An integral inequality. 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.
 A: If $f$ is an entire function of exponential type $< \pi$ you might have some luck. Indeed, by Theorem 10.6.4 of Boas "Entire functions" (for the first inequality)
$$
\int_{\mathbb{R}} |f(x)|^m dx \leq C \sum_{\ell \in \mathbb{Z}} |f(\ell)|^{m}
\leq C \bigg ( \sum_{\ell \in \mathbb{Z}} |f(\ell)| \bigg )^{m}
\leq C' \bigg ( \int_{\mathbb{R}} |f(x)| dx \bigg )^{m}
$$
The second inequality is trivial, and the last inequality follows essentially by sub-harmonicity or Bernstein's inequality, but you could also appeal to Boas, Theorem 6.7.15. Finally note that the restriction for $f$ to be of exponential type $< \pi$ can be replaced by ``finite exponential type'', because if $f(z)$ is of exponential type $B$ then $f(z/B)$ is of exponential type $1 < \pi$, and you can repeat the above argument for the dilate $f(z/B)$. 
Finally you'll notice that I've integrated over the whole line. You can get the restriction to $[0,T]$, by multiplying $f$ against an entire smoothing, for example you could multiply by a dilated Fejer kernel $(\sin (x/T)/ (x/T))^2$. Much sharper cut-off's are of course possible (for one thing you can take a higher power than $2$ on the Fejer kernel, but one could do better than that if necessary). 
A: Let $a(t) = \displaystyle t^{-\frac{2p}{p+1}}$. Since $0 < \dfrac{2p}{p+1} < 1$, $a(t)$ is integrable on $(0,T)$, but $a(t)^{\frac{p+1}{2p}} = t^{-1}$ is not.
