# Generalization of Jensen's inequality for integrals?

Jensen's inequality for sums says that for $f$ convex, $$f\left(\sum_1^n \alpha_i x_i\right)\leq \sum_1^n \alpha_i f(x_i), \,\,\,\,\text{for } \sum_1^n \alpha_i = 1.$$ I have read that a generalization of it is that $$f\left(\frac{\sum_1^n \alpha_i x_i}{\sum_1^n \alpha_i}\right)\leq \frac{\sum_1^n \alpha_i f(x_i)}{\sum_1^n \alpha_i}.$$

Now I read Jensen's inequality for integrals: for $f,g$ functions, $f$ convex, $g$ and $f\circ g$ integrable, we have $$f\left(\int_0^1g(x) \, dx\right)\leq \int_0^1 (f\circ g)(x)\, dx.$$

What is the analogous generalization of this to an interval other than $[0,1]$? I don't think it's just dividing by the length of the interval...

• maybe you can find someting useful here: math.stackexchange.com/questions/171599/…. Jul 23, 2013 at 2:15
• It's right there on Wikipedia: en.wikipedia.org/wiki/… Jul 23, 2013 at 2:52
• Shoot, that's embarrassing. Sorry. Should I delete the question? Jul 23, 2013 at 2:58

• Right, and we can then correct for it afterward by dividing by $\mu ([a,b])$, for $\mu$ the standard one, right? Jul 23, 2013 at 2:39
I believe we can prove that in fact it should be the length of the interval by using Jensen's finite equality on the Riemann sums: $f\left(\frac{\sum_1^n \alpha_i x_i}{\sum_1^n \alpha_i}\right)\leq \frac{\sum_1^n \alpha_i f(x_i)}{\sum_1^n \alpha_i},$ where $x_i = g(t^*_j)$ and $\alpha_i=t_{i+1}-t_i$.
@Zach has also mentioned that it can be shown by just choosing another measure $\nu$ such that $\nu([a,b])=1$.