(edited for more clarity)

For a given function $f$, which is continuous, and $a < b$ real numbers, I need to make an estimation of the type $ \Bigg| \frac{\int_a^b f(t) (-t)dt}{\int_a^b f(t)dt} \Bigg| \leq M $. That is to say, I want to find the minimal $M$ with that property.

(1) Are there any tricks for estimating this for such general $f$?

(2) I could further assume for my application that $f$ is positive... then my best result is that $a \leq M \leq b$, because $f(t)a \leq f(t)t \leq f(t) b \quad \forall t \in [a,b]$ Does somebody have an idea how to improve these bounds?

  • $\begingroup$ Is $f$ a polynomial function? $\endgroup$ – ghosts_in_the_code Jun 3 '16 at 12:34
  • $\begingroup$ No, it is just an arbitrary continous function $\endgroup$ – jenna Jun 3 '16 at 12:36
  • $\begingroup$ If $f$ isn't positive, the LHS can very well be unbounded (take any odd function in $[-a,a]$). $\endgroup$ – Yves Daoust Jun 9 '16 at 10:34
  • $\begingroup$ If you can bound $f$ below and above with functions that you can integrate, you can get tighter estimates. But I don't think you can find the minimal $M$ without solving the exact integrals. $\endgroup$ – Yves Daoust Jun 9 '16 at 10:37
  • $\begingroup$ If I do find functions s.t. $g(x) \leq f(x) \leq h(x)$ on $[a,b]$, is it valid to say that $\Bigg|\frac{\int_a^b f(t) (-t) dt}{\int_a^b f(t) dt} \Bigg| \leq \Bigg| \frac{\int_a^b h(t) (-t) \ dt }{\int_a^b g(t) dt} \Bigg|$ ? $\endgroup$ – jenna Jun 9 '16 at 12:03

Well the estimation $$(b-a)\cdot \min_{x\in [a,b]}\{f(x)\}\leq \int_a^b f(x)dx \leq (b-a)\cdot \max_{x\in [a,b]}\{f(x)\}$$

is an exact estimation (i.e., we can find functions for which both inequalities become equalities), so with no other info about $f$, it's the best you can do...

  • $\begingroup$ Ok, yes. I was just hoping that I could multiply out something, such that a lot cancels. But apparently not. $\endgroup$ – jenna Jun 6 '16 at 6:56
  • $\begingroup$ Mh, is this enough to say anything about the inequation in the OP ? $\endgroup$ – Yves Daoust Jun 9 '16 at 10:33
  • $\begingroup$ I think the answer was referring to something in my original post, but I have updated it. Using the mean value theorem does not yield enough for what I am trying to do... $\endgroup$ – jenna Jun 9 '16 at 12:08

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