Given $a, T>0$, by Holder's inequality we have $$ \int_T^{T+a}f(s)ds\leq \left(\int_0^{T+a}f(s)^2ds\right)^{1/2}\cdot\sqrt{a}. $$ Do we have similar result if we replace $x^2$ by some convex function? That is, let $\Psi$ be a convex function such that $\Psi(x)/x\to\infty$ as $x\to\infty$, do we have $$ \int_T^{T+a}f(s)ds\leq \int_0^T\Psi(f(s))ds\cdot\Phi(a), $$ where $\Phi(a)$ goes to 0 as $a\to0$?

3$\begingroup$ Jensen's inequality. $\endgroup$ – Jochen Wengenroth Jul 1 '19 at 17:23

$\begingroup$ Could you please provide more details? Jensen's inequality seems not work. Thanks. $\endgroup$ – Wenguang Zhao Jul 1 '19 at 20:31

$\begingroup$ Just a typo report: in the last line of your question it should be $\Phi(a)$, not $Phi(a)$. $\endgroup$ – Daniele Tampieri Jul 1 '19 at 20:45

3$\begingroup$ is everything correct with the limits of integration in right hand side guys? Should not they be $\int_T^{T+a}$ in both inequalities? $\endgroup$ – Fedor Petrov Jul 1 '19 at 20:56
Yes, this is known as Young's (or Hölder's) inequality for Orlicz spaces, and one should take for $\Phi$ the (Legendre) convex conjugate $$ \Phi(y)=\Psi^*(y)=\sup_x \{ xy\Psi(x) \} $$