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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...

Eric Auld
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    maybe you can find someting useful here: http://math.stackexchange.com/questions/171599/jensens-inequality-for-integrals. – sigmatau Jul 23 '13 at 02:15
  • It's right there on Wikipedia: http://en.wikipedia.org/wiki/Jensen%27s_inequality#Measure-theoretic_and_probabilistic_form – tomasz Jul 23 '13 at 02:52
  • Shoot, that's embarrassing. Sorry. Should I delete the question? – Eric Auld Jul 23 '13 at 02:58

2 Answers2

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The generalization should just be the fact that it hold for other measures than the lebesgue measure. The generalization for sums just says that you can work with weighted sums. This weighting just becomes some other measure than the uniform one in the continuous setting.

Zach L.
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  • Right, and we can then correct for it afterward by dividing by $\mu ([a,b])$, for $\mu$ the standard one, right? – Eric Auld Jul 23 '13 at 02:39
  • Well, you need them to be probability measures, though. Otherwise, try constant functions. – tomasz Jul 23 '13 at 02:51
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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 $\nu([a,b])=1$.

Eric Auld
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