Does $\lim_{p \to \infty}\biggl(\int_0^T \Bigl(\int_\Omega |f|^{\frac{3p}{1+2p}}\Bigr)^{\!\frac{1+2p}{3}}\biggr)^{\!\frac 1p} < \infty$ exist?

Question

Let $f \in L^2(0,T;L^2(\Omega))$ on a bounded set $\Omega$. Does the following limit exist?

$$\lim_{p \to \infty}\biggl(\int_0^T \Bigl(\int_\Omega |f|^{\frac{3p}{1+2p}}\Bigr)^{\!\frac{1+2p}{3}}\biggr)^{\!\frac 1p} < \infty?$$

I don't know how to bring the factor of $\frac 1p$ inside.. Jensen's inequality doesn't help as the function is not convex.

Answer 1

We can rewrite the integral as $$\int_0^T \left(\int_\Omega |f|^{\frac{3p}{1+2p}}\right)^{\!\frac{1+2p}{3}} = \int_0^T \|f(t)\|_{L^{\frac{3p}{1+2p}}(\Omega)}^{p}dt = \|f\|_{L^p(0,T;L^{\frac{3p}{1+2p}}(\Omega))}^p.$$ The limit $p\to\infty$ exists if and only if $f\in L^\infty(0,T;L^{3/2}(\Omega))$. Try to find a counter-example.