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Convex sets have property that a line between any two points, lies in the set. This is stated algebraically here as:
Given an affine space E, ... a subset V of E is convex if for any two points $a, b ∈ V,$ we have $c ∈ V$ for every point $c = (1 − λ)a + λb,$ with $0 ≤ λ ≤ 1 (λ ∈ R).$
I.e., any point (in the convex set) $c,$ that lies on the line joining $a, b;$ can be stated as a linear combination of points $a, b.$

But, there is an alternative definition of the convex function, which is expressed by stating that any point on the line joining any points, on a convex function, lies at least at a height (i.e., the $y-$ value) equal to that of the point on the convex function.
This is stated as, in Definition 3.3.8:
Let $A$ be a nonempty convex subset of $A_n.$ A function, $f : A → R,$ is convex if $f ((1 − λ)a + λb) ≤ (1 − λ)f (a) + λf (b)$ for all $a, b ∈ A$ and for all $λ ∈ [0, 1].$

I intuitively can see the dependence of the first definition on the second, but cannot prove it theoretically; though the first concerns with convex set; and the second with convex functions.

So, request help on proving the dependence of the first definition, on the second.
And if possible, also to prove the vice-versa; i.e. to prove the dependence of the second definition, on the first one.

jiten
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  • These are definitions of different objects: a convex function is not the same thing as a convex set. Logically speaking, there does not need to be any dependence between the definitions - though we might hope that there is some relationship, if the terminology is well chosen. (In this case, it is. There is indeed a relationship between the two definitions - a function $f\colon A \to \mathbb R$ is convex if and only if the set ${(x,y) \in A \times \mathbb R : y \ge f(x)}$ is convex. Maybe you are asking for a proof of that?) – Misha Lavrov Aug 10 '24 at 03:41
  • @MishaLavrov Request more details, though it seems that implicitly am asking the same proof. But, details are requested. – jiten Aug 10 '24 at 04:40
  • There's this question, for instance: https://math.stackexchange.com/questions/925835/a-function-is-convex-if-and-only-if-its-epigraph-is-convex – Misha Lavrov Aug 10 '24 at 04:47
  • @MishaLavrov Based on your first response: Is it possible to have no link between the two definitions, given above? If yes, can an example be provided for that. That would serve to elucidate. My intuition was that, in order that the first property is satisfied, the second is a must; and vice-versa; though with out a formal proof. This can be seen with any simple convex region, defined by a set of convex functions. Am taking multiple convex functions, in order to have a practical convex region, which is bounded, as say formed by an intersection of linear combinations. It arises in LP regularly. – jiten Aug 10 '24 at 10:16
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    It makes no sense to say "if the first property is satisfied, the second must be" because they are talking about different things! A function is not a set. If I give you a convex function $f$, and ask "does $f$ satisfy the second definition?" the answer is - "no, because $f$ is not a set". You first have to construct a set out of $f$ somehow - the epigraph is one way (which does give a convex set), and defining a convex set using several inequalities with convex functions is another way (which can work, depending on which way the inequalities go). – Misha Lavrov Aug 10 '24 at 16:18
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    Similarly, if I give you a convex set $S$, you are asking "does $S$ satisfy the first definition?" and the answer is, again, "no, because $S$ is not a function". You have to construct a function out of $S$ before you can ask if it is convex. (For example, the function $f$ defined by $f(x) = 0$ if $x \in S$ and $f(x) = +\infty$ if $x \notin S$ will always be a convex function, though it is not very interesting.) Note that given a set $S$ there will not, in general, be a collection of inequalities defining it - so you cannot ask about whether those inequalities use convex functions or not. – Misha Lavrov Aug 10 '24 at 16:23
  • Convex functions must be defined on domains that are convex sets, since you cannot talk about whether $f((1 - \lambda) a + \lambda b)$ is above or below a line unless $(1 - \lambda) a + \lambda b$ is in the domain of $f$. Also $f$ is convex if and only if the region above the graph of $f$ is a convex set. But these are just connections between the two concepts, not one somehow depending on the other. – Paul Sinclair Aug 12 '24 at 15:27

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