I'm reading a proof of Theorem 2.20 in Barbu's textbook Convexity and Optimization in Banach Spaces.
Proposition 2.20 Any convex, proper and lower-semicontinuous function is
bounded from below by an affine function.
The proof is given below. There is a positive constant $-\alpha \varepsilon$ in inequality $\color{blue}{{(*)}}$. Could you please explain how the author gets rid off this constant?
Let $f: X \rightarrow]-\infty,+\infty]$ be any convex and lower-semicontinuous function on $X, f \not \equiv+\infty$. As already seen, the epigraph epi $f$ of $f$ is a proper convex and closed subset of product space $X \times \mathbb{R}$. If $x_{0} \in \operatorname{Dom}(f)$, then $\left(x_{0}, f\left(x_{0}\right)-\varepsilon\right) \bar{\in}$ epi $f$ for every $\varepsilon>0$. Thus, using the Hahn-Banach theorem (see Corollary $1.45$ ), there exists $u \in(X \times \mathbb{R})^{*}$ such that
$$
\sup _{(x, t) \in \text { epi } f} u(x, t)<u\left(x_{0}, f\left(x_{0}\right)-\varepsilon\right) .
$$
Identifying the dual space $(X \times \mathbb{R})^{*}$ with $X^{*} \times \mathbb{R}$, we may infer that there exist $x_{0}^{*} \in X^{*}$ and $\alpha \in \mathbb{R}$, not both zero, such that
$$
\sup _{(x, t) \in \text { epi } f}\left\{x_{0}^{*}(x)+t \alpha\right\}<x_{0}^{*}\left(x_{0}\right)+\alpha\left(f\left(x_{0}\right)- \color{blue}{\varepsilon}\right) . \quad \quad \color{blue}{{(*)}}
$$
We observe that $\alpha \neq 0$ and must be negative, since $\left(x_{0}, f\left(x_{0}\right)+n\right) \in$ epi $f$ for every $n \in \mathbb{N}$. On the other hand, $(x, f(x)) \in$ epi $f$ for every $x \in \operatorname{Dom}(f)$. Thus,
$$
x_{0}^{*}(x)+\alpha f(x) \leq x_{0}^{*}\left(x_{0}\right)+\alpha f\left(x_{0}\right), \quad \forall x \in \operatorname{Dom}(f)
$$
or
$$
f(x) \geq-\frac{1}{\alpha} x_{0}^{*}(x)+\frac{1}{\alpha} x_{0}^{*}\left(x_{0}\right)+f\left(x_{0}\right), \quad \forall x \in \operatorname{Dom}(f),
$$
but the function in the right-hand side is affine, as claimed.
Best Answer
The proof is flawed, but can be fixed easily. Indeed, by keeping the $\varepsilon$, we arrive at $$ f(x) \geq-\frac{1}{\alpha} x_{0}^{*}(x)+\frac{1}{\alpha} x_{0}^{*}\left(x_{0}\right)+f\left(x_{0}\right) - \varepsilon, \quad \forall x \in \operatorname{Dom}(f). $$ The right hand side is an affine function and we are done.
I would like to mention, that the final line in the proof cannot hold. Note that in the beginning of the proof, $x_0$ is an arbitrary element in $\operatorname{Dom}(f)$. Then, the proof constructs $(x_0^*, \alpha)$ such that the last line holds. This last line is just $-\alpha^{-1} x_0^* \in \partial f(x_0)$. However, it is easy to provide an example with $x_0 \in \operatorname{Dom}(f)$ but $\partial f(x_0) = \emptyset$.