convexity and a non-decreasing function?

I need to check that this function is non-decreasing:

$$f(x) = \frac{g(x)}{x},$$

where $g(x)=\log[\text{E}(|X|^x)]$ is the $x$'th absolute moment of random variable $X$ which is proved that is a convex and non-decreasing function of $x$; $x>0$.

• This is false in general: consider $g(x)=x+1$. Then $f(x) = 1 + 1/x$ which is decreasing. – amakelov Apr 14 '17 at 8:11
• But $x + 1$ is not strictly convex. – badjohn Apr 14 '17 at 8:19
• nobody said it has to be. – amakelov Apr 14 '17 at 8:22
• Thanks @amakelov. But, I think it is just a convex function not affine. – Amin Apr 14 '17 at 8:23
• @amakelov True but there may have been a mistake. As it happens, the question has been edited but in a different way. – badjohn Apr 14 '17 at 8:23

It suffices to show that $e^{f(x)}$ is increasing, and $$e^{f(x)} = \mathbb{E}[|X|^x]^{1/x}$$ So now we want to show that $\mathbb{E}[|X|^x]^{1/x} \geq \mathbb{E}[|X|^y]^{1/y}$ whenever $x\geq y$, which follows from the power mean inequality.
Edit: to prove the power mean inequality in this case, we can use Holder's inequality: namely, it tells us that with $p=x/y>1$ and $q=\frac{p}{p-1}$ we have $$\mathbb{E}[(|X|^y)^p]^{1/p}\mathbb{E}[1^q] \geq \mathbb{E}[|X|^y]$$ which decodes to what we want.
• I am wondering is there another way using the direct usage of convexity and being non-decreasing of $g(x)$ for proving this? – Amin Apr 14 '17 at 8:51