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I was hoping someone could help me prove the statement that a concave function of a linear function is concave. Thank you in advance.

user133993
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  • Are you able to write the statement you want to prove in symbols? – Karolis Juodelė Mar 29 '14 at 16:00
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    Let $x$ be a bunch of points, $A(x)$ a weighted average of $x$, $L$ a linear function, and $C$ a concave one. Then $C(L(A(x)))=C(A(L))\geq A(C(L(x)))$. The first equality is because $L$ is linear and $A$ is a linear combination, so they commute. The inequality is the definition of concavity, that the value of the function at an average of some points is not less than the same average of the values of the function at those points. – OR. Mar 29 '14 at 16:04

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Let us see: assume that

$$ 0<t<1\implies g(tx+(1-t)y) \ge tg(x)+(1-t)g(y)\\ L(tx+sy) = tL(x) + L(sy) $$then:

$$ 0<t<1\implies g\circ L(tx+(1-t)y) = g(L(tx+(1-t)y)) \\= g(tL(x)+(1-t)L(y)))\\\ge tg(L(x))+(1-t)g(L(y)) = tg\circ L(x)+(1-t)g\circ L(y) $$

mookid
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Let $c(x)$ be concave and $l(x)$ linear. You want to show that $c[l(x)]$ is concave. Square brackets only for clarity. All you have to do is check the definition: $$c[l(tx + (1-t)y)] \geq tc[l(x)] + (1-t)c[l(y)]$$

Do you see what $l(tx + (1-t)y)$ equals?

Karolis Juodelė
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