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How to proof that a function f:R->R is convex if and only if dom(f) is convex and for each a, b, c in it's domain that are $a<b<c$, we have:

Determinant of matrix: $$ \begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ f(a) & f(b) &f(c) \end{vmatrix}\ge 0. $$

Determinant is:

$$ bf(c)-cf(b)+cf(a)-af(c)+af(b)‐bf(a) >= 0$$

Then:

$$ f(a)(c-b) + f(b)(a-c) + f(c)(b-a) >=0$$

Then according to the a<b<c, we can say:

$$ f(a)(c-b) + f(c)(b-a) >= f(b)(c-a)$$ [edited]

So I went until here but I don't know how to connect this to the Jensen's inequality to prove that f is convex.

shsh
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1 Answers1

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You need almost no more steps to go beyond there. The definition of convexity of a function includes the following: $$ f(\theta x+(1-\theta)y)\le \theta f(x)+(1-\theta)f(y) $$ Now, try to rearrange the last inequality you obtained with $x=a$, $y=c$ and a proper choice of $\theta$.

Mostafa Ayaz
  • 31,924
  • Thanks, my main problem is with f( t x + (1-t) y ). If I choose t= (c-b/c-a), then the right side of the Jensen's inequality is reached, but the left side still remains...like I have f(b)(a-c/c-a) = -f(b) in the left. And I still don't know how to convert this to f( t a + (1-t) c). – shsh Nov 19 '20 at 15:11
  • You're welcome. A slight mistake in your answer is $$f(a)(c-b) + f(b)(a-c) + f(c)(b-a) >=0\implies f(a)(c-b) +f(c)(b-a) \ge f(b)(a-c) $$while it should be$$f(a)(c-b) +f(c)(b-a) \ge f(b)(c-a)$$ – Mostafa Ayaz Nov 19 '20 at 15:39
  • Oh, thank you ! – shsh Nov 19 '20 at 16:24
  • Good luck.${}{}{}{}$ – Mostafa Ayaz Nov 19 '20 at 16:24