Theorem: Existence of Continuous Affine Minorant of Lower Semicontinuous Proper Convex Function with Prescribed Point Evaluation

Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\R$.
Let $f : X \to \overline \R$ be a lower semicontinuous proper convex function.
Let $\Dom f$ be the effective domain of $f$.
Let $x \in \Dom f$.
Let $\alpha \in \R$ such that:
:$\alpha < \map f x$
Then there exists a continuous affine functional $g : X \to \overline \R$ such that:
:$\map g x = \alpha$
and:
:$\map g y < \map f y$ for each $y \in X$.

Proof:
Let $\struct {X \times \R}^\ast$ be the topological dual of the direct product $X \times \R$. 
Let $\map {\operatorname {epi} } f \subseteq X \times \R$ be the epigraph of $f$.
Since $\map f x < \infty$, there exists $\beta \in \R$ such that $\map f x < \beta$.
Hence $\tuple {x, \beta} \in \map {\operatorname {epi} } f$, showing that $\map {\operatorname {epi} } f \ne \O$.
Since $f$ is lower semicontinuous, we have that $\map {\operatorname {epi} } f$ is closed from Characterization of Lower Semicontinuity.
Since $f$ is convex, we have that $\map {\operatorname {epi} } f$ is convex from Function is Convex iff Epigraph is Convex.
From Finite Topological Space is Compact, $\set {\tuple {x, \alpha} }$ is compact.
From Singleton is Convex Set, $\set {\tuple {x, \alpha} }$ is convex.
Finally, since $\alpha < \map f x$, we do not have $\map f x \le \alpha$ and hence $\tuple {x, \alpha} \not \in \map {\operatorname {epi} } f$.
From Hahn-Banach Separation Theorem: Hausdorff Locally Convex Space: Real Case: Compact Convex Set and Closed Convex Set, there exists $\phi \in \paren {X \times \R}^\ast$ such that:
:$\ds \map \phi {x, \alpha} < \inf_{\tuple {y, \beta} \in \map {\operatorname {epi} } f} \map \phi {y, \beta}$
Let $\epsilon > 0$ such that:
:$\ds \map \phi {x, \alpha} + \epsilon < \inf_{\tuple {y, \beta} \in \map {\operatorname {epi} } f} \map \phi {y, \beta}$
From Topological Dual Space of Product of Topological Vector Spaces: Corollary, there exists a continuous linear functional $g : X \to \R$ and $\lambda \in \R$ such that:
:$\map g x + \lambda \alpha + \epsilon < \map g y + \lambda \beta$ for each $\tuple {y, \beta} \in \map {\operatorname {epi} } f$.
{{AimForCont}} $\lambda < 0$.
If $\tuple {y, \beta} \in \map {\operatorname {epi} } f$, then we have:
:$\map f y \le \beta < \gamma$
for all $\gamma > \beta$.
Hence $\tuple {y, \gamma} \in \map {\operatorname {epi} } f$ for all $\gamma > \map f x$. 
Then we can take $\beta \to \infty$ in:
:$\map g x + \lambda \alpha + \epsilon < \map g y + \lambda \beta$
to obtain:
:$\map g x + \lambda \alpha = -\infty$
a contradiction. 
Hence we have $\lambda \ge 0$.
We now rule out $\lambda = 0$.
Suppose that:
:$\map g x + \epsilon < \map g y$ for each $\tuple {y, \beta} \in \map {\operatorname {epi} } f$.
We have $\tuple {x, \map f x} \in \map {\operatorname {epi} } f$, and hence:
:$\map g x + \epsilon < \map g x$
a contradiction.
Hence we must have $\lambda > 0$.
Now, define $h : X \to \R$ by:
:$\ds \map h z = \frac {\map g {x - z} } \lambda + \alpha$
for each $z \in X$.
We have $\map h x = \alpha$. 
If $z \in \Dom f$, then we have $\map f z < \infty$ and hence $\tuple {z, \map f z} \in \map {\operatorname {epi} } f$.
Then we have:
:$\ds \map g x + \lambda \alpha + \epsilon < \map g z + \lambda \map f z$
so that:
:$\ds \map g {x - z} + \lambda \alpha + \epsilon < \lambda \map f z$
giving:
:$\ds \frac {\map g {x - z} } \lambda + \alpha + \frac \epsilon \lambda < \map f z$
That is:
:$\ds \map h z + \frac \epsilon \lambda < \map f z$
so that $\map h z < \map f z$. 
If $z \not \in \Dom f$, then we have $\map f z = \infty$.
Since $\map h z < \infty$, we then have $\map h z < \map f z$.
Hence $\map h z = \alpha$ and $\map h z < \map f z$ for all $z \in X$.
Hence $h$ is our desired continuous affine functional.
{{qed}}