22
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ take any two points on the graph... where would be the line joining those two points present? $\endgroup$
    – user87543
    Dec 18, 2013 at 12:06

2 Answers 2

28
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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
    Dec 16, 2014 at 20:15
  • 3
    $\begingroup$ 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. $\endgroup$ 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
    Dec 18, 2014 at 7:03
11
$\begingroup$

Linear function is both convex and concave. You may be interested in this page.

$\endgroup$
2
  • $\begingroup$ 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
    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
    Nov 5, 2016 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.