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I was wondering if linear function is convex or concave? For example f(x)=x, is function whose second derivate is 0 so we cant tell anything using this criteria. Can someone help?

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    $\begingroup$ take any two points on the graph... where would be the line joining those two points present? $\endgroup$
    – user87543
    Commented Dec 18, 2013 at 12:06

2 Answers 2

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A linear function is both. Use this definition of convexity:

For any two points $x_1$ and $x_2$ $$\forall a \in [0,1] \quad f(ax_1 + (1-a)x_2) \leq af(x_1) + (1-a)f(x_2)$$

Flip inequality for concave. Do you see why linear is both?

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  • $\begingroup$ I don't get it. Linear is supposed to be f(cx1+bx2) = cf(x1) + bf(x2) where c and b are real numbers and x1 and x2 are elements of the domain/I/interval/whatever right? The definition of convex and concave use a and 1-a which only covers numbers in [0,1] so how are we extending this to all real numbers from just [0,1]? $\endgroup$
    – BCLC
    Commented Dec 16, 2014 at 20:15
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    $\begingroup$ Exactly my point. To prove convexity, you simply pick the coefficients as mentioned above. Convexity requires that coefficints be in $[0,1]$. Linearity covers this and far more!! That too with equality. $\endgroup$ Commented Dec 18, 2014 at 3:27
  • $\begingroup$ Hahaha sorry. I misread the question. Linear is convex and concave but the link to the answer below says iff $\endgroup$
    – BCLC
    Commented Dec 18, 2014 at 7:03
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Linear function is both convex and concave. You may be interested in this page.

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  • $\begingroup$ I don't get it. Linear is supposed to be f(ax1+bx2) = af(x1) + bf(x2) where a and b are real numbers and x1 and x2 are elements of the domain/I/interval/whatever right? The definition of convex and concave uses $\lambda$ and 1-$\lambda$ which only cover numbers in [0,1] so how are we extending this to all real numbers from just [0,1]? $\endgroup$
    – BCLC
    Commented Dec 16, 2014 at 20:14
  • $\begingroup$ I don't think you need to extend to all real numbers. You already know that it's true for all real numbers which implies it's also true for [0, 1]. $\endgroup$
    – confused00
    Commented Nov 5, 2016 at 14:21

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