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 $a\to0$?
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3$\begingroup$ Jensen's inequality. $\endgroup$– Jochen WengenrothCommented Jul 1, 2019 at 17:23
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$\begingroup$ Could you please provide more details? Jensen's inequality seems not work. Thanks. $\endgroup$– Wenguang ZhaoCommented Jul 1, 2019 at 20:31
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$\begingroup$ Just a typo report: in the last line of your question it should be $\Phi(a)$, not $Phi(a)$. $\endgroup$– Daniele TampieriCommented Jul 1, 2019 at 20:45
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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 PetrovCommented Jul 1, 2019 at 20:56
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1 Answer
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Yes, this is known as Young's (or Hölder's) inequality for Orlicz spaces, and one should take for $\Phi$ the (Legendre) convex conjugate $$ \Phi(y)=\Psi^*(y)=\sup_x \{ xy-\Psi(x) \} $$
answered Jul 2, 2019 at 9:23