- #1
amirmath
- 8
- 0
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.
$$
\int_{0}^{T}\Bigr(a(t
)\Bigr)^{\frac{p+1}{2p}}dt\leq C\Bigr(\int_{0}^{T}a(t)dt\Bigr)^{\frac{p+1}{2p}}
$$
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.