Suppose I have a real-valued function $f \in L^{\infty}([0,2\pi)^2)$. Suppose also that, if $\| f \|_{p}$ denotes the $L^{p}([0,2\pi)^2)$ norm of $f$, then the following bounds are known:
$$\displaystyle C_{1}r^{1/2} \leqslant \|f\|_{1} \leqslant \|f\|_{2} \leqslant C_{2}r^{1/2},$$
where $r \gg 1$. I would like to know if the following is possible:
$$\displaystyle \|f\|_{4} \leqslant C r^{1/4}.$$
This seems counter-intuitive as I expected the norms to have the same or greater bound. Moreover, by the generalised Holder's inequality, this seems impossible, since then we would have
$$\displaystyle C_{1}^{2}r \leqslant \|f\|_{2}^{2} = \|f \cdot f \cdot 1 \cdot 1\|_{1} \leqslant \|f\|_{4}\|f\|_{4}\|1\|_{4}\|1\|_{4} \leqslant Cr^{1/2},$$
but the left hand side is bounded below by $r$, which indicates that my claimed upper bound for $\|f\|_4$ is impossible, and the above implies that $r^{1/2}$ is a lower bound for $\|f\|_4$ -- right?
There is no upper bound on $L^4$ norm in terms of the $L^2$ norm. Generally, on a finite measure space $L^p$ controls $L^q$ for $p>q$, but not for $p<q$.
As a concrete example, consider $f = \epsilon^{-1/2}\chi_E$ where $E$ has measure $\epsilon$. Then $\|f\|_2=1$ but $\|f\|_4 = \epsilon^{-1/4}$ which can be arbitrarily large.