How to prove the convexity of $f$ if the strict epigraph of $f$ is convex

Question

I have trouble to prove the equivalence of the following two definitions of convex function. For convenience, I list them as follows:

Def-A. Let $f : I\to \mathbb{R}$ be a function, where $I$ is (and from now on) an interval in $\mathbb{R}.$ We call $f$ is convex on $I,$ if for every pair of $x_1, x_2\in I$ with $x_1\neq x_2,$ and for all $\lambda \in (0,1),$ we have \begin{gather*} f\big(\lambda x_1+(1-\lambda)x_2\big)\leq \lambda f(x_1)+(1-\lambda )f(x_2). \end{gather*}

Def-B. Let $f: I\to \mathbb{R},$ we call the set \begin{gather*} \text{epi}(f):=\{(x,y)\in\mathbb{R}^2\mid x\in I, y\geq f(x)\} \end{gather*} the epigraph of $f.$

Def-C. Let $f: I\to\mathbb{R},$ we call the set \begin{gather*} \text{epi}_s(f):=\{(x,y)\in\mathbb{R}^2\mid x\in I, y> f(x)\} \end{gather*} the strict epigraph of $f.$

As usual, we have known that $f: I\to\mathbb{R}$ is convex on $I$ if and only if the epigraph of $f,$ that is $\text{epi}(f)$ is convex in $\mathbb{R}^2.$

But recently we I read Giaquinta and Modica's book Mathemtical Analysis, Functions of one variable, I found that in Page 226, they called the strict epigraph of $f$ here (see Def-C above) the epigraph of $f,$ that is to say, they call $\text{epi}_s(f)$ the epigraph of $f,$ and claimed that \begin{gather*}\tag{$\star$} \text{$f$ is convex on $I,$ if and only if $\text{epi}_s (f) $ is convex.} \end{gather*} Then, I tried to prove $(\star).$ I have finished proof of that $\text{epi}_s(f)$ is convex, if $f$ is convex on $I.$ But when I tried to prove the converse, I was stuck!

Then, my question is, how to prove that $f$ is convex on $I,$ provided $\text{epi}_s(f)$ is convex in $\mathbb{R}^2?$

Answer 1

I am adapting the proof from Convexity and Optimization in Banach Spaces from my own interpretation.

Since $\text{epi} f$ is strictly convex, this means for the pair of points $(x,\alpha), (y,\beta) \in \text{epi} f$ we have the condition $$f(z) < t\alpha + (1-t)\beta,$$

where $t\in[0,1]$ and $z = tx + (1-t)y.$ We are required to show that $f(z) \leq tf(x) + (1-t)f(y)$, but we will instead assume on the contrary that we have $$f(z) > tf(x) + (1-t)f(y).$$

Consider the set $$\mathcal{A} = \{ (1-t)\alpha + t\beta: (x,\alpha),(y, \beta) \in \text{epi}(f)\}.$$

This set is nonempty and bounded below (this follows from the fact neither $f(x)$ or $f(y)$ can be equal to $+\infty$), thus $\mathcal{A}$ admits an infinium.

Therefore the chain of inequality shows,

$$f(z) \leq \inf \mathcal{A} = (1-t)f(x) + tf(y) <f(z).$$

This leads to a contradiction, so we conclude our original assumption is incorrect.

Answer 2

Proposition: $$ \text{epi}(f)=\bigcap_{\epsilon>0}\bigl(\text{epi}_S(f)-(0,\epsilon)\bigr). $$ Proof: A simple observation $$ f(x)\le y\quad\Leftrightarrow\quad\forall\epsilon>0\colon \ f(x)<y+\epsilon $$ leads directly to $$ (x,y)\in\text{epi}(f)\quad \Leftrightarrow\quad \forall\epsilon>0\colon \ (x,y+\epsilon)\in\text{epi}_S(f)\quad \Leftrightarrow\quad $$ $$ \quad \Leftrightarrow\quad \forall\epsilon>0\colon \ (x,y)\in\text{epi}_S(f)-(0,\epsilon)\quad \Leftrightarrow\quad (x,y)\in\bigcap_{\epsilon>0}\bigl(\text{epi}_S(f)-(0,\epsilon)\bigr). $$

Now if $\text{epi}_S(f)$ is convex then the translations $\text{epi}_S(f)-(0,\epsilon)$ are convex as well and, hence, $\text{epi}(f)$ is convex as an itersection of convex sets.