Is the Integral Inequality Possible to Prove for Certain Parameters?

[amirmath]

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.

 

[fresh_42]

The crucial point is whether $C$ may depend on $T$ or not. E.g. $p=\frac{2}{3}\, , \,a(t)=t$.

 

[math771]

Yes, it is possible to show that the inequality holds for $p\in(\frac{1}{2},1)$. First, we can rewrite the left-hand side of the inequality as
$$
\int_{0}^{T}\Bigr(a(t)\Bigr)^{\frac{p+1}{2p}}dt = \int_{0}^{T}\Bigr(a(t)\Bigr)^{\frac{1}{2}}a(t)^{\frac{p}{2p}}dt
$$
Then, we can use the Cauchy-Schwarz inequality to obtain
$$
\int_{0}^{T}\Bigr(a(t)\Bigr)^{\frac{1}{2}}a(t)^{\frac{p}{2p}}dt \leq \Bigr(\int_{0}^{T}a(t)dt\Bigr)^{\frac{1}{2}}\Bigr(\int_{0}^{T}a(t)^{\frac{p}{p-1}}dt\Bigr)^{\frac{p-1}{2p}}
$$
Since $p\in(\frac{1}{2},1)$, we have $\frac{p}{p-1} > 1$, which means that the second term on the right-hand side is integrable. Thus, we can apply Holder's inequality again to obtain
$$
\Bigr(\int_{0}^{T}a(t)dt\Bigr)^{\frac{1}{2}}\Bigr(\int_{0}^{T}a(t)^{\frac{p}{p-1}}dt\Bigr)^{\frac{p-1}{2p}} \leq \Bigr(\int_{0}^{T}a(t)dt\Bigr)^{\frac{1}{2}}\Bigr(\int_{0}^{T}a(t)dt\Bigr)^{\frac{p-1}{2p}} = \Bigr(\int_{0}^{T}a(t)dt\Bigr)^{\frac{p+1}{2p}}
$$
Therefore, combining all of these steps, we have
$$
\int_{0}^{T}\Bigr(a(t)\Bigr)^{\frac{p+1}{2p}}dt \leq \Bigr(\int_{0}^{