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I am struggling with this algebraic development and wonder if any of you can help me out. Would be very nice if you please could explain your "strategy" when doing so as well.

First of all:

$\xi = (1 - \theta)x_1 + \theta x_2$

I cannot understand how the right side is equal to the left side.

$(1- \theta) f(x_1) + \theta f(x_2) - f((1-\theta)x_1 + \theta x_2)$ $ =$ $(1-\theta)[f(x_1-f(\xi)] + \theta[f(x_2)-f(\xi)]$

This is part of a proof of convex functions and the proof has more steps beyond this which will incorporate the mean value theorem. I think I can manage the next steps if I get some help with this first one :)

Thank you very much!

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1 Answer

up vote 1 down vote accepted

Open up parentheses (and use square/round ones to make, perhaps, things clearer):

$$(1- \theta) f(x_1) + \theta f(x_2) - f[(1-\theta)x_1 + \theta x_2] \stackrel{?}=(1-\theta)[f(x_1)-f(\xi)] + \theta[f(x_2)-f(\xi)]$$

putting, as you did, $\,\xi=(1-\theta)x_1+\theta x_2\,$ , on the LHS we have:

$$f(x_1)-\theta f(x_1)+\theta f(x_2)-f(\xi)$$

and on the RHS we have

$$f(x_1)-f(\xi)-\theta f(x_1)+\theta f(\xi)+\theta f(x_2)-\theta f(\xi)=f(x_1)-\theta f(x_1)+\theta f(x_2)-f(\xi)$$

and both sides are equal.

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Thank you very much for your answer! –  Lukas Arvidsson Nov 5 '12 at 6:39
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