3
$\begingroup$

Let $f$ be integrable on $[0, t]$ , $t≥0$ then prove that $$\exp\left(\int_0^t f(s)ds \right) \le 1+ \int_0^t e^{\max(1,t)f(s)}ds$$

This clearly smells like Jensen's inequality where we instead have, $$\exp\left(\frac1t\int_0^t f(s)ds \right) \le \frac1t \int_0^t e^{f(s)}ds$$ I don't see a special function I can plug into the Jensen inequality to get the result.

Does anyone knows how the aforementioned inequality could be derived from Jensen inequality?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.