An inequality-Is it possible?

Question

I want to know that is it possible to show that $$ \int_{0}^{T}\Bigr(a(t )\Bigr)^{\frac{p+1}{2p}}dt\leq C\Bigr(\int_{0}^{T}a(t)dt\Bigr)^{\frac{p+1}{2p}} $$ for some $C>0$ where $a(t)>0$ and integrable on $(0,T)$ and $p\in(\frac{1}{2},1)$. It is worth noting that this range for $p$ yields $\frac{p+1}{2p}>1$. In the case $p>1$ we have $\frac{p+1}{2p}<1$ and the Holder's inequality can be applied to obtain the result immediately. But in the first case I don't know the validity of the inequality.

Answer 1

No, this is not possible. Given any $q>1$, we can find a positive function on $[0,T]$ for which $\int_0^T a(t)\,dt=1$ but $\int_0^T a(t)^q\,dt=1$ can be as large as we want, or even infinite.

For example, pick $\alpha\in(1/q,1)$ and consider $a(t)= t^{-\alpha}$: then $\int_0^T a(t)\,dt=T^{1-\alpha}/(1-\alpha)$ while $\int_0^T a(t)\,dt$ diverges.

You can make $a$ bounded by truncating it -- that is, taking $\min(a,M)$ -- and do other things to arrange a counterexample with desired properties.

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There is a subject of reverse Hölder inequalities (not to be confused with Hölder inequality for $p<1$), which concerns the classes of functions for which such inequalities hold. Here is a sample paper from this area.