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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.

Analyst
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2 Answers2

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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 $\partial f(x_0) = \emptyset$.

gerw
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This a late posting but here is a result that holds for locally convex vector spaces of which a Banach space being an example.

Theorem: Let $X$ be a real locally convex topological space and suppose $f:X\rightarrow\mathbb{R}\cup\{+\infty\}$ is a proper lower semicontinuous convex function. For any $(x,\alpha)\in X\times\mathbb{R}$, if $\alpha<f(x)$, then there exits a continuous affine function $g$ such that $\alpha\leq g(x)$ (with equality if $x\in\operatorname{dom}(f)$) and $g(y)<f(y)$ for all $y\in X$.

Proof: By assumption $(x,\alpha)\notin\operatorname{epi}(f)$. Being $f$ l.s.c. and convex, $\operatorname{epi}(f)$ is a closed subset in $X\times\mathbb{R}$. Since $X\times \mathbb{R}$ (with the product topology) is also locally convex and so, by the Hahn-Banach separation theorem there is $(v,\lambda)\in (X\times\mathbb{R})^*=X^*\times\mathbb{R}$ and $\varepsilon>0$ such that $$v(x)+\lambda\alpha+\varepsilon<v(y)+\lambda\beta\qquad\forall(y,\beta)\in\operatorname{epi}(f).\tag{0}\label{zero}$$ By taking $\beta\rightarrow\infty$ we conclude that $\lambda\geq0$.

If $\lambda>0$ then $$g(y):=\frac1\lambda v(x-y)<g(y)+\frac{\varepsilon}{\lambda}<\beta\tag{1}\label{one}$$ whenever $(y,\beta)\in\operatorname{epi}(f)$. Hence, $g$ is a continuous affine function satisfying $g(x)=\alpha$ and $$g(y)<f(y)\qquad\forall y\in X$$

If $x\in\operatorname{dom}(f)$, that is $f(x)\in\mathbb{R}$, then $(x,f(x))\in\operatorname{epi}(f)$ and so, from \eqref{zero} $\lambda>0$ and so, the continuous affine function $g$ define in \eqref{one} satisfies the desired properties.

It remains to consider the case where $f(y)=\infty$ and $\lambda=0$. Then \eqref{zero} reads $$v(x-y)\leq-\varepsilon\qquad \forall y\in \operatorname{dom}(f)\tag{2}\label{two}$$ Fix $y_0\in\operatorname{dom}(f)$. The first part of this proof implies that there exists a continuous affine function $\phi$ such that $\phi(y_0)=f(x_0)-1$ and $\phi(y)<f(y)$ for all $y\in X$. For $c>0$ define $$g_c(y):=c\Big(v(x-y)+\frac{\varepsilon}{2}\Big) + \phi(y)$$ This is a continuous affine function and by \eqref{two}, $$g_c(y)<\phi(y)<f(y)\qquad\forall y\in X.$$ At $y=x$, $g_c(x)=\frac{c\varepsilon}{2}+f(y_0)-1$. Take $c^*>0$ large enough so that $g_{c^*}(x)\geq\alpha$. The continuous affine function $g_{c^*}$ satisfies the desired conditions.

Final comment:

The local convexity assumption is used only to guarantee that $\operatorname{epi}(f)$ and the compact set $\{(x,\alpha)\}$ can be separated by a continuous linear functional on $X\times\mathbb{R}$. All linear functionals $\phi$ on $X\times \mathbb{R}$ are uniquely expressed as $\phi=(v,\lambda)$ where $v\in X^*$ and $\lambda\in\mathbb{R}$ and $\phi(y,\alpha)=v(y)+\lambda\alpha$.

Mittens
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