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Suppose you have: $a_0 + a_1x_1 + \cdots + a_nx_n \leq b$ how would I show that this is a convex function?

  • would it be enough to show that it's second derivate with respect to any $x_i$ would be equal to zero and thus the function is both convex and concave?
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    yes I would like to prove in general that a multi-variable linear function is convex – 2017-01-22
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    But what is a function you are asking for? Actually we have the inequality defining so-called half-space. Of course, this is a convex set. The DeepSea'a answer below treats convexity of this set. – 2017-01-22
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    overall I am trying to prove that a set of variables that satisfy a constraint is convex, I can prove that the set is convex if the function is convex I just dont know how to show that this function is convex – 2017-01-22

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hint: I guess that your sense of convex here is for the domain rather than the function. So let $A=(x_1,x_2,...,x_n)$ and $B=(y_1,y_2,...,y_n)$ be $2$ solution points of the above inequality, you wish to show any point of the segment $\bar{AB}$ is also a solution point. So let $P = At+ (1-t)B$ with $0 \le t \le 1$. We have: $a_0 + \displaystyle \sum_{i=1}^n \left(a_i(tx_i+(1-t)y_i)\right)= a_0 + t\displaystyle \sum_{i=1}^n a_ix_i + (1-t)\displaystyle \sum_{i=1}^n a_iy_i\le a_0 + t(b-a_0) + (1-t)(b-a_0) = a_0+b-a_0 = b$. Thus $P$ is a solution of the inequality and you're done-which means the set of solutions is convex ( I think this means the function is convex in your sense )

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    like I know what the definition of a convex set is but I don't know how to prove that AB satisfies this which is why I asked on here – 2017-01-22
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    see my edited post and tell me if you can take it . – 2017-01-22
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    yes thank you that helps out alot :) – 2017-01-22