My Proof goes as follows: Let $ \Phi : X \to \mathbb{R}^{n}$ be a convex function with $ X \subseteq \mathbb{R}^{n}$ a convex open set.

We can (I think) extend $ \Phi$ to a convex function $ \Phi _{\text{ext}}: \mathbb{R}^{n} \to \mathbb{R}$ by defining

\begin{equation} \Phi_{\text{ext}}(x)=\Phi(x) \text{ if } x \in X \text{ and } \Phi_{\text{ext}}(x)= \infty \text{ if } x \notin X \end{equation} . By the proposition stated above, we know that for $ p \in X$ , $ \partial \Phi_{\text{ext}}(p)$ is non-empty, closed and compact. We also know that $ \partial \Phi_{\text{ext}}(p)= \partial \Phi(p)$ holds, since \begin{gather} \partial \Phi_{\text{ext}}(p)=\{ y \in \mathbb{R}^{n} \ | \ \langle y, q-p \rangle \leq \Phi_{\text{ext}}(q) – \Phi_{\text{ext}}(p) \ \forall q \in \mathbb{R}^{n} \} \ = \{ y \in \mathbb{R}^{n} \ | \ \langle y, q-p \rangle \leq \Phi_{\text{ext}}(q) – \Phi(p) \ \forall q \in X \} \cap \{ y \in \mathbb{R^{n}} \ | \ \langle y, \rangle \leq \Phi_{\text{ext}}(q) – \Phi(p) \ \forall q \in \mathbb{R}^{n} \setminus X \} \ =\{ y \in \mathbb{R}^{n} \ | \ \langle y, q-p \rangle \leq \Phi_{\text{ext}}(q) – \Phi(p) \ \forall q \in X \} \cap \mathbb{R}^{n} \ =\{ y \in \mathbb{R}^{n} \ | \ \langle y, q-p \rangle \leq \Phi(q) – \Phi(p) \ \forall q \in X \} \ = \partial \Phi(p) \end{gather}

In particular, the fact that $ \partial \Phi(p)$ is non empyt, closed and compact follows from the fact that $ \partial \Phi_{\text{ext}}(p)$ fulfills the property.

Where my doubt lies: My doubt lies in the notion of $ \infty$ and the fact that $ \phi_{\text{ext}}$ is convex on the whole $ \mathbb{R}^{n}$ . To be more specific, it is clear that $ \Phi_{\text{ext}}$ is convex on $ X$ since $ \Phi $ is. But note that,for $ x \in X$ and $ y \in \mathbb{R}^{n} \setminus X$ it might happen that $ \alpha x +(1-\alpha)y \notin X$ , $ \alpha \in [0,1]$ which implies that $ \Phi_{\text{ext}}(\alpha x +(1-\alpha)y )= \infty$ . In particluar, we have

\begin{equation} \infty=\Phi_{\text{ext}}(\alpha x +(1-\alpha)y ) , \end{equation}

and

\begin{equation} \infty=\alpha \Phi_{\text{ext}}(x)+(1-\alpha)\Phi_{\text{ext}}(y). \end{equation}

Hence, in order for $ \Phi_{\text{ext}}$ to be convex, it must satisfy $ \infty \leq \infty$ , which is not well defined.

Question: How can we make the above argument more regorous? I was thinking about taking an upper bound on the value of $ \Phi$ instead of $ \infty$ and repeat the argument, but $ \Phi$ might be unboudnend on $ X$ , since $ X$ is open. (Note for $ X$ compact, continuity of $ X$ would imply a finite upper bound and we would be fine).