How to prove the convexity of $f$ if the strict epigraph of $f$ is convex 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 $\mathbb{R}^2?$ 
 A: 
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.
A: 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. 
