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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 $\mathbb{R}^2?$

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  • $\begingroup$ I may be missing something (indeed, very likely) but I see no difference between the two epigraph definitions B and C. $\endgroup$ – Michael Grant Nov 22 '14 at 5:30
  • $\begingroup$ @MichaelGrant, Sorry for my careless, I have corrected. $\endgroup$ – azc Nov 22 '14 at 6:01
  • $\begingroup$ @Nameless, Yes, you can use, if you can show the equivalence of the convexity of $\text{epi}_s(f)$ with the strict convexity of $f.$ $\endgroup$ – azc Nov 22 '14 at 6:40
  • $\begingroup$ @azc, I'll come back tomorrow and check. $\endgroup$ – IAmNoOne Nov 22 '14 at 7:05
  • $\begingroup$ Are you able to prove (or assume) the standard result---the convexity of f implies the convexity of the non-strict epigraph? If so, then your challenge reduces to showing that the convexity of the non-strict epigraph implies the convexity of the strict epigraph. For most convex functions, if not all, the strict epigraph is just the relative interior of the epigraph, and the relative interior operation preserves convexity. $\endgroup$ – Michael Grant Nov 22 '14 at 13:17
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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.

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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.

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