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In my introductory class to calculus, we were given this statement and we were supposed to find a proof of it.

Function is convex on a closed interval <a,b>$\iff $ if given three random numbers in the interval, which satisfy this inequality: $x_1<x_2<x_3$; determinant of $$ \left( \begin{matrix} x_1 & f(x_1) & 1 \\ x_2 & f(x_2) & 1 \\ x_3 & f(x_3) & 1 \end{matrix} \right) $$ is positive.

I do not know how to approach this problem; I figured out that the determinant of the matrix is equal to $$x_1(f(x_2)-f(x_3)) + x_2(f(x_3)-f(x_1)) + x_3(f(x_1)-f(x_2))$$ Still, I don't see how this helps me prove that this is always positive for all convex functions. I do not want anyone to solve this for me, I just think I need a nudge in the right direction.

ampersander
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1 Answers1

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This is the shoestring formula for the signed area of the "triangle" formed by the points $(x_i, f(x_i))$, $i = 1, 2, 3$. (I put "triangle" in quotes in case these points lie on a straight line). What does convexity of the function imply about this signed area?

Rivers McForge
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  • That looks very promising, thank you! I am just unsure about why the sign changes when we go counterclockwise versus clockwise. The wikipedia page just states this fact, but it does not explain why this happens. – ampersander Mar 09 '21 at 18:36
  • Going clockwise instead of counterclockwise is equivalent to swapping the top and bottom rows of this determinant, which has the effect of multiplying the determinant by $-1$. – Rivers McForge Mar 10 '21 at 23:39