Context : I wish to show that if $X$ is a real normed space, $f : X \longrightarrow \overline{\mathbb{R}}$ is convex lower semi continuous and proper, then forall $(x_0,t_0) \in X \times \mathbb R$ such that $t_0 < f(x_0)$, there exists $\varphi = \varphi_{x_0,t_0} : X \longrightarrow \mathbb R$ affine continuous such that
$\forall x \in X,\quad \varphi(x) \leq f(x)$
$\varphi(x_0) \geq t_0$
This allows one to approximate $f$ by affine continuous functions, namely $f$ is the pointwise supremum of such functions.
Question : How to construct such a minorant when $x_0 \notin dom(f)$?
Knowledge :
$\underline{x_0 \in dom(f)}$ We can do as in Ekeland and Temam's Convex analysis book. Since $f$ is convex LSC its epigraph is convex closed, the set $\{ (x_0,t_0 )\}$ does not meet the epigraph and is compact convex. Apply the first Hahn-Banach geometric theorem to get $\varphi \in X'$, $\alpha,\lambda \in \mathbb R$ and $\epsilon > 0$ such that $$ \forall (x,t) \in epi(f),\quad \varphi x_0 + \lambda t_0 \leq \alpha < \alpha + \epsilon \leq \varphi x + \lambda t. \tag{HB} $$ Since $x_0 \in dom(f)$ we have $(x_0,f(x_0)) \in epi(f)$ and applying the last inequality to this element, remembering that $t_0 < f(x_0)$, yields that $\lambda > 0$. This means that forall $x \in dom(f)$, $$ f(x) \geq \frac{\alpha - \varphi x}{\lambda} $$ so call $\varphi_{x_0,t_0}$ the affine continuous functionnal at the right hand side, it solves the question.
$\underline{x_0 \notin dom(f),\quad \lambda > 0}$, we can introduce the same objects $\varphi,\alpha,\lambda,\epsilon$ and the sign of $\lambda$ yields the same seeked properties of $\frac{\alpha - \varphi x}{\lambda}$.
$\underline{x_0 \notin dom(f),\quad \lambda = 0}$, here (HB) tells us that $$ \forall x \in dom(f),\quad \varphi x_0 \leq \alpha < \alpha + \epsilon \leq \varphi x. $$ Observe that $\alpha + \epsilon - \varphi$ is non negative of $dom(f)$ and positive at $x_0$. Apply the case "$x_0 \in dom(f)$" to introduce $\overline \varphi$ continuous affine minorant of $f$ (the domain is non empty), it is possible to find $c > 0$ large enough such that $$ c(\alpha + \epsilon - \varphi) + \overline \varphi $$ works. Indeed the only condition that this candidate may not meet is the value at $t_0$, and large enough $c$ does the job because $(\alpha + \epsilon - \varphi)(x_0) > 0$.
The problem is when $x_0 \notin dom(f)$ and $\lambda < 0$, here (HB) has the wrong inequality when we wish to isolate $t$. Therefore I think the problem reduces to prove that $\lambda$ cannot be negative. I don't see how to contradict that.
More precise question : How to prove that when applying Hahn-Banach (HB) we cannot find $\lambda < 0$?
In (HB) fix $x\in dom \ f$ (which exists). Then $\lambda$ cannot be negative, because (HB) is valid for arbitrarily large $t$. Or phrased differently: if $\lambda<0$ then letting $t\to \infty$ in (HB) gives a contradiction.