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Suppose $g$ is increasing and continuous. Does it follow that $G(x) = \int_0^xg(y)dy$ is convex?

Clearly $G'$ is increasing and continuous, and $G''\geq 0$ exists a.e., but I don't see how this implies convexity.

nullUser
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1 Answers1

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$ G'$ increasing is enough.

To show it:

$$x<y<z\implies \frac{G(y)-G(x)}{y-x} = G'(c_{x,y})\le G'(c_{y,z})=\frac{G(z)-G(y)}{z-y} $$because of the mean value theorem, with $x<c_{x,y}<y<c_{y,z}<z$.

Then $$ (z-y+y-x)G(y) \le (y-x)G(z) + (z-y)G(x)\\ G(y)\le \frac{y-x}{z-x} G(z)+ \frac{z-y}{z-x} G(x) $$then you have the convexity:

Let $t\in (0,1)$, define $y=(z-x)t+x = (1-t)x+tz\in (x,z)$.

Then $$ G((1-t)x+tz)=G(y)\le \frac{y-x}{z-x} G(z)+ \frac{z-y}{z-x} G(x) = t G(z)+ (1-t) G(x). $$

mookid
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