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I am trying to prove the following:

Suppose $\phi:[0,\infty) \to \mathbb{R}$ a continuous, bounded and integrable function and $\psi:[0,\infty) \to \mathbb{R}$ is a continuous, bounded above by 1, strictly decreasing, positive function.

Then, there exists a $\xi \in [0,\infty)$ such that $\int_0^{\infty} \psi(x)\phi(x)dx = \psi(\xi)\int_0^{\infty}\phi(x)dx$

I was able to show that there exists $\xi$ such that $\mid \int_0^{\infty} \psi(x)\phi(x)dx \mid = \psi(\xi)\int_0^{\infty}\mid \phi(x) \mid dx$ using the inequality $\mid\int_0^{\infty} \phi(x) dx \mid < \int_0^{\infty}\mid \phi(x)\mid dx$ and the intermediate value theorem. Any ideas how to proceed from here?

user385459
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  • Are you sure $\psi$ is positive and not $\phi$? The positive one should be the one staying in the integral. – Harnak Apr 29 '17 at 17:13
  • Yes, i think this is what makes it different to the mean value theorem. I have tried it with a few examples and it seems to hold but i can't come up with a general proof! – user385459 Apr 29 '17 at 17:14
  • Uh, ok, then the proof I had in mind won't work :) – Harnak Apr 29 '17 at 17:20
  • I think the proof you had in mind is exactly what i said i was able to show !:) – user385459 Apr 29 '17 at 17:21

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