# Holder inequality with respect to convex function

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 $$a\to0$$?

• Jensen's inequality. – Jochen Wengenroth Jul 1 at 17:23
• Could you please provide more details? Jensen's inequality seems not work. Thanks. – Wenguang Zhao Jul 1 at 20:31
• Just a typo report: in the last line of your question it should be $\Phi(a)$, not $Phi(a)$. – Daniele Tampieri Jul 1 at 20:45
• is everything correct with the limits of integration in right hand side guys? Should not they be $\int_T^{T+a}$ in both inequalities? – Fedor Petrov Jul 1 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 $$\Phi(y)=\Psi^*(y)=\sup_x \{ xy-\Psi(x) \}$$