An integral inequality

  • Thread starter amirmath
  • Start date
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

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

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